Answer to Question #185816 in Mechanics | Relativity for Atiqa

Question

A particle of mass m0 moving with velocity u in the first frame. The second frame is moving

at velocity v with respect to the first frame, along the x-axis, and the particle has a velocity u

´as seen in the second frame. Showing that momentum and energy are the components of a

4-vector is to write the momentum and energy of a particle in terms of its real mass and

velocity in one Lorentz frame and then use the Lorentz transformation for velocity (that is,

the velocity addition law) to and the form of the momentum and energy of the particle in a

second Lorentz frame.

Answer 1

The velocity 4-vector can be defined as,

$v_\mu=\dfrac{\partial x_\mu}{\partial\tau}=\gamma\dfrac{\partial x_\mu}{\partial t}=\gamma\dfrac{\partial }{\partial t}(ct,\vec{x})=\gamma(c,\dfrac{\partial \vec{x}}{\partial t})=(\gamma c,\vec{v})$

So we see that the time-component of the usual velocity vector is  $\gamma c$  and we have the velocity 4-vector

$v_\mu=\gamma (c,\vec{v})$

where $\gamma$  is for the transformation from the rest frame to whatever frame we are defining $v_\mu$  in.

We can dot the velocity 4-vector into itself.

$v_\mu v_\mu=\gamma^2(-c^2+v^2)=\dfrac{-c^2+v^2}{1+\beta^2}=-c^2\dfrac{1-\beta^2}{1-\beta^2}=-c^2$

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  no matter what the velocity is. To be consistent with non-relativistic equations we will define the momentum.

$p_\mu=mv_\mu=\gamma(mc,m\vec{v})$

If we identify the time component as above, $\dfrac{E}{c}=\gamma mc^2$ , we have the relation

$E=\gamma mc^2$

Inserting the energy into the momentum 4-vector. Start with the energy equation from above

$E=\sqrt{(mc^2)^2+(pc)^2}=mc^2\sqrt{1+(\dfrac{pc}{mc^2})^2}$

$\approx mc^2(1+\dfrac{1}{2}(\dfrac{pc}{mc^2})^2)=mc^2+\dfrac{p^2}{2m}$

Lorentz Velocity Transformation

$u_x'=\dfrac{u_x-v}{1-u_x\dfrac{v}{c^2}}$

As per the data provided in question,

$u_x=u$

$v=v$

$u'_x=u'$

Rewriting the above equation,

$u'=\dfrac{u-v}{1-u\dfrac{v}{c^2}}$