1

See, the energy of a photon is given out by $E = pc = hv$ why don't we substitute for $p$ in $E ^2= p^2 c^2 + m^2 c^4$ by putting $p = \gamma mv$ and then get a value for $m$ (which will be $0$ for a photon) and therefore rendering the equation to $E = 0$. I am just getting a bit confused.

EDIT

Since, $p= \frac{E}{c}$ can we conclude that the momentum for red light will be lesser than that for blue light as red light has less $E$ ?

Qmechanic
  • 201,751
rahulgarg12342
  • 2,469
  • 2
    You cannot conclude that $p=0$ because $m=0$. Notice that $\gamma\rightarrow\infty$ for $v\rightarrow c$. Therefore, $m\gamma$ is undetermined. – jordix Apr 16 '14 at 15:48
  • Related: http://physics.stackexchange.com/q/3541/2451 and links therein. – Qmechanic Apr 16 '14 at 18:14

3 Answers3

6

That's because the relation $p=\gamma mv$ doesn't hold universally. As you just showed yourself, using this relation for a photon would lead to a contradiction because the energy of a photon isn't zero.

A heuristic way of seeing why this relation won't hold for a photon is by recognizing that $$p=\gamma mv =m\frac{d x}{d\tau}$$ but a photon doesn't experience any proper time, so it wouldn't seem like a sensible idea to proceed in this fashion.

Danu
  • 16,302
5

You can understand the expression by attempting the limit for $v\rightarrow c$ and $m\rightarrow0$. Notice that $\gamma\rightarrow\infty$ when $v\rightarrow c$. Therefore $m v \gamma$ is an undetermination of the form $0\cdot\infty$. From your expressions, you cannot say that $E=0$. The limit does not exist, and this implies that this expression is not valid for a massless particle.

Actually, if I can go a bit further, you can wonder what would happen if we relax one of the two properties: being massless and travelling at $c$. If the particle is not massless, then $E=m c^2 \gamma$ enforces the fact that it cannot travel at $v=c$, or otherwise $\gamma\rightarrow\infty$ and the particle would have infinite energy. Also, if it does not travel at $v=c$, then it cannot be massless, or otherwise it would not have energy. The conclusion of this would be that only massless particles can travel at $v=c$, and particles that travel at $v=c$ must be necessarily massless. In other words, a particle is massless if and only if it travels at $v=c$, which is actually a fact.

jordix
  • 221
3

by putting $p=\gamma mv$ and then get a value for $m$ (which will be 0 for a photon) and therefore rendering the equation to $E=0$

First, let's write this out in full (in 1D)

$$p = \frac{m v}{\sqrt{1 - \frac{v^2}{c^2}}} $$

Then, solve for $m$

$$m = p\frac{\sqrt{1 - \frac{v^2}{c^2}}}{v}$$

Now, holding $p$ constant, see that the limit of $m$ as $v \rightarrow c$ is zero

$$\lim_{v \rightarrow c}m = 0$$

So, a massless particle can have non-zero momentum and thus, non-zero $E$ when $v = c$.

Alfred Centauri
  • 59,560
  • Nice answer but according to the statement that "a massless particle can have non-zero momentum", won't it contradict the equation $p =\gamma mv$ itself. It will state that the momentum does not even depend on mass. – rahulgarg12342 Apr 16 '14 at 16:55
  • @rahulgarg12342, I don't see a contradiction. If the invariant mass is non-zero, the momentum is proportional to the invariant mass. If the invariant mass is zero, the four-momentum is null, and the (three-) momentum is proportional to the energy. – Alfred Centauri Apr 16 '14 at 17:05
  • Please answer the question after edit. – rahulgarg12342 Apr 16 '14 at 17:11
  • @rahulgarg12342, the momentum for a photon is $p = \frac{E}{c} = \frac{h\nu}{c} = \frac{h}{\lambda}$ so yes, a photon of red light will have less momentum than a photon of blue light. – Alfred Centauri Apr 16 '14 at 17:20
  • This is the last of my doubts. Please also tell me that why do we have to make $m$ the subject in the equation. Is there a direct logical explanation for why shouldn't $p$ be affected by zero mass. Because if we put $m = 0$ in $p = \gamma mv$ then we directly get $p = 0$ but why not otherwise. It is the same equation. – rahulgarg12342 Apr 16 '14 at 17:24
  • 1
    @rahulgarg12342, as others have mentioned, setting the invariant mass $m=0$ gives $p=0$ for any $v < c$. However, if $v = c$, $\gamma = \frac{1}{\sqrt{1- \frac{v^2}{c^2}}}$ is not defined thus, the equation $p = \gamma m v$ cannot be applied for the case $m=0, v=c$. However, one can rearrange the equation as I did and see that, in fact, $p$ can be non-zero for the case $m=0, v=c$ – Alfred Centauri Apr 16 '14 at 17:52
  • @rahulgarg12342 what Alfred is doing is looking at the limit as $v\rightarrow c$. We see that $m$ tends to zero as you approach $v=c$. This is some information that you can't extract by simply putting $v=c$ in blindly. – Danu Apr 16 '14 at 17:53