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I have studied from Griffiths, the relativistic form of momentum is $$p = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} m_0v$$

Now when I evaluate the momentum for photon, I just insert $v=c$ and $m_0=0$ and I get $p= 0/0$. How does it make sense?

Can you tell me that where I am wrong?

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    $\begingroup$ possible duplicate of If photons have no mass, how can they have momentum? $\endgroup$
    – John Rennie
    Commented Jun 16, 2014 at 13:04
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    $\begingroup$ You have to take the double limit $m \to 0$ and $v \to 1$ (I'm using $c=1$ units), don't simply set $m = 0$ and $v = 1$. You know how the the mass $m$ is bound to energy and momentum from the energy-momentum relation. So, in $p = mv/\sqrt{1 - v^2}$ replace $m$ by $\sqrt{E^2 - p^2}$ and let $v$ be $1$, in this way you will recover $p = E$, but starting from energy-momentum relation and setting $m = 0$ is much simpler. $\endgroup$
    – giordano
    Commented Jun 16, 2014 at 13:48
  • $\begingroup$ can you plz explain without setting c=1. from the equation $\vec{p}=\frac{m \vec{v}}{\sqrt{1-\frac{v^2}{c^2}}}$ to E =pc please? $\endgroup$
    – zero_field
    Commented Jun 16, 2014 at 15:52

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You should consider a particle with some finite energy $E$ and use that constraint to take the $v\rightarrow c$ limit.

With Lorentz factor $\gamma = 1/\sqrt{1-v^2/c^2}$, the relativistic total energy is $E = \gamma mc^2$. Therefore, $p/E = v/c^2$. With the particular case of $v = c$, it follows that $E = pc$.

Although really, you should simply consider $E = pc$ for massless particles to be more fundamental. The general relation is $(mc^2)^2 = E^2 - (pc)^2$, which corresponds to the the norm-squared of the four-momentum vector in relativity.

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  • $\begingroup$ I understand how to get this form, but $p = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} mv$ is it not valid for photon momentum ? why not? $\endgroup$
    – zero_field
    Commented Jun 16, 2014 at 12:22
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    $\begingroup$ @zero_field: it's not valid because photons have zero mass and $0/0$ doesn't mean anything. However, having the constraint of some particular energy enables you to take the limit in a way that gets the correct result--with the numerator and denominator changing in a consistent way. $\endgroup$
    – Stan Liou
    Commented Jun 16, 2014 at 12:33
  • $\begingroup$ I have found that form in Griffiths, so why cant i use that for photon? $\endgroup$
    – zero_field
    Commented Jun 16, 2014 at 12:49
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    $\begingroup$ @zero_field: You have just discovered why yourself in the question you asked. Why is is the fact that it is found in Griffiths relevant? Which Griffiths, anyway? (If it's Introduction to Electrodynamics, then he discusses just this issue for photons in section 10.2.3 in the 1981 ed.--sorry, I don't have the newer one on-hand right now.) $\endgroup$
    – Stan Liou
    Commented Jun 16, 2014 at 12:55
  • $\begingroup$ Yes it is Introduction to Electrodynamics, page no 509 and the topic is, relativistic energy and momentum. $\endgroup$
    – zero_field
    Commented Jun 16, 2014 at 13:07

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