# Thread: Speed of light in GR

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Originally Posted by Hetman
It is rather impossible - in principle.
It's certainly not impossible. Add up the angles of a triangle, if you get 180° the curvature is flat, if you get something else you have either some positive or negative curvature.

And finally, it is possible to calculate the speed that I mentioned at the beginning of a subject?
Sure, use pythagoras, just remember that it doesn't really mean anything physical, it's just a coordinate thingy.

ETA: i see you want it for any general angle, the formula then is

2. I wonder whether, (or not), the choice of co-ordinate system, might result in different physical predictions ? ...

For example .. the X-Ray spectrum from a Schwarzschild coordinate or Hilbert version X-Ray Binary type BH, might predict a different physical X-Ray spectra from a version which still has a physical singularity at r=0 (eg: a Kruskal co-ordinate system) ?

Direct matter collisions with a 'hard' singularity surface, should produce higher energy collisions, and would thus be different from those emanating within the vicinity of an EH?

Regards

3. Originally Posted by Hetman
What is the speed of light according to the Schwarzschild metric, but in any direction, not only radial or tangential separately?

I mean the dependence of c on the angle, which is measured relative to the radius, like this:

+---------> c
..+ / - angle
....+
-----M
Well, let's see. Let's say the x axis lies along the radial direction and a hovering observer emits a pulse of light locally at c along the angle θ from the x axis toward the mass. The pulse travels a very short distance so that we can ignore the gravity gradient and our measurements remain local. According to the local observer, then, the pulse travels a distance d in a time t, ending up at coordinates x = (cos θ) d = (cos θ) c t, y = (sin θ) d = (sin θ) c t, y being revolved to include the z axis. To the distant observer using regular Schwarzschild coordinates, the distance the pulse travels along y is the same, the distance travelled along x is contracted to x' = sqrt(1 - r_s / r) x, the time that passes is t' = t / sqrt(1 - r_s / r), and there is no simultaneity difference, where r is the distance of the hovering observer from the mass according to the distant observer. So the distant observer measures a speed for the light pulse at r in the gravitational field using the locally measured angle θ of

c' = d' / t' = sqrt[(1 - r_s / r) x^2 + y^2] / [t / sqrt(1 - r_s / r)]

= sqrt[(1 - r_s / r) (cos θ)^2 c t + (sin θ)^2 c t] / [t / sqrt(1 - r_s / r)]

= c sqrt[(1 - r_s / r) (cos θ)^2 + (sin θ)^2] sqrt(1 - r_s / r)

= c sqrt[(1 - r_s / r) (cos θ)^2 + 1 - (cos θ)^2] sqrt(1 - r_s / r)

= c sqrt[1 - (cos θ)^2 (r_s / r)] sqrt(1 - r_s / r)

The angle that the pulse travels according to the distant observer is

cos θ' = x' / d' = sqrt(1 - r_s / r) (cos θ) / sqrt[(1 - r_s / r) (cos θ)^2 + (sin θ)^2]

= sqrt(1 - r_s / r) (cos θ) / sqrt[1 - (cos θ)^2 (r_s / r)]

Solving for θ, we get

[1 - (cos θ)^2 (r_s / r)] (cos θ')^2 = (1 - r_s / r) (cos θ)^2

(cos θ)^2 [1 - r_s / r + (cos θ')^2 (r_s / r)] = (cos θ')^2

(cos θ) = (cos θ') / sqrt[1 - r_s / r + (cos θ')^2 (r_s / r)]

And substituting that into the equation to gain the coordinate speed c' in terms of the angle θ' the distant observer measures, we have

c' = c sqrt(1 - (cos θ')^2 (r_s / r) / [1 - r_s / r + (cos θ')^2 (r_s / r)]) sqrt(1 - r_s / r)

= c sqrt([1 - r_s / r + (cos θ')^2 (r_s / r)] - (cos θ')^2 (r_s / r)) sqrt(1 - r_s / r) / sqrt[1 - r_s / r + (cos θ')^2 (r_s / r)]

= c (1 - r_s / r) / sqrt[1 - r_s / r + (cos θ')^2 (r_s / r)]

= c (1 - r_s / r) / sqrt[1 - (sin θ')^2 (r_s / r)]
Last edited by grav; 2012-Apr-13 at 03:50 AM.

