# Speed of light not an invariant in GR

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• Pyter
In summary, Einstein states that the ##c## is not an invariant in GR, and that from (70) and (73), it stems that $$\gamma = \sqrt{ -\frac {g_{44}}{g_{22}} },$$ where ##\gamma## is ##|c| <= 1## (in relativistic units), and =1 in a Lorentz frame when ##g_{44} = -1## and ##g_{22} = 1##. Then Einstein proceed to compute the light ray curvature using the partial derivative of ##\gamma##. Questions:If the computed |c| in a frame subjected to gravity is generally <=
Pyter
TL;DR Summary
Speed light modulo changes to allow ligth ray bending in non-inertial frames.
Hi all,
I need help understanding the light ray bending in the original GR 1916 paper, Die Grundlagen....
First of all, Einstein states the ##c## is not an invariant in GR.
In fact, from (70) and (73), it stems that $$\gamma = \sqrt{ -\frac {g_{44}}{g_{22}} },$$ where ##\gamma## is ##|c| <= 1## (in relativistic units), and =1 in a Lorentz frame when ##g_{44} = -1## and ##g_{22} = 1##.
Then Einstein proceed to compute the light ray curvature using the partial derivative of ##\gamma##.
Questions:
1. If the computed ##|c|## in a frame subjected to gravity is generally ##<= 1##, how come when an observer measures it, it's always 1?
2. Since the ##g_{**}## are not unitary in an accelerated frame not subjected to gravity, this means that the invariance of ##c## doesn't apply also in this case, i.e. mass-less flat space?
3. Can you explain the following equations where E. computes the total curvature, in particular $$curvature = -\frac {\partial {\gamma}} {\partial n}$$? He talks about the Huygens principle, but that formula doesn't seem a direct consequence.

Pyter said:
First of all, Einstein states the##c## is not an invariant in GR.
The precise statement in GR is that the speed of light is ##c## in all local inertial frames, and that is consistent with what Einstein is saying here.

The qualifier about local inertial frames is often not stated in discussions of general relativity because by the time you get to studying GR you're presumed to understand the issues around coordinate velocities and non-inertial frames. You do not want to take on GR and curved spacetime until you have a clear understanding of these issues; the thread that @PeroK linked above would be a good start.

vanhees71
Pyter said:
Summary:: Speed light modulo changes to allow ligth ray bending in non-inertial frames.

If the computed |c| in a frame subjected to gravity is generally <=1, how come when an observer measures it, it's always 1?
The second postulate says that light moves at c in an inertial frame. In non-inertial frames the coordinate speed of light is arbitrary and can even be much greater than c. In GR there are no global inertial frames, only local ones.

What is invariant is that light always travels on null geodesics. In a local inertial frame null geodesics are the paths traveling at c. But even frames where the coordinate speed of light is not c agree that it is a null geodesic. So the invariant fact that generalizes the second postulate is that light travels on null geodesics.

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cianfa72, meekerdb, hutchphd and 3 others
In the paper
$$\gamma=\sqrt{(\frac{dx_1}{dx_4})^2+(\frac{dx_2}{dx_4})^2+(\frac{dx_3}{dx_4})^2}$$
is speed of light which pass tangential to the sun. So your equation means light passing nearer the sun slower its speed is. So the light bend down to the sun slightly. You know "slower" is the observation who use system time ##x_4##. Local time which does not coincide with ##x_4## goes slower, the nearer to the sun. These changes cancel so that local observation of light speed is c in any place.

meekerdb and Pyter
Pyter said:
Can you explain the following equations where E. computes the total curvature, in particular $$curvature = -\frac {\partial {\gamma}} {\partial n}$$? He talks about the Huygens principle, but that formula doesn't seem a direct consequence.
Einstein used Huygens principle to calculate the deflection of light. This calculation is similar to how you calculate the deflection of light due to diffraction in optics. Diffraction occurs because the speed of light changes if you go from one medium to another. In the presence of gravity, the coordinate speed of light becomes coordinate-dependent. So, using Huygens principle, if you integrate all the wavefronts with the changing speed of light, you get deflection.

Some time ago I saw a textbook on GR which performed the same calculation. Nowadays, most textbooks use a different approach. Einstein's approach can be found here,

https://www.mathpages.com/rr/s8-09/8-09.htm

where the peculiar factor of 2 is also explained. That factor turns out to be quite subtle!

Pyter, Dale, vanhees71 and 1 other person
Thanks to all for the replies, it's much clearer now.

The source of the misunderstanding was also that Einstein, in his Relativity: The Special and General Theory, states that:

[90]
In the second place our result shows that, according to the general theory of relativity, the law of the constancy of the velocity of light in vacuo, which constitutes one of the two fundamental assumptions in the special theory of relativity and to which we have already frequently referred, cannot claim any unlimited validity. A curvature of rays of light can only take place when the velocity of propagation of light varies with position. Now we might think that as a consequence of this, the special theory of relativity and with it the whole theory of relativity would be laid in the dust. But in reality this is not the case. We can only conclude that the special theory of relativity cannot claim an unlimited domain of validity; its results hold only so long as we are able to disregard the influences of gravitational fields on the phenomena (e.g. of light).

