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photon through mass delay

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.

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  • $\begingroup$ I am not following why there is a problem with momentum. With a mass, space is not flat. Without even considering time, the distance from A to B is larger than it would have been. So one would expect the photon to take longer. $\endgroup$
    – mmesser314
    Dec 29 '16 at 15:39
  • $\begingroup$ The problem arise especially because of the time delay. $\endgroup$
    – Misuser
    Dec 30 '16 at 1:45
  • $\begingroup$ If we consider the analogue situation with dielectric instead of gravitational mass, the only way the momentum is conserved is to have an (Abraham) momentum given to the dielectric as the photon pass through it. So if we have a time delay t, the dielectric is pushed with distance d=tE/(cM). But this can not be applied to the gravitational mass without antigravity of the photon. $\endgroup$
    – Misuser
    Dec 30 '16 at 1:56
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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).

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  • $\begingroup$ I understand that momentum and energy is not changed by the delay but the delay of momentum translates to some movement of some mass and if the delay is different for different path (straight vs curved) we can translate this difference in some displacement of mass left when we return the photon to the start possition. Which means, as crazy as it sounds, reactionless drive. $\endgroup$
    – Misuser
    Jan 12 '17 at 5:00
  • $\begingroup$ No, don't think so. In fact impossible. If you have two masses one gains the other will loose. Now, I can't off the cuff analyze if the motion has enough asymmetry that it will radiate gravitational waves, if so eventually it'll stop. Of course also if any gas etc around to slow it down. $\endgroup$
    – Bob Bee
    Jan 12 '17 at 5:09
  • $\begingroup$ From my understanding of situation, we must have a gravitational wave carrying momentum when photon pass through a variation in gravitational potential. The type of momentum radiated away depends on the gradient of the gravitational field ecountered by the photon. When the photon enters the gravity well it radiates away a "push" and when the photon exit the gravity well it radiates away a "pull" (relative to the dirrection of photon). So we will have a "+" and "-" momentum radiated away which will carry a displacement equal with the momentum of the photon times the (Shapiro) delay time. $\endgroup$
    – Misuser
    Jan 12 '17 at 8:19
  • $\begingroup$ Look, if the photon/planet system emit gravitational waves in the postulated motion of the photon trough 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 thrust 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. More in next $\endgroup$
    – Bob Bee
    Jan 13 '17 at 2:13
  • $\begingroup$ 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. So, forget about it. But, worse, the photon-planet system has no quadrupole moment, only dipole, and thus cannot radiate. General Relativity doesn't allow monopole or dipole radiation. So, again, forget about it. 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. But too small to measure, forget about it. We did detect real grav radiation from 2 black holes SEE NEXT FOR MORE $\endgroup$
    – Bob Bee
    Jan 13 '17 at 2:19
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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.

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  • $\begingroup$ Yes, the momentum is unchanged when the photon get to the point B, but what happen with the entity that caused the delay of the momentum to happen ? Should it not be moved by the momentum delay ? I mean, if we apply a momentum P to a mas M for a time t, we get a distance d = t*P/M. Or, the momentum P give the mass M a velocity v = P/M and if this momentum is applied a time t (the delay time), the mass should be left moved to a new possition at d distance. $\endgroup$
    – Misuser
    Dec 30 '16 at 2:07
  • $\begingroup$ @Misuser: no the planet won't be moved. The easy way to see this is that the equations of motion are symmetrical i.e. the motion follows the same path in reverse from B to A. The symmetry means the position of the planet can't be displaced. $\endgroup$ Dec 30 '16 at 6:19
  • $\begingroup$ You say we apply a momentum P to a mas M for a time t but that doesn't make sense. We can apply a force but we can't apply momentum. However force is the rate of change of momentum, which is maybe where you're getting mixed up. $\endgroup$ Dec 30 '16 at 6:21
  • $\begingroup$ I want to say something like this : If we ignore the Shapiro delay, or let's say it's zero, in order for us to have the photon delayed some time t, the photon must hit the mass, give the mass some speed v, move along with the mass the time t and after that the photon can continue on his path (when photon leave mass it will get back his momentum and the mass stop moving). In this situation the mass will be moved with d = tv but v = P/M =E/(cM) so the momentum P in time t change the possition of mass M with d distance. And this is what i want to say by "momentum applied for time t". $\endgroup$
    – Misuser
    Dec 30 '16 at 6:55
  • $\begingroup$ @ John Rennie : I can agree with the planet not to be moved when photon go back to point A but the delayed momentum must do something to somebody on a single path. If the photon go A->B and then B->A the effect of momentum delay is canceled out but on a single path (A->B) it is not cancelled. $\endgroup$
    – Misuser
    Dec 30 '16 at 7:02

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