# Bending of light in a gravitational field

• B
I have a few conceptual issues following a standard thought experiment to argue why light bends in a gravitational field and I'm hoping I can clear them up here.

Consider an observer in a lift in free-fall in a uniform gravitational field and an observer at rest in the uniform gravitational field. Suppose that a laser is fired from a fixed height on the inside of the lift from one side to the other. According to the equivalence principle, inside the lift the laws of special relativity apply and the observer in the lift will observe the laser beam to travel from one side of the lift to the other in a straight line (at a fixed height). However, the stationary observer outside the lift will observe the lift to accelerate downwards at some constant acceleration ##g##.

Now, I understand it up to this point, but doesn't one have to invoke the principle of (general) relativity, i.e. that all observers are equivalent, in order to complete the argument? To me this seems to make sense, and (personally) I would complete the argument as follows:

... Now, if the stationary observer (outside of the lift) were to observe the laser beam to propagate in a straight line then it would arrive at the other side of the lift at a different height to that observed by the observer inside the lift. This, however, would distinguish between reference frames, violating the principle of (general) relativity. In order for the principle of (general) relativity to hold we must therefore conclude that the (external) stationary observer observes the light to arrive at the other side of the lift at the same height as that recorded by the observer inside of the lift. As the lift is accelerating downwards at ##g## relative to the (external) stationary observer, this implies that the laser beam must follow a curved path. Thus light "bends" in the presence of a uniform gravitational field.

Would this argument be correct at all? (I feel I may be over thinking the issue!).

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PeterDonis
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doesn't one have to invoke the principle of (general) relativity, i.e. that all observers are equivalent, in order to complete the argument?
No. All one has to invoke is that all observers must agree on invariants. That doesn't require that all observers are equivalent (which is not really a correct statement of the general principle of relativity anyway).

In this case, the invariant is the location on the wall of the lift where the laser beam arrives. We could make it more concrete by placing a photo eye at the proper height, according to the observer free-falling with the lift, to detect the beam. Then we can just invoke the requirement that the stationary observer (the one who feels acceleration and is at rest in the gravitational field) must agree that the beam hits the photo eye. That doesn't require this observer to be equivalent in all respects to the other one; obviously he's not since he feels acceleration and the other observer doesn't. But they both agree on invariants like whether the beam hits the photo eye.

if the stationary observer (outside of the lift) were to observe the laser beam to propagate in a straight line then it would arrive at the other side of the lift at a different height to that observed by the observer inside the lift. This, however, would distinguish between reference frames
No, it would be impossible, because the beam can't hit the photo eye according to one observer but not hit it according to another. All observers must agree on something like that.

That doesn't require that all observers are equivalent (which is not really a correct statement of the general principle of relativity anyway).
Is a correct statement of the general principle of relativity that a reference frame in a homogeneous gravitational field is equivalent to a uniformly accelerating reference frame?

In this case, the invariant is the location on the wall of the lift where the laser beam arrives. We could make it more concrete by placing a photo eye at the proper height, according to the observer free-falling with the lift, to detect the beam. Then we can just invoke the requirement that the stationary observer (the one who feels acceleration and is at rest in the gravitational field) must agree that the beam hits the photo eye. That doesn't require this observer to be equivalent in all respects to the other one; obviously he's not since he feels acceleration and the other observer doesn't. But they both agree on invariants like whether the beam hits the photo eye.
Ah ok. So the point is that the detector fixed to the point on the elevator wall is frame independent and hence should be observed to be the same for all observers?
Would it be correct to say that, if the observer in the elevator observes the light beam hitting the detector at the fixed location on the opposite side of the elevator then these are coincident events in the frame of the elevator (i.e. the space-time locations of the detector and the light beam are coincident), and coincident events are frame independent, thus the observer at rest (outside of the elevator) in the gravitational field will also observe the light beam to hit the detector, hence the light beam must bend in the gravitational field?!

No, it would be impossible, because the beam can't hit the photo eye according to one observer but not hit it according to another. All observers must agree on something like that.
Is this because physics (such as a light beam hitting a detector) should be observer independent?!

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PAllen
2019 Award
Is a correct statement of the general principle of relativity that a reference frame in a homogeneous gravitational field is equivalent to a uniformly accelerating reference frame?
No, that is one variant of the principle of equivalence.

No, that is one variant of the principle of equivalence.
I thought that the general principle of relativity was the statement that all equations describing physical phenomena are "form invariant", i.e. they are of the same form in all frames of reference?!

A.T.
I thought that the general principle of relativity was the statement that all equations describing physical phenomena are "form invariant", i.e. they are of the same form in all frames of reference?!
All inertial frames.

All inertial frames.
I thought that general principle of relativity extended from inertial frames to include non-inertial frames?!

PeterDonis
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Is a correct statement of the general principle of relativity that a reference frame in a homogeneous gravitational field is equivalent to a uniformly accelerating reference frame?
No. There are several issues here.

First, as PAllen said (and you appear to agree in a later post), you are confusing "the general principle of relativity" with the equivalence principle. The general principle of relativity says that the laws of physics must take the same form in any coordinate system.

