The Momentum-Energy 4-Vector

It is obviously important it determine how Energy and Momentum transform in Special Relativity. A reasonable guess is that momentum is a 3-vector conjugate to position, so we need to find what the fourth component is to make a 4-vector. We again have the problem of the speed of light not being equal to one in our units. The answer, which we will derive below, is that the Momentum-Energy 4-vector is

\bgroup\color{black}$ \displaystyle p_\mu=\left({E\over c}, p_x, p_y, p_z\right)$\egroup
where the choice of where to put the \bgroup\color{black}$ c$\egroup could be made by dimensional analysis.

The dot product with itself is

\bgroup\color{black}$\displaystyle p_\mu p_\mu =-{E^2\over c^2}+p_x^2+p_y^2+p_z^2=-{E^2\over c^2}+p^2 $\egroup

This quantity should be a Lorentz scalar, which we will call \bgroup\color{black}$ -m^2c^2$\egroup, and we get the equation.

\bgroup\color{black}$\displaystyle p_\mu p_\mu =-{E^2\over c^2}+p^2=-m^2c^2 $\egroup

Multiplying by \bgroup\color{black}$ c^2$\egroup and rearranging.

\bgroup\color{black}$\displaystyle E^2=p^2c^2+m^2c^4 $\egroup

Again the problem of \bgroup\color{black}$ c\neq 1$\egroup is vexing but we get the basic Energy equation of Special relativity.
\bgroup\color{black}$ \displaystyle E=\sqrt{(mc^2)^2+(pc)^2}$\egroup
We understand this as the rest energy \bgroup\color{black}$ mc^2$\egroup added in quadrature with \bgroup\color{black}$ pc$\egroup. For a particle at rest we get the rest energy equation.

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

Of course any 4-vector transforms like a 4-vector so we have the transformation equations for momentum

\bgroup\color{black}$ \displaystyle p'_x=\gamma\left(p_x-\beta {E\over c}\right)$\egroup
\bgroup\color{black}$ \displaystyle p'_y=p_y$\egroup
\bgroup\color{black}$ \displaystyle p'_z=p_z$\egroup
\bgroup\color{black}$ \displaystyle E'=\gamma\left(E-\beta p_xc\right)$\egroup

Lets start in the rest frame and do a transformation.

  $\displaystyle p_\mu=(mc,0,0,0)$    
  $\displaystyle {E'\over c}=\gamma mc$    
  $\displaystyle E'=\gamma mc^2$    
  $\displaystyle p'_x=-\beta\gamma mc= -\beta E'$    

If we consider a boost in the minus \bgroup\color{black}$ x$\egroup direction to have the particle moving in the plus \bgroup\color{black}$ x$\egroup direction afterward, then the boost transformation gives.
\bgroup\color{black}$ \displaystyle E=\gamma mc^2$\egroup
\bgroup\color{black}$ \displaystyle pc=\beta E$\egroup
These are very useful relations for many kinematic calculations.

Jim Branson 2012-10-21