How is this definition compatible with the derivation from special relativity?
Express total energy as function of the momentum E(p) = (m^2*c^4 + p^2*c^2)^0.5 and do a Taylor expansion for momentum p_0. You'll arrive at E(p_0) = mc^2 + p_0^2/(2m /* <-- here is the factor 1/2 coming in */) + p_0^4/(8m^3*c^2) + ... The first term is the rest energy, the second term is the kinetic energy, higher ones are relativistic corrections to that. Setting kinetic energy arbitrarily to p^2/m (= mv^2 i.e. without the factor 1/2) breaks that.
Thanks. I admit that my previous article was in error but this new article uses a different approach to the problem. It shows that the current equation was never meant to replace the previous one. Coriolis clearly wrote that he used vis viva = ½mv² for purposes of convenience.
Edit: I'm grateful for our previous exchange. You pointed out new places and ideas that were useful to my research.
In my opinion, Ek = mv² contradicts special relativity. The maximum energy is E = mc². There can be no doubt about it. And it is not rest energy. It is obviously kinetic energy. But that's just me.
There is no contradiction, because the 2 "energies" from the 2 formulas are 2 different quantities.
The energy is not a primitive physical quantity. It is a quantity that is defined.
The definition of energy is not unique, it can contain an arbitrary multiplicative factor without changing anything in physics, as long as the same definition is used everywhere.
If the kinetic energy is defined as mv^2, everything is fine, but then the mechanical work is 2Fd and the total energy from the special relativity is 2mc^2.
If the kinetic energy is defined in the usual way, all the values of energy or work are halved.
There is nothing mysterious about this. It is just a matter of convention.
In the beginning, the kinetic energy was defined as mv^2, because it was the simplest formula.
After more complex mechanical problems began to be studied and solved, requiring differential and integral calculus, the definition was modified to be that from today, because there are more formulas that become simpler than formulas that become more complex, like the direct computation of the kinetic energy from mass and velocity.
The same has happened with the Coulomb law. Initially the electric force was defined as the product of charges divided by the square of the radius, because that was the simplest possible formula.
Later, it was realized that making the Coulomb formula more complex, by defining the force as the product of charges divided by the area of the sphere centered on one charge, many other formulas become simpler, so the old definition was changed.
The change of the kinetic energy formula was a good change, because computing a half of mv^2 is a negligible complication, while a large number of formulas containing integrals or derivatives become simpler.
The ½ comes from an integration. Whenever you see a square term in an equation, there's often a corresponding ½.
Yes, we could "simplify" equations by getting rid of the ½ and squashing it into the units, but at that point, all of the stuff makes less sense.
I do physics with kids, and often show area (usually of a triangle) as the distance traveled with constant acceleration, the energy of something, the accumulated debt or what-not. The source of the ½ is very obvious -- we're looking at a triangle and not a rectangle.
If you define kinetic energy as mv^2 it basically means that you redefine the word "energy" to mean double energy (as we currently definine it). It could work just as well, but we would have to have the factor 2 in all of the other equations that define any kind of energy or work so the "double energy" is conserved. The author alread gives such doubled equation for potential energy Epd=2mgh. You'd need to do the same for every other equation that involves any kind of energy in physics.
I wonder how author feels about 1/2 factor in equation that gives distance traveled by moving with constant acceleration. Should it be just at^2 ? This would redefine distance to be double distance and we would have to consider what it means for acceleration and velocity.
All these 1/2 come straight from integration of linear functions. To get rid of them we'd have to patch math results and since math holds this would mean factors of 2 pop up around everywhere else.
Thanks for the comment. I see your point but I don't think it's a matter of definition. There is no doubt that the correct equation is Ek = mv². Since velocity v² = 2ad, it follows that mv² = 2mad = 2Fd. The work done is the acceleration of the body which is perfectly expressed in mv². The distance traveled is a result of the acceleration not the cause of it. Only the change in velocity matters. At least, that is my take on it.
But the work done is Fd so 2Fd is double the work, and hence mv^2 is double energy.
Why do you think work is given by 2Fd? Do you want it to be that? Do you want to define work as two times the force times the distance the force acted over?
You can, if you like, but you'll be plagued by a factor of 2 in every place where you will talk about any work.
