# Curved Space-Time and the Speed of Light

1. Jun 9, 2010

### Anamitra

The speed of light in vacuum is expected to be a universal constant. Now let us review the fact in the light of the following situation:
We consider three different points A,B and C in curved space-time(in vacuum).Observers stationed at A and B see a light ray flashing across an small spatial interval at C.
Now the length of the spatial interval at C is same for the two observers. But what about the time interval?.The coordinate separation for the two time intervals is the same for both the observers.But the physical separations may by different because of the differences in the values of g(00) at A and B. We must take note of two facts very carefully:
1)The observers have theirs clocks at the points A and B and the physical intervals measured by these clocks are dependent on the values of g(00) at these points.
2) The length of the spatial interval at C as measured by A and B does not depend on the values of g(mu,nu) at A and B. It depends on the value of g(mu,nu) at C only.

This implies that the speed of light as measured by the two observers should be different.
The uploaded file discusses the above problem.

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2. Jun 9, 2010

### Passionflower

What determines the spatial length of a (large) distance in curved spacetimes is subject to the type of measurement and interpretation. Consequently the speed of light may not necessarily be viewed as constant for those measurements.

But locally the speed of light is always c.

3. Jun 9, 2010

### starthaus

Several things:

-your derivations are incorrect
-it is well known that coordinate speed of light is not constant in GR (it is in SR)
-you could prove that to yourself by using the Schwarzschild and respectively Minkowski metrics correctly and by making $$ds=0$$
-the local speed of light is constant (as Passionflower already posted)

4. Jun 9, 2010

### JesseM

Well, in SR it is also not constant if you use a non-inertial coordinate system like Rindler coordinates, and as you and Passionflower noted, in GR it is constant in a locally inertial frame (and only local frames can qualify as 'inertial' in GR). So, it might be conceptually simplest to say that the speed of light is constant in inertial coordinate systems, but not necessarily constant in non-inertial coordinate systems (although in some curved spacetimes it is possible to find non-inertial coordinate systems where it happens to be constant, like Kruskal-Szekeres coordinates in the Schwarzschild spacetime).

5. Jun 9, 2010

### Anamitra

I have simply assumed that the physical speed of light is constant in curved space-time and then I have tried to contradict it.
I have also highlighted in the uploaded article that the speed of light is locally constant.

Now to make my posting clear I consider two points A and B, A being far away from a dense object and B being close to it.Now it is true indeed that observers at A and B measure the speed of light to be 'c' locally. Now the observer at A may be interested in knowing the speed of a light ray as it emerges from B.
Speed of light at B as observed from A= {Spatial separation at B}/{sqrt{g(00)} at A.dt}
Speed of light at B as observed at B ie, c ={Spatial separation at B}/{sqrt{g(00)} at B.dt}
Speed of light as B as observed from A= c sqrt{(g(00) at B)/(g(00) at A)}

Again the observer at B may want to know the speed of light as it emerges from A

Speed of light at A as observed by B={Spatial separation at A}/{sqrt{g(00)} at B.dt}
Speed of light at A as observed at A ie, c ={Spatial separation at A}/{sqrt{g(00)} at A.dt}
Speed of light at B as observed from A= c sqrt{(g(00)}at A)/(g(00)at B)}
The above considerations do have a physical significance.These issues have been seriously discussed in the uploaded file.

Last edited: Jun 9, 2010
6. Jun 9, 2010

### JesseM

But you can't define the speed "in curved space-time" without picking a coordinate system on that spacetime--for any given spacetime there are an infinite number of different possible coordinate systems you could choose, for example in the Schwarzschild spacetime around a nonrotating uncharged black hole you could use Schwarzschild coordinates, Eddington-Finkelstein coordinates, Kruskal-Szekeres coordinates, etc. (see the bottom part of this page for a diagram of each). Do you agree all notion of "speed" is coordinate-dependent, and on any given spacetime there are an infinite number of different possible coordinate systems where the Einstein Field Equations would hold?

7. Jun 9, 2010

### starthaus

Well, the assumption is wrong and so is the math in your article.

Not if both the physics and the math are wrong.

8. Jun 9, 2010

### Anamitra

The fact remains that we can always break the speed barrier.Just think of a light ray shooting out from the vicinity of a blackhole(from beyond the Schwarzschild Radius). Let a person observe it in a direction away from the blackhole. He should see it moving with a speed greater than 'c'. To understand this let us denote the position of the person by A.We consider another position B further away from the blackhole.

Speed of light at B as observed from A= c sqrt{(g(00) at B)/(g(00) at A)}

For a spherical body the above quantity is greater than 'c'.

This should affect the time of travel of a light ray as it goes out of a spherical body. Such calculations do become relevant even to the rays coming out of the sun.Observers near the sun and near the earth should calculate different values of the time of passage of light rays between them!Strictly speaking we should not use a constant value of c for such calculations.These considerations become more relevant if we consider two high densit spherical bodies.

