証明 E=mc2!

The correct formula is

圖片參考：http://imgcld.yimg.com/8/n/HA05726829/o/701011080004413873409800.jpg

where m is the rest mass (mass of the object when stationary), p is the momentum and c is the speed of light.

This is a result of Lorentz symmetry. Two inertial reference frames O and O' (think of two observers with a relative velocity) are related by the Lorentz transformation

圖片參考：http://imgcld.yimg.com/8/n/HA05726829/o/701011080004413873409811.jpg

where (x,y,z,t) are space-time coordinates of frame O, (x',y',z',t') are coordinates of O' and u is the relative velocity of the observers O and O'. This can be rewritten as a matrix equation

圖片參考：http://imgcld.yimg.com/8/n/HA05726829/o/701011080004413873409812.jpg

The matrix (Lambda) representing the Lorentz transformation preserves the space-time metric (eta), i.e.

圖片參考：http://imgcld.yimg.com/8/n/HA05726829/o/701011080004413873409823.jpg

[In general, Lorentz transformations are matrices (Lambda) that satisfy the above equation. And these are the only linear transformations that preserves the speed of light.]

Since momentum-energy represents (infinitesimal) translation in space-time, we require momentum-energy to transform also under Lorentz transformation

圖片參考：http://imgcld.yimg.com/8/n/HA05726829/o/701011080004413873409824.jpg

[Alternatively, this is also because a plane wave exp[i hbar (Et - px)] should be invariant under Lorentz transformation, i.e. Et - px = E't' - p'x'.]

Let P = (px,py,pz,E/t) and P'=(px',py'pz'E'/c) be the 4-momenta for frame O and O'. Plug in the previous equation, we have

圖片參考：http://imgcld.yimg.com/8/n/HA05726829/o/701011080004413873409835.jpg

This means this quantity is a constant independent from reference frame, and we define the mass m by

圖片參考：http://imgcld.yimg.com/8/n/HA05726829/o/701011080004413873409836.jpg


Next, we will show that the mass m coincide with the inertial mass in non-relativistic limit. In the small velocity limit (p<<mc), we Taylor expand

圖片參考：http://imgcld.yimg.com/8/n/HA05726829/o/701011080004413873409800.jpg

in power series of p/mc,

圖片參考：http://imgcld.yimg.com/8/n/HA05726829/o/701011080004413873409848.jpg

So we see the first term mc^2 is the rest energy (when p=0),. The second term p^2/2m agrees the non-relativistic kinetic energy, and therefore the mass m corresponds the non-relativistic inertial mass. Higher order terms are relativistic corrections.

To deduce the momentum-velocity relation, we use the Hamiltonian equation

圖片參考：http://imgcld.yimg.com/8/n/HA05726829/o/701011080004413873409859.jpg

Substituting into the energy equation, we have

圖片參考：http://imgcld.yimg.com/8/n/HA05726829/o/7010110800044138734098510.jpg

where M(v) is the relativistic inertial mass in velocity v.

Please refer to an earlier answer in this forum by the following link:

http://hk.knowledge.yahoo.com/question/question?qid=7010092500282