# Proof for momentum Lorentz Transformation

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1. Feb 21, 2016

### little neutrino

2. Feb 22, 2016

### TeethWhitener

Which part of this step don't you understand? Can you show exactly where you get stuck?

3. Feb 23, 2016

### little neutrino

I tried substituting the Lorentz transformation for u'
ux' = (ux-v)/(1-uxv/c2)
into the LHS
and simplifying it from there, but I couldn't derive the expression on the RHS :(

4. Feb 23, 2016

### PeroK

Yes, it's quite difficult algebra to derive this identity. You have three gamma factors and a velocity transformation formula to manage. You just have to keep working at it.

There is a quicker way using the properties of 4-vectors and the proper time of the particle if you are comfortable with those?

Actually, I've just noticed a very quick way simply using the transformation of the time coordinates $t$ and $t'$.

Let me know if you are interested in the quick way!

Last edited: Feb 23, 2016
5. Feb 23, 2016

### TeethWhitener

Maybe this is a trivial point, but you have to use the Lorentz transformed expressions for $u'_y$ and $u'_z$ as well as $u'_x$. Start with $u'^2=u'^2_x+u'^2_y+u'^2_z$ in the LHS. As PeroK says, there's a lot of messy algebra to work through, but it should work out.

Well, I'm certainly interested if OP isn't.

6. Feb 23, 2016

### PeroK

The relationship between the proper time of the particle and the two coordinate times is:
$\frac{dt}{d\tau} = \gamma, \frac{dt'}{d\tau} = \gamma '$ (1)

Differentiate the transformation for time wrt proper time:

$t' = \gamma_V(t - Vx/c^2)$

$\frac{dt'}{d\tau} = \gamma_V(\gamma - V \gamma v_x/c^2) = \gamma_V\gamma (1- Vv_x/c^2)$

Then, using (1) we have:

$\gamma ' = \gamma_V\gamma (1- Vv_x/c^2)$

In fact, you can derive the energy-momentum transformation much more easily by differentiating the components of position wrt proper time of the particle.

7. Feb 23, 2016

### TeethWhitener

Wow, that is pretty.

8. Feb 23, 2016

### robphy

With rapidities (i.e. trigonometry), $\theta'=\theta-\theta_V$,
\begin{align*}\cosh\theta' &=\cosh(\theta-\theta_V)\\ &=\cosh\theta\cosh\theta_V-\sinh\theta\sinh\theta_V\\ &=\cosh\theta\cosh\theta_V(1-\tanh\theta\tanh\theta_V)\\ \gamma_{u'} &=\gamma_{u}\gamma_{V}(1-uV) \end{align*}

9. Feb 27, 2016

### little neutrino

Sflr I was caught up with some stuff this week :/ I just figured out the algebraic manipulation after substitution of u' ; I realised that I had to observe the final expression more closely instead of blindly manipulating the algebra. Thanks so much for your help! :)

10. Feb 27, 2016

### little neutrino

Yup, I finally worked through the algebra. Thanks! :)

11. Feb 27, 2016

### little neutrino

Woah that's sweet XD Thanks for the alternative solution!