Last edited: Jun 9, 2010
9. Jun 9, 2010

### starthaus

No, it is easy to prove your repeated claim wrong.

Start with the Schwarzschild metric describing radial motion ONLY:

$$ds^2=(1-2m/r)dt^2-\frac{dr^2}{1-2m/r}$$

For light, $$ds=0$$ so:

$$\frac{dr}{dt}=1-2m/r$$

10. Jun 9, 2010

### Anamitra

Your calculation is quite interesting.It has two important features.
1)The person is standing at the same place where he is measuring the speed of light .The metric coefficients of dt$$^{2}$$ and dr$$^{2}$$ have the same values for r.Now let the Person stand at one point and measure the speed of light at some other point in the radial direction. The values of r in the two metric coefficients would be different now!That will give you results that are more interesting. You might like to use my calculations.
2) Interesting enough, you have calculated the coordinate speed of light and not the physical speed

I believe that you would try to understand my calculations now

11. Jun 10, 2010

### starthaus

Sure, I calculated the coordinate speed of light. What you call "physical speed of light" (i.e. the local speed of light) has the value "c'. Always.

I already did, they are dead wrong. This is the third time I'm telling you this.

12. Jun 10, 2010

### Anamitra

We consider the formula:

ds^2=(1-2m/r)dt^2 - (1-2m/r)^(-1) dr^2 --------------------- (1)

For a light ray indeed ds =0
But can we write ds=0 if the observer is stationed at one point(say A),with the clock in his hand, and the light ray flashes past another(say B)?Is it correct to assume that equation (1) should hold true for such a situation? Would it be reasonable to write

ds^2=(1-2m/r1)dt^2 - (1-2m/r2)^(-1) dr^2

with ds=0 for a light ray?

Of course for each of the two points we may write locally equation (1).

For an observer standing at A and the light ray flashing past B, I am suggesting a method here:

Speed of light at B as seen from A= (spatial interval at B)/(sqroot{g(0,0) at A}).dt
Speed of light at B as seen from B,ie,c=(spatial interval at B)/(sqroot{g(00) at B}).dt

[In the above formulas by" spatial interval" I have meant the physical separation. By "dt" is meant the coordinate separation of time]
Therefore ,

Speed of light at B as seen from A= c (sqroot{g(00) at B})/(sqroot{g(00) at A})

The right hand side of the equation may exceed c if

g(00) at B> g(00) at A

In case there is any mistake in my suggested method, I am ready to learn from others,if the error is pointed out in a specific way.
[It is to carefully noted that I have never once denied the fact that locally the speed of light remains "c"]

13. Jun 10, 2010

### JesseM

Anamitra, are you going to answer the questions I asked above? If you agree with the above statements, then you should see why it's not surprising that you can show the speed of light is other than c in a non-inertial coordinate system, this would be true in special relativity where there is no spacetime curvature and no gravity as well.

14. Jun 10, 2010

### Anamitra

Let us examine the situation with reference to an accelerating frame---one which is accelerating with respect to an inertial one.Let us assume for the simplicity of analysis that the acceleration is taking place along the x-x' direction. The primed frame is the accelerating one.It is accelerating along the positive direction of the x-axis(assumed).The primed frame is a non-inertial one but it is equivalent to an inertial one with gravity acting along the negative direction of the x'-axis.Now if this is true the speed of light should locally be "c" in the primed frame also.For non-local points(in the primed frame) you may try using my "suggested method"[from my last thread]

15. Jun 10, 2010

### JesseM

If you're using "locally" in the standard way of referring to a local inertial frame, then saying "locally ... in the primed frame" is meaningless, since the primed frame is a non-inertial frame so measuring the local velocity at any given point on the light's world line wouldn't involve the primed frame's coordinates at all. If by "locally" you just mean the instantaneous velocity in the primed frame, then it's not true that the instantaneous velocity would always be c. For example, suppose the x,t frame is an inertial frame in SR, and the accelerating frame x',t' is defined by the following coordinate transformation:

x' = x - at^2 (where a is some acceleration like a = 0.1 light-seconds/second^2)
t' = t

So the inverse transformation must be:

x = x' + at'^2
t = t'

Now suppose you have a beam of light moving at c in the original inertial frame, so x(t) = ct. In this case we can use the inverse transformation to show that for this beam it must be true that x' + at'^2 = ct', so x'(t') = ct' - at'^2, therefore the instantaneous velocity as a function of time must be dx'/dt' = c - 2at'.
How can a "point" be non-local? Again, you need to explain what you mean by local, it doesn't seem to correspond to how physicists use this word.

16. Jun 10, 2010

### starthaus

No, it isn't reasonable. You don't get to make up your own brand of physics.

17. Jun 11, 2010

### Anamitra

I have been misinterpreted by Jesse.I have simply tried to replace the accelerating frame by an inertial frame with gravity acting along the negative direction of the x'-axis.We can always think in terms of local inertial frames in the replaced inertial frame with gravity acting on it.