The equivalence principle, which is what I think you are actually thinking of, says that a small enough patch of any spacetime looks flat. Or, to put it another way, in a small enough patch of any spacetime, all of the laws of physics take the same form as they do in special relativity, i.e., in flat spacetime. This means that in a small enough patch of any spacetime, there is no "gravity"; it doesn't appear, because it doesn't appear in SR--there is no "gravity" in flat spacetime. But this principle holds regardless of how you choose a "reference frame".

A particular special case of the equivalence principle is the comparison between two observers who both feel the same proper acceleration: one who is uniformly accelerating in flat spacetime, the other who is "hovering" at a constant altitude in a static gravitational field, or more precisely in the curved spacetime surrounding a static massive object. In a small enough patch of spacetime around a given event on each of these observers' worldlines, all of the laws of physics look the same, since in both cases they take the same form as in SR. That means that, if each observer restricts all of his observations and measurements to such a small patch of spacetime around an event on his worldline, he can't tell which of the two situations he is in--uniform acceleration in flat spacetime or hovering in a static gravitational field. But the restriction to a small patch of spacetime is crucial: by making measurements over a large enough distance or over a long enough period of time, the observers can tell which situation they are in.

So the point is that the detector fixed to the point on the elevator wall is frame independent and hence should be observed to be the same for all observers?
The detector detecting the laser beam is frame independent and must be observed to be the same for all observers. The position of the detector is not the same for all observers; only the event of the detector detecting the laser beam is.

All inertial frames.
Not if we are talking about the general principle of relativity; that holds for all valid coordinate charts, whether they are local inertial charts or not.

I thought that general principle of relativity extended from inertial frames to include non-inertial frames?!
It does. See above.

First, as PAllen said (and you appear to agree in a later post), you are confusing "the general principle of relativity" with the equivalence principle. The general principle of relativity says that the laws of physics must take the same form in any coordinate system.
Yes, apologies for that confusion. I realised afterwards that I'd muddled them up. Thanks for the clarification though!

The detector detecting the laser beam is frame independent and must be observed to be the same for all observers. The position of the detector is not the same for all observers; only the event of the detector detecting the laser beam is.
Is it correct to say that the light beam hitting the detector at the fixed location on the opposite side of the elevator and the space-time location of the detector are coincident events in the frame of the elevator (i.e. the space-time locations of the detector and the light beam are coincident), and coincident events are frame independent, thus the observer at rest (outside of the elevator) in the gravitational field will also observe the light beam to hit the detector, hence the light beam must bend in the gravitational field?!

vanhees71
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Well, as the many words show, the equivalence principle is a shaky ground to discuss. It's a great heuristics to argue for Einstein's General Relativity, and it was the key ansatz for Einstein to formulate this most beautiful theory, but now that we have this theory it's simpler to argue using it.

First of all we have to specify what a "light beam" is. It's ray optics and thus the eikonal approximation of the free electromagnetic field in the curved spacetime of GR. The final conclusion of this ideas simply is that you can define the "light beam" as the trajectory of a fictitious massless particle in curved spacetime. As long as there are no nother interactions than gravitation, this trajectory is a light-like geodesic in spacetime and can be calculated given the pseudometric describing this spacetime (e.g., the Schwarzschild solution as a good proxy for our sun's gravitational field),
$$\frac{\mathrm{d}^2 q^{\mu}}{\mathrm{d} \lambda^2}+{\Gamma^{\mu}}_{\rho \sigma} \frac{\mathrm{d} q^{\rho}}{\mathrm{d} \lambda} \frac{\mathrm{d} q^{\sigma}}{\mathrm{d} \lambda}=0,$$
to be solved under the contraint
$$g_{\mu \nu} \frac{\mathrm{d} q^{\mu}}{\mathrm{d} \lambda}\frac{\mathrm{d} q^{\nu}}{\mathrm{d} \lambda} =0.$$
##\lambda## is an arbitrary (affine) parameter of the geodesic.

This equation can be analytically solved for the Schwarzschild metric and lead to Einstein's prediction of how much a light beam is deviating from a straight line in space (!), and this was verified by Eddington's solar-eclipse expedition in 1919, which made Einstein the first super-media star of science in history.

Battlemage!
PeterDonis
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the space-time locations of the detector and the light beam are coincident
Yes. Or to put it another way, the worldlines of the detector and the light beam intersect.

coincident events are frame independent
A better way to put it is that intersection of worldlines at a particular event is frame independent. The word "event" denotes a point in spacetime, so the intersection of the worldlines of the detector and the light beam is a single event, not "coincident events".

thus the observer at rest (outside of the elevator) in the gravitational field will also observe the light beam to hit the detector, hence the light beam must bend in the gravitational field?!
More precisely, the light beam must bend relative to the observer at rest in the field.

PeterDonis
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now that we have this theory it's simpler to argue using it.
Please note that I have adjusted the level of this thread to "B", based on the OP's apparent level of knowledge.

Well, as the many words show, the equivalence principle is a shaky ground to discuss. It's a great heuristics to argue for Einstein's General Relativity, and it was the key ansatz for Einstein to formulate this most beautiful theory, but now that we have this theory it's simpler to argue using it.

