So some flat earth arguments are actually correct if general relativity is correct, namely that gravity is an illusion and that the real reason we are stuck to the earth is that the earth is accelerating toward us at 9.8 m/s^2
I don't understand it. How can it accelerate towards anyone on its surface? Where does it get energy to accelerate? We generally need to burn fuel to accelerate something in the space.
This amounts to a confusion over notation: "proper acceleration" (e.g. as measured by an accelerometer) vs "coordinate acceleration" (the acceleration an observer observes an object to be undergoing).
The acceleration an observer sees you undergoing is the same as the inherent "proper" acceleration you're undergoing, minus the acceleration of their coordinate frame with respect to yours. For me to stay still with respect to you, if you're in a frame that is accelerating away from me, I need some proper acceleration to catch up and counteract the fact that our frames are diverging. But if spacetime is curved, your frame probably is accelerating relative to mine - c.f. the example on the Earth's surface, where our frames inexorably accelerate towards each other as we move parallel to each other. So for me to stay still with respect to you, I need to have some proper acceleration to balance out the coordinate acceleration derived from the fact that our frames are moving in a curved space.
My understanding of General Relativity[1] is that mass distorts space-time, so an object traveling in a "straight line" through distorted space-time will curve with that distortion.
If the object's velocity isn't enough to traverse the curved space-time, it will move toward the center of the mass generating the distortion and fall out of the sky.
If the object is traveling quickly enough, it can continue traversing the distorted space-time and orbit that mass.
If the object is traveling even more quickly, it will traverse the distorted space-time and continue on without orbiting the mass.
In all three cases, from the perspective of the object traversing the distorted space-time, it continues to travel in a straight line, as it's the space-time that's distorted.
A (flawed) analogy would be riding a bicycle between the peaks of two identically sized hills. Starting at the top of the first hill, you coast down increasing your velocity.
Once you reach the bottom of the first hill and head up the second, your velocity decreases.
If your velocity at the bottom of the first hill is too small, you'll go up the second hill and as your velocity reaches zero, you roll back toward the bottom of the hill.
You will pick up velocity and then roll back up the first hill, then down again, then back up the second, etc. until you end up stopped at the bottom of the hill. This is akin to falling to the center of the distorting mass.
If your velocity is high enough to carry you back up to the top of the second hill and then stop, you'll roll back down and get to the bottom with the same velocity you had coming down the first hill. You'll then oscillate between the tops of both hills. This is akin to orbiting the mass.
If your velocity at the bottom of the first hill is enough to carry you past the top of the second hill, you'll just keep going after reaching the top of the second hill. That's akin to flying by the mass.
It's a flawed analogy, because in a curved space-time the directional portion of the motion vector doesn't change.
As John Wheeler[0] simplified it:
"Mass tells space-time how to curve, and space-time tells mass how to move."
This is about relativity. We're going straight in space-time, but space-time itself is curved because of the heavy mass nearby (the Earth). This is visualized by the rocket curving towards the planet in the video, and the bent sheet experiment where the balls spiral towards the center.
So we're curving in towards the center of the mass of the Earth, but the reason we don't end up in the core of the Earth is because the surface stops us. The Earth is "pushing" us away from the center, and that's the acceleration. It's accelerating you off your straight line path, and this is the deviation from the geodesic.
I thought that the Earth curved spacetime, and since an inertial observer approaching the earth would follow a straight line through curved spacetime, they'd appear to be following a curved line through space towards the earth.
What's confusing me here is the notion that when two objects collide, they accelerate into each other. Why and how is force constantly applied after the collision? My intuition is falling down here, and none of the resources I've looked at so far have explained why the acceleration happens.
1st sentence is correct and is the same as what I said, sorry if I wasn't clear.
Remember that spacetime = space + time dimensions. The object is always travelling through time, and the curvature of spacetime is converting some of that speed through time into speed through space. That's what you perceive as motion (caused by gravity).
Time and space are linked together. The faster you go through space, the "slower" you go through time (as in you experience it slower). This is very measurable and even used to alter timings for satellites GPS readings. You can take an atomic clock on a plane and age slower than someone who just stayed on the ground.
So the spacetime curvature is continuously converting some of your temporal motion into spatial motion, until that's stopped by the surface of the Earth which is constantly "accelerating" to stop you from going further.
As to why we always move through time, that's beyond my understanding at this point but it's a fundamental axiom of physics.
Yeah, sorry, I was restating what I understood you had said, in case I'd misunderstood any part of it, and then explaining where I was lost.
Your explanation makes sense, and it sounds like I'd need to understand the maths behind relativity to be able to really understand how objects behave in spacetime.
I do not understand how you can have acceleration without changing position (at 10:06). Acceleration is the derivative of speed, which is the derivative of position change. If the position change is zero, how can the acceleration be non-zero?
Position is not an absolute notion: you need to answer "position with respect to what?".
If the thing you're measuring position against is also accelerating, then you need to apply some acceleration of your own to stay still with respect to it.
The terms you want to look up are "proper acceleration" and "coordinate acceleration". The curvature of spacetime means the thing I'm measuring position against is moving relative to me (c.f. the example of two people walking in parallel across the Earth, nevertheless eventually meeting: the curvature means that even though neither of them is measuring an acceleration, nevertheless they are accelerating towards each other), so I need to have some internal ("proper") acceleration of my own to counteract the fact that our geodesics are moving away from each other.
Your position in spacetime is changing. You're going straight in spacetime, but spacetime is curved by the mass of the Earth so you're following that curve into the center.
The surface of the Earth keeps you from actually falling in, and is therefore pushing you away or upwards from the center. This is the acceleration acting on your straight line path through curved spacetime. This is the deviation from your geodesic.
How would periodic “free-fall” motion look in this setup? E.g. a point mass orbiting around a body, or a point mass oscillating back and forth in a 1D gravity well.
In the left-hand image, an orbit would be a horizontal line, because it's a constant distance. So it's a mirror of the time axis, but translated upwards in space. It would be exactly the same axis-mirroring translation in the other images. So, importantly, it would not be a line in the right-hand image.
The concept of a constant distance orbit doesn't make much sense in 1D. A horizontal line in this model wouldn't indicate an orbit, but rather a completely stationary object. It would then make sense for the world line to curved because it must be accelerating in order to resist the attraction of the body it's near.
By definition, the world line of a stable orbit would be a line in curved spacetime.
This has some visualizations of space time curvature where you can get a sense of the lines particles take in different circumstances:
http://www.relativitet.se/Webtheses/lic.pdf
I tried to explain it with words, but I guess the images are worth more than I could write..
With two space dimensions and one time dimension (2+1), an orbit in Newtonian physics is a helix. In general relativity that same helical path would be a straight line in an interestingly curved spacetime.
Likewise I guess for the 1D gravity well case, where a sine wave would become a straight line in spacetime.
True gravitational force is something that can’t be transformed away by an arbitrary choice of frame (even an accelerating one).
As a brief example, consider two objects in downwards free fall toward the centre of some massive object. Since they head towards the centre, in a free falling frame the two objects actually get closer to each other until they collide as they reach the centre.
This is known as the tidal effect of gravity and is the actual physical content of general relativity. This effect can be shown to be obtained by an appropriate curvature of spacetime which itself can be shown to be related to the stress-energy of matter inhabiting spacetime.
"If you and a friend started walking straight north, both at the equator but a long distance apart, you would gradually get closer to each other until you collided at the north pole."
I always thought that was a nice way to drop one dimension down to get the intuition. To the metaphorical 2D ant they see two friends attracted to/falling towards each other, but they are going in a straight line on a curved surface and there are no forces at play.
It's been some time since I studied these things, but I believe the post was trying to illustrate that the geodesic lines in spacetime created by Earth's gravity field can be visualized as straight lines after a nonlinear change of the coordinates system. [1] To simplify, these are the lines along which particles move when no outside force is exerted on them--again, considering gravity to be a distortion of spacetime and not a force, very much like the well-known metaphor of steel ball on a rubber sheet, which would curve marbles towards the "well" it's created in the sheet. But you're right that once the marbles and the steel ball get to a point where one of the gravitational fields cannot be ignored, this framework becomes less useful.
I take your point, but I think the author's main point was that because objects in free-fall merely follow spacetime geodesics, it makes calling gravity a "force" a little bogus, at least compared to the other forces. Tidal effects don't change that; tidal effects mean the spacetime curvature "over here" is different than the spacetime curvature "over there", which means the principle of equivalence isn't true in a global sense. But objects still follow spacetime geodesics, which is a concept that's hard to reconcile with the notion of a "force."
That "perspective" can be extended to other forces. E.g.: one could say that an electron taking a complex spiralling path through a magnetic field on Earth is merely following a "gravity+EM geodesic" and is actually in free fall the whole time.
I'm super layman in all this, but I think the word force here refers to the Newton's force. Which experimentally was shown to not properly predict the effect of gravity, where as seeing gravity as a distortion of spacetime instead of as a acceleration force applied to the objects, did predict accurately the trajectory in experiments.
You can call both a force in the generic dictionary definition of force. Because obviously it's crazy that something can be so powerful as to distort spacetime itself. But gravity wouldn't be a force in the Newton sense of being something that affects the acceleration of an object.
If you choose to view the universe in one particular coordinate system (e.g. one centred on your eyes), then you'll see some things mysteriously happening. For example, you might find that if you release something from your hand, it will mysteriously move towards your feet. Eventually you'll realise that a bunch of things can cause things to mysteriously move, and you'll start using the word "force" to describe "tendency to mysteriously move".
But some of these forces are not observed by other people who are watching; they arise as by-products of the fact that you are the one doing the measuring. For example, if you're falling in a lift, you won't observe the force that causes a ball to fall towards your feet; but I, standing on the ground outside the lift, will observe that actually you and the ball and the lift are subject to this force.
In fact there's an underlying reality which we can't directly observe but which we can infer from the paths of objects. That reality is a curved spacetime, not a flat one (as it appears to be). This curvature means that "straight line" is actually not quite what you're used to; things follow straight lines, but those straight lines don't look straight to us, because of the underlying space's curvature: we can't see the whole of spacetime, only small segments of it, so we can't see enough to get a proper sense of the curvature. But since our limited frames of reference have their own notion of "straight line" which is incompatible with that of the global spacetime, we observe a mysterious tendency of things to deviate from what we think is the straight line they should be following.
So gravity "is a force": it's a mysterious tendency of things to move, because we are limited in what we can see and our own observation frames are subtly incompatible with the global structure. But it's also "not a force": if we were somehow able to take a fully global view of the universe, there would be no mysterious movement, only a huge number of things moving at exactly the same speed in perfect straight lines through a curved spacetime. (Assuming General Relativity is 100% accurate.)
I saw this and the related Veritasium video, but am still scratching my head about something. Does this mean that away from all gravity a body does not experience any time? ie a person on a spacecraft stopped in intergalactic space would not age relative to persons on planets?
>How do you get away from all gravity? Aren't you subject to gravity from all other matter in the universe at all times?
Yes. But the effect of the distortion of space-time that we call "gravity" is subject to the inverse square law[0].
This means that, as Newton described:
"The gravitational attraction force between two point masses is directly proportional to the product of their masses and inversely proportional to the square of their separation distance. The force is always attractive and acts along the line joining them"
As such, while objects are affected by the distortion of space-time, the effect is diminished (but not eliminated) by increased distance from the mass that's distorting space-time.
Yes, you did. However, it depends on how you define "getting away" from it.
At some point, the effect is so small that it either can't be measured or even if it can, the effect is so small that any impact is irrelevant in practical terms.
If that's the case, you're effectively "getting away" from it.
I believe (from my limited knowledge from Interstellar) that it’s the opposite: higher gravity environments move slower relative to lower gravity environments, which enabled the “go down to high gravity planet for a couple hours and come back and it’s ten years later” plot point in the movie. So if you’re in intergalactic space, time is actually passing very quickly in the higher gravity environment of a galaxy.
But a person’s experience should always be the same, no matter what reference they are in. They will perceive time as passing at the same rate in all environments, but it will be different from people in other environments.
Caveat: IANA physicist, and this is not physics advice.
I don't think that is quite right. If you travelled near the speed of light and then slowed down again it would look (from your perspective) like you jumped 10 years into the future. So high gravity environments "moving slower" shouldn't result in a jump to the future upon exit.
Instead I think what is happening is that massive objects actually stretch the fabric of spacetime somehow so that the closer you are to the object, the slower you travel through both space and time. And the more massive the object, the more stretched space and time become as you get closer to it.
Hence if you go near a very massive object, from an outside observer it looks like you are frozen in time because time is so stretched it takes forever for you to move through it.
As you mentioned though, from any given frame of reference time will always feel the same. 1 second will always feel like 1 second.
Ok then in that case I think of it as massive objects tend to compress space the closer you get to them. So though your velocity is constant, you are travelling through more space the closer you get to a massive object (since the space is now compressed) which means you are travelling through less time (in order to maintain same velocity).
> So if you’re in intergalactic space, time is actually passing very quickly in the higher gravity environment of a galaxy.
Not "very". Look up the actual equation, it's a really easy one. AFAIR for a standard stellar black hole, you'd need to be within meters from the event horizon to get any substantial time difference.
Thus the implication that Interstellar's Gargantua was an SMBH, i.e. probably the center of it's galaxy.