It's way easier to keep this factor of 1/2 just in definition of kinetic energy instead.
You can name mv^2 kinetic energy, and 2Fd work. But then you can name (mv^2)/2 kinetic half-energy and Fd half-work and physicist will still prefer to do calculations for half-work and half-energy because there will be less places where this factor of 2 will crop up in calculations.
Coriolis was wrong. The work done has nothing to do with distance. It only has to do with velocity. Accelerating the body from 0 to a final velocity is the work. This is what force does. If you know the distance and the acceleration, you can use them to calculate the final velocity with the well-known kinematics equation v² = 2ad. That is all. Likewise, if you know the elapsed time and the acceleration you can use the formula v = at to get the final velocity. It's simple, really.
Probably about many things in his life as we all are.
> The work done has nothing to do with distance. It only has to do with velocity.
The thing that has to do only with the velocity we call kinetic energy.
Work is something else.
> Accelerating the body from 0 to a final velocity is the work.
It's one example of work. You can also consider for example a force acting upon a body that travels with constant velocity of v over a distance d. And this force would still do the work equal to Fd. You might ask, how is it possible that this body that the force is acting upon doesn't have an acceleration? Because at the same time this body might have interactions with other bodies through different forces and do work on them.
As example of this imagine a body that moves directly up on Earth with constant velocity v, because the force that does the work on it is directed up and equal exactly mg. That force does the work on this body, but none of it has anything to do with velocity and it goes directly into potential energy.
Regardless of what kind of motion the body uses to travel distance d, if it's under the influence of the force F during that, the equation holds that work done by the force F equals Fd.
This allows us to do calculations for various kinds of energies and multiple acting forces, just adding the work that each force does and changes in every kind of energy the body has.
> If you know the distance and the acceleration, you can use them to calculate the final velocity with the well-known kinematics equation v² = 2ad.
This equation is not some core law of the universe. It just comes from assuming we have a movement under influence of a single force of a body that at t=0 started at rest. It comes from v=at and d=(at^2)/2 and those come from integrals over time of a done once and twice.
This equation you really like is just a result of how we define speed and acceleration simply applied to one kind of motion. It has nothing to do with forces or energies. It just deals with displacement and its first and second derivatives (velocity and acceleration).
> Likewise, if you know the elapsed time and the acceleration you can use the formula v = at to get the final velocity. It's simple, really.
Yes. It's simple to the point of not being especially interesting.
What's interesting is that resulting formula deltaEk = Fd is more general. It works regardless of whether the motion that causes the change in kinetic energy has constant acceleration or not. And you can use it when multiple forces act upon this body and multiple kinds of energies are involved, for example deltaEp + deltaEk = Fd + Ge + Hi (where deltaEp is a change in potential energy, deltaEk is a change in kinetic energy, F, G, H are the forces and d, e, i are the displacement of the body while the forces F, G and H acted upon the body respectively).
You are not wrong writing mv^2 = 2Fd and we might have called mv^2 energy but the we would have to call 2Fd work, and we really prefer to call Fd work because that factor 2 would crop up everywhere in physics where there are any energies and works considered (you already saw it in your version of the potential energy equation Ep=2mgh, and there are so many places in physics where it would need to show up, basically any place where there are energies and fields involved). So it's just more convenient to have it in the denominator of kinetic energy than literally everywhere else.
> This equation is not some core law of the universe.
> It's simple to the point of not being especially interesting.
Either you don't realize what you're saying or you are playing a game of deception.
1. This equation is indeed a core law of the universe.
2. It is extremely interesting because it has a specific meaning. It expresses the kinetic energy of a massive body in motion.
3. More specifically, it means that E = mc² does not represent what Einstein claimed it did. It represents the maximum kinetic energy that a massive body can have. This is why it is extremely interesting.
I'm not sure who downvoted you. I can't downvote comments that are responses to my comments.
I don't believe person who did this was motivated by cowardice. More likely annoyance or exhaustion. Or just wanted to express that he thinks you are wrong but wouldn't bother with writing a comment.
> Either you don't realize what you're saying or you are playing a game of deception.
I'm sorry you feel that way. I'm just trying to help you understand source of those equations.
> 1. This equation is indeed a core law of the universe.
Let me show you step by step where the equation v² = 2ad comes form.
