
Einstein's E=mc^2 is only for stationary masses - chimpino
https://www.fxsolver.com/browse/formulas/Energy+-+Momentum+relation
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frutiger
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.

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BeeBoBub
For non-accelerating masses, or more precisely within an inertial reference
frame

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Fjolsvith
Relatively speaking?

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fjsolwmv
Are arbitrary textbook physics formulas HN worthy?

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

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jdsully
I think a hacker would find this interesting. I know I do.

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gumby
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.

~~~
exabyte
it's still valuable information to know that it goes a layer deeper

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dublin
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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soundpuppy
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?