It's not that the absence of gravitational fields slows time up relative to observers in more massive reference frames, it's that the presence of gravitational fields speeds up time relative to observers in other reference frames.
You would age about the same as you would in microgravity. You could get closer and closer approximations by going into Earth orbit, solar orbit, and galactic orbit. Each approximation has less gravity, so time would pass slightly slower relative to an Earthly observer at each step, but the effect diminishes.
I understand, but the corollary of time appearing to speed up close to gravity well (article explains this is not a field per se) is that it would appear to slow down further away. Is there a minimum speed of time relative to earth?
Somewhere I read that a free fall parabola does not even take into account earth's curvature. Although I cannot remember, what kind of function describes the reference system specific ``path, as a function of time''.
Could this be named more correctly:
Gravity is not a force – free-fall hyperbolas are straight lines in spacetime (timhutton.github.io)
?
The planet's curvature is not very relevant to the article, but, yes, if an object is in a free fall at a slow speed (I mean non-orbital) its trajectory is:
- parabola if you assume flat&infinite ground,
- ellipsis if you assume a spherical planet (an ellipsis is crossing the planet's surface).
Of course, a very very eccentric ellipse approximates a parabola quite well. You only start to notice errors when dealing with trajectories dozens of miles long (on Earth).
I heard an interesting question at one point: "how come, when you throw a ball up on Earth, the parabola is so strongly curved? Spacetime is nearly flat, so how can a straight line become such a steep parabola?"
I'll answer this question as I understand it, but I only took four lectures of General Relativity before I gave it up in favour of computability and logic, so if there is a more intuitive and/or less wrong answer out there, please correct me.
Intuitive answer: the curve is indeed very gentle, and (e.g.) light will be deflected only very slightly by the curvature; but the ball is moving for a couple of seconds, and that's an eternity. On human scales, the time dimension is much "bigger" than the space dimensions (we're quite big in the time dimension and quite small in the spatial dimensions); the ball moves only a small distance through space but a very large distance through time, amounting to a big distance in spacetime, and so the slight curvature has a bigger effect than you might expect.
I'm not aware of any game engines that simulate general relativity / 4D spacetime.
I think what you may be noticing is that, as you reduce the tick frequency of a newtonian physics simulation, parabolas become less accurate as integration error accumulates.
>I heard an interesting question at one point: "how come, when you throw a ball up on Earth, the parabola is so strongly curved? Spacetime is nearly flat, so how can a straight line become such a steep parabola?"
Air resistance, wind, and horizonal acceleration. Over long vertical distances, these perturbations in the x-axis cause an arc. Nothing to do with general relativity.
When you're tossing a ball into the air by hand, gravity is going to have a far more dominant effect on things than air resistance and friction. Things still fall on the moon...
It’s something of a philosophical question. We tend to think of distant galaxy’s as if we are viewing them via some FTL means with a single consistent now. NASA for example tracks distant Mars probes like that rather than marking timing based on when the signal was received.
Alternatively, you can think of everything that could impact you as something of a now light cone. The second view has the universe existing as a 3D surface in 4D space time which means objects have a temporal width for each observer. That can be a really useful mathematical model.
your size in X, Y, Z, is how much space you occupy at once, i.e., at a fixed point on the time axis. But how would we even make sense of the notion of "size" on the time axis? How much time we take up for a fixed value in one or more of the X, Y, and Z axes?
If you consider your body has extent in time, then when you move between two points in space, your body is a long worm connecting those two points. So your size in some way is related to the distance you travel in your life. But even without time, a person's size isn't really well defined anyway. Is a person wider if he stretches his arms out sideways?
> The creatures can see where each star has been and where it is going, so that the heavens are filled with rarefied, luminous spaghetti. And Tralfamadorians don’t see human beings as two-legged creatures, either. They see them as great millepedes—“with babies’ legs at one end and old people’s legs at the other,” says Billy Pilgrim.
Interesting way to phrase the question, the answer is... depends on relative motion. The notion of simultaneity which is the technical term for what you describe as "being flat" in the time-direction is not an absolute thing in relativity. With respect to your own reference system you are essentially "flat" in the time direction [1], but with respect to someone that moves at a certain speed with respect to you you have certain size in the time direction.
Simply picture an extended object in the x direction, Lorentz transformations are going to rotate it slightly in the time direction causing it to become extended in the time direction.
[1] If you take it to the extreme you may consider that different parts of your body move with respect to each other and thus they are not in the same reference frame. In which case, not even with respect to yourself are totally "flat" in the time direction.
Under special relativity, everyone and everything moves at a constant speed `c` through spacetime. If you feel like you're not moving, it's because all your speed is being put towards travelling faster through time. Conversely, if you manage to move very fast through space, the world around you will appear to speed up, because you've had to trade off some of your forward travel through time so as to travel in space; the rest of the world is moving forward in time faster than you are.
So you can change your acceleration through the time dimension of spacetime, by dint of changing your acceleration in the spatial ones.
And at the same time, in your new frame of reference, you're still moving at exactly c through the time and 0 through space. But your time axis is no longer parallel with the time axis of the rest of the world.
I think it does if they do not occupy the exact same space. It's only objective that they have the same velocity from their point of view (but then parallel lines concept hardly makes any sense).
I mean the whole way looking at it seems wrong to me. There is no "rest of the world" in relativity. Assigning some objective vector to everybody doesn't work. These only make sense from some specific point of view.
By the question you asked now I'm assuming, you meant "but hey, without GR..", but even without GR, ignoring that the universe is expanding, assuming flat space time etc. If the universe consisted just of 3 bodies, 1 being you and 2 being rest of the world, then the way of thinking you described still doesn't make sense in context of relativity and may lead to some confusion (apart from it being, to me, incoherent in context of special relativity).
But maybe I'm missing something from your picture, I'm happy to read and learn.
Special relativity definitely won't be a problem. Everyone see timings and speeds a little differently, but everyone can also do the calculation from anyone else's point of view. Objects that are in the same reference frame are objectively in the same reference frame in special relativity; everyone agrees. The reference frame is determined only by velocity. And the distortion is determined only by the observer's velocity, so both objects will have the same distortion.
You can certainly imagine a scenario where two objects measure their mutual distance as being constant in time. You can also imagine other scenarios, but I'm asking to imagine the scenario where two objects do measure their distance to be D and then measure it again and it's still D, and measure it again and again and it's still D. That's just the definition of standing still with respect of each other, and when you plot a space time diagram, their space curve is parallel (because their spacial distance doesn't change).
Special relativity doesn't make that scenario impossible. It doesn't force things to move. It just describes what happens when things do move (through spacetime).
I've always intuitively understood this to be the reason why it would take infinite force to achieve light speed for a massive object. When we apply a physical force, it is applied in the spatial axes, so it is always perpendicular to the time axis. Acceleration is just rotating some magnitude of your fixed velocity vector out of the time axis and into the spatial axes. When your spatial velocity is apparently zero, then the component of force that is perpendicular to your velocity is large, so you achieve a large deflection. But as you rotate velocity out of time and into space, it becomes more perpendicular to time, so any force applied perpendicularly to time is now more parallel to your velocity, having a smaller component perpendicular to one's velocity. You can't rotate a vector with a parallel force.
This is also why you can't travel backwards in time through just acceleration. There is no way to impart a force perpendicular to your velocity vector when it is already perpendicular to time, giving you no way to rotate the vector to have a component that points backward in time.
So I've always wondered, whether general relativity allows for forces parallel to time, and we just don't know of any mechanism to actually do so, or if it does not cover such cases because we have no mechanism, or if it disallows it entirely.
This is a useful way to think about it, but you have to keep in mind that (even flat) spacetime is not Euclidean but Minkowskian: in the distance metric, time has an opposite sign to the spatial dimensions. So when you "rotate" a four-vector, it actually follows the surface of a hyperboloid rather than a sphere, which means that rotation has a discontinuity at Θ = π/4 (in normalized units) and the vector escapes to infinity! Only massless objects can travel "at the speed of π/4", everything else can only approach but never reach that speed.
I still don’t understand the graphs at the link, but this intuitively makes sense to me. Thank you - I now have a new, apparently accurate mental model of relativity.
The word “dimension” in this context is overloaded. We think of space being three dimensions but really it’s only one - velocity relative to a specific reference frame. Thinking of it this way, the word “spacetime” makes sense; it’s a two-dimensional system: “spatial velocity” (S) on one axis and “temporal velocity” (T) on another. Both velocities are always measured against a reference frame, and their sum is c (c=S+T).
This would mean that time travel is impossible not because of a “speed limit”, but because c is a dimensionless physical constant.
Whether something is dimensionless is pure convention. By making c dimensionless and, for even more convenience, setting its value to 1, you can measure distance in seconds!
Off topic, but in reality, the trajectory of a ball thrown on Earth is not a parabola, but an ellipse [1]:
> under the laws of gravity, a parabola is an impossible shape for an object that's gravitationally bound to the Earth. The math simply doesn't work out. If we could design a precise enough experiment, we'd measure that projectiles on Earth make tiny deviations from the predicted parabolic path we all derived in class: microscopic on the scale of a human, but still significant. Instead, objects thrown on Earth trace out an elliptical orbit similar to the Moon.
Good point; this becomes more obvious if you imagine throwing the ball up and then immediately collapsing all the mass of the Earth into a single point at the centre. What path does the ball follow now? It's probably following a path we would more usually call an "orbit", and it sure looks a lot like an ellipse. Now just put the mass of the Earth back where it was, and notice that the ball hits the ground before it can trace out very much of its orbit.
Despite both being conic sections, cutting up an ellipse won't yield you parabolas. An ellipse has two focal points to which the sum of the distances is constant, while a parabola has a focal point and a directrix line to which the difference of the distances is constantly 0. Two different things.
More symmetrically, Every comic section has one focus and one directrix (and a semi-major axis for scale), eccentricity is the ratio of distance. Ellipse has eccentricity less than 1, and a parabola has eccentricity equal to 1.
A good point to make, but if one is going to be that pedantic one shouldn't call it "impossible". A parabola is possible with the right wind, air resistance, gravity from nearby mountains, etc, and in practice those have a larger impact on almost any suborbital projectile than the difference that turns the trajectory from a parabola to an elliptical arc.
It's a parabola in a uniform gravity field, an ellipse in a circular gravity field coming from a point mass.
So if you want to be really pedantic, it's never an ellipse because the Earth is not a point mass. It would be equivalent to a point mass if the Earth were a perfect sphere of uniform density, but it isn't. In reality it's a potato like mass blob that's approximated by what geodesists call the "geoid". So in order of approximations the path of a ball thrown on earth is a parabola -> ellipse -> numerical integration of 6-dof initial conditions and spherical harmonics approximation of the earth gravity field.
> it's never an ellipse because the Earth is not a point mass.
At least classically, a sphere is indistinguishable (gravitationally) from a point mass while you're outside it. The earth is pretty sphere-ish, locally speaking.
There are very few stable orbits close to the lunar surface. Basically a couple of polar orbits with very specific parameters, and that's it.
The rest get so perturbed by gravitational anomalies that they fall out of orbit after a few months or years--faster than low Earth orbit where there is atmospheric drag!
Yeah, the moon's gravitational field is quite lumpy compared to Earth's, due to its smaller size and the way it is believed to have formed. Plus you can orbit much closer to the surface due to the lack of atmosphere, so the lumps are steeper.
And a ball thrown in the air follows a parabola, locally speaking. At least, you're better off correcting for air resistance before you sub in the ellipse equation.
Spherical mass can be replaced with a point mass. Earth however is not spherical (biggest deviation is polar flattening). And even then, Einstein's model, unlike Newton's, says it's not an ellipse even for a point mass.
So, moral of the story - we have to be not too pedantic.
> Spherical mass can be replaced with a point mass
I'm not a physicist, but that doesn't sound right. One example is that if you are inside of the sphere, at least some of the mass will be pulling you away from the point at its center. I think what you mean is that a point mass closely approximates a spherical mass, but the degree to which that is true becomes less and less the closer you get to the center. I don't think you have successfully contradicted what the person you responded to said.
In Newtonian physics, a sphere and point mass are exactly interchangeable as long as you are not inside the sphere. If you are outside the sphere, the equivalence is exact, regardless of distance.
Proving this is a classic problem in undergraduate physics.
And that brings fluid dynamics into the picture, which produces discrepancies orders of magnitude greater than the parabolic/elliptical/etc. distinction!
Thanks! Shame I never took undergraduate physics but now that you've rung my bell I think we may have discussed this in high school. What an unintuitive result that being even a meter under ground breaks what is, up to that point, a fine model.
The way the result works is that if you are a meter underground, it is equivalent to standing on top of a sphere with the top meter of mass removed.
Or equivalently, if you are inside a shell (sphere outside, hollow sphere inside) then the gravitational effect is zero. This can be explained by analogy with light which also follows the inverse square law--changing your position inside a hollow sphere does not change the fact that you see the sphere all around you. Same reason that there is no electrical field inside a hollow conductive object.
Check out gauss law for electromagnetism and gravity. It says that total flux through closed surface is proportional to strength of field sources inside the surface. Flux from outside sources cancels out.
You can use this law to see what's the gravity fields as you move under ground and in other very symmetrical cases.
