Einstein's E=mc^2 is only for stationary masses
fxsolver.com
fxsolver.com
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.He title didn't even copy the formula correctly, a + is missing.