# Showing that Energy-momentum relation is invariant

Tags:
1. Oct 5, 2014

### Rubber Ducky

1. The problem statement, all variables and given/known data

A particle of mass $m$ is moving in the $+x$-direction with speed $u$ and has momentum $p$ and energy $E$ in the frame $S$.

(a) If $S'$ is moving at speed $v$, find the momentum $p'$ and energy $E'$ in the $S'$ frame.
(b) Note that $E' \neq E$ and $p' \neq p$, but show that $(E')^2-(p')^2c^2=E^2-p^2c^2$

2. Relevant equations

$E=p^2c^2+m^2c^4$
$E'=\gamma mc^2=\frac{mc^2}{\sqrt{1-\frac{(u')^2}{c^2}}}$
$p'=\gamma mu'=\frac{mu'}{\sqrt{1-\frac{(u')^2}{c^2}}}$
$u'=\frac{u-v}{1-\frac{uv}{c^2}}$

3. The attempt at a solution

My prof says we need to use (a) to derive (b). It seems like it'd be easier a different way, but whatever. My strategy has been to show that $(E')^2-(p')^2c^2=m^2c^4$, which would complete the derivation.

I tried playing around with some algebra and got $(E')^2=\frac{m^2c^4(c^2-uv)^2}{(c^2-u^2)(c^2-v^2)}$ and $(p')^2=\frac{m^2(u-v)^2}{(1-\frac{uv}{c^2})^2(c^2-u^2)(c^2-v^2)}$

I got a little overwhelmed with the algebra at this point. I'd like to know if this is right so far, at least, before trying anything else.

2. Oct 5, 2014

### vela

Staff Emeritus
For part (a), you should be able to show that
\begin{eqnarray*}
E' &= \gamma(E - \beta pc) \\
p'c &= \gamma(pc - \beta E)
\end{eqnarray*} where $\beta = v/c$ and $\gamma = 1/\sqrt{1-\beta^2}$.

Your expression for $E'^2$ looks okay. Divide both the numerator and denominator by $c^4$ and take the square root. You should be able to show the first equation above. Just make sure you keep track of the three $\gamma$'s you have in this problem.

Last edited: Oct 6, 2014
3. Oct 6, 2014

### Rubber Ducky

Thanks very much for the help. It worked like a charm - the last part of the derivation that I ended up using was pretty neat. My only worry is that my prof will wonder how I thought to express $E'$ and $cp'$ in that way.

4. Oct 6, 2014

### vela

Staff Emeritus
Look up the Lorentz transformation. You might already be familiar with it applied to x and t. It applies to all four-vectors.

5. Oct 6, 2014

### Rubber Ducky

Interesting - we've covered the Lorentz transformation, but not four-vectors. My class gives a brief overview of SR, but maybe I'll see it in future.

I have another question that I'm editing into the OP now - It's somewhat related, but I didn't want to flood the section with threads.

EDIT: Woops, guess it's been too long to edit. See you in the next thread XD