- #1
quZz
- 125
- 1
Could someone explain to me, what does [tex]E=mc^2[/tex] mean? I've got a feeling that this famous relation is just a definition of m.
First, if we use an old version
[tex]E=mc^2\:\:(1)[/tex]
where E is total energy and m is relativistic mass, then is there a way to measure m and E independently so that this equation becomes a physical law?
In fact, if we are dealing with a free moving particle than [tex]\textbf{p} = E\textbf{v}/c^2 [/tex]. Now if we want to folow Newton and define [tex]\textbf{p} = m\textbf{v} [/tex] we get (1).
In particle interactions it is 4-momentum [tex]p = (E/c,\:\textbf{p})[/tex] that is conserved.
If we try to weigh smth then (as it follows from GR) it is [tex]T_{\mu \nu}[/tex] that matters. Actually the gravitational force acting on a moving particle can be different depending on it's moving direction. In the simplest case only [tex]T_{00}[/tex] survive, but again it is energy density.
Second, if we use a modern one
[tex]E_0=mc^2\:\:(2)[/tex]
with E being rest energy, then it follows right from the definition of 4-momentum norm
[tex]mc \equiv \sqrt{(E/c)^2-\textbf{p}^2}[/tex].
If we recall Einstein's derivation of (2), it was based on the fact that c.m. of a closed system is not accelerating, but in SR c.m. is defined... again from energies of particles.
So please explain to me, what is physical about (1) or (2) or both?
First, if we use an old version
[tex]E=mc^2\:\:(1)[/tex]
where E is total energy and m is relativistic mass, then is there a way to measure m and E independently so that this equation becomes a physical law?
In fact, if we are dealing with a free moving particle than [tex]\textbf{p} = E\textbf{v}/c^2 [/tex]. Now if we want to folow Newton and define [tex]\textbf{p} = m\textbf{v} [/tex] we get (1).
In particle interactions it is 4-momentum [tex]p = (E/c,\:\textbf{p})[/tex] that is conserved.
If we try to weigh smth then (as it follows from GR) it is [tex]T_{\mu \nu}[/tex] that matters. Actually the gravitational force acting on a moving particle can be different depending on it's moving direction. In the simplest case only [tex]T_{00}[/tex] survive, but again it is energy density.
Second, if we use a modern one
[tex]E_0=mc^2\:\:(2)[/tex]
with E being rest energy, then it follows right from the definition of 4-momentum norm
[tex]mc \equiv \sqrt{(E/c)^2-\textbf{p}^2}[/tex].
If we recall Einstein's derivation of (2), it was based on the fact that c.m. of a closed system is not accelerating, but in SR c.m. is defined... again from energies of particles.
So please explain to me, what is physical about (1) or (2) or both?