>What an unintuitive result that being even a meter under ground breaks what is, up to that point, a fine model.
It doesn't break it at all. The meter above you can be treated as a hollow shell, which surprisingly has zero net pull, and the solid sphere below can be treated as a point mass just as before.
Just remember these two facts, each provable with a simple integral calculation, usually done in high school physics or freshman college physics: a uniform sphere has the same gravitational pull on an object as a point mass at it's center, and the net gravitational pull on an object inside a spherical shell is zero.
This all works under perfect spheres, uniform (at the spherical shell level at least) density... There are other cases it works, but this simple case is the basic idea.
The intuition we came up with when we had to solve this issue in undergrad physics was interesting.
For each point of any given distance, you can find a disk of points at the plane of the same distance whose gravitational pull will be equivalent to a single point at their center.
You do this for all distances x, and you will find an equivalent rod going from the attracted objected to the center of the sphere, of a non-uniform but symmetrical density.
We find that the density is linearly correlated to the area of the corresponding section, and we then take the closer half of the rod, multiply the density by 1/r^2, and find that it is constant!
For the far half of the rod, we cannot do this, so instead we divide it into two halves of equal pull, then divide those to halves and so on, and find that it this reduces to the equivalent of a point.
Now that we have two points in-line with the object, we find the point with the same total mass that exerts and equivalent force, and lo and behold it is at the center.
I'm sure there is a much simpler way to intuit it, though :)
> if you are inside of the sphere, at least some of the mass will be pulling you away from the point at its center
If the mass is spherically symmetric, this will not be the case; all of the Newtonian forces from the masses further away from the center than you are will cancel out. This is called the "shell theorem", and it turns out to hold even in General Relativity.
I don’t think this is correct. I think the shell theorem says that the gravitational forces cancel if you are on the inside of a hollow sphere and all mass is on the surface. A perfect sphere of uniform density would not meet the shell theorem assumptions.
> I think the shell theorem says that the gravitational forces cancel if you are on the inside of a hollow sphere and all mass is on the surface.
No, it's stronger than that. It says that any spherically symmetric distribution of matter outside a certain radius exerts no "gravitational force" on anything inside that radius, whether there is matter inside that radius or not.
What if my distance to one point on the shell was zero? Would I not feel infinite acceleration towards the massive point on the shell that I was infinitely close to?
> What if my distance to one point on the shell was zero?
Then you are not inside the shell, you are on it. The shell theorem only applies if you are inside the shell.
> Would I not feel infinite acceleration towards the massive point on the shell that I was infinitely close to?
In General Relativity, matter is not viewed as point particles. It is viewed as a continuum, described by the stress-energy tensor. This is one of those cases where the difference shows up.
I think how gravity works at really short distances like 10^(-50)m is still a mystery as we don't have a consistent quantum theory of gravity that works there.
But as you get close to stuff, say two atoms, the electrostatic and other forces are far greater than the gravitational ones.
Any point on the shell, whether the shell has zero thickness or not, has zero mass, but a finite mass density is assigned to it. Any finite mass is spread out, so you cannot have zero distance to enough of its points to feel infinite force.
You and pdonis are correct but you are saying different things. Inside a perfect solid sphere, you would of course feel force everywhere except at the center, but this force would only be due to the mass that is nearer to the center than you.
In relativity, there is no such thing as a "uniform gravity field", if by that you mean a field where the "acceleration due to gravity" is the same everywhere. The closest you can come is the "gravity field" inside a rocket accelerating in a straight line in empty space, where the acceleration felt by the crew is constant. That kind of "gravity field" has an "acceleration due to gravity" that decreases linearly with height.
Look up the Bell Spaceship Paradox. In relativity, two spatially separated objects (or two ends of a spatially extended single object) that have the same proper acceleration in the same direction do not stay at rest relative to each other; they move apart, as seen in each of their own frames. This is different from the behavior predicted by Newtonian mechanics. In order to have the two objects (or two ends of a single extended object) remain at rest relative to each other, the one in front has to have a smaller proper acceleration than the one in back. (Rindler coordinates are often used to describe this case.)
> So if you want to be really pedantic, it's never an ellipse because the Earth is not a point mass.
This turns out to be not pedantic but very important if you're guiding an ICBM. And when landing on the Moon, Apollo had to deal with irregularities in the Moon's gravity due to mass concentrations, called mascons.
(If you're interested in missile guidance, take a look at the book Inventing Accuracy. Among other things, it discusses some of the efforts to map the Earth's gravity field to increase missile accuracy for Trident and Minuteman missiles. I knew a physicist who worked on this.)
(Ah, yes, pedantry. It has its own gravity. You can't tell me it's not a force. I can't resist its pull.)
It wouldn't be an ellipse even if Earth were a point mass. The gravity of the Moon and Sun, the gravitational lumpiness in the sky, has the same effect as the gravitational lumpiness underground. The combined result may be no closer to an ellipse than it is to a parabola.
But on low earth orbit, the L2 term of earth's oblateness dominates by an order of magnitude compared to the moon and the sun, and the rest of the planets are negligible.
Source is a table in the first chapters of Fundamentals of Astrodynamics.
> "It would be equivalent to a point mass if the Earth were a perfect sphere of uniform density"
Would it? Even of uniform density, the mass would not the at the core, only the center of mass would be. Most of the mass is actually closer to the surface. I'm no physicist, but I'd imagine that would have quite some impact[0] on any trajectory that crosses the surface.
And for any ball thrown at human speeds, I'd expect Earth's gravity would be much closer to a uniform gravitational field than one from a point mass.
[0] Pun not originally intended but I'm quite happy with it now.
Based on what I've seen over the years I've been around here, I'm quite certain that's a primary motivator for a non-trivial number of commenters here.
Ellipse, parabola, hyperbola and circles are all the same thing (called a comic). The only difference is if it includes or goes through the point at infinity. The relevant branch of mathematics is called projective geometry.
I think of it as being because the ball has mass and photons don't. So Newton's Gm1m2/r^2 = 0 for photons, and you have to use Einstein to measure the "force" on a photon.
But because massive objects also cannot move as fast as photons, we're probably both saying the same thing from two different perspectives.
The curvature does indeed depend on the ball's mass. The contribution to spacetime curvature the ball's mass brings to the table is extremely small compared to that of the earth, and the contribution that comes from the (massless) photon's energy[0] is smaller still, but they both influence the spacetime curvature.
[0] Photons bend spacetime according to GR, although as far as I know this has not been proven experimentally. We'll probably need a theory of quantum gravity to be sure.
I’d think that the curvature of spacetime due to the ball’s mass wouldn’t affect the ball’s trajectory, for similar reasons as why the electric field from a charged particle doesn’t move the particle itself.
I've always wanted to know why a ball doesn't follow a beam of light if they are both following straight lines in spacetime.
But even more important, if light beams are reversible under relativity (reflected off a mirror they will backtrack the same path) then light can not enter a black hole because its reversed path could allow it a way out. But then there's that whole thing of objects falling in appear to slow and stop as they approach the event horizon, so maybe light doesnt enter after all.
My conclusion is that you cant really understand it without serious study of under someone who already gets it.
They both follow straight lines but unless you have a really really really strong arm, the baseball is “moving” across far more time than the ray of light. So if it “experiences” much more gravitational effect from the larger swath of space time it covers.
Using quotes since terms are more metaphorical than exact.
The difference is that light travels through space, but not time (similar to how a vertical line does not travel the x axis, only the y axis). The ball travels through both space and time.
The faster you go, the less you travel through time. Thus, if the ball were travelling at the speed of light, it would not travel through time either and would follow the same path as light.
> The difference is that light travels through space, but not time
This is a common pop science statement, but it's not correct. A correct statement is that the concept of "speed through spacetime", which is what has to be split into "speed through space" and "speed through time" in the pop science statement, does not apply to a light ray.
In more technical language, the tangent vector to the light ray's worldline is not a unit vector, it's a null vector, and the concept of "speed through spacetime" only makes sense for a worldline whose tangent vector is a unit vector.
> From the point of view of a photon, no time elapses between its the origin and destination endpoints.
No, this is not correct. The correct statement is that the concept of "elapsed time" does not apply to a photon; it only applies to timelike worldlines, not null worldlines.
To put it another way, if your statement were true, it would mean that the origin and destination events were the same point in spacetime. But they're not; they're distinct points in spacetime. Which means that, since the spacetime interval along the worldline is zero, you can't use the interval to distinguish points on the photon's worldline. And the concept of "elapsed time" requires that you be able to do that. So the concept of "elapsed time" can't be used for a photon.
It's impossible for an object with mass to go the speed of light. But you can observe what happens as you approach it:
Let's say I put you in a spaceship and accelerate you to 50% the speed of light toward the sun. From an inertial viewer's perspective you are travelling toward the sun at half the speed of light and it takes you ~16 minutes to crash into the sun. But from your perspective it only took ~14 minutes to crash into the sun[0].
Repeat the experiment except I accelerate you to .99c. From an inertial viewer's perspective you are travelling toward the sun at nearly speed of light and it takes you ~8 minutes to crash into the sun. But from your perspective it only took ~1 minute to crash into the sun.
Repeat the experiment except I accelerate you to .999c. From an inertial viewer's perspective you are travelling toward the sun at nearly speed of light and it takes you ~8 minutes to crash into the sun. But from your perspective it only took 20 seconds to crash into the sun.
Repeat the experiment except I accelerate you to .9999c. From an inertial viewer's perspective you are travelling toward the sun at nearly speed of light and it takes you ~8 minutes to crash into the sun. But from your perspective it only took 6 seconds to crash into the sun.
Repeat the experiment except I accelerate you to .99999c. From an inertial viewer's perspective you are travelling toward the sun at nearly speed of light and it takes you ~8 minutes to crash into the sun. But from your perspective it only took 2 seconds to crash into the sun.
See what's happening? As you approach the speed of light, the amount of time that elapses until you reach your destination approaches zero. So from an inertial observer's point of view, time has completely frozen for travelers approaching light speed.
Yes, but you cannot extrapolate from this to say that the time lapse for a photon would be zero. A photon is not the limit of objects with mass going closer and closer to the speed of light, because "closer and closer to the speed of light" is frame-dependent, but a photon's speed being c is not. I can find an inertial frame in which each of your objects is at rest, and in that frame, you are the one who is "close to c" (in the opposite direction). But that doesn't mean your elapsed time approaches zero. By contrast, it is impossible to find any frame in which a photon is at rest. The two types of objects are fundamentally different.
In more technical language, the action of Lorentz transformations on photons is fundamentally different from their action on timelike objects. So it is simply not valid to view photons as some sort of limit "as speed approaches c" of timelike objects.
But is there any physical way to distinguish these fundamentally different situations? If not, then perhaps the fundamentalness of it is just an artifact of the formulation.
I'm thinking of solar neutrinos which, for a while, we weren't sure if they were massless or not. We had to observe them experiencing a duration of time to conclude they were massive. If we didn't find that, maybe it was just an even shorter duration, not the absence of one and we would never be able to tell the difference.
> is there any physical way to distinguish these fundamentally different situations?
Are you asking if there is a way to distinguish a timelike object from a lightlike object? Of course there is. The fact that, for something that has a very, very small invariant mass, it might be practically difficult does not change the fundamental principle.
Also note that the reason it was difficult, for example, to tell whether neutrinos have mass or not is that we can't just do the obvious and straightforward thing and find an inertial frame in which they are at rest (by, for example, taking a rocket and accelerating it in the direction of a neutrino to see if we can bring it to rest relative to the rocket). So we have to resort to indirect methods. But, again, that's a practical limitation that doesn't change the fundamental principle.
It still doesn't sound physically distinct any more than distinguishing any continuous quantity as being zero or nonzero. If we measure something that looks like 0, we can't be sure if it's just below the sensitivity of our instruments.
For neutrinos, even if we accelerated an rocket and somehow checked if a neutrino was at rest relative to it, we might find that it's not. That means we won't know if we need more speed or if it's impossible. I suppose it's a bit easier than that because we only have to accelerate the rocket fast enough that the neutrino's speed becomes measurably less than c, rather than 0. But still, what if we can't even get it to go fast enough for that? No way to prove that it's travelling at c, it seems.
I'd like to add that even photons have a nonzero upper bound to their possible rest mass. At least they used to. Is there any way, in principle, to show that it's exactly zero, and thus falls into this distinct category?
With neutrinos we might find that it's not but it'd be impossible to catch a photon as it would always have the same speed of c in our reference frame.
> I'd like to add that even photons have a nonzero upper bound to their possible rest mass. At least they used to.
Not sure what you're talking about, their momentum?
No object with mass can reach the speed of light and we know they're travelling at that exact speed.
How do we know they're travelling at exactly c? That's my concern. Last I heard, a couple of decades ago, physicists would occasionally measure a new maximum possible rest-mass for photons. It would be very tiny, of course, but they couldn't say it's exactly zero.
> How do we know they're travelling at exactly c? That's my concern.
We don't, strictly speaking. The measurements you refer to aren't even measuring the speed of photons. They're measuring their rest mass.
> physicists would occasionally measure a new maximum possible rest-mass for photons. It would be very tiny, of course, but they couldn't say it's exactly zero.