Let's consider an object moving in any way. It may have some mass, it may be massless, it doesn't matter.
Let's call the distance it moved so far d. Because object moves, d is a smooth function of time so we might be interested with its first and second derivative. Let's call the first derivative v and second derivative a. Those might be constant or might be also some functions of time.
Then let's say we are interested in a very specific type of motion. A motion where at t=0 first derivative was equal zero, and second derivative is constant through the whole motion.
So we have second derivative equal some constant value a. Because integration is the opposite of derivation to calculate v we need to integrate a over time t. This is a purely mathematical operation. And gives us v = at
Then we want to calculate the function d itself for this specific motion. To do that we need to integrate over time again and we get d = at²/2 This again is purely mathematical operation that doesn't rely on any connection to the world we live in.
So we have:
d = at²/2
v = at
when we calculate t from the second equation and put it in the first one we get:
d = av²/2a²
by rearranging we get:
v² = 2ad
All of the above is true and pure math. Even if universe didn't exist or worked completely differently (for example if real objects didn't really move in smooth motions but teleported instead) this equation and its derivation would still exist, it just would not reflect reality.
So this equation comes from pure math and a concept of how movement might work, defined by purely mathematical means. It doesn't have to describe our universe and existence of this equation doesn't depend on the existence of our universe or how movement actually works in our universe. So it's not a core law of universe, just a mathematical consequence of considering one specific type of motion that we think might happen in our universe.
> It is extremely interesting because it has a specific meaning. It expresses the kinetic energy of a massive body in motion.
Yes, but that's all. There are many other forms of energy. And what's way more interesting, that we get from considering kinetic energy, is the concept of energy itself, that it can come in many forms and can get transferred from one form to another and that transfer is mediated by work done by forces acting on moving bodies over some distances.
> 3. More specifically, it means that E = mc² does not represent what Einstein claimed it did. It represents the maximum kinetic energy that a massive body can have. This is why it is extremely interesting.
Absolutely not. Kinetic energy defined as mV² (with 1/2 or without) is only the approximation what the kinetic energy of a massive body is when it travels at slow speeds. When the speeds get near c best approximation of kinetic energy that we know is
Ek = (1/sqrt(1-v²/c²) - 1)mc²
It's not immediately obvious but this equation can be approximated by mv²/2 for small v (by using Maclaurin series expansion of part of it and taking only first terms).
This better equation again is purely mathematical result derived by considering of how adding movements must work if there is such a thing as the maximum speed of movement (that we called c).
You can see that this equation tends to infinity as v approaches c. So there's no maximum kinetic energy. You can pump energy into the moving body by applying the force to it to do the work regardless of what speed the body has already. Buy pumping in more energy you just bring the speed closer to c but you never reach it no matter how much yo...
21 comments
[ 2.9 ms ] story [ 68.9 ms ] threadExpress total energy as function of the momentum E(p) = (m^2*c^4 + p^2*c^2)^0.5 and do a Taylor expansion for momentum p_0. You'll arrive at E(p_0) = mc^2 + p_0^2/(2m /* <-- here is the factor 1/2 coming in */) + p_0^4/(8m^3*c^2) + ... The first term is the rest energy, the second term is the kinetic energy, higher ones are relativistic corrections to that. Setting kinetic energy arbitrarily to p^2/m (= mv^2 i.e. without the factor 1/2) breaks that.
My exchange with the author is at https://news.ycombinator.com/item?id=27974055 .
Edit: I'm grateful for our previous exchange. You pointed out new places and ideas that were useful to my research.
The energy is not a primitive physical quantity. It is a quantity that is defined.
The definition of energy is not unique, it can contain an arbitrary multiplicative factor without changing anything in physics, as long as the same definition is used everywhere.
If the kinetic energy is defined as mv^2, everything is fine, but then the mechanical work is 2Fd and the total energy from the special relativity is 2mc^2.
If the kinetic energy is defined in the usual way, all the values of energy or work are halved.
There is nothing mysterious about this. It is just a matter of convention.
In the beginning, the kinetic energy was defined as mv^2, because it was the simplest formula.
After more complex mechanical problems began to be studied and solved, requiring differential and integral calculus, the definition was modified to be that from today, because there are more formulas that become simpler than formulas that become more complex, like the direct computation of the kinetic energy from mass and velocity.
