# Shouldn't the momentum of a photon, and hence its energy, become 0 if it's trapped inside two perfectly reflecting walls?

A photon has no mass, but it does have momentum. So, let's say, we cause a photon to constantly reflect back and forth two perfectly reflecting walls.

Since momentum is a vector, it should become zero and hence, the photon would cease to exist (no momentum, no energy.) So... what happens to the photon?

ALSO, I had another question:

I know, I know, this is a pretty triggering question to a lot of people, but my teachers seems to stress on the fact that a photon has relativistic mass but no rest mass.

I get the rest mass part of the equation, but don't understand how it even has relativistic mass when it doesn't even have a rest mass to begin with? According to the formula for relativistic mass, don't we require a rest mass to begin with?

• How should it's momentum become zero? It will still have momentum whenever you measure it.
– Max
Dec 4, 2020 at 16:25
• Ah, yes! At any instant, yes, but isn't the OVERALL momentum zero? Dec 4, 2020 at 16:28
• Yes but that is like saying if I run to somewhere and back my overall velocity is zero. Or even better, my overall momentum is zero, so I cease to exist
– Max
Dec 4, 2020 at 16:30
• The overall momentum won't be zero, the photon still has momentum at any given moment. If you were to measure the average over some time period you would find it averages to zero momentum though if it's just confined to bouncing between two reflective walls. Dec 4, 2020 at 16:32
• If I bounce a ball between my hand and the ground, does the ball stop moving? Dec 4, 2020 at 17:21

The photon reflecting between walls can be compared to any other particle bouncing to and fro between walls: such a particle has a non-zero momentum at any instant, but zero momentum when averaged over time. There is, let's acknowledge, something of a difference because we can regard the photon's state as a simultaneous combination of the two momentum states (called a superposition) but the point about momentum still stands. It exchanges momentum with the wall at each reflection, thus exerting a pressure.

As you will find repeatedly urged on this site (by experts in relativity at least), it is best not to invoke the concept known as "relativistic mass". Rather, each particle has an energy $$E$$ and a momentum $$p$$, and the combination $$E^2 - p^2 c^2$$ is an important property of the particle, because its value does not depend on which inertial frame of reference is used to define the energy and the momentum at any given moment in the "life" of the particle (the strict terminology is "at any given event".) With this in mind, let's introduce a letter to stand for this combination of energy and momentum. The way it is done is to use $$m$$ but define it this way: $$E^2 - p^2 c^2 = m^2 c^4.$$ This $$m$$ is called the rest mass of the particle. We then find there are two cases that arise: either $$m=0$$ or $$m \ne 0$$. If $$m=0$$ then we have something like a photon or a pulse of light. It will be travelling at the speed of light. If $$m \ne 0$$ then we have something like a thing made of matter, or a particle such as electron or proton, and it will be travelling at less than the speed of light relative to any frame of reference.

For the second case ($$m \ne 0$$) and for this case only, it is then found that $$E = \frac{1}{\sqrt{1 - v^2/c^2}} m c^2$$ and $$p = \frac{1}{\sqrt{1 - v^2/c^2}} m v.$$

Thus everything is fully and precisely set out and there is no need to introduce anything called "relativistic mass". The reason why it gets introduced is that people wish to point out the similarity between the above formula for momentum and the Newtonian formula $$p = m v$$, so they combined the $$v^2$$ part with the $$m$$ to make $$p = \left( \frac{1}{\sqrt{1 - v^2/c^2}} m \right) \times v.$$ But we could equally well think of it as $$p = m \times \left( \frac{v}{\sqrt{1 - v^2/c^2}} \right) .$$

• Makes perfect sense, thank you so much! Dec 5, 2020 at 16:32

I recommend avoiding "relativistic mass", it's not helpful. Worse: it seems to imply that there is an absolute rest frame so that moving near $$c$$ is somehow special because you have diverging mass. (You don't, you're moving near $$c$$ right now, in some frame...and nothing is different...that's "relativity").

I also recommend not thinking about the frame of a photon. PBS Spacetime, Fermilab, and so on yt videos are full of "but a photon doesn't experience time" questions. There is no inertial frame moving at $$c$$, and thinking there is only leads to confusion.

However: you can make progress with affine-parameter, $$\xi$$. Suppose a particle travels on a path $$P(\tau)$$ (parametrized by its proper-time, $$\tau$$). The 4-velocity is:

$$u^{\mu} = \frac{dP}{d\tau} = \frac{dx^{\mu}}{d\tau}$$

from that, you can define 4-momentum:

$$p^{\mu} = mu^{\mu} = m\frac{dx^{\mu}}{d\tau} = \frac{dx^{\mu}}{d\xi}$$

where:

$$\xi \equiv \tau/m$$

is a renormalized version of proper time (aka: the affine parameter).

Now for a massive particle moving slower than light:

$$u^{\mu} = \frac 1 {\sqrt{1-\frac{v^2}{c^2}}}(c, \vec v)$$

and

$$p^{\mu}=mu^{\mu} = \frac 1 {\sqrt{1-\frac{v^2}{c^2}}}(mc, m\vec v) =(E/c, \vec p)$$

is all good, but if $$||v||=c$$, the 4-velocity equation breaks, and you are stuck.

Now you apply the affine parameter ($$\tau, m\rightarrow 0$$, with $$\tau/m$$ fixed), because it allows you to get 4-momentum by skipping the 4-velocity step:

$$p^{\mu} = \frac{x^{\mu}}{d\xi} = (||\vec p||, \vec p)$$

which becomes, in quantum mechanics:

$$p^{\mu} = (\hbar\omega/c, \hbar\vec k)$$

with

$$\omega = c||\vec k||$$

So that's what you do at $$v=c$$. It's good to know, but it will not help in understanding any of the classic relativity paradoxes.

The notion that a photon has relativistic mass is deeply wrong. One related concern that pops up a lot is, "if a photon has no mass, how can it have momentum?". It is entirely based on the Newtonian concept of momentum.

If you look at a massive particle, the momentum per unit energy is:

$$\frac p E = \frac{\gamma mv}{\gamma mc^2} = \frac v {c^2}\rightarrow \frac 1 c$$

where the arrow is as $$\vec v \rightarrow c$$.

Meanwhile, for a photon:

$$\frac p E = \frac{\hbar k}{\hbar \omega} = \frac k{\omega} = \frac 1 c$$

The point is, mass just slows you down (literally), and makes it harder to get momentum out of energy. This is in complete contrast to our Newtonian intuition, which says:

$$p = \sqrt{2mE}$$