Based on just those measurements, no. The most they can say is that the photon rest mass is zero to within some error bar, and the size of the error bar keeps getting smaller. (The current error bar, IIRC, is 10^-52 grams, or about 24 orders of magnitude smaller than the electron mass.)
However, we have a ton of indirect evidence that photons are massless; the most extensive body of such evidence is all the evidence for the gauge invariance of electromagnetism. If photons had a nonzero rest mass, that would break electromagnetic gauge invariance. So photons having a nonzero rest mass would be a huge issue for our current theories, in the way that neutrinos having a nonzero rest mass would not; there is no important symmetry coresponding to electromagnetic gauge invariance that is broken by neutrinos having a nonzero rest mass.
If you try what I described with a light ray, it will be moving away from you at c no matter how much you accelerate in its direction.
If you try it with a massive object, even a neutrino with a very, very tiny invariant mass, that will not be the case; its speed relative to you will decrease as you accelerate after it, eventually to zero.
There is no continuum between those two possibilities; they are distinct and discrete. The only continuum is in the latter case, where the final speed of the object relative to you will depend continuously on how long you accelerate.
> even if we accelerated an rocket and somehow checked if a neutrino was at rest relative to it, we might find that it's not. That means we won't know if we need more speed or if it's impossible
Yes, you will know, because you will know if the neutrino's speed relative to you has decreased or not. If it has, it's possible to bring it to rest relative to you. If it hasn't, it's not. See above.
> I suppose it's a bit easier than that because we only have to accelerate the rocket fast enough that the neutrino's speed becomes measurably less than c, rather than 0.
Exactly.
> But still, what if we can't even get it to go fast enough for that?
That's basically the position we are in now: we have no way of building a rocket or other device that can accelerate after a neutrino long enough to tell whether its speed relative to the rocket is measurably decreasing. So we have to resort to indirect measurements. But as I said before, that doesn't change the principle.
> even photons have a nonzero upper bound to their possible rest mass
Yes, because, as I said, practically speaking we can't run the obvious and straightforward experiment I described, to confirm that a photon moves away from you at c no matter how much you accelerate after it. So we have to resort to indirect measurements, like trying to measure its invariant mass by other means. But that doesn't change the principle.
> Yes, you will know, because you will know if the neutrino's speed relative to you has decreased or not. If it has, it's possible to bring it to rest relative to you. If it hasn't, it's not. See above.
I don't think this quite works because of relativistic addition of velocities. Naively, it seems to. For example, if the object was travelling at 0.999999c relative to you (appears to be 1.0c according to your limited instruments), then you accelerate to 0.50c in its direction, you'd see its speed reduce to 0.50c (same 2s.f. instrument), which would clearly prove it's not massless. But velocities don't add like that relativistically and I think you'd still see it as travelling at 1.0c because it only decreased a tiny amount, below what you instrument can detect. If you use a more precise instrument or a faster rocket, you might measure it as 1.00000c but then you still won't know if it's exactly c or a smidgen less.
Maybe I've got my relativistic velocity addition wrong? But it still looks like the same measurement problem as trying to prove a classical object has a speed of exactly 0, which can't be done no matter how accurate our instruments are.
> velocities don't add like that relativistically and I think you'd still see it as travelling at 1.0c
Not indefinitely. Sure, if you accelerated for a short enough time in the direction of the neutrino, you might still be within your measurement error and so not have learned anything. But that just means you need to accelerate for a longer time. For any given measurement accuracy, you will be able to calculate how long you need to accelerate, by your clock, to definitely distinguish the two cases. Relativistic velocity addition does not change that fact. All it changes is the details of that calculation; yes, for a given measurement accuracy, you need to accelerate for a longer time, by your clock, to definitely distinguish the cases than you would if velocity addition were linear. But that doesn't mean relativistic velocity addition makes it impossible to distinguish the cases at all, ever. It doesn't.
So if your accuracy is, say, 1 part in 100,000, you wouldn't be able to see the difference. But with an accuracy of 1 part in 500,000, you would, even though you wouldn't have been able to see the difference before the acceleration.
Also, suppose you accelerated for a second increment equal to the first; you would get
As you can see, the differences in velocity grow fairly quickly for each equal increment of acceleration; the growth is not at all linear. And, as I said in my other post just now, for any given measurement accuracy, it would be simple to calculate how much acceleration you would need to be able to distinguish moving at exactly c from moving at 0.999999c (or any other speed less than c that you choose) to that accuracy.
>> In more technical language, the action of Lorentz transformations on photons is fundamentally different from their action on timelike objects.
I don't believe that, and have never heard it before. There are many ways in which light actually behaves just like particles with mass traveling at speed c. It has to or conservation of momentum is violated.
Show me an actual textbook or peer-reviewed paper Tyson has written where he makes this claim. Pop science videos don't count. (Tyson is by no means the only one; Brian Greene is notorious for the same thing.)
You won't be able to because there aren't any. No scientist who talks about a photon "experiencing zero time" in informal contexts will try it in a textbook or paper. That's because they know that if they did, other scientists would call them out on it, so they confine such claims to contexts where there are no other experts so there's nobody to call bullshit.
Another point is that if this concept were actually scientifically useful, somebody would be using it in a textbook or peer-reviewed paper. The fact that nobody is is a huge clue that the concept is not scientifically useful. It's only useful for selling pop science books or getting views of pop science videos, where, again, there are no other experts around.
Dude, this is basic theory at the high school level. Here's how it works: spacetime is 4-dimensional, and a vector in spacetime is always c units long. You can restrict your travel solely to X, Y, Z, or time if you like. If you do that, the other three components are going to be zero.
Photons put it all into the X, Y, and Z components, leaving nothing for the t component. They experience a change of position in space, but not in time. What's so hard to grasp about this?
Another point is that if this concept were actually scientifically useful, somebody would be using it in a textbook or peer-reviewed paper.
Seems that a fellow named Maxwell got a lot of mileage out of the concept, even if he didn't know what was really going on.
> Dude, this is basic theory at the high school level.
No, Tyson's claim is not "basic theory at the high school level". It is a particular interpretation of a theory (Special Relativity) that does not work, for the reasons I gave.
> Photons put it all into the X, Y, and Z components, leaving nothing for the t component.
Wrong. The spacetime vector that describes a photon's trajectory does have a t component.
> Seems that a fellow named Maxwell got a lot of mileage out of the concept
Dude, if you think the concept Tyson described is the same as the concept that Maxwell got a lot of mileage out of, then you are the one who needs to learn more about "how it works".
Again, take it up with Tyson and others with doctoral-level credentials who've made a career out of explaining these subjects to the unwashed laity. I'm not one of those people. Nobody posting on Hacker News is, as far as can be discerned.
If you and others in the thread feel that these popular authorities are spreading misinformation or using inappropriate analogies, doesn't it behoove you to raise an objection with them directly? Or perhaps with the appropriate faculty committees at their institutions?
> If you and others in the thread feel that these popular authorities are spreading misinformation or using inappropriate analogies, doesn't it behoove you to raise an objection with them directly?
If they want to make money by getting "the unwashed laity" to buy their books, why should I stop them? I simply don't buy them myself. If other people want to get told comforting nonsense, that's their problem. Caveat lector.
> Or perhaps with the appropriate faculty committees at their institutions?
Which would be pointless and absurd, since, as I have already said, the claims in question are not being made in textbooks and peer-reviewed papers.
> I've always wanted to know why a ball doesn't follow a beam of light if they are both following straight lines in spacetime.
Because they're following different straight lines in spacetime. Roughly speaking, if you pick a point in space and a particular direction in space from that point, there is a continuous infinity of possible straight lines in spacetime that point in that direction in space. One endpoint of that continuous infinity is the worldline of a light ray. The ball's path is somewhere in the middle of that continuous infinity.
It all depends on your frame of reference. From the ligth beam's view nothing happens, it just falls into the black hole. In fact the light beam never experiences time as it travels with the speed of causality.
From the reference of the outside observer the light emitted at the event horizon will forever try to leave it, but as spacetime itself casdades into the hole at the speed of light (at the horizon) this light will get redshifted until you can't see it anymore. This doesn't mean that the light you shoot into the hole never reaches the singularity. It just means that the light emitted at the event horizon will struggle forever to get out of the insane warp. Think about this: if you're walking on a conveyor belt with a constant speed `n` in the opposite direction and the belt itself is moving with a constant speed `n` you'll never make progress. This is what happens at the event horizon.
Light beams are not reversible. If you use a mirror they won't travel back in time, they will just change course. They will never go back in time.
> I've always wanted to know why a ball doesn't follow a beam of light if they are both following straight lines in spacetime.
Because light is much faster. Throw a ball at the speed of light and you will see exactly the same path.
Lets say you shoot rifles with different calibers. Each time the bullet takes one second to hit the ground but the faster bullets travel longer distances. Light is so fast it escapes from the planet but if the planet were big enough even light would hit the ground like a bullet.
So if light were traveling through "slow glass" where it's speed through the medium was significantly slowed down, we would see it go in a parabola like the ball?
For example, one of the first observational confirmations of relativity was being able to observe starlight near the sun during an eclipse being slightly out of place because it had been bent by the sun’s gravity.
I never studied this stuff, so the genius of your intuitive explanation is appreciated.
My intuitive response to your intuitive explanation: This ball is moving through spacetime relative to the earth, which is in turn also moving through spacetime relative to the sun - and so light is being deflected off of this ball at each point in its position in spacetime relative to the sun for much longer than light is being deflected off of this ball at each point in its position in spacetime relative to the earth - have I got that right?
Light is a hitman
Perpetual driveby shooter
Never missing
As we dance through the night
On human scales, the time dimension is much "bigger" than the space dimensions...
This is really interesting, and it made me wonder how to convert between space and time. I mean, one meter up is equivalent in magnitude to one meter forward, is equivalent to one meter to the right. Is _c_ the conversion between space and time? In other words, is 300 million meters equivalent in magnitude to one second of time?
It is bigger only because you travel slowly in the spacial dimensions. You always travel thorugh spacetime with a constant speed (the speed of light). What happens is that you're usually going with 460 m/s (as Earth revolves around the Sun) and this is not really comparable to your `t` speed in the x/y/z/t coordinate system. So when you are still your speed is something like 230/230/0/299.791.998.
It is much easier to think in two dimensions than three. If you picture the ball following a parabolic trajectory in space then you are really thinking in three dimensions here. It might be easier to think of throwing the ball straight up, after which it eventually falls straight back down. NOW plot this in two dimensions with one of the dimensions being time. As the original comment says, the scaling of the time dimension in seconds is not a good comparison to meters. Physicist like units where the speed of light is 1 (one). In these units, in our two dimensional space-time graph, the ball travels a very far distance while the change in our space dimension is still small. It is very close to straight line.
It’s also why I despise the popular portrayal of space time curvature. It looks at space in isolation rather than space time as a whole, and provides no intuition as to why objects traveling at different speeds follow different trajectories.
FWIW I think that in general it is better to just teach people that gravity is an acceleration in classical spacetime (as opposed to a force or curvature). It is simply too hard to create intuition for laymen around minkowskian spacetime, and even harder for curved minkowskian spacetime.
I think treating gravity as an acceleration is too abstract for many high-school students. For all the on-the-Earth problems, you'd have to analyze them in an accelerating reference frame. That means Newton's laws and the equations of motion don't apply. You'd have to use special linear acceleration equations. An inertial reference frame is simpler and more general. Gravity fits that OK if you consider it to be a force distributed in proportion to mass.
Think about this. You're always travelling through spacetime with a constant speed (the speed of causality == the speed of light). This is your 4 vector because it has 4 components: x, y, z, t. Therefore the faster you go in one dimension (say, x), the slower you go in the other dimensions. The dimension that's affected the most by the warping of spacetime is time (t) in most cases, because you're moving much slower in the other 3. As you go faster your path becomes less warped. This is why the faster your ball is, the less curved its path look like. This also explains why objects with zero x/y/z speed fall straight towards the object that causes the warping: only t remains from your 4 vector so you're essentially moving through spacetime with the speed of causality (light) --> you fall straight down.
So this is the core problem with all of the general relativity materials that model it as a rubber sheet causing curvature in spacetime. They always model it with focus on _spatial_ curvature: which is totally able to model an orbit or a hyperbolic trajectory as a geodesic, but it totally cannot model "throwing a ball up" since the geodesic for throwing a ball up is just a straight line.
The important thing is that gravitation is a distortion in space-_time_, which is way trickier to model as a rubber sheet because you end up with one dimension of space and one of time. If you distort _those_ (also, they don't distort quite like a ball-in-a-rubber-sheet), you can get the results of a ball being thrown up. It's also possible to visualize this for 2 spatial dimensions with a distorted 3d space, but tricky.
I'm a bit confused, how does a 2D graph with a curved X axis work?
Does this just mean to say that if we saw things distorted with an outward curve, then something that is moving in a curve would look to be moving in a straight line?
Gravity is not a force. The surface of the Earth is moving up to the object in free-fall at an acceleration of 9.8 m/s^2. The force pushing the surface, and the pressurized atmospheric shell, upward is a result of the processes occurring within the Earth (likely, in particular, those within the the core).
How is it not a force though? Regardless of curvature, a ball starts moving if you let it go without applying any force. Curvature alone can't account for that could it?
