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I am aware of the relativistic equation:

$E^2 = (pc)^2 + (mc^2)^2$

And if we are dealing with a massless particle then $E = pc$

However I am doing some work in Astrophysics and have been told that the momentum of an ultra relativistic electron is $p =\frac E c$

I am confused as to why this is so seeing as though an electron does have a mass.

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    $\begingroup$ Hint: when $E\gg m$, the mass doesn't matter, and is only a small correction. $\endgroup$ – probably_someone Oct 24 '18 at 15:10
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    $\begingroup$ It is a valid approximation of course, electrons never reach velocity c. $\endgroup$ – anna v Oct 24 '18 at 15:14
  • $\begingroup$ You might find this helpful: Ultrarelativistic limit $\endgroup$ – Alfred Centauri Oct 24 '18 at 20:50
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That is just an approximation. Of course electrons have mass, but for an ultrarelativistic electron you have that $p \gg m \implies \frac {m^2} {p^2} \approx 0$, so it is reasonable to make such an approximation. Explicitly you have:

$$E^2=p^2c^2+m^2c^4=p^2(c^2+ \frac {m^2}{p^2}) \approx p^2c^2 \implies p \approx \frac E c$$

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As others have said, $p=\frac{E}{c}$ is an approximation in the ultra-relativistic case.

I will make this more explicit.

$p$ and $E$ are related to the spatial and temporal components of a 4-momentum vector $\tilde P$.
In terms of rapidities ($\theta$, where $v=c\tanh\theta$ for a timelike 4-momentum),
we have $p=mc\sinh\theta$ and $E=mc^2\cosh\theta$.
So, for all timelike 4-momenta, $$\frac{p}{E}=\frac{1}{c}\tanh\theta=\frac{1}{c}\left(\frac{v}{c}\right),$$ or equivalently, $$p=\frac{E}{c}\tanh\theta=\frac{E}{c}\left(\frac{v}{c}\right),$$

As $v\rightarrow c$ but never reaching $c$ (that is, as $\theta\rightarrow\infty$) [while keeping $m$ fixed], $$p \rightarrow \frac{E}{c}.$$

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If you stipulate that the electron is ultrarelativistic then you are stipulating that its kinetic energy is far greater than its invariant energy $m_ec^2$.

Write the equation for the energy of the electron and assume $pc\gg m_ec^2$ so that we can approximate the radical as so

$$E = pc\,\sqrt{1 + \left(\frac{(m_ec^2)}{(pc)}\right)^2}\approx pc\left(1 + \frac{1}{2}\left(\frac{m_ec^2}{pc}\right)^2\right) = pc\left(1 + \frac{1}{2}\left(\frac{c}{\gamma v_e}\right)^2\right)$$

Now recall that $\gamma v_e$ becomes arbitrarily large as $v_e$ approaches $c$ and so, in the ultrarelativistic limit

$$\lim_{v_e \rightarrow c} E = pc$$

For example, if $v_e = 0.99999\,c$, then $\gamma = \frac{1}{\sqrt{1 - (0.99999)^2}} \approx 223.6$ and then

$$E = pc\,(1.00001000015)$$

and so this ultrarelativistic electron's momentum is well approximated as $p = \frac{E}{c}$

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  • $\begingroup$ In my answer, my (pc/E) is exactly (v/c)... for all v. So, (1/0.99999)=1.0000100001.... which is obtained without using approximations involving $\gamma$ [but maybe one is interested in $\gamma$ for other reasons]. $\endgroup$ – robphy Oct 24 '18 at 22:18
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If a particle is ultrarelativistic it means that its momentum is way bigger than its mass. So in the equation $$E^2=(pc)^2+(mc^2)^2$$ The second term is negligible and so you can approximate E=pc. The electron mass is very small (~0.5 MeV/c^2) so the ultrarelativistic regime is reached pretty soon. To convince yourself you can compute the energy of an electron with p=10 GeV considering or not its mass. You will see that including the mass in the calculation makes a barely visible difference at those energies.

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