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

$$p_\mu=\left({E\over c}, p_x, p_y, p_z\right)$$

where the choice of where to put the $c$ could be made by dimensional analysis.

The dot product with itself is

$$p_\mu p_\mu =-{E^2\over c^2}+p_x^2+p_y^2+p_z^2=-{E^2\over c^2}+p^2$$

This quantity should be a Lorentz scalar, which we will call $-m^2c^2$, and we get the equation.

$$p_\mu p_\mu =-{E^2\over c^2}+p^2=-m^2c^2$$

Multiplying by $c^2$ and rearranging.

$$E^2=p^2c^2+m^2c^4$$

Again the problem of $c\neq 1$ is vexing but we get the basic Energy equation of Special relativity.

$$E=\sqrt{(mc^2)^2+(pc)^2}$$

We understand this as the rest energy $mc^2$ added in quadrature with $pc$. For a particle at rest we get the rest energy equation.

$$E=mc^2$$

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

$$p'_x=\gamma\left(p_x-\beta {E\over c}\right)$$

$$p'_y=p_y$$

$$p'_z=p_z$$

$$E'=\gamma\left(E-\beta p_xc\right)$$

Lets start in the rest frame and do a transformation.

$$p_\mu=(mc,0,0,0)$$

$${E'\over c}=\gamma mc$$

$$E'=\gamma mc^2$$

$$p'_x=-\beta\gamma mc= -\beta E'$$

If we consider a boost in the minus $x$ direction to have the particle moving in the plus $x$ direction afterward, then the boost transformation gives.

$$E=\gamma mc^2$$

$$pc=\beta E$$

These are very useful relations for many kinematic calculations.