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Re:Einstein's E=mc^2 is only for stationary masses by frutiger on Sunday April 28, 2019 @05:03PM (#19773658)
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.
Re:Einstein's E=mc^2 is only for stationary masses by BeeBoBub on Sunday April 28, 2019 @02:50PM (#19772894)
For non-accelerating masses, or more precisely within an inertial reference frame
Re:Einstein's E=mc^2 is only for stationary masses by Fjolsvith on Monday April 29, 2019 @10:12AM (#19778477)
Relatively speaking?
Re:Einstein's E=mc^2 is only for stationary masses by fjsolwmv on Sunday April 28, 2019 @03:04PM (#19772999)
Are arbitrary textbook physics formulas HN worthy?

He title didn't even copy the formula correctly, a + is missing.

Re:Einstein's E=mc^2 is only for stationary masses by N_trglctc_joe on Sunday April 28, 2019 @05:56PM (#19773955)
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.
Re:Einstein's E=mc^2 is only for stationary masses by jdsully on Sunday April 28, 2019 @03:50PM (#19773281)
I think a hacker would find this interesting. I know I do.
Re:Einstein's E=mc^2 is only for stationary masses by gumby on Sunday April 28, 2019 @06:03PM (#19773997)
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.
Re:Einstein's E=mc^2 is only for stationary masses by exabyte on Sunday April 28, 2019 @10:12PM (#19775312)
it's still valuable information to know that it goes a layer deeper
Re:Einstein's E=mc^2 is only for stationary masses by dublin on Tuesday April 30, 2019 @10:33AM (#19788239)
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...
Re:Einstein's E=mc^2 is only for stationary masses by soundpuppy on Sunday April 28, 2019 @05:13PM (#19773719)
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?