4. Since it's engendering significant discussion, I've moved the thread from Q&A to S&T, with an expiring redirect from the former to the latter.

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Originally Posted by Selfsim
I wonder whether, (or not), the choice of co-ordinate system, might result in different physical predictions ? ...
No.

At least, not if you're working from GR (and haven't made mistakes).

That's rather the point; what any observer sees is independent of whatever coordinate system she uses.

In our everyday experience we know this intuitively; the two cars collided, whether we say it was in Main St just before the intersection with New Ave, or {insert lat-long values here}, or in front of the (town's one and only) pub.

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Originally Posted by caveman1917
It's certainly not impossible. Add up the angles of a triangle, if you get 180° the curvature is flat, if you get something else you have either some positive or negative curvature.
After all, it's the curvature with respect to the flat space - Euclidean or Minkowski.

Originally Posted by caveman1917
Sure, use pythagoras, just remember that it doesn't really mean anything physical, it's just a coordinate thingy.

ETA: i see you want it for any general angle, the formula then is
Why it does not mean anything physical?
I think it is just the speed of light relative to a flat space.

In the curved space we assume: c = const, and now we change the distance, respectively, which gives non-zero curvature.

These are two mutually exclusive interpretations: c variable and flat space, or c = const and non-zero curvature.

Which interpretation is correct?
This can be verified by measuring the anisotropy of lightspeed.

What is the curvature of space - according to the Schwarzschild metric and Eddington?
If they are equal, then automatically the speed of light must also be equal, and according to both equally anisotropic.

7. Originally Posted by Hetman
After all, it's the curvature with respect to the flat space - Euclidean or Minkowski.

Why it does not mean anything physical?
I think it is just the speed of light relative to a flat space.

In the curved space we assume: c = const, and now we change the distance, respectively, which gives non-zero curvature.

These are two mutually exclusive interpretations: c variable and flat space, or c = const and non-zero curvature.

Which interpretation is correct?
This can be verified by measuring the anisotropy of lightspeed.

What is the curvature of space - according to the Schwarzschild metric and Eddington?
If they are equal, then automatically the speed of light must also be equal, and according to both equally anisotropic.
With Schwarzschild coordinates, we just measure distance along the plane that the light travels with the mass at the origin. With Eddington coordinates, we would expand the distance along the radial direction, like an extra dimension, very much like the light is falling down into a gravity well with the mass at the bottom. So Schwarzschild only measures the distance travelled along the plane at the top of the gravity well, those which coincide vertically with the location of a pulse within the well, while Eddington measures the distance travelled along the walls of the well as the light travels along them, which will be greater, the radial dimension of those walls being such that they correspond to the tangential speed of light measured by a distant observer.

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Originally Posted by Hetman
Why it does not mean anything physical?
I think it is just the speed of light relative to a flat space.

In the curved space we assume: c = const, and now we change the distance, respectively, which gives non-zero curvature.

These are two mutually exclusive interpretations: c variable and flat space, or c = const and non-zero curvature.

Which interpretation is correct?
This can be verified by measuring the anisotropy of lightspeed.

What is the curvature of space - according to the Schwarzschild metric and Eddington?
If they are equal, then automatically the speed of light must also be equal, and according to both equally anisotropic.
Let me try another way. Suppose you and i are standing in flat space and you have a light bulb. Call the event where you switch on the light bulb E1, and the event where the first photon reaches my position E2. The metric in a standard basis being . Suppose you are at the origin and i am 1 unit of distance along the positive x-axis. Suppose the photon takes 1 unit of time to go from you to me. So the photon did indeed follow a null-path .

Now suppose i do a coordinate change , i'm halving the unit of distance. So now the metric takes the form . You can see that the coordinate speed of light changed. However this doesn't mean anything physical. We can do the same calculation. We give the photon one unit of time to travel to my location, which is now 2 units of distance from you (we halved our unit of distance), so . As you can see the "nullness" of the path didn't change, as publius said earlier.

