Proof for Lorentz Transformation of Momentum: Step Explained

[little neutrino]

Hi. In the attached proof for Lorentz transformation for momentum http://www.colorado.edu/physics/phys2170/phys2170_sp07/downloads/lorentz_transformation_E_p.pdf, there is this step that I don't understand:

1/√1-u'²/c² = γ(1-vu_x/c²)/√1-u²/c²

Can someone explain how they derived this? Thanks! :)

 

[TeethWhitener]

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

 

[little neutrino]

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

 

[PeroK]

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

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!

 

[TeethWhitener]

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

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.

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

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

 

[PeroK]

Spoiler

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.

 

[TeethWhitener]

Wow, that is pretty.

 

[robphy]

With rapidities (i.e. trigonometry), \theta'=\theta-\theta_V,
\begin{align*}\cosh\theta&#039;<br /> &amp;=\cosh(\theta-\theta_V)\\<br /> &amp;=\cosh\theta\cosh\theta_V-\sinh\theta\sinh\theta_V\\<br /> &amp;=\cosh\theta\cosh\theta_V(1-\tanh\theta\tanh\theta_V)\\<br /> \gamma_{u&#039;}<br /> &amp;=\gamma_{u}\gamma_{V}(1-uV)<br /> \end{align*}

 

[little neutrino]

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!

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

 

[little neutrino]

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.

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

 

[little neutrino]

Spoiler

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.

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