I came here to ask something along these lines. Sometimes I'm baffled by what percolates up to the first few pages of HN. I've seen more than a couple Wikipedia pages about random scientific trivia, and that sort of thing just doesn't feel up to snuff.
Somewhat more interesting is that (E, px, py, pz) form a proper 4-vector, which means under rotations in 4-space its length does not change.
To see this, first note that in flat 4-space of our universe (known as Minkowski space), experiments show that our metric is diag(1, -1, -1, -1), thus giving the length as
E^2 - px^2 - py^2 - pz^2
This should be a scalar, i.e. a constant that does not change under transformations. So we label it m^2, the rest mass of the object. From there it follows that
E^2 = m^2 + px^2 + py^2 + pz^2
You may be wondering where all the 'c's went. For exposition I chose units where the speed of light is 1. We can easily reconstruct where the 'c's should go by looking at the relevant units:
E: m * v * v
p: m * v
m: m
Thus we need to multiply masses by c^2 and momenta by c:
E = (mc^2)^2 + (px^2 + py^2 + pz^2)c^2
Compare with perhaps a more familiar Euclidean 2-space where the metric is diag(1, 1) and the length is given by the familiar Pythagorean theorem. In such a space we can explore the set of transformations which preserve the length, and see they are of the form
(cos x, -sin x)
(sin x, cos x)
In Minkowski space, the set of transformations can be categorized into familiar rotations in 3-space and so-called boosts in 4-space where some space dimension is rotated into some time dimension (or vice versa). This gives rise to the famous observations of time dilation and length contraction when considering relativistic speeds.
So how do we simulate the fallout and mushroom cloud of a tera-ton nuclear fusion explosion detonating while travelling at mach 300 in the stratosphere? What would that look like?
Seems like anybody who understands that equation already knows it's a derivation from special relativity. And anybody who doesn't probably doesn't know what an inertial reference frame is, so probably doesn't care.
Actually, it's not just stationary masses, but it does apply only to things in linear motion - rotation is specifically excluded, as (IIRC) he makes clear in the appendix to his book on relativity...
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[ 3.3 ms ] story [ 37.0 ms ] threadHe title didn't even copy the formula correctly, a + is missing.
To see this, first note that in flat 4-space of our universe (known as Minkowski space), experiments show that our metric is diag(1, -1, -1, -1), thus giving the length as
This should be a scalar, i.e. a constant that does not change under transformations. So we label it m^2, the rest mass of the object. From there it follows that You may be wondering where all the 'c's went. For exposition I chose units where the speed of light is 1. We can easily reconstruct where the 'c's should go by looking at the relevant units: Thus we need to multiply masses by c^2 and momenta by c: Compare with perhaps a more familiar Euclidean 2-space where the metric is diag(1, 1) and the length is given by the familiar Pythagorean theorem. In such a space we can explore the set of transformations which preserve the length, and see they are of the form In Minkowski space, the set of transformations can be categorized into familiar rotations in 3-space and so-called boosts in 4-space where some space dimension is rotated into some time dimension (or vice versa). This gives rise to the famous observations of time dilation and length contraction when considering relativistic speeds.