In general, it is convenient to assume that forces exist. However, the model presented here shows that you can instead dispense of the need for there to be a force and show that the apparent acceleration is just the object moving along its curved path in spacetime. You could consider it as if the curvature itself making it look like there are forces.
An object at rest is still travelling through time. In fact it continues to travel at the same speed through spacetime when you let it go. It's just that spacetime is distorted such that the straight line takes it across space too, as seen in the inertial frame on the right.
The falling object doesn't accelerate. You, standing on the ground are the one that's accelerating. You see the object as accelerating but that's an illusion due to frames of reference.
As evidence: which object feels a force on it?
You can feel the force the ground continually pushes up at you. The ground is accelerating you up. The falling object is completely idle in its inertial frame and feels nothing.
Ok so that makes sense, but if we are accelerating by standing on the ground, isn't that implying that we are continuously increasing in energy? Couldn't that be harvested for a perpetual motion machine of some sort?
First, "it all adds up to normality". Second, hm, I'm not sure. Well, the glass on the table next to me is "accelerating", but in my frame of reference (which is the same as that of the glass) it is also stationary...?
If I drop a ball above the glass, and we consider things from the frame of reference where the ball is stationary, which (neglecting air resistance) is an inertial reference frame. In this frame, the glass is accelerating ball-wards , because of the table pushing on it, and, this force is applied over a distance as the glass approaches the ball.
So, yes, it seems that the KE of the glass should be increasing in the reference frame of the ball.
Hm, I'm confused.
I suppose if we consider a geodesic representing a periodic orbit around the earth, that the positions and velocities of various objects on earth will be in a cyclic pattern?
The layman descriptions are probably incomplete or rough abstractions missing nuance. Reminds me of the idea of a rubber membrane with balls on it representing the curvature of spacetime explaining how gravity works - I've always disliked that example because you still need plain old gravity to pull the balls into the membrane.
He addresses that basic science makes this confusing because there's a curvature term that is usually left out of acceleration equations which balances out when you're accelerating along your curvature (e.g. against gravity), instead of along your spatial coordinates.
It acts like an acceleration, rather than a force.
Imagine a bowling ball and a marble at rest from your perspective, and an apparatus with a pair of pneumatic guns that eject pistons with the same precisely-calibrated amount of force. You arrange the guns so that they will hit the marble and the bowling ball at the same moment, and you trigger them together. The marble and the bowling ball are hit at the same time by the same amount of force. The marble, being much lighter takes off much faster than the bowling ball.
Now take the same marble and bowling ball to a place a mile above the moon (so that there is no confounding atmosphere to complicate things) and release them next to one another at the same moment.
If gravity is a force, then we should expect that the marble will fall much faster than the bowling ball, because the same force is acting on two different masses; the lighter mass should be accelerated more, just as it was when the source of the force was the pneumatic gun. F = ma, after all. If the force is the same and the mass is less, then the acceleration must be more.
That's not what happens, though. The marble and the bowling ball fall together at the same accelerating rate.
Gravity acts like an acceleration, not a force.
Lo these many years ago when I was an undergraduate physics student, my advisor told me that we should say "the force due to gravity", not "the force of gravity".
In every day colloquial speech it doesn't matter, of course.
Mm, just like we have wave-particle duality, one can also speak of some sort of “force-coordinate duality”, as it were. Just as we let the sine function be linear for small angles, quasiparticles with forces in Cartesian space are wonderful abstractions. Good physics – especially applied physics – is about handling this balance between precision and mathematical/computational simplicity. Mind, research physics can be the opposite; a lot of nuclear involves (often special) relativity as combined with QM. Until we have a unifird theory, however, the graviton and its force ain’t dead yet.
I'm unconvinced that treating spacetime as a quantum field is the right approach -- just a gut feeling -- given gravity "feels" so much different than the quantum theories and general relativity is just so beautifully geometric.
I'd be very sad if we had to lose that and deal with gravitons.
I'm happy they didn't, making it a great video for me as well :)
Not that I'm afraid of math, I just don't feel like I need to see equations when it's a concept that is being explained rather than how to numerically calculate something. Math helps me as much as a code implementation does to explain the concept of a variable: I can observe its behavior but it doesn't necessarily teach me the concept.
Gravitational attraction is a function of both objects, so the force between them is not a constant, same as for the electron.
Also, what if you have a charge so high that the electrical-escape velocity exceeds the speed of light? A strange kind of black hole?
I ask these things because I do not understand why gravity is called "not a force", while electromagnetism is, when I see no real difference between how particles act.
Gravitational attraction is a function of both objects, so the force between them is not a constant, same as for the electron.
The way that varies is always in proportion to the object's inertial mass though, which has the result that the path traced by an any object with the same starting position and velocity in a given gravitational field is going to be the same.
Contrast this with charge, which can vary independently of inertial mass. This has the result that the paths traced by objects with the same starting position and velocity in a given electrical field will vary.
This also means there is no fixed "electrical escape velocity". More massive objects with the same charge can escape with a lower velocity.
the link mentions that it's not proven that intertial mass and gravitational mass are the same, they are related. If gravitation mass contributes to the majority of intertial mass it would look like they're the same.
But let's just say they are the same, then gravity and "spacial" inertia are intertwined. Gravity is only special if intertia is also considered special. Inertia seems special because it's explainable in terms of motion in space. There might be other kinds of motion in other dimensions that could explain the randomness of quantum space.
The model shown is for 1D space and 1D time, and this 2D spacetime is curved, remaining 2D. Can you though show us a 3D model of curved 2D spacetime? Maybe as a rotatable, scaleable 3D scene, where more complex phenomena, like planetary motion around a star, would be possible to show?
My enlightening moment about general relativity: apples do not fall on the ground, instead, the earth is inflating, and the inflation of the earth is accelerating at 9.8 m/s^2. Eventually, the ground catches the apple.
Of course, you are going to tell me that the earth is not inflating, obviously, because it is still the same size after so many years.
But here is the trick: the earth is inflating at the same rate as spacetime contracts. If the earth didn't inflate, the contraction of spacetime would have collapsed it into a black hole.
Note: It is related to Einstein's elevator thought experiment. Here, the inflating earth replaces the rocket powered elevator.
Note 2: If the idea of an inflating earth bothers you, I suggest you start considering that the earth is flat, seriously! Flat Earthers took Einstein's thought experiment quite literally and consider the Earth to be a disk that is continuously accelerated upwards. And in fact, if free fall trajectories were parabolic, that would be the correct explanations. In reality, because the earth is not flat, free fall trajectories are elliptic, though it is only apparent on a large scale.
Yes - and it's observable in every rocket launch. When an astronaut is pushed back against their seat, they feel artificial gravity caused by the exhaust behind them inflating faster than the Earth normally does.
The thing is, I watched the French version and I didn't know an English version existed, that's why I didn't post it here, French speakers are, I believe, a minority.
And BTW, the French channel has an 8 part explanation of the maths behind general relativity that is the best I have ever seen. It is on a level above most pop science video since it actually shows the equations, tensors, etc... but the explanations are actually quite accessible.
That the big idea behind Einstein field equations: energy (and mass because E=mc2) curve spacetime and spacetime curvature affects the energy fluctuations (and the way things move).
Because both terms of the equation affect one another, solving it is complicated but here, the result is that spacetime contracts to the center of the earth.
I'm not sure how that hypothesis works out with things like black holes and gravitational lenses, friction, and a lot of other established physics. Somehow I figure people that believe in the expansion hypothesis will have some kind of workaround for those.
Some interesting alternative viewpoints I've seen:
Matter continuously destroys spacetime at its location, sucking in the fabric of the universe.
The Universe isn't expanding; matter is shrinking. Light isn't redshifted on the way to us, it's just that our sensors are getting smaller relative to the unmodified wavelength of light.
I'm curious, how does the "gravity is not a force" viewpoint relate to the hypothesized graviton particle [1]?
Are they incompatible viewpoints, or just different perspectives on the same thing? (E.g. are gravitons hypothesized to disappear depending on frame of reference?)
I'm assuming they're incompatible (that we need the theory of everything [2] to reconcile them) but would love to know if there's something I'm missing.
It is likely that both the "fabric of spacetime" idea and the Graviton are parts of the whole picture, like with particle-wave duality. In the end, particles, waves, and spacetime are really just abstractions on top of the true, possibly unknowable rules of the universe.
As you may know, one of the great problems in physics is to unite relativity and quantum mechanics. It happens to be that the graviton is a concept from QM and "gravity is not a force" is a concept that lives in the theory of relativity.
Yeah, I always found the assumption that QM was more correct than GR to be a bit odd. One of the outcomes of the geometric approach to gravity is the total lack of anti-gravity: there is simply no such thing as an anti-geodesic. The lesson of quantum field theory to me was that observables are operators on fields and not simple scalars evolving in time. General relativity is quite similar in that respect.
In quantum physics the particles are excitations of a particular field. The gravitons correspond to (excitations in) the curvature of space-time. The other force bosons correspond to (excitations in) the curvature of other fields, such as the electromagnetic field.
Interestingly the Yang-Mills equations (used in quantum mechanics) and the Einstein-Hilbert action (used in general relativity) are pretty much identical if you use general enough mathematics.
Speaking of reducing the dimensions of physical reality, a physics professor shared a very interesting video saying we should be thinking of photons and other massless objects as existing in 2d "areatime" as opposed to 3d "spacetime." I'm wondering if a revolution in simplifying the math behind physics is brewing.
Can second this. I was worried that course would be over my head since I never took physics in college, but he does a great job prepping you to understand the end of the course.
How come on half of the earth's sphere people don't get ejected out into outer space while the other half of the sphere it pushes onto the people giving them sense of acceleration?
If earth is pushing equally on all of the people around the globe how come it doesn't swell up and burst?
And how do you explain force of attraction weakening the higher in altitude you go away from earth?
450 comments
[ 3.5 ms ] story [ 372 ms ] threadSo some flat earth arguments are actually correct if general relativity is correct, namely that gravity is an illusion and that the real reason we are stuck to the earth is that the earth is accelerating toward us at 9.8 m/s^2
I don't really understand why, that was just the explanation Derek gave.
The acceleration an observer sees you undergoing is the same as the inherent "proper" acceleration you're undergoing, minus the acceleration of their coordinate frame with respect to yours. For me to stay still with respect to you, if you're in a frame that is accelerating away from me, I need some proper acceleration to catch up and counteract the fact that our frames are diverging. But if spacetime is curved, your frame probably is accelerating relative to mine - c.f. the example on the Earth's surface, where our frames inexorably accelerate towards each other as we move parallel to each other. So for me to stay still with respect to you, I need to have some proper acceleration to balance out the coordinate acceleration derived from the fact that our frames are moving in a curved space.
If the object's velocity isn't enough to traverse the curved space-time, it will move toward the center of the mass generating the distortion and fall out of the sky.
If the object is traveling quickly enough, it can continue traversing the distorted space-time and orbit that mass.
If the object is traveling even more quickly, it will traverse the distorted space-time and continue on without orbiting the mass.
In all three cases, from the perspective of the object traversing the distorted space-time, it continues to travel in a straight line, as it's the space-time that's distorted.
A (flawed) analogy would be riding a bicycle between the peaks of two identically sized hills. Starting at the top of the first hill, you coast down increasing your velocity.
Once you reach the bottom of the first hill and head up the second, your velocity decreases.
If your velocity at the bottom of the first hill is too small, you'll go up the second hill and as your velocity reaches zero, you roll back toward the bottom of the hill.
You will pick up velocity and then roll back up the first hill, then down again, then back up the second, etc. until you end up stopped at the bottom of the hill. This is akin to falling to the center of the distorting mass.
If your velocity is high enough to carry you back up to the top of the second hill and then stop, you'll roll back down and get to the bottom with the same velocity you had coming down the first hill. You'll then oscillate between the tops of both hills. This is akin to orbiting the mass.
If your velocity at the bottom of the first hill is enough to carry you past the top of the second hill, you'll just keep going after reaching the top of the second hill. That's akin to flying by the mass.
It's a flawed analogy, because in a curved space-time the directional portion of the motion vector doesn't change.
As John Wheeler[0] simplified it: "Mass tells space-time how to curve, and space-time tells mass how to move."
[0] https://en.wikipedia.org/wiki/John_Archibald_Wheeler
[1] https://en.wikipedia.org/wiki/General_relativity
So we're curving in towards the center of the mass of the Earth, but the reason we don't end up in the core of the Earth is because the surface stops us. The Earth is "pushing" us away from the center, and that's the acceleration. It's accelerating you off your straight line path, and this is the deviation from the geodesic.
What's confusing me here is the notion that when two objects collide, they accelerate into each other. Why and how is force constantly applied after the collision? My intuition is falling down here, and none of the resources I've looked at so far have explained why the acceleration happens.
Remember that spacetime = space + time dimensions. The object is always travelling through time, and the curvature of spacetime is converting some of that speed through time into speed through space. That's what you perceive as motion (caused by gravity).
Time and space are linked together. The faster you go through space, the "slower" you go through time (as in you experience it slower). This is very measurable and even used to alter timings for satellites GPS readings. You can take an atomic clock on a plane and age slower than someone who just stayed on the ground.
So the spacetime curvature is continuously converting some of your temporal motion into spatial motion, until that's stopped by the surface of the Earth which is constantly "accelerating" to stop you from going further.
