Deriving the Momentum-Energy 4-Vector

The problem we have is how to take a time derivative if the time is the component of a 4-vector. We need some kind of scalar time to make sense of the equations we know and love. A well defined time, that does not need to be transformed, is the time in the rest frame of the particle. We call this the proper time \bgroup\color{black}$ \tau$\egroup. We will make use of it here, but later just try to rewrite our equations so that they are covariant in 4 dimensions.

The velocity 4-vector can be defined as.

\bgroup\color{black}$\displaystyle v_\mu={\partial x_\mu\over\partial\tau}=\gamm...
...vec{x})=\gamma (c,{\partial \vec{x}\over\partial t})=(\gamma c,\vec{v}) $\egroup

So we see that the time-component of the usual velocity vector is \bgroup\color{black}$ \gamma c$\egroup and we have the velocity 4-vector
\bgroup\color{black}$ \displaystyle v_\mu=\gamma (c,\vec{v})$\egroup
where \bgroup\color{black}$ \gamma$\egroup is for the transformation from the rest frame to whatever frame we are defining \bgroup\color{black}$ v_\mu$\egroup in.

We can dot the velocity 4-vector into itself.

\bgroup\color{black}$\displaystyle v_\mu v_\mu=\gamma^2(-c^2+v^2)={-c^2+v^2\over 1-\beta^2}=-c^2{1-\beta^2\over 1-\beta^2}=-c^2 $\egroup

This is certainly a scalar. It is an example of the problem that many 4D ``lengths'' are not very useful. That is, the magnitude of the velocity vector is \bgroup\color{black}$ c$\egroup no matter what the velocity is.

To be consistent with non-relativistic equations we will define the momentum.

\bgroup\color{black}$\displaystyle p_\mu=mv_\mu =\gamma (mc,m\vec{v}) $\egroup

If we identify the time component as above, \bgroup\color{black}$ {E\over c}=\gamma mc$\egroup, we have the relation

\bgroup\color{black}$ \displaystyle E=\gamma mc^2$\egroup

which looks similar to the rest energy equation but actually is true in any frame.

A crucial test of this ``derived'' 4-vector is whether it gives the right physics in the non-relativistic limit. We did have some choice to make when inserting the energy into the momentum 4-vector. Start with the energy equation from above.

$\displaystyle E=\sqrt{(mc^2)^2+(pc)^2}=mc^2\sqrt{1+\left({pc\over mc^2}\right)^...
...t)^2\right) = \left(mc^2+{1\over 2}{p^2c^2\over mc^2}\right)=mc^2+{p^2\over 2m}$    

This is the correct non-relativistic limit. The total energy is, in the non-relativistic limit, the rest energy \bgroup\color{black}$ mc^2$\egroup plus the kinetic energy \bgroup\color{black}$ {1\over 2}mv^2$\egroup. Normally, we ignore the rest energy as being unchangeable. It does change in nuclear interactions for example.

Jim Branson 2012-10-21