What is physical is the actual result. For example i might use the arrival of the photon as a trigger for some physical device, and we see that in both coordinatizations the same actual result would be achieved (the photon arrives when it should). Whatever value (or isotropy) the coordinate speed of light might have is irrelevant, what is relevant is whether a given path is null (ie traverseable by light) or not, which is invariant.

This was an easy example, but the same thing happens with schwarzschild and other coordinate charts of the schwarzschild metric. They change the coordinate speeds of light, but also the used time and distance coordinates, in such way that the physical results (does the photon get there or not?) is preserved.

9. Originally Posted by Nereid
No.

At least, not if you're working from GR (and haven't made mistakes).

That's rather the point; what any observer sees is independent of whatever coordinate system she uses.

In our everyday experience we know this intuitively; the two cars collided, whether we say it was in Main St just before the intersection with New Ave, or {insert lat-long values here}, or in front of the (town's one and only) pub.
Hi Nereid (thanks for your post);
Yep … agreed .. naked singularities (implied in what I said), don't appear/disappear with changing co-ordinate systems. An EH still exists in other co-ordinate systems .. the difference being that the EH is non-singular.
With rotating BHs, its might not be quite so clear-cut .. a 'rotator' could become a naked singularity, and the collisions should then become visible, but this would be due to increasing angular momentum.

Regards

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Originally Posted by Selfsim
Hi Nereid (thanks for your post);
Yep … agreed .. naked singularities (implied in what I said), don't appear/disappear with changing co-ordinate systems. An EH still exists in other co-ordinate systems .. the difference being that the EH is non-singular.
Thanks; I missed the part about the naked singularity being implied.

A black hole event horizon is not the same sort of thing as two cars colliding!

If you are on a moving train, sitting in a seat, you are motionless with respect to the train. I, being outside and sitting in a seat on the train platform, see you moving.

Event horizons are the same sort of thing: their nature depends upon the observer. An infalling observer sees an event horizon ... but it is always in front of them, while you - being well away and not falling in - disagree (caveats apply).

With rotating BHs, its might not be quite so clear-cut .. a 'rotator' could become a naked singularity, and the collisions should then become visible, but this would be due to increasing angular momentum.

Regards
Would you clarify please? What do you mean by "a 'rotator' could become a naked singularity"? And what sort of "collision" do you have in mind?

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Originally Posted by grav
With Schwarzschild coordinates, we just measure distance along the plane that the light travels with the mass at the origin. With Eddington coordinates, we would expand the distance along the radial direction, like an extra dimension, very much like the light is falling down into a gravity well with the mass at the bottom. So Schwarzschild only measures the distance travelled along the plane at the top of the gravity well, those which coincide vertically with the location of a pulse within the well, while Eddington measures the distance travelled along the walls of the well as the light travels along them, which will be greater, the radial dimension of those walls being such that they correspond to the tangential speed of light measured by a distant observer.
I understand, but this is some kind of misunderstanding.
In practice, we use the units of measure, which are omitted in the equations.

Of course, it is not difficult to imagine a situation where the object's length is measured in inches and the height in meters, though I never met with anything like that.

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Originally Posted by caveman1917
Let me try another way. Suppose you and i are standing in flat space and you have a light bulb. Call the event where you switch on the light bulb E1, and the event where the first photon reaches my position E2. The metric in a standard basis being . Suppose you are at the origin and i am 1 unit of distance along the positive x-axis. Suppose the photon takes 1 unit of time to go from you to me. So the photon did indeed follow a null-path .

Now suppose i do a coordinate change , i'm halving the unit of distance. So now the metric takes the form . You can see that the coordinate speed of light changed. However this doesn't mean anything physical. We can do the same calculation. We give the photon one unit of time to travel to my location, which is now 2 units of distance from you (we halved our unit of distance), so . As you can see the "nullness" of the path didn't change, as publius said earlier.

