Impossible dilemma about Shapiro delay and momentum conservation

Question

I'm having an impossible dilemma, with no right answer. Starting with the Shapiro delay from the General Relativity theory, I can't understand how the conservation of momentum applies to it.

The problem is like this: If I launch a photon from point $A$ to point $B$, the photon goes with light speed $c$ and gets there in time $t = (B-A)/c$. If I have a mass somewhere on the photon trajectory, the photon arrives later at the point $B$, or, it arrives at a closer point $C$ in the same time as before. So, we have for the case with the mass near photon trajectory, time $t = (C-A)/c$. (Here I've taken for simplicity a mass with a tunnel in it for the photon.)

Now, if the photon arrives later at point $B$ because of the Shapiro delay, the momentum of the photon is delayed. This delay should be apparent on the movement of the mass. Meaning, the mass should move.

The question is: which direction will the mass move when the photon passes through it, forward (+), backward (-), or will it stay put? If the mass moves forward, momentum is conserved. But the photon has antigravity. If the mass moves backward, the photon is gravitationally attractive but the conservation of momentum is violated. If the mass stays put, we violate the conservation of momentum and we have an antigravity photon.

It seems to me like an impossible situation. Any thoughts?

Note - the problem originated after the EM drive hype.

Answer 1

The photon starts with some energy $E=h\nu$. As it falls into the gravity well of the planet it is blue shifted, and then as it climbs back out of the gravity well of the planet it is red shifted again. The end result is that when the photon reached us its original energy is unchanged i.e. it is still $E=h\nu$.

The momentum of the photon is:

$$ p =\frac{h}{\lambda} = \frac{h\nu}{c} = \frac{E}{c} $$

And since the energy of the photon is unchanged by its passage through the gravity well that means the momentum is unchanged as well. Since the original momentum of the photon is unchanged when it reaches us there need be no momentum change of the planet so there is no dilemma.

As you say in the question, the fall into the gravitational well and the climb back out cause the travel time to be longer than if the planet wasn't there and the photon arrives with a delay. However this does not affect the momentum of the photon when we receive it.

An interesting question is whether the planet moves during the photon transit. Suppose instead of a photon we had a large mass passing through the planet (from left to right on your diagram). As the mass approaches from the left the planet will accelerate leftwards towards it due to the mutual gravitational attraction. Then as the mass passes through and moves away to the right the planet will accelerate to the right. These two accelerations will cancel out leaving the momentum of both the planet and the mass unchanged.

The interesting question is whether this also happens with light. In practice the oscillation in the planet's momentum as the light passed through would be far too small to measure, but in principle it must be there. This would imply there is a gravitational attraction between the planet and the light.

Answer 2

In proper distance and proper time you will see the Shapiro delay, and the distance is also larger by the same amount. The speed of light remains c. That explains your concern

If we use coordinate time and distance yes the speed of light can be different, but coordinate speed of light means nothing invariant, it's just an artifact of the coordinate system selected.

You can see this in pg. 10 of the following, which explains and calculates the effect.

http://onlyspacetime.com/Chapter_2.pdf

You can also see the graphic for the measured Shapiro delay (1 or 2 pages before) where you can see the delay is largest for the smaller impact parameter, i.e., when Venus was in (almost) superior conjunction with the Sun (that was the Shapiro measurement, Earth to Venus and back, of course they couldn't drill a hole through the Sun like you did). The delay is real, due to gravity (so, for instance, it is much bigger than the effect of the curved path if the light is skimming the surface of the Sun or not too far away) even apart from a curved path.

So there is no problem with momentum either, or energy.

Added as Edit from Comments below

If the photon/planet system emit gravitational waves in the postulated motion of the photon through the center of the planet ( big if, probably no, see below), then a small amount of momentum, and energy is carried away by the GW waves. Those will be lost by the 2 body system, and the oscillation of the photon through the planet will decrease in distance. No mystery if so, the energy and momentum they lose will go to infinity in GW waves. And if the photon looses energy and momentum (if one also the other), the freq will decrease, and it won't oscillate as far.

But, it'll be too small to measure, maybe 40 or so orders of magnitude smaller than the EM wave from an electron doing the same. But, worse, the photon-planet system has no quadrupole moment, only dipole (the angular momentum, and of course monopole, the mass), and thus cannot radiate. General Relativity doesn't allow monopole or dipole radiation – equivalently a spin two field can only radiate spin, or angular momentum, greater than 2-1 = 1, ie spin 2's, ie gravitons (for electromagnetic waves of spin 1, it has to be angular momentum greater than zero, ie, spin 1's, ie, photons).

The only way would be if the photon was orbiting the planet, I think that is a quadruple moment, and I think it's changing. Then they radiate. A changing quadrupole moment will radiate. But it still would be too small to measure – we did barely detect real gravitational radiation from 2 black holes. The two black holes were orbiting each other, and that was a changing quadrupole moment. From a total of 65 (or so) solar masses we got about 3 solar masses of radiation when the grav field was 'super' strong near their horizons. In your case it would undetectable (when further away the effect was too small and we were not able to detect it).