First of all we have to specify what a "light beam" is. It's ray optics and thus the eikonal approximation of the free electromagnetic field in the curved spacetime of GR. The final conclusion of this ideas simply is that you can define the "light beam" as the trajectory of a fictitious massless particle in curved spacetime. As long as there are no nother interactions than gravitation, this trajectory is a light-like geodesic in spacetime and can be calculated given the pseudometric describing this spacetime (e.g., the Schwarzschild solution as a good proxy for our sun's gravitational field),
$$\frac{\mathrm{d}^2 q^{\mu}}{\mathrm{d} \lambda^2}+{\Gamma^{\mu}}_{\rho \sigma} \frac{\mathrm{d} q^{\rho}}{\mathrm{d} \lambda} \frac{\mathrm{d} q^{\sigma}}{\mathrm{d} \lambda}=0,$$
to be solved under the contraint
$$g_{\mu \nu} \frac{\mathrm{d} q^{\mu}}{\mathrm{d} \lambda}\frac{\mathrm{d} q^{\nu}}{\mathrm{d} \lambda} =0.$$
##\lambda## is an arbitrary (affine) parameter of the geodesic.

This equation can be analytically solved for the Schwarzschild metric and lead to Einstein's prediction of how much a light beam is deviating from a straight line in space (!), and this was verified by Eddington's solar-eclipse expedition in 1919, which made Einstein the first super-media star of science in history.
Thanks for the details. I understand that one can show that massless particles obey the geodesic equation you quote and hence the path of a photon in a gravitational field will appear to be curved relative to a stationary observer in said gravitational field. I was really hoping to try and understand the thought experiment that is often used (before introducing the mathematics) for why light must bend (relative to a stationary observer) in a gravitational field due to the equivalence principle?!

Yes. Or to put it another way, the worldlines of the detector and the light beam intersect.

A better way to put it is that intersection of worldlines at a particular event is frame independent. The word "event" denotes a point in spacetime, so the intersection of the worldlines of the detector and the light beam is a single event, not "coincident events".

More precisely, the light beam must bend relative to the observer at rest in the field.
OK cool, I think I understand it a bit better now.
So would it be correct to say that in general, although two different observers will assign different temporal and spatial coordinates to a given physical phenomenon, e.g. a light beam hitting a detector, they will both agree that the phenomenon occurred (it is observer independent), that is, they will both agree that the light beam hit the detector. It would be absurd if the light beam hit the detector according to one observer but not another.

PeterDonis
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would it be correct to say that in general, although two different observers will assign different temporal and spatial coordinates to a given physical phenomenon, e.g. a light beam hitting a detector, they will both agree that the phenomenon occurred (it is observer independent), that is, they will both agree that the light beam hit the detector. It would be absurd if the light beam hit the detector according to one observer but not another.
Yes, this is what we already said.

Yes, this is what we already said.
OK great, thanks for your help!

By the way, what is the correct terminology when referring to such frame invariant physical occurrences (for want of a better description), since the term event seems to be exclusively reserved for its space-time location?! Does one simply say that physics is observer independent?!

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Yes, this is what we already said.
OK great, thanks for your help!

By the way, what is the correct terminology when referring to such frame invariant physical occurrences (for want of a better description), since the term event seems to be exclusively reserved for its space-time location?! Does one simply say that physics is observer independent?!
...Or would it be better to say that physical processes/phenomena are observer independent (they possess an objective reality)?! (Apologies, I think I may be over thinking the whole thing at this point).

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PeterDonis
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what is the correct terminology when referring to such frame invariant physical occurrences (for want of a better description), since the term event seems to be exclusively reserved for its space-time location?
"Event" can refer to a point in spacetime or the frame-independent occurrence that happens at that point. Since the frame-independent occurrence is how we identify points in spacetime, the first meaning depends on the second anyway.

PeterDonis
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someone recently published an article deriving and promoting it

https://arxiv.org/pdf/1503.01970v1.pdf

I don't see anything at all talking about "linear curvature" or any of the claims relating to it that you have made in this thread. Nor do I see a justification for them in the other thread. Accordingly, I am deleting the entire subthread relating to these claims.

"Event" can refer to a point in spacetime or the frame-independent occurrence that happens at that point. Since the frame-independent occurrence is how we identify points in spacetime, the first meaning depends on the second anyway.
OK, thanks for the the clarification! So space-time points are themselves observer independent (absolute), i.e. they exist independently of any observer (coordinate chart)?!
Would it be correct at all to say that physical processes/phenomena are observer independent (that is, they possess an objective reality)?! So if a physical process is observed to occur in one frame of reference, then it will occur in all other frames of reference (i.e. all observers will agree that the given physical process occurred).
Apologies, I think I may be over thinking the whole thing at this point!

PeterDonis
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2019 Award
I think I may be over thinking the whole thing at this point!
Yes, you are. You're just restating what has already been said.

Yes, you are. You're just restating what has already been said.
OK fair enough, I'll stop rambling. Thanks for all your help and merry Christmas!