What is physical is the actual result. For example i might use the arrival of the photon as a trigger for some physical device, and we see that in both coordinatizations the same actual result would be achieved (the photon arrives when it should). Whatever value (or isotropy) the coordinate speed of light might have is irrelevant, what is relevant is whether a given path is null (ie traverseable by light) or not, which is invariant.

This was an easy example, but the same thing happens with schwarzschild and other coordinate charts of the schwarzschild metric. They change the coordinate speeds of light, but also the used time and distance coordinates, in such way that the physical results (does the photon get there or not?) is preserved.
You are talking about something like this:
we have the equation of the circle x^2 + y^2 = 1
what is written in polar coordinates: r = 1;

We change the name of r to x, then we get: x = 1 - vertical line?

This is a consequence of losing the meaning of the symbols in the equations.
Not only the units of measurement, but also a reference system, in which was measured, determined the size and coordinates of bodies, and the boundary conditions.

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Originally Posted by Hetman
You are talking about something like this:
we have the equation of the circle x^2 + y^2 = 1
what is written in polar coordinates: r = 1;

We change the name of r to x, then we get: x = 1 - vertical line?

This is a consequence of losing the meaning of the symbols in the equations.
Not only the units of measurement, but also a reference system, in which was measured, determined the size and coordinates of bodies, and the boundary conditions.
But that is somewhat the point. The "r" coordinate in the schwarzschild chart does not represent proper radial distance, that's why the coordinate light speed is anisotropic in those coordinates. If you use proper light speed about any local observer in the schwarzschild metric, it is isotropic.

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Originally Posted by caveman1917
But that is somewhat the point. The "r" coordinate in the schwarzschild chart does not represent proper radial distance, that's why the coordinate light speed is anisotropic in those coordinates. If you use proper light speed about any local observer in the schwarzschild metric, it is isotropic.
This is only the other interpretation, which I mentioned: deform the space in such a way to keep c

Anisotropy of light, or space in the second interpretation, disappears only when we put: m = 0.
Then you get a space with zero curvature, ie Minkowski, which is precisely isotropic.

The point is that gravity is inherently anisotropic, one can even say that this anisotropy is just gravity!

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Originally Posted by Hetman
This is only the other interpretation, which I mentioned: deform the space in such a way to keep c

Anisotropy of light, or space in the second interpretation, disappears only when we put: m = 0.
Then you get a space with zero curvature, ie Minkowski, which is precisely isotropic.

The point is that gravity is inherently anisotropic, one can even say that this anisotropy is just gravity!
It doesn't have anything to do with gravity, only with a choice of coordinates. You can choose coordinates in Minkowski in which it is anisotropic, and coordinates in Schwarzschild in which it is isotropic.

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c' = c(1 - r_s / r) / sqrt[1 - (sinθ)^2 (r_s / r)];

r^2 = x^2 + R^2 => cosθ = x/r;

From this we get Shapiro delay:

It will be too small about x/r, which converges to 1 for x >> R;
double, and back and forth, in total we get less of: 4GM/c^3 = 2rs/c = 2*3km/300000km/s = 20us.

Quite a significant reduction - about 10%, and probably in relation to the measurement, if I'm not mistaken.

17. Originally Posted by Nereid
Originally Posted by Selfsim
With rotating BHs, its might not be quite so clear-cut .. a 'rotator' could become a naked singularity, and the collisions should then become visible, but this would be due to increasing angular momentum.
Would you clarify please? What do you mean by "a 'rotator' could become a naked singularity
Well, in a rotating BH, the size of the EH is a function of the angular momentum, (inversely proportional). I guess in theory, if the angular momentum is large enough, a rotating BH can have a naked singularity, in which case, we should then be able to observe some effects from the sudden deceleration of particles as they collide with, (or imminently collide with) the surface (??)
(Crudely speaking, and in theory, that is).

Cheers

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I don't know enough about rotating black holes to say how the
size of event horizon relates to the angular momentum, so I'm
not going to say anything about that ... but ... In Schwarzschild
black holes, everything is pulled apart in the radial direction by
the gravitational tide. Nothing in free-fall will ever collide with
anything else close to the center. Angular momentum can't
prevent matter from falling farther once it is inside the event
horizon, so there cannot be a surface to collide with.

