Find total energy and momentum of an moving electron in a rest frame

Question

I have an electron moving with speed $u'$ in a frame $S'$ moving with speed $v'$ relative to a rest frame $S$.

How do I find the total energy and momentum of the electron in the rest frame $S$?

I thought the equations were:

$E_{total} = \gamma \times mc^2$
$p = \gamma \times mv$

But, that doesn't look right... Could someone point out to me what is wrong here?

Answer 1

I would simply calculate the velocity of the electron in frame S. You can either sit down and work it out using the Lorentz transforms, or just use the relativistic velocity addition formula.

Answer 2

Your equations are correct you just have to be careful about their meaning.

In an inertial frame $S$, the energy and momentum of a particle of mass $m$ are given by
$$E_S = \frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}},\quad \text{and} \quad \vec{p}_S=\frac{m \vec{v}}{\sqrt{1-\frac{v^2}{c^2}}}$$ where $v$ is the speed of the particle relative to $S$.

If you can work out the velocity of the particle relative to $S$, your formulae will give you the correct answer. However, you have only been given $v^\prime$, the velocity with respect to $S^\prime$. You can compute $v$ using the velocity addition formula.

Answer 3

The velocity-addition formulae should indeed give the correct answer. Here's what happens if you work it out using vector Lorentz transformations.

We know that the four-velocity of the electron is $S'$ frame is $\tilde u = W(ce_t' + ue_x')$, where $W=(1-[u/c]^2)^{-1/2}$. Some notes on notation: $\tilde u$ denotes the four-vector while $u$ is just the three-velocity. I'm calling this Lorentz factor $W$ to avoid confusion with the Lorentz factor of the boost (which will just be $\gamma$). So once again, $u$ goes with $W$ the way $v$ goes with $\gamma$.

Now then, the boost between the $S,S'$ frames tells us that $e_t' = \gamma( e_t + v e_x/c)$ and $e_x' = \gamma(e_x + v e_t/c)$. We can be very mechanical here and just plug these expressions back into our original four-velocity. The result is

$$\tilde u = W(c[\gamma(e_t + ve_x/c)] + u[\gamma(e_x + ve_t/c)])=W\gamma([c+uv/c]e_t+\gamma[u+v]e_x)$$

Take $\tilde p = m_e \tilde u$, and you're done. The components fall out as

$$E/c = m_e W\gamma(c + uv/c), \quad p = m_e W\gamma(u + v)$$

In the limiting case that the boost velocity $v=0$, then $\gamma=1$ and we get $E/c=m_e W c \implies E = m_e W c^2$, which checks, and $p = m_e W u$, which also checks. Similarly, if $u=0$, then $W=1$, the correct limits hold, too. It's always a good idea to check limits; doing so helped me find a mistake in this calculation which I've since corrected!

Answer 4

Method 1

Taking all motion to be along the x/x' axis, the energy-momentum 4-vector (we'll only use 2 components since this is a 1 dimension problem spatially) in the primed frame is $$ \left( \begin{array}{c} E'/c\\ p_x'\\ \end{array} \right) = m \ \gamma\left(u'\right) \left( \begin{array}{c} c\\ u'\\ \end{array} \right), $$ where $\gamma\left(u\right) = 1/\sqrt{1-\left(u/c\right)^2}$. Now just Lorentz transform to get the energy-momentum 4-vector in the unprimed frame: $$ \begin{eqnarray} \left( \begin{array}{c} E/c\\ p_x\\ \end{array} \right) &=& \gamma\left(v'\right) \begin{pmatrix} 1 & v'/c \\ v'/c & 1 \end{pmatrix} \left( \begin{array}{c} E'/c\\ p_x'\\ \end{array} \right) \\ &=& m \ \gamma\left(v'\right)\gamma\left(u'\right) \begin{pmatrix} 1 & v'/c \\ v'/c & 1 \end{pmatrix} \left( \begin{array}{c} c\\ u'\\ \end{array} \right) \\ &=& m \ \gamma\left(v'\right)\gamma\left(u'\right) \left( \begin{array}{c} c + u' v' / c\\ v' + u' \\ \end{array} \right). \\ \end{eqnarray} $$ So, $$ E = m c^2 \ \gamma\left(v'\right)\gamma\left(u'\right) \left(1 + u' v' / c^2\right) $$ and $$ p_x = m \ \gamma\left(v'\right)\gamma\left(u'\right) \left(v' + u'\right). $$ Method 2

Using the relativistic velocity addition rule, the velocity in the unprimed frame is $$ u = \frac{u' + v'}{1 + u' v' / c^2}. $$ Noting that $$ \begin{eqnarray} \gamma\left(u\right) &=& \left[1-\frac{1}{c^2}\left(\frac{u' + v'}{1 + u' v' / c^2}\right)^2\right]^{-1/2} \\ &=& \frac{1+u' v' / c^2}{\sqrt{1 + \left(u' v' / c^2\right)^2 - \left(u'/c\right)^2 - \left(v'/c\right)^2}} \\ &=& \gamma\left(v'\right)\gamma\left(u'\right) \left(1 + u' v' / c^2\right), \end{eqnarray} $$ the energy in the unprimed frame is $$ E = \gamma\left(u\right) m c^2 = m c^2 \ \gamma\left(v'\right)\gamma\left(u'\right) \left(1 + u' v' / c^2\right), $$ and the momentum in the unprimed frame is $$ p_x = \gamma\left(u\right) m u = m \ \gamma\left(v'\right)\gamma\left(u'\right) \left(1 + u' v' / c^2\right) \left(\frac{u' + v'}{1 + u' v' / c^2}\right) = m \ \gamma\left(v'\right)\gamma\left(u'\right) \left(v' + u'\right), $$ which are in agreement with Method 1.