The same has happened with the Coulomb law. Initially the electric force was defined as the product of charges divided by the square of the radius, because that was the simplest possible formula.
Later, it was realized that making the Coulomb formula more complex, by defining the force as the product of charges divided by the area of the sphere centered on one charge, many other formulas become simpler, so the old definition was changed.
The change of the kinetic energy formula was a good change, because computing a half of mv^2 is a negligible complication, while a large number of formulas containing integrals or derivatives become simpler.
The ½ comes from an integration. Whenever you see a square term in an equation, there's often a corresponding ½.
Yes, we could "simplify" equations by getting rid of the ½ and squashing it into the units, but at that point, all of the stuff makes less sense.
I do physics with kids, and often show area (usually of a triangle) as the distance traveled with constant acceleration, the energy of something, the accumulated debt or what-not. The source of the ½ is very obvious -- we're looking at a triangle and not a rectangle.
The equation should be W = mv² = 2Fd. Why? because v² = 2 a*d.
There's a pattern of factorial terms you see with successive integrations: 1/1, 1/2, 1/6, 1/24, 1/n!,
Multiplying by 2 changes this to 2, 1, 1/3, 1/12, 2/n!.
I view that as a clear loser, not a winner.
I wonder how author feels about 1/2 factor in equation that gives distance traveled by moving with constant acceleration. Should it be just at^2 ? This would redefine distance to be double distance and we would have to consider what it means for acceleration and velocity.
All these 1/2 come straight from integration of linear functions. To get rid of them we'd have to patch math results and since math holds this would mean factors of 2 pop up around everywhere else.
Why do you think work is given by 2Fd? Do you want it to be that? Do you want to define work as two times the force times the distance the force acted over?
You can, if you like, but you'll be plagued by a factor of 2 in every place where you will talk about any work.
It's way easier to keep this factor of 1/2 just in definition of kinetic energy instead.
You can name mv^2 kinetic energy, and 2Fd work. But then you can name (mv^2)/2 kinetic half-energy and Fd half-work and physicist will still prefer to do calculations for half-work and half-energy because there will be less places where this factor of 2 will crop up in calculations.
Probably about many things in his life as we all are.
> The work done has nothing to do with distance. It only has to do with velocity.
The thing that has to do only with the velocity we call kinetic energy.
Work is something else.
> Accelerating the body from 0 to a final velocity is the work.
It's one example of work. You can also consider for example a force acting upon a body that travels with constant velocity of v over a distance d. And this force would still do the work equal to Fd. You might ask, how is it possible that this body that the force is acting upon doesn't have an acceleration? Because at the same time this body might have interactions with other bodies through different forces and do work on them.
As example of this imagine a body that moves directly up on Earth with constant velocity v, because the force that does the work on it is directed up and equal exactly mg. That force does the work on this body, but none of it has anything to do with velocity and it goes directly into potential energy.
Regardless of what kind of motion the body uses to travel distance d, if it's under the influence of the force F during that, the equation holds that work done by the force F equals Fd.
This allows us to do calculations for various kinds of energies and multiple acting forces, just adding the work that each force does and changes in every kind of energy the body has.
> If you know the distance and the acceleration, you can use them to calculate the final velocity with the well-known kinematics equation v² = 2ad.
This equation is not some core law of the universe. It just comes from assuming we have a movement under influence of a single force of a body that at t=0 started at rest. It comes from v=at and d=(at^2)/2 and those come from integrals over time of a done once and twice.
This equation you really like is just a result of how we define speed and acceleration simply applied to one kind of motion. It has nothing to do with forces or energies. It just deals with displacement and its first and second derivatives (velocity and acceleration).
> Likewise, if you know the elapsed time and the acceleration you can use the formula v = at to get the final velocity. It's simple, really.
Yes. It's simple to the point of not being especially interesting.
What's interesting is that resulting formula deltaEk = Fd is more general. It works regardless of whether the motion that causes the change in kinetic energy has constant acceleration or not. And you can use it when multiple forces act upon this body and multiple kinds of energies are involved, for example deltaEp + deltaEk = Fd + Ge + Hi (where deltaEp is a change in potential energy, deltaEk is a change in kinetic energy, F, G, H are the forces and d, e, i are the displacement of the body while the forces F, G and H acted upon the body respectively).