My last statement above was sloppy. Caveman1917 makes a

-- Jeff, in Minneapolis
Last edited by Jeff Root; 2012-Apr-16 at 05:16 PM. Reason: add correction & link

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The horizon is just in the center.

The confusion with the horizon are based on the fact that the radial coordinate r in the Schwarzschild metric can not run from 0 due to the spherical curvature.

20. Originally Posted by Hetman
c' = c(1 - r_s / r) / sqrt[1 - (sinθ)^2 (r_s / r)];

r^2 = x^2 + R^2 => cosθ = x/r;
Right, if we consider only the delay for a straight line path of the light along an angle coming from the mass where R is the closest point of approach, with no gravitational lensing involved, which should be close enough for R>>r_s.

From this we get Shapiro delay:
I'm not sure what you integrated here, though. We are integrating for the path of the light along x, so we want to integrate for changing x while keeping the rest constant, but r changes as x becomes greater while R remains constant, so we want to change that to r = sqrt(R^2 + x^2), gaining

delay = (G M / c^3) ∫ dx [(1 + x^2 / (x^2 + R^2)] / sqrt(x^2 + R^2)

= (G M / c^3) [2 log(2 (sqrt(R^2 + x^2) + x)] - x / sqrt(x^2 + R^2) from x = 0 to x = sqrt(r^2 - R^2), which becomes

2 ln(2 R) for x = 0 and 2 ln(2 (r + sqrt(r^2 - R^2)) - sqrt(r^2 - R^2) / r for x = sqrt(r^2 - R^2), giving

delay = (G M / c^3) [2 ln(2 (r + sqrt(r^2 - R^2)) - sqrt(1 - (R/r)^2) - 2 ln(2 R)]

= (G M / c^3) [2 ln((r / R) (1 + sqrt(1 - (R/r)^2))) - sqrt(1 - (R/r)^2)]

or twice that from -r to r.

EDIT - Okay, I think I see what you did. You integrated this way, then just switched back to the total distance travelled along x where x = sqrt(r^2 - R^2), gaining

delay = (G M / c^3) [2 ln((r / R) (1 + x / r)) - x / r]

but you just didn't subtract the x = 0 part of the original integration.

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I just calculated the indefinite integral.

In order to calculate the delay, you should insert the limits of integration.

For transmission: Venus - Earth or Mars-Earth, we get 20 microseconds less.

22. Originally Posted by Hetman
I just calculated the indefinite integral.

In order to calculate the delay, you should insert the limits of integration.

For transmission: Venus - Earth or Mars-Earth, we get 20 microseconds less.
I'm not sure what you are saying here. The precise delay would just depend upon the precise values for the distances you are using, the exact distances during their orbits of each of the planets from the sun when the experiment is performed and how close the light passes the sun.

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Originally Posted by grav
I'm not sure what you are saying here. The precise delay would just depend upon the precise values for the distances you are using, the exact distances during their orbits of each of the planets from the sun when the experiment is performed and how close the light passes the sun.
For large distances in comparison with R will be about 20 microseconds less - in the transmission to a planet behind the Sun (there and back).

x/r -> 1.

With the increase of R (impact parameter), this will change -
maybe this is a sine wave in the Pioneer anomaly.

24. Okay, for a pulse sent past the sun to another planet and back, we have

delay = 2 (G M / c^3) [2 ln((r_e / R) (1 + sqrt(1 - (R / r_e)^2))) + 2 ln((r_p / R) (1 + sqrt(1 - (R / r_p)^2))) - sqrt(1 - (R / r_e)^2) - sqrt(1 - (R / r_p)^2)]

where r_e and r_p are the distances from the sun to Earth and the planet, and to first approximation sqrt(1 - (R / r)^2) reduces to unity, the difference being neglible, so we get

delay = 2 (G M / c^3) [2 ln((r_e / R) (2)) + 2 ln((r_p / R) (2)) - 2]

= 4 (G M / c^3) [ln(4 (r_e / R) (r_p / R)) - 1]

If we just approximate with 4 (G M / c^3) ln(4 (r_e / R) (r_p / R)) taking only the first part of the equation, then this will be 4 (G M / c^3) or about 20 microseconds greater for any planet with distance from the sun r>>R, right, large enough to make a difference, so not precise enough. This is just for the straight line path of light, though, so if you're only seeing the first part of the equation somewhere and not the last, it is either an imprecise approximation or perhaps gravitational lensing has been included in the calculation and makes up the difference.