As to why we always move through time, that's beyond my understanding at this point but it's a fundamental axiom of physics.
Your explanation makes sense, and it sounds like I'd need to understand the maths behind relativity to be able to really understand how objects behave in spacetime.
If the thing you're measuring position against is also accelerating, then you need to apply some acceleration of your own to stay still with respect to it.
The terms you want to look up are "proper acceleration" and "coordinate acceleration". The curvature of spacetime means the thing I'm measuring position against is moving relative to me (c.f. the example of two people walking in parallel across the Earth, nevertheless eventually meeting: the curvature means that even though neither of them is measuring an acceleration, nevertheless they are accelerating towards each other), so I need to have some internal ("proper") acceleration of my own to counteract the fact that our geodesics are moving away from each other.
The surface of the Earth keeps you from actually falling in, and is therefore pushing you away or upwards from the center. This is the acceleration acting on your straight line path through curved spacetime. This is the deviation from your geodesic.
By definition, the world line of a stable orbit would be a line in curved spacetime.
I tried to explain it with words, but I guess the images are worth more than I could write..
Likewise I guess for the 1D gravity well case, where a sine wave would become a straight line in spacetime.
As a brief example, consider two objects in downwards free fall toward the centre of some massive object. Since they head towards the centre, in a free falling frame the two objects actually get closer to each other until they collide as they reach the centre.
This is known as the tidal effect of gravity and is the actual physical content of general relativity. This effect can be shown to be obtained by an appropriate curvature of spacetime which itself can be shown to be related to the stress-energy of matter inhabiting spacetime.
I always thought that was a nice way to drop one dimension down to get the intuition. To the metaphorical 2D ant they see two friends attracted to/falling towards each other, but they are going in a straight line on a curved surface and there are no forces at play.
[1] https://en.wikipedia.org/wiki/Geodesic
You can call both a force in the generic dictionary definition of force. Because obviously it's crazy that something can be so powerful as to distort spacetime itself. But gravity wouldn't be a force in the Newton sense of being something that affects the acceleration of an object.
But some of these forces are not observed by other people who are watching; they arise as by-products of the fact that you are the one doing the measuring. For example, if you're falling in a lift, you won't observe the force that causes a ball to fall towards your feet; but I, standing on the ground outside the lift, will observe that actually you and the ball and the lift are subject to this force.
In fact there's an underlying reality which we can't directly observe but which we can infer from the paths of objects. That reality is a curved spacetime, not a flat one (as it appears to be). This curvature means that "straight line" is actually not quite what you're used to; things follow straight lines, but those straight lines don't look straight to us, because of the underlying space's curvature: we can't see the whole of spacetime, only small segments of it, so we can't see enough to get a proper sense of the curvature. But since our limited frames of reference have their own notion of "straight line" which is incompatible with that of the global spacetime, we observe a mysterious tendency of things to deviate from what we think is the straight line they should be following.
So gravity "is a force": it's a mysterious tendency of things to move, because we are limited in what we can see and our own observation frames are subtly incompatible with the global structure. But it's also "not a force": if we were somehow able to take a fully global view of the universe, there would be no mysterious movement, only a huge number of things moving at exactly the same speed in perfect straight lines through a curved spacetime. (Assuming General Relativity is 100% accurate.)
Yes. But the effect of the distortion of space-time that we call "gravity" is subject to the inverse square law[0].
This means that, as Newton described: "The gravitational attraction force between two point masses is directly proportional to the product of their masses and inversely proportional to the square of their separation distance. The force is always attractive and acts along the line joining them"
As such, while objects are affected by the distortion of space-time, the effect is diminished (but not eliminated) by increased distance from the mass that's distorting space-time.
Or at least that's how I understand it.
[0] https://en.wikipedia.org/wiki/Inverse-square_law
So... you can't get away from it. That's what I said isn't it?
At some point, the effect is so small that it either can't be measured or even if it can, the effect is so small that any impact is irrelevant in practical terms.
If that's the case, you're effectively "getting away" from it.
I guess it's a matter of perspective.
But a person’s experience should always be the same, no matter what reference they are in. They will perceive time as passing at the same rate in all environments, but it will be different from people in other environments.
Caveat: IANA physicist, and this is not physics advice.
Instead I think what is happening is that massive objects actually stretch the fabric of spacetime somehow so that the closer you are to the object, the slower you travel through both space and time. And the more massive the object, the more stretched space and time become as you get closer to it.
Hence if you go near a very massive object, from an outside observer it looks like you are frozen in time because time is so stretched it takes forever for you to move through it.
As you mentioned though, from any given frame of reference time will always feel the same. 1 second will always feel like 1 second.
No. You are always traveling at a constant velocity through spacetime.
[0] Spacetime: https://en.wikipedia.org/wiki/Spacetime
Not "very". Look up the actual equation, it's a really easy one. AFAIR for a standard stellar black hole, you'd need to be within meters from the event horizon to get any substantial time difference.
Thus the implication that Interstellar's Gargantua was an SMBH, i.e. probably the center of it's galaxy.
You would age about the same as you would in microgravity. You could get closer and closer approximations by going into Earth orbit, solar orbit, and galactic orbit. Each approximation has less gravity, so time would pass slightly slower relative to an Earthly observer at each step, but the effect diminishes.
Could this be named more correctly: Gravity is not a force – free-fall hyperbolas are straight lines in spacetime (timhutton.github.io) ?
- parabola if you assume flat&infinite ground,
- ellipsis if you assume a spherical planet (an ellipsis is crossing the planet's surface).
This is Newton, not GR.
I'll answer this question as I understand it, but I only took four lectures of General Relativity before I gave it up in favour of computability and logic, so if there is a more intuitive and/or less wrong answer out there, please correct me.
Intuitive answer: the curve is indeed very gentle, and (e.g.) light will be deflected only very slightly by the curvature; but the ball is moving for a couple of seconds, and that's an eternity. On human scales, the time dimension is much "bigger" than the space dimensions (we're quite big in the time dimension and quite small in the spatial dimensions); the ball moves only a small distance through space but a very large distance through time, amounting to a big distance in spacetime, and so the slight curvature has a bigger effect than you might expect.
I think what you may be noticing is that, as you reduce the tick frequency of a newtonian physics simulation, parabolas become less accurate as integration error accumulates.
"A Slower Speed of Light" http://gamelab.mit.edu/games/a-slower-speed-of-light/
"Velocity Raptor" https://testtubegames.com/velocityraptor.html
Air resistance, wind, and horizonal acceleration. Over long vertical distances, these perturbations in the x-axis cause an arc. Nothing to do with general relativity.
Alternatively, you can think of everything that could impact you as something of a now light cone. The second view has the universe existing as a 3D surface in 4D space time which means objects have a temporal width for each observer. That can be a really useful mathematical model.
PS: Edited the above several times for clarity.
That's just another way of describing the total time you exist.
—Kurt Vonnegut, Slaughterhouse-Five
299,792,458m is for the distance traveled by light in one second. The -1m is for the distance between your feet.
PBS spacetime did an episode about this recently.
The best way to picture it is via Minkowski diagrams, you can find some neat visualizations in the wikipedia article: https://en.wikipedia.org/wiki/Relativity_of_simultaneity
Simply picture an extended object in the x direction, Lorentz transformations are going to rotate it slightly in the time direction causing it to become extended in the time direction.
[1] If you take it to the extreme you may consider that different parts of your body move with respect to each other and thus they are not in the same reference frame. In which case, not even with respect to yourself are totally "flat" in the time direction.
So you can change your acceleration through the time dimension of spacetime, by dint of changing your acceleration in the spatial ones.
I mean the whole way looking at it seems wrong to me. There is no "rest of the world" in relativity. Assigning some objective vector to everybody doesn't work. These only make sense from some specific point of view.
By the question you asked now I'm assuming, you meant "but hey, without GR..", but even without GR, ignoring that the universe is expanding, assuming flat space time etc. If the universe consisted just of 3 bodies, 1 being you and 2 being rest of the world, then the way of thinking you described still doesn't make sense in context of relativity and may lead to some confusion (apart from it being, to me, incoherent in context of special relativity).
But maybe I'm missing something from your picture, I'm happy to read and learn.
GR I'm less familiar with.
You can certainly imagine a scenario where two objects measure their mutual distance as being constant in time. You can also imagine other scenarios, but I'm asking to imagine the scenario where two objects do measure their distance to be D and then measure it again and it's still D, and measure it again and again and it's still D. That's just the definition of standing still with respect of each other, and when you plot a space time diagram, their space curve is parallel (because their spacial distance doesn't change).
Special relativity doesn't make that scenario impossible. It doesn't force things to move. It just describes what happens when things do move (through spacetime).
This is also why you can't travel backwards in time through just acceleration. There is no way to impart a force perpendicular to your velocity vector when it is already perpendicular to time, giving you no way to rotate the vector to have a component that points backward in time.
So I've always wondered, whether general relativity allows for forces parallel to time, and we just don't know of any mechanism to actually do so, or if it does not cover such cases because we have no mechanism, or if it disallows it entirely.
The word “dimension” in this context is overloaded. We think of space being three dimensions but really it’s only one - velocity relative to a specific reference frame. Thinking of it this way, the word “spacetime” makes sense; it’s a two-dimensional system: “spatial velocity” (S) on one axis and “temporal velocity” (T) on another. Both velocities are always measured against a reference frame, and their sum is c (c=S+T).
This would mean that time travel is impossible not because of a “speed limit”, but because c is a dimensionless physical constant.
I suspect that the answer is that this is a false comparison.
> under the laws of gravity, a parabola is an impossible shape for an object that's gravitationally bound to the Earth. The math simply doesn't work out. If we could design a precise enough experiment, we'd measure that projectiles on Earth make tiny deviations from the predicted parabolic path we all derived in class: microscopic on the scale of a human, but still significant. Instead, objects thrown on Earth trace out an elliptical orbit similar to the Moon.
[1] https://www.forbes.com/sites/startswithabang/2020/03/12/we-a...
Vi Hart has a nice explanation: https://www.youtube.com/watch?v=v-pyuaThp-c
So if you want to be really pedantic, it's never an ellipse because the Earth is not a point mass. It would be equivalent to a point mass if the Earth were a perfect sphere of uniform density, but it isn't. In reality it's a potato like mass blob that's approximated by what geodesists call the "geoid". So in order of approximations the path of a ball thrown on earth is a parabola -> ellipse -> numerical integration of 6-dof initial conditions and spherical harmonics approximation of the earth gravity field.
At least classically, a sphere is indistinguishable (gravitationally) from a point mass while you're outside it. The earth is pretty sphere-ish, locally speaking.
Kinda. There are mountains, they have their own gravitational attraction, and it can be measured... even in the 18th century! https://en.m.wikipedia.org/wiki/Schiehallion_experiment
It all depends on how pedantic one wants to be :)
The rest get so perturbed by gravitational anomalies that they fall out of orbit after a few months or years--faster than low Earth orbit where there is atmospheric drag!
https://www.lpi.usra.edu/lunar/documents/NTRS/collection2/NA...
So, moral of the story - we have to be not too pedantic.
I'm not a physicist, but that doesn't sound right. One example is that if you are inside of the sphere, at least some of the mass will be pulling you away from the point at its center. I think what you mean is that a point mass closely approximates a spherical mass, but the degree to which that is true becomes less and less the closer you get to the center. I don't think you have successfully contradicted what the person you responded to said.
Proving this is a classic problem in undergraduate physics.
Or equivalently, if you are inside a shell (sphere outside, hollow sphere inside) then the gravitational effect is zero. This can be explained by analogy with light which also follows the inverse square law--changing your position inside a hollow sphere does not change the fact that you see the sphere all around you. Same reason that there is no electrical field inside a hollow conductive object.
You can use this law to see what's the gravity fields as you move under ground and in other very symmetrical cases.
It doesn't break it at all. The meter above you can be treated as a hollow shell, which surprisingly has zero net pull, and the solid sphere below can be treated as a point mass just as before.
Just remember these two facts, each provable with a simple integral calculation, usually done in high school physics or freshman college physics: a uniform sphere has the same gravitational pull on an object as a point mass at it's center, and the net gravitational pull on an object inside a spherical shell is zero.
This all works under perfect spheres, uniform (at the spherical shell level at least) density... There are other cases it works, but this simple case is the basic idea.
For each point of any given distance, you can find a disk of points at the plane of the same distance whose gravitational pull will be equivalent to a single point at their center.
You do this for all distances x, and you will find an equivalent rod going from the attracted objected to the center of the sphere, of a non-uniform but symmetrical density.
We find that the density is linearly correlated to the area of the corresponding section, and we then take the closer half of the rod, multiply the density by 1/r^2, and find that it is constant!
For the far half of the rod, we cannot do this, so instead we divide it into two halves of equal pull, then divide those to halves and so on, and find that it this reduces to the equivalent of a point.
Now that we have two points in-line with the object, we find the point with the same total mass that exerts and equivalent force, and lo and behold it is at the center.
I'm sure there is a much simpler way to intuit it, though :)
If the mass is spherically symmetric, this will not be the case; all of the Newtonian forces from the masses further away from the center than you are will cancel out. This is called the "shell theorem", and it turns out to hold even in General Relativity.
