# Homework Help: Relative energy and momentum

1. Feb 11, 2013

### PirateFan308

1. The problem statement, all variables and given/known data
A particle of rest mass $m_0$ is moving the S-frame at a velocity of $u=(c/2, c/2, c/2)$. Calculate:

a) its energy E and its momentum p in the S-frame

b) E' and p' in the S'-frame moving at $v=(c/2, 0,0)$ relative to S

2. Relevant equations

$v_x = \frac{v_x-u}{\sqrt{1-(v_xu)/c^2}}$

$E= \gamma m_0 c^2$

$p = \gamma m_0 u$

3. The attempt at a solution
I was given the solutions, I just have difficulties understanding. The solution to part a) is fairly simple, simply calculating $\gamma = \frac{1}{\sqrt{1-(u/c)^2}}$ where $(u/c)^2 = 3/4$. Then $E= \gamma m_0 c^2 = 2m_0c^2$ and $p = \gamma m_0 u = 2m_0(c/2, c/2, c/2) = (m_0c, m_0c, m_0c)$

It's part b) that's giving me problems. The solution given is as follows:

LT for E and p. $\gamma(v) = \frac{1}{(1-1/4)^{1/2}} = 2/ \sqrt{3}$. $E' = \gamma (v) (E-vp_x) = (2/ \sqrt{3})(2-1/2)m_0c^2 = \sqrt{3}m_0c^2$

$p_x' = \gamma(v)(p_x - vE/c^2) = 2/ \sqrt{3} (m_0c - c/2*2m_0) = 0$. So then since $p_y'=p_y=m_0c$ and $p_z'=p_z=m_0c$, we get $p' = (0, m_0c, m_0c)$.

The following is how I thought the question should be solved.

Note that $v_x' = 0, v_y' = c/2, v_z' = c/2$ so that $u = (0, c/2, c/2)$. Then we would find $\gamma = \frac{1}{\sqrt{1-(u^2/c^2)}}$ where $u^2 = (c/2)^2+(c/2)^2 = 1/2c^2$ so that $\gamma = \sqrt{2}$.

Then using we find that $E= \gamma m_0 c^2 = \sqrt{2} m_0 c^2$.

We also get $p = \gamma m_0 u = \sqrt{2} m_0 (0, c/2, c/2) = (0, m_0c/\sqrt{2}, m_0c/ \sqrt{2})$.

I understand the logic of the given solution for momentum, but I don't understand where I made the mistake in my solution. Also, does anyone know how the two equations for relative energy and momentum were derived? I hadn't seen these two equations before I looked at the solutions.

I'm completely at a loss for the given solution for energy ... I'm confused why exactly we used $\gamma (v)$, when shouldn't we generally use the relative velocity of the particle with respect to that particular frame (so take $v=(0,c/2,c/2)$ with respect to the S'-frame)??

Any help would be appreciated. Thanks!

Last edited: Feb 12, 2013
2. Feb 12, 2013

### cepheid

Staff Emeritus
You can't just add velocities to transform between reference frames. Velocities don't add that way in special relativity. So it's NOT true that u' = u - v.

In more detail: it looks like you tried to used the Galilean rule for the addition of velocities (which is wrong and only holds approximately true at low speeds), when you should use the special relativistic rule. Check it out:

Hmm, I'm not quite sure where you believe I added velocities. I used Lorentz transformations to find that $v_x'=0, v_y=c/2, v_z=c/2$. Where $v_x' = \frac{v_x-u}{1-\frac{uv_x}{c^2}} = \frac{c/2-c/2}{1-\frac{(c/2)(c/2)}{c^2}} = 0$ and $v_y' = v_z = \frac{v_y-u}{1-\frac{uv_x}{c^2}} = \frac{c/2-0}{1-\frac{(0)(c/2)}{c^2}} = c/2$. It just occurred to me now that instead of using $u=(u_x,u_y,u_z)$, should I instead use $u=||u||$?
For the $u^2=(c/2)^2+(c/2)^2$, I was attempting to find the norm of u = $\sqrt{u_x^2+u_y^2+u_z^2}$ where each u_x, u_y, and u_z were all in the same frame (S'-frame).