25. Ahah. Here is a link where the delay is calculated. It is done by a somewhat different process but you can see at the bottom that they gain the same result. They then reduce it further with

delay = 4 (G M / c^3) [ln(4 (r_e / R) (r_p / R)) - 1]

= 4 (G M / c^3) [ln(4 (r_e / R) (r_p / R) - ln(e)]

= 4 (G M / c^3) ln(4 (r_e / R) (r_p / R) / e)

Maybe you just missed the division by e in the log wherever you found this result.

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Originally Posted by grav
If we just approximate with 4 (G M / c^3) ln(4 (r_e / R) (r_p / R)) taking only the first part of the equation, then this will be 4 (G M / c^3) or about 20 microseconds greater for any planet with distance from the sun r>>R, right, large enough to make a difference, so not precise enough. This is just for the straight line path of light, though, so if you're only seeing the first part of the equation somewhere and not the last, it is either an imprecise approximation or perhaps gravitational lensing has been included in the calculation and makes up the difference.
The path along the arc is just a few meters longer.
Error 20 microseconds remains intact.

Shapiro measured: 240us with an error of 3%.
x_v = 108, x_e = 150, R = 0.7 [million km]

ln (4r_e.r_v / R ^ 2) - 1 = 10.8

dt = 4GM/c^3 * 10.8 = 20us * 10.8 = 216us;
240 - 216 = 24us, so much is missing.

Mars: x_m = 228;
ln (4 r_e r_m / R^2) - 1 = 11.54

20us * 11.54 = 230.8us
Used an incorrect approximation: c '= c(1 - rs / r)... and probably measured in accordance with this.

27. Originally Posted by Hetman
The path along the arc is just a few meters longer.
Error 20 microseconds remains intact.

Shapiro measured: 240us with an error of 3%.
x_v = 108, x_e = 150, R = 0.7 [million km]

ln (4r_e.r_v / R ^ 2) - 1 = 10.8

dt = 4GM/c^3 * 10.8 = 20us * 10.8 = 216us;
240 - 216 = 24us, so much is missing.

Mars: x_m = 228;
ln (4 r_e r_m / R^2) - 1 = 11.54

20us * 11.54 = 230.8us
Used an incorrect approximation: c '= c(1 - rs / r)... and probably measured in accordance with this.

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Originally Posted by grav
Of course. You can find these numbers and formulas in most textbooks on GR.

Here also is the simplified formula - ignoring anisotropy.

29. Originally Posted by Hetman
The horizon is just in the center.

The confusion with the horizon are based on the fact that the radial coordinate r in the Schwarzschild metric can not run from 0 due to the spherical curvature.
Hello Hetman;
Well, what I'm exploring here goes somewhat beyond the abstract modelling viewpoint. I get the uneasy feeling that not everything going on here, can easily be explained away by simple coordinate shifting.

One has to remember that BHs exist for sound physical reasons. (They have a physical origin in the first place, no ?)

As a consequence, EHs must also not exist purely because of an artifact of the coordinate system used. The same reasoning applies for the rest of the physical properties of a BH, (unless one is prepared to abandon the reality of these principles as well).

I think a physical explanation beckons, and seems wanting here.

Regards

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Originally Posted by caveman1917
You don't need the first if you have the third and you don't need the third if you have the first.
As long as you ignore that the universe is expanding and that this expansion is accelerating.

It's been a while since I read Einstein's "Relativity". In the book, he mentions that we must abandon the idea that the speed of light is constant when considering gravity. His argument, I think, was that if the velocity of light was constant, it would not be able to change direction. That tangential deflection is like an acceleration.

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