A solid ball (filled sphere) is just a union of many shells (hollow spheres), so the theorem still applies.
en.wikipedia.org/wiki/Shell_theorem
No, it's stronger than that. It says that any spherically symmetric distribution of matter outside a certain radius exerts no "gravitational force" on anything inside that radius, whether there is matter inside that radius or not.
Then you are not inside the shell, you are on it. The shell theorem only applies if you are inside the shell.
> Would I not feel infinite acceleration towards the massive point on the shell that I was infinitely close to?
In General Relativity, matter is not viewed as point particles. It is viewed as a continuum, described by the stress-energy tensor. This is one of those cases where the difference shows up.
But as you get close to stuff, say two atoms, the electrostatic and other forces are far greater than the gravitational ones.
Even if you consider the point mass to move because of the mutual attraction?
Assuming a uniform+spherical earth. And also a uniform+spherical ball I supposed.
In relativity, there is no such thing as a "uniform gravity field", if by that you mean a field where the "acceleration due to gravity" is the same everywhere. The closest you can come is the "gravity field" inside a rocket accelerating in a straight line in empty space, where the acceleration felt by the crew is constant. That kind of "gravity field" has an "acceleration due to gravity" that decreases linearly with height.
Look up the Bell Spaceship Paradox. In relativity, two spatially separated objects (or two ends of a spatially extended single object) that have the same proper acceleration in the same direction do not stay at rest relative to each other; they move apart, as seen in each of their own frames. This is different from the behavior predicted by Newtonian mechanics. In order to have the two objects (or two ends of a single extended object) remain at rest relative to each other, the one in front has to have a smaller proper acceleration than the one in back. (Rindler coordinates are often used to describe this case.)
Yes, it is certainly not intuitive. That's why Bell took the time to write a paper about it years ago.
This article might help:
https://www.physicsforums.com/insights/what-is-the-bell-spac...
This turns out to be not pedantic but very important if you're guiding an ICBM. And when landing on the Moon, Apollo had to deal with irregularities in the Moon's gravity due to mass concentrations, called mascons.
(If you're interested in missile guidance, take a look at the book Inventing Accuracy. Among other things, it discusses some of the efforts to map the Earth's gravity field to increase missile accuracy for Trident and Minuteman missiles. I knew a physicist who worked on this.)
It wouldn't be an ellipse even if Earth were a point mass. The gravity of the Moon and Sun, the gravitational lumpiness in the sky, has the same effect as the gravitational lumpiness underground. The combined result may be no closer to an ellipse than it is to a parabola.
Source is a table in the first chapters of Fundamentals of Astrodynamics.
It's not a force. From the right frame of mind[0], it's just a straight line.
[0] Substituting for frame of reference.
ICBMs were never used to toss nukes at people.
Pretty sure their existence saved the world from WW3.
> Surely there's something else we could be doing instead?
The application most useful in everyway life is GPS.
Would it? Even of uniform density, the mass would not the at the core, only the center of mass would be. Most of the mass is actually closer to the surface. I'm no physicist, but I'd imagine that would have quite some impact[0] on any trajectory that crosses the surface.
And for any ball thrown at human speeds, I'd expect Earth's gravity would be much closer to a uniform gravitational field than one from a point mass.
[0] Pun not originally intended but I'm quite happy with it now.
Based on what I've seen over the years I've been around here, I'm quite certain that's a primary motivator for a non-trivial number of commenters here.
But because massive objects also cannot move as fast as photons, we're probably both saying the same thing from two different perspectives.
A more energetic particle of the same mass will have a very slightly higher gravitational attraction.
[0] Photons bend spacetime according to GR, although as far as I know this has not been proven experimentally. We'll probably need a theory of quantum gravity to be sure.
But even more important, if light beams are reversible under relativity (reflected off a mirror they will backtrack the same path) then light can not enter a black hole because its reversed path could allow it a way out. But then there's that whole thing of objects falling in appear to slow and stop as they approach the event horizon, so maybe light doesnt enter after all.
My conclusion is that you cant really understand it without serious study of under someone who already gets it.
The line is only straight in spacetime, not in space.
* photons
* spherical, massless cows in a vacuum
Using quotes since terms are more metaphorical than exact.
The faster you go, the less you travel through time. Thus, if the ball were travelling at the speed of light, it would not travel through time either and would follow the same path as light.
This is a common pop science statement, but it's not correct. A correct statement is that the concept of "speed through spacetime", which is what has to be split into "speed through space" and "speed through time" in the pop science statement, does not apply to a light ray.
In more technical language, the tangent vector to the light ray's worldline is not a unit vector, it's a null vector, and the concept of "speed through spacetime" only makes sense for a worldline whose tangent vector is a unit vector.
No, this is not correct. The correct statement is that the concept of "elapsed time" does not apply to a photon; it only applies to timelike worldlines, not null worldlines.
To put it another way, if your statement were true, it would mean that the origin and destination events were the same point in spacetime. But they're not; they're distinct points in spacetime. Which means that, since the spacetime interval along the worldline is zero, you can't use the interval to distinguish points on the photon's worldline. And the concept of "elapsed time" requires that you be able to do that. So the concept of "elapsed time" can't be used for a photon.
Let's say I put you in a spaceship and accelerate you to 50% the speed of light toward the sun. From an inertial viewer's perspective you are travelling toward the sun at half the speed of light and it takes you ~16 minutes to crash into the sun. But from your perspective it only took ~14 minutes to crash into the sun[0].
Repeat the experiment except I accelerate you to .99c. From an inertial viewer's perspective you are travelling toward the sun at nearly speed of light and it takes you ~8 minutes to crash into the sun. But from your perspective it only took ~1 minute to crash into the sun.
Repeat the experiment except I accelerate you to .999c. From an inertial viewer's perspective you are travelling toward the sun at nearly speed of light and it takes you ~8 minutes to crash into the sun. But from your perspective it only took 20 seconds to crash into the sun.
Repeat the experiment except I accelerate you to .9999c. From an inertial viewer's perspective you are travelling toward the sun at nearly speed of light and it takes you ~8 minutes to crash into the sun. But from your perspective it only took 6 seconds to crash into the sun.
Repeat the experiment except I accelerate you to .99999c. From an inertial viewer's perspective you are travelling toward the sun at nearly speed of light and it takes you ~8 minutes to crash into the sun. But from your perspective it only took 2 seconds to crash into the sun.
See what's happening? As you approach the speed of light, the amount of time that elapses until you reach your destination approaches zero. So from an inertial observer's point of view, time has completely frozen for travelers approaching light speed.
[0] Using time dilation formula from this page: https://www.phy.olemiss.edu/HEP/QuarkNet/time.html
Yes, but you cannot extrapolate from this to say that the time lapse for a photon would be zero. A photon is not the limit of objects with mass going closer and closer to the speed of light, because "closer and closer to the speed of light" is frame-dependent, but a photon's speed being c is not. I can find an inertial frame in which each of your objects is at rest, and in that frame, you are the one who is "close to c" (in the opposite direction). But that doesn't mean your elapsed time approaches zero. By contrast, it is impossible to find any frame in which a photon is at rest. The two types of objects are fundamentally different.
In more technical language, the action of Lorentz transformations on photons is fundamentally different from their action on timelike objects. So it is simply not valid to view photons as some sort of limit "as speed approaches c" of timelike objects.
I'm thinking of solar neutrinos which, for a while, we weren't sure if they were massless or not. We had to observe them experiencing a duration of time to conclude they were massive. If we didn't find that, maybe it was just an even shorter duration, not the absence of one and we would never be able to tell the difference.
Are you asking if there is a way to distinguish a timelike object from a lightlike object? Of course there is. The fact that, for something that has a very, very small invariant mass, it might be practically difficult does not change the fundamental principle.
Also note that the reason it was difficult, for example, to tell whether neutrinos have mass or not is that we can't just do the obvious and straightforward thing and find an inertial frame in which they are at rest (by, for example, taking a rocket and accelerating it in the direction of a neutrino to see if we can bring it to rest relative to the rocket). So we have to resort to indirect methods. But, again, that's a practical limitation that doesn't change the fundamental principle.
For neutrinos, even if we accelerated an rocket and somehow checked if a neutrino was at rest relative to it, we might find that it's not. That means we won't know if we need more speed or if it's impossible. I suppose it's a bit easier than that because we only have to accelerate the rocket fast enough that the neutrino's speed becomes measurably less than c, rather than 0. But still, what if we can't even get it to go fast enough for that? No way to prove that it's travelling at c, it seems.
I'd like to add that even photons have a nonzero upper bound to their possible rest mass. At least they used to. Is there any way, in principle, to show that it's exactly zero, and thus falls into this distinct category?
> I'd like to add that even photons have a nonzero upper bound to their possible rest mass. At least they used to. Not sure what you're talking about, their momentum? No object with mass can reach the speed of light and we know they're travelling at that exact speed.
We don't, strictly speaking. The measurements you refer to aren't even measuring the speed of photons. They're measuring their rest mass.
> physicists would occasionally measure a new maximum possible rest-mass for photons. It would be very tiny, of course, but they couldn't say it's exactly zero.
Based on just those measurements, no. The most they can say is that the photon rest mass is zero to within some error bar, and the size of the error bar keeps getting smaller. (The current error bar, IIRC, is 10^-52 grams, or about 24 orders of magnitude smaller than the electron mass.)
However, we have a ton of indirect evidence that photons are massless; the most extensive body of such evidence is all the evidence for the gauge invariance of electromagnetism. If photons had a nonzero rest mass, that would break electromagnetic gauge invariance. So photons having a nonzero rest mass would be a huge issue for our current theories, in the way that neutrinos having a nonzero rest mass would not; there is no important symmetry coresponding to electromagnetic gauge invariance that is broken by neutrinos having a nonzero rest mass.
If you try what I described with a light ray, it will be moving away from you at c no matter how much you accelerate in its direction.
If you try it with a massive object, even a neutrino with a very, very tiny invariant mass, that will not be the case; its speed relative to you will decrease as you accelerate after it, eventually to zero.
There is no continuum between those two possibilities; they are distinct and discrete. The only continuum is in the latter case, where the final speed of the object relative to you will depend continuously on how long you accelerate.
> even if we accelerated an rocket and somehow checked if a neutrino was at rest relative to it, we might find that it's not. That means we won't know if we need more speed or if it's impossible
Yes, you will know, because you will know if the neutrino's speed relative to you has decreased or not. If it has, it's possible to bring it to rest relative to you. If it hasn't, it's not. See above.
> I suppose it's a bit easier than that because we only have to accelerate the rocket fast enough that the neutrino's speed becomes measurably less than c, rather than 0.
Exactly.
> But still, what if we can't even get it to go fast enough for that?
That's basically the position we are in now: we have no way of building a rocket or other device that can accelerate after a neutrino long enough to tell whether its speed relative to the rocket is measurably decreasing. So we have to resort to indirect measurements. But as I said before, that doesn't change the principle.
> even photons have a nonzero upper bound to their possible rest mass
Yes, because, as I said, practically speaking we can't run the obvious and straightforward experiment I described, to confirm that a photon moves away from you at c no matter how much you accelerate after it. So we have to resort to indirect measurements, like trying to measure its invariant mass by other means. But that doesn't change the principle.
I don't think this quite works because of relativistic addition of velocities. Naively, it seems to. For example, if the object was travelling at 0.999999c relative to you (appears to be 1.0c according to your limited instruments), then you accelerate to 0.50c in its direction, you'd see its speed reduce to 0.50c (same 2s.f. instrument), which would clearly prove it's not massless. But velocities don't add like that relativistically and I think you'd still see it as travelling at 1.0c because it only decreased a tiny amount, below what you instrument can detect. If you use a more precise instrument or a faster rocket, you might measure it as 1.00000c but then you still won't know if it's exactly c or a smidgen less.
Maybe I've got my relativistic velocity addition wrong? But it still looks like the same measurement problem as trying to prove a classical object has a speed of exactly 0, which can't be done no matter how accurate our instruments are.
Not indefinitely. Sure, if you accelerated for a short enough time in the direction of the neutrino, you might still be within your measurement error and so not have learned anything. But that just means you need to accelerate for a longer time. For any given measurement accuracy, you will be able to calculate how long you need to accelerate, by your clock, to definitely distinguish the two cases. Relativistic velocity addition does not change that fact. All it changes is the details of that calculation; yes, for a given measurement accuracy, you need to accelerate for a longer time, by your clock, to definitely distinguish the cases than you would if velocity addition were linear. But that doesn't mean relativistic velocity addition makes it impossible to distinguish the cases at all, ever. It doesn't.
For the specific case you give, that depends on what accuracy you are assuming. The relativistic velocity addition would be:
v_new = (0.999999 - 0.5) / (1 - 0.999999 * 0.5) = 0.999997.
So if your accuracy is, say, 1 part in 100,000, you wouldn't be able to see the difference. But with an accuracy of 1 part in 500,000, you would, even though you wouldn't have been able to see the difference before the acceleration.
Also, suppose you accelerated for a second increment equal to the first; you would get
v_new = (0.999997 - 0.5) / (1 - 0.999999 * 0.5) = 0.999991.
And one more increment:
v_new = (0.999991 - 0.5) / (1 - 0.999999 * 0.5) = 0.999973.