You are not wrong writing mv^2 = 2Fd and we might have called mv^2 energy but the we would have to call 2Fd work, and we really prefer to call Fd work because that factor 2 would crop up everywhere in physics where there are any energies and works considered (you already saw it in your version of the potential energy equation Ep=2mgh, and there are so many places in physics where it would need to show up, basically any place where there are energies and fields involved). So it's just more convenient to have it in the denominator of kinetic energy than literally everywhere else.
> It's simple to the point of not being especially interesting.
Either you don't realize what you're saying or you are playing a game of deception.
1. This equation is indeed a core law of the universe.
2. It is extremely interesting because it has a specific meaning. It expresses the kinetic energy of a massive body in motion.
3. More specifically, it means that E = mc² does not represent what Einstein claimed it did. It represents the maximum kinetic energy that a massive body can have. This is why it is extremely interesting.
Thank you for the exchange.
Edit: I forgot to laugh. hahahaha...HAHAHAHA...hahahaha
I don't believe person who did this was motivated by cowardice. More likely annoyance or exhaustion. Or just wanted to express that he thinks you are wrong but wouldn't bother with writing a comment.
Let's stick to math and physics though.
To understand it you'll need to have ability to calculate derivatives and integrals.
I'm sorry you feel that way. I'm just trying to help you understand source of those equations.
> 1. This equation is indeed a core law of the universe.
Let me show you step by step where the equation v² = 2ad comes form.
Let's consider an object moving in any way. It may have some mass, it may be massless, it doesn't matter.
Let's call the distance it moved so far d. Because object moves, d is a smooth function of time so we might be interested with its first and second derivative. Let's call the first derivative v and second derivative a. Those might be constant or might be also some functions of time.
Then let's say we are interested in a very specific type of motion. A motion where at t=0 first derivative was equal zero, and second derivative is constant through the whole motion.
So we have second derivative equal some constant value a. Because integration is the opposite of derivation to calculate v we need to integrate a over time t. This is a purely mathematical operation. And gives us v = at Then we want to calculate the function d itself for this specific motion. To do that we need to integrate over time again and we get d = at²/2 This again is purely mathematical operation that doesn't rely on any connection to the world we live in.
So we have:
d = at²/2
v = at
when we calculate t from the second equation and put it in the first one we get:
d = av²/2a²
by rearranging we get:
v² = 2ad
All of the above is true and pure math. Even if universe didn't exist or worked completely differently (for example if real objects didn't really move in smooth motions but teleported instead) this equation and its derivation would still exist, it just would not reflect reality.
So this equation comes from pure math and a concept of how movement might work, defined by purely mathematical means. It doesn't have to describe our universe and existence of this equation doesn't depend on the existence of our universe or how movement actually works in our universe. So it's not a core law of universe, just a mathematical consequence of considering one specific type of motion that we think might happen in our universe.
> It is extremely interesting because it has a specific meaning. It expresses the kinetic energy of a massive body in motion.
Yes, but that's all. There are many other forms of energy. And what's way more interesting, that we get from considering kinetic energy, is the concept of energy itself, that it can come in many forms and can get transferred from one form to another and that transfer is mediated by work done by forces acting on moving bodies over some distances.
> 3. More specifically, it means that E = mc² does not represent what Einstein claimed it did. It represents the maximum kinetic energy that a massive body can have. This is why it is extremely interesting.
Absolutely not. Kinetic energy defined as mV² (with 1/2 or without) is only the approximation what the kinetic energy of a massive body is when it travels at slow speeds. When the speeds get near c best approximation of kinetic energy that we know is
Ek = (1/sqrt(1-v²/c²) - 1)mc²
It's not immediately obvious but this equation can be approximated by mv²/2 for small v (by using Maclaurin series expansion of part of it and taking only first terms).
This better equation again is purely mathematical result derived by considering of how adding movements must work if there is such a thing as the maximum speed of movement (that we called c).
You can see that this equation tends to infinity as v approaches c. So there's no maximum kinetic energy. You can pump energy into the moving body by applying the force to it to do the work regardless of what speed the body has already. Buy pumping in more energy you just bring the speed closer to c but you never reach it no matter how much yo...