Where it's not clear at all that, in the GR case, he refers to coordinate velocity .

Pyter said:
We can only conclude that the special theory of relativity cannot claim an unlimited domain of validity; its results hold only so long as we are able to disregard the influences of gravitational fields on the phenomena (e.g. of light).
Einstein stated equivalence principle. Frame of reference of free falling observer is local IFR. You go anywhere holding wearable light speed measurement machine with you, switch it on jumping into a pitfall. You measure light speed is c. Your friend next to you on the ground would also measure light speed c with his own machine, i.e. in GR local speed of light is constant c.

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vanhees71
@anuttarasammyak but my friend on the ground is not in an IFR. Why should he measure c = 1 then?

Pyter said:
@anuttarasammyak but my friend on the ground is not in an IFR. Why should he measure c = 1 then?

A better way of phrasing this question is that your friend on the ground is not in free fall. He is still "in an IFR"; he just isn't at rest in one.

anuttarasammyak said:
Your friend next to you on the ground would also measure light speed c with his own machine

Careful. Observers with nonzero proper acceleration cannot be assumed to always measure the speed of light to be ##c##, because they are not at rest in an IFR, and the strictly correct version of the second postulate of SR requires that the observer measuring the speed of light be at rest in an IFR.

The way this point is usually finessed is to say that the measurement in question is not actually being made by the observer with nonzero proper acceleration, but by a second observer in free fall (zero proper acceleration) who is momentarily comoving with the first observer during the measurement.

cianfa72 and Pyter
Pyter said:
but my friend on the ground is not in an IFR. Why should he measure c = 1 then?
You are right to ask, because @anuttarasammyak's wording here was rather careless.

However, you have just tripped over another wording-related hazard: that phrase "not in an IFR" is dangerously misleading. Frames (whether inertial are not) are not things that we are "in", such that free-faller might be in one and I in another, so the issue is not that free-faller is "in" an inertial frame and ground guy is not. Frames are things that we use to assign coordinates to events and then to calculate things like coordinate velocities; thus everything is "in" whatever frame we choose.
Yes, you'll hear people saying things like "in my frame" all the time... but that's because the more precise "doing the calculations using the coordinates assigned by the frame I find most convenient, which is one in which my spatial coordinates are constant and which is to very good approximation inertial" is a mouthful. No one is going to write that out every time.

A more accurate statement of what is going on is something like: There is a local inertial reference frame in which free-faller is at rest; if we choose to use the coordinates assigned by that frame, we will find that the speed of light is ##c##. There is also a local inertial frame in which ground-guy is at rest and there is a downwards gravitational force; if we choose to use the coordinates assigned by that frame we will find that the speed of light is ##c##. The second is the way we usually think about experiments we perform in a lab in a building on the surface of the earth.

And do note that in both cases the inertialness of the frame is an approximation. That's implied by the word "local" - sufficiently careful measurements over a large enough distance will show curvature effects.

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Pyter and Dale
Nugatory said:
There is also a local inertial frame in which ground-guy is momentarily at rest

To be clear, the qualifier I put in bold above should be added, to make clear the key difference between this case and the free-faller case.

cianfa72
PeterDonis said:
The way this point is usually finessed is to say that the measurement in question is not actually being made by the observer with nonzero proper acceleration, but by a second observer in free fall (zero proper acceleration) who is momentarily comoving with the first observer during the measurement.

Momentarily comoving in the sense that their world-lines intersect at just one point?

Pyter said:
Momentarily comoving in the sense that their world-lines intersect at just one point?
Not just intersect, but are tangent. i.e. matching velocities. But yes, at one point.

So there's no way to measure c < 1 or even > 1 in any frame, IFR or not?

How come when we see the light bent by a physical lens, we agree that it's (non-coordinate) velocity has decreased due to ##\mu <> \mu_0## and/or ## \epsilon <> \epsilon_0## in the medium, but when we see it bent by a gravitational lens we assume that its velocity has always been 1?

Pyter said:
So there's no way to measure c < 1 or even > 1 in any frame, IFR or not?
Sure you can. The easiest example is a rotating reference frame.

@Dale like the Earth?

Sure, but you can make your frame rotate any speed you like. You are not restricted to the earth’s rotation.

At a rotation of about 1 Hz I think that the moon is moving faster than c, and light is going faster than that in the direction of rotation.

@Dale but you're talking about Moon's and light's coordinate velocities, not "physical", measured velocities.

If what you measure is coordinate velocity, shouldn't there be an anisotropy in c measured in the direction of rotation and ortogonally to it, detected by a Michelson-Morley-like experiment?

Pyter said:
@Dale but you're talking about Moon's and light's coordinate velocities, not "physical", measured velocities.
Yes, that is what you were talking about from the beginning.

@Dale ok, but my question #16 was about measured velocities, not "computed" ones like the coordinate velocities.