As you can see, the differences in velocity grow fairly quickly for each equal increment of acceleration; the growth is not at all linear. And, as I said in my other post just now, for any given measurement accuracy, it would be simple to calculate how much acceleration you would need to be able to distinguish moving at exactly c from moving at 0.999999c (or any other speed less than c that you choose) to that accuracy.
I don't believe that, and have never heard it before. There are many ways in which light actually behaves just like particles with mass traveling at speed c. It has to or conservation of momentum is violated.
https://www.youtube.com/watch?v=5ELA3ReWQJY
Show me an actual textbook or peer-reviewed paper Tyson has written where he makes this claim. Pop science videos don't count. (Tyson is by no means the only one; Brian Greene is notorious for the same thing.)
You won't be able to because there aren't any. No scientist who talks about a photon "experiencing zero time" in informal contexts will try it in a textbook or paper. That's because they know that if they did, other scientists would call them out on it, so they confine such claims to contexts where there are no other experts so there's nobody to call bullshit.
Another point is that if this concept were actually scientifically useful, somebody would be using it in a textbook or peer-reviewed paper. The fact that nobody is is a huge clue that the concept is not scientifically useful. It's only useful for selling pop science books or getting views of pop science videos, where, again, there are no other experts around.
Photons put it all into the X, Y, and Z components, leaving nothing for the t component. They experience a change of position in space, but not in time. What's so hard to grasp about this?
Another point is that if this concept were actually scientifically useful, somebody would be using it in a textbook or peer-reviewed paper.
Seems that a fellow named Maxwell got a lot of mileage out of the concept, even if he didn't know what was really going on.
No, Tyson's claim is not "basic theory at the high school level". It is a particular interpretation of a theory (Special Relativity) that does not work, for the reasons I gave.
> Photons put it all into the X, Y, and Z components, leaving nothing for the t component.
Wrong. The spacetime vector that describes a photon's trajectory does have a t component.
> Seems that a fellow named Maxwell got a lot of mileage out of the concept
Dude, if you think the concept Tyson described is the same as the concept that Maxwell got a lot of mileage out of, then you are the one who needs to learn more about "how it works".
Sounds interesting. Where can I read more about this t component?
However, this comment has significantly raised my estimate of the probability that you are just trolling.
So you are completely incorrect insofar as what you are saying is physically nonsense.
If you and others in the thread feel that these popular authorities are spreading misinformation or using inappropriate analogies, doesn't it behoove you to raise an objection with them directly? Or perhaps with the appropriate faculty committees at their institutions?
If they want to make money by getting "the unwashed laity" to buy their books, why should I stop them? I simply don't buy them myself. If other people want to get told comforting nonsense, that's their problem. Caveat lector.
> Or perhaps with the appropriate faculty committees at their institutions?
Which would be pointless and absurd, since, as I have already said, the claims in question are not being made in textbooks and peer-reviewed papers.
I suspect you are trolling.
You can win an internet argument anywhere, but if that's all you want to achieve here, then maybe stop trying to debate physics.
Because they're following different straight lines in spacetime. Roughly speaking, if you pick a point in space and a particular direction in space from that point, there is a continuous infinity of possible straight lines in spacetime that point in that direction in space. One endpoint of that continuous infinity is the worldline of a light ray. The ball's path is somewhere in the middle of that continuous infinity.
From the reference of the outside observer the light emitted at the event horizon will forever try to leave it, but as spacetime itself casdades into the hole at the speed of light (at the horizon) this light will get redshifted until you can't see it anymore. This doesn't mean that the light you shoot into the hole never reaches the singularity. It just means that the light emitted at the event horizon will struggle forever to get out of the insane warp. Think about this: if you're walking on a conveyor belt with a constant speed `n` in the opposite direction and the belt itself is moving with a constant speed `n` you'll never make progress. This is what happens at the event horizon.
Light beams are not reversible. If you use a mirror they won't travel back in time, they will just change course. They will never go back in time.
Because light is much faster. Throw a ball at the speed of light and you will see exactly the same path.
Lets say you shoot rifles with different calibers. Each time the bullet takes one second to hit the ground but the faster bullets travel longer distances. Light is so fast it escapes from the planet but if the planet were big enough even light would hit the ground like a bullet.
That is a valid path for the light to follow, hypothetically.
But if that's what the light was doing, you wouldn't see it. You can only see things when light enters your eyes.
My intuitive response to your intuitive explanation: This ball is moving through spacetime relative to the earth, which is in turn also moving through spacetime relative to the sun - and so light is being deflected off of this ball at each point in its position in spacetime relative to the sun for much longer than light is being deflected off of this ball at each point in its position in spacetime relative to the earth - have I got that right?
Light is a hitman Perpetual driveby shooter Never missing As we dance through the night
Probably not, because "the sun" appears in your response. If you're watching a ball move on Earth, the sun is an irrelevant variable.
This is really interesting, and it made me wonder how to convert between space and time. I mean, one meter up is equivalent in magnitude to one meter forward, is equivalent to one meter to the right. Is _c_ the conversion between space and time? In other words, is 300 million meters equivalent in magnitude to one second of time?
Your answer is basically the one given in an early chapter of Misner, Thorne, and Wheeler, which is one of the classic General Relativity textbooks.
It’s also why I despise the popular portrayal of space time curvature. It looks at space in isolation rather than space time as a whole, and provides no intuition as to why objects traveling at different speeds follow different trajectories.
FWIW I think that in general it is better to just teach people that gravity is an acceleration in classical spacetime (as opposed to a force or curvature). It is simply too hard to create intuition for laymen around minkowskian spacetime, and even harder for curved minkowskian spacetime.
The important thing is that gravitation is a distortion in space-_time_, which is way trickier to model as a rubber sheet because you end up with one dimension of space and one of time. If you distort _those_ (also, they don't distort quite like a ball-in-a-rubber-sheet), you can get the results of a ball being thrown up. It's also possible to visualize this for 2 spatial dimensions with a distorted 3d space, but tricky.
Does this just mean to say that if we saw things distorted with an outward curve, then something that is moving in a curve would look to be moving in a straight line?
And what's the point of such an observation?
[0] https://www.youtube.com/watch?v=XRr1kaXKBsU&t=504 (Veritasium on Youtube: Why Gravity is NOT a Force @ 8:24)
https://youtu.be/XRr1kaXKBsU
As evidence: which object feels a force on it?
You can feel the force the ground continually pushes up at you. The ground is accelerating you up. The falling object is completely idle in its inertial frame and feels nothing.
If I drop a ball above the glass, and we consider things from the frame of reference where the ball is stationary, which (neglecting air resistance) is an inertial reference frame. In this frame, the glass is accelerating ball-wards , because of the table pushing on it, and, this force is applied over a distance as the glass approaches the ball. So, yes, it seems that the KE of the glass should be increasing in the reference frame of the ball.
Hm, I'm confused.
I suppose if we consider a geodesic representing a periodic orbit around the earth, that the positions and velocities of various objects on earth will be in a cyclic pattern?
See Veritasium's video at 10m17s:
https://youtu.be/XRr1kaXKBsU?t=617
He addresses that basic science makes this confusing because there's a curvature term that is usually left out of acceleration equations which balances out when you're accelerating along your curvature (e.g. against gravity), instead of along your spatial coordinates.
Imagine a bowling ball and a marble at rest from your perspective, and an apparatus with a pair of pneumatic guns that eject pistons with the same precisely-calibrated amount of force. You arrange the guns so that they will hit the marble and the bowling ball at the same moment, and you trigger them together. The marble and the bowling ball are hit at the same time by the same amount of force. The marble, being much lighter takes off much faster than the bowling ball.
Now take the same marble and bowling ball to a place a mile above the moon (so that there is no confounding atmosphere to complicate things) and release them next to one another at the same moment.
If gravity is a force, then we should expect that the marble will fall much faster than the bowling ball, because the same force is acting on two different masses; the lighter mass should be accelerated more, just as it was when the source of the force was the pneumatic gun. F = ma, after all. If the force is the same and the mass is less, then the acceleration must be more.
That's not what happens, though. The marble and the bowling ball fall together at the same accelerating rate.
Gravity acts like an acceleration, not a force.
Lo these many years ago when I was an undergraduate physics student, my advisor told me that we should say "the force due to gravity", not "the force of gravity".
In every day colloquial speech it doesn't matter, of course.
An artifact of rotating coordinate frames in one, the curvature of spacetime in the other
https://www.diffen.com/difference/Centrifugal_Force_vs_Centr...
Gravity arises similarly, by being in a non inertial frame induced by mass energy
https://en.m.wikipedia.org/wiki/Fictitious_force
This was the key insight behind Einstein attempting to get general relativity working.
I'd be very sad if we had to lose that and deal with gravitons.
Not that I'm afraid of math, I just don't feel like I need to see equations when it's a concept that is being explained rather than how to numerically calculate something. Math helps me as much as a code implementation does to explain the concept of a variable: I can observe its behavior but it doesn't necessarily teach me the concept.
[1] https://youtu.be/Xc4xYacTu-E
Would that mean electromagnetism is not a force either?
What is the difference between the two that makes one a force and one not a force?
The indicator that gravity is special is the fact that 'inertial mass' and 'gravitational mass' are the same thing.
That's undecidable and gravity is not special in that way
https://physics.stackexchange.com/questions/390540/what-is-t...
> the gravitational field and acceleration are inductive pairs (similar to the electromagnetic field and electric current.)
How so? You can have different electric charge to inertial mass ratios, but that is not so for gravitational mass to inertial mass.
Also, what if you have a charge so high that the electrical-escape velocity exceeds the speed of light? A strange kind of black hole?
I ask these things because I do not understand why gravity is called "not a force", while electromagnetism is, when I see no real difference between how particles act.
The way that varies is always in proportion to the object's inertial mass though, which has the result that the path traced by an any object with the same starting position and velocity in a given gravitational field is going to be the same.
Contrast this with charge, which can vary independently of inertial mass. This has the result that the paths traced by objects with the same starting position and velocity in a given electrical field will vary.
This also means there is no fixed "electrical escape velocity". More massive objects with the same charge can escape with a lower velocity.
But let's just say they are the same, then gravity and "spacial" inertia are intertwined. Gravity is only special if intertia is also considered special. Inertia seems special because it's explainable in terms of motion in space. There might be other kinds of motion in other dimensions that could explain the randomness of quantum space.
The model shown is for 1D space and 1D time, and this 2D spacetime is curved, remaining 2D. Can you though show us a 3D model of curved 2D spacetime? Maybe as a rotatable, scaleable 3D scene, where more complex phenomena, like planetary motion around a star, would be possible to show?
Of course, you are going to tell me that the earth is not inflating, obviously, because it is still the same size after so many years.
But here is the trick: the earth is inflating at the same rate as spacetime contracts. If the earth didn't inflate, the contraction of spacetime would have collapsed it into a black hole.
Note: It is related to Einstein's elevator thought experiment. Here, the inflating earth replaces the rocket powered elevator.
Note 2: If the idea of an inflating earth bothers you, I suggest you start considering that the earth is flat, seriously! Flat Earthers took Einstein's thought experiment quite literally and consider the Earth to be a disk that is continuously accelerated upwards. And in fact, if free fall trajectories were parabolic, that would be the correct explanations. In reality, because the earth is not flat, free fall trajectories are elliptic, though it is only apparent on a large scale.
The thing is, I watched the French version and I didn't know an English version existed, that's why I didn't post it here, French speakers are, I believe, a minority.
And BTW, the French channel has an 8 part explanation of the maths behind general relativity that is the best I have ever seen. It is on a level above most pop science video since it actually shows the equations, tensors, etc... but the explanations are actually quite accessible.
https://www.youtube.com/watch?v=sg2BBldgKpo
Because both terms of the equation affect one another, solving it is complicated but here, the result is that spacetime contracts to the center of the earth.
Matter continuously destroys spacetime at its location, sucking in the fabric of the universe.
The Universe isn't expanding; matter is shrinking. Light isn't redshifted on the way to us, it's just that our sensors are getting smaller relative to the unmodified wavelength of light.
Indeed; light is blue-shifted as its source is on its way to us (and is red-shifted as its source is on its way away from us).
Is there any way to distinguish these? Surely it depends on who's point of view you're looking from
The apple doesn't curve space-time as much as the earth does.
Are they incompatible viewpoints, or just different perspectives on the same thing? (E.g. are gravitons hypothesized to disappear depending on frame of reference?)
I'm assuming they're incompatible (that we need the theory of everything [2] to reconcile them) but would love to know if there's something I'm missing.
[1] https://en.wikipedia.org/wiki/Graviton
[2] https://en.wikipedia.org/wiki/Theory_of_everything
Interestingly the Yang-Mills equations (used in quantum mechanics) and the Einstein-Hilbert action (used in general relativity) are pretty much identical if you use general enough mathematics.
Play with the accelaration to see some spooky seemingly faster-than-light movement.
but the real question is: can you counter your own arguments? Because that's where absolutely fascinating science happens.
https://www.youtube.com/watch?v=cb_PCuv0vcY&feature=emb_logo
https://www.audible.com/pd/Black-Holes-Tides-and-Curved-Spac...
If earth is pushing equally on all of the people around the globe how come it doesn't swell up and burst?
And how do you explain force of attraction weakening the higher in altitude you go away from earth?