Pyter said:
but you're talking about Moon and c coordinate velocities, not "physical", measured velocities.
All velocities are coordinate velocities, there’s no other kind. What you’re calling a “physical” velocity is a coordinate velocity calculated using an inertial reference frame.

cianfa72, Abhishek11235 and PeroK
Dale said:
At a rotation of about 1 Hz I think that the moon is moving faster than c, and light is going faster than that in the direction of rotation.
Are you sure about "light is going faster than that in the direction of rotation"? I was under the impression that, for the equivalence principle, the rotating frame should be equivalent (at least locally) to a static frame in a gravitational field, so you should see the light bending phenomenon and consequently compute a coordinate velocity c < 1.

Nugatory said:
All velocities are coordinate velocities, there’s no other kind. What you’re calling a “physical” velocity is a coordinate velocity calculated using an inertial reference frame.
According to this post, measured velocity is ##ds/d\tau##, which should be an invariant in any FR, non-IFRs included.

Pyter said:
@Dale ok, but my question #16 was about measured velocities, not "computed" ones like the coordinate velocities.
Then you need to specify the measurement procedure. Perhaps a bar of known length with a clock and light source on one end and a mirror on the other (in vacuum of course)

Pyter said:
Are you sure about "light is going faster than that in the direction of rotation"? I was under the impression that, for the equivalence principle, the rotating frame should be equivalent (at least locally) to a static frame in a gravitational field, so you should see the light bending phenomenon and consequently compute a coordinate velocity c < 1.
100% sure. This isn’t how the equivalence principle works. Please decide which topic you want to address and stick to it. These changes in topic are making it difficult for me to follow

Dale said:
Then you need to specify the measurement procedure. Perhaps a bar of known length with a clock and light source on one end and a mirror on the other (in vacuum of course)

Even a Michelson-Morley experiment would do.
Always according to this post, the measured velocity ##ds/d\tau## depends on the light beam direction (the equations in the post are for a radially-propagating beam). If we compare the velocities of two beams, one radial and the other tangential, there should be a difference, detected by the MM experiment.

Dale said:
100% sure. This isn’t how the equivalence principle works. Please decide which topic you want to address and stick to it. These changes in topic are making it difficult for me to follow
I'm sticking to the topic of the velocity of light in non-inertial FRs.

Pyter said:
Even a Michelson-Morley experiment would do
The MMX doesn’t measure the speed of light, just its isotropy.

Pyter said:
I'm sticking to the topic of the velocity of light in non-inertial FRs.
So my first post today holds.

PeroK
So the linear velocity of an object "computed" from an angular rotation (##= \omega r##) is just "fictitious", has not the same physical meaning of the linear velocity v of an object moving of rectilinear motion wrt to my FR, which we all know can't exceed c.

Dale said:
The MMX doesn’t measure the speed of light, just its isotropy.
If we send two orthogonal beams along two same-length arms, if their velocity is different they should form an interference pattern when they're reflected back.
That was the original MMX purpose, to detect if c was different alongside different directions, to prove or disprove the ether hypothesis.
To show that c <> 1 in at least one direction, the appearance of the interference pattern should be enough.

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Pyter said:
If we send two orthogonal beams along two same-length arms, if their velocity is different they should form an interference pattern when they're reflected back.
That was the original MMX purpose, to detect if c was different alongside different directions, to prove or disprove the ether hypothesis.
To show that c <> 1 in at least one direction, the appearance of the interference pattern should be enough.
What is your question?

@PeroK as per your post in the other thread, the measured velocity ##ds/d\tau## of a light beam should vary according to its propagation direction?

PeroK
Pyter said:
If we send two orthogonal beams along two same-length arms, if their velocity is different they should form an interference pattern when they're reflected back.
That was the original MMX purpose, to detect if c was different alongside different directions, to prove or disprove the ether hypothesis.
To show that c <> 1 in at least one direction, the appearance of the interference pattern should be enough.
Sure. If it is anisotropic then it will not be c in at least one direction. So if you do one arm vertical and the other arm horizontal then you will get an interference pattern.

Pyter said:
According to this post, measured velocity is ##ds/d\tau##, which should be an invariant in any FR, non-IFRs included.
Careful... did you read and understand the post immediately before that one, also by @PeroK?

What is ##\mathrm{d} \tau## for a "light beam"? In the naive photon picture you can consider a light beam as described as the trajectory of a massless particle, and thus if no other interactions than gravity act on it, it moves along a null-geodesic, and there is no unique intrinsic measure of time, while for massive particles you can always go to a local inertial frame, where the particle is at rest and where the time is given by the proper time of the particle, which is a natural scalar world-line parameter). For a null geodesic you can still have an affine parameter, but it has no natural scale attached to it, i.e., you can just multiply it with any positive constant without changing anything about the physics.

All one can say is that in a local inertial reference (LIF) frame a locally measured speed of light is always ##c## (no matter whether how the light source is moving at this moment wrt. this LIF, as in special relativity, according to the weak equivalence principle).

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