Again Jesse has used some simple rules of transformation (in her mathematical formulations) which are correct only in the classical context.Such applications are very much doubtful in Relativistic applications.

18. Jun 11, 2010

### JesseM

If you're using a local inertial frame, then do you have any response to my first comment? Namely:
Do you disagree, and think it makes sense for you to say "Now if this is true the speed of light should locally be 'c' in the primed frame also" despite the fact that you had earlier said "The primed frame is the accelerating one"? How can you say the speed of light is c "in the primed frame" if you're actually not using the coordinates of the primed frame to measure speed at all, but rather using a local inertial frame?
Not true, the equations of general relativity will hold under all smooth coordinate transformations due to the principle of "diffeomorphism invariance", see http://www.aei.mpg.de/einsteinOnline/en/spotlights/background_independence/index.html [Broken]), but it's still one of an infinite number of allowable coordinate transformations where the Einstein Field Equations will hold thanks to diffeomorphism invariance. If you wish to use a coordinate transformation that would be a more "natural" one to use for an accelerating observer in SR like Rindler coordinates, the math would be more complicated but you'd still find that the instantaneous velocity of light dx'/dt' would be something different than c.

Last edited by a moderator: May 4, 2017
19. Jun 11, 2010

### Staff: Mentor

Hi Anamitra,

Geometrically speaking, the relative speed of two objects at a given pair of events is the angle between their worldlines (tangent vectors) at those events. However, in a curved space there is no unique way to compare the angle at two distant points. The reason for this is that in a curved space the result of parallel transporting a vector depends on the path taken.

So in GR it does not make any sense at all to talk about how a distant observer measures speed without specifying the path you are taking for parallel transporting the vector. Also, once you do specify that path the speed determined has no physical significance since it is not unique. Thus, in GR the only speed which has any physical meaning is the speed determined by a local observer, which is always c for light.

20. Jun 11, 2010

### Anamitra

When I said "Now if this is true the speed of light should locally be 'c' in the primed frame also", I meant the the inertial frame(with gravity acting on it) that replaces the primed frame.I should have made the point more explicit.

"How can you say the speed of light is c "in the primed frame" if you're actually not using the coordinates of the primed frame to measure speed at all, but rather using a local inertial frame?"---Jesse

If the primed frame and the inertial frame (with gravity acting on it) are equivalent my assertion is correct indeed.

"If you wish to use a coordinate transformation that would be a more "natural" one to use for an accelerating observer in SR like Rindler coordinates, the math would be more complicated but you'd still find that the instantaneous velocity of light dx'/dt' would be something different than c."----Jesse

Now one of the basic aims of General Relativity is to formulate results independent of the coordinate systems. In fact the interval ds$$^{2}$$ is always independent of the nature of the coordinate system you choose to suit your convenience.It is totally immaterial as to how the frames are moving,accelerating etc,etc.The basic properties of the metric do not change.Now one important property of these metrics is that locally they are diagonalizable so that we may think in terms of inertial frames of reference in a local manner.Given that, I must say that you cannot change the speed of light locally.
Accepting that property/restriction we may proceed in manner I have suggested in my postings.

Last edited: Jun 11, 2010
21. Jun 11, 2010

### JesseM

But the equivalence principle does not say a non-inertial coordinate system can be "equivalent" to an inertial one--I'm not sure what that would even mean. Rather it says that in a curved spacetime, in the neighborhood of any point you can pick a locally inertial coordinate system where the laws of physics will take the same form that they do in an inertial coordinate system in flat spacetime. The speed of light is not c in the primed frame, i.e. dx'/dt' for a light ray can be something other than c.
So do you agree that when you found the speed of light could be other than c, you were finding its coordinate speed in a non-inertial coordinate system, not its speed in a locally inertial coordinate system? And do you agree it's completely unsurprising that the coordinate speed of light can be other than c in a non-inertial coordinate system, since this would even be true for non-inertial coordinate systems in flat SR spacetime (like Rindler coordinates, or like the simple coordinate system I defined in post #15)?

22. Jun 11, 2010

### Anamitra

From the above formula it is clear that dx'/dt' exceeds the value of 'c ' if 'a' is negative and t' is positive. The speed barrier gets broken but from a non-inertial frame of reference.
[This should be true if your considerations are correct]

But in the problem that I have posted you are standing in a "local inertial frame of reference" in curved space-time and watching a ray of light coming from another "local inertial frame" of reference at a distance. In such a situation you observe differences in the speed of light.

The two problems are categorically different.

23. Jun 11, 2010

### Staff: Mentor

As I explained above, the measurement of speed at a distance simply does not make sense in a curved spacetime.

24. Jun 11, 2010

### Anamitra

If I am standing at A and I want to know the speed of light as it comes from a particular direction[as it emerges from another distant point B] the relative should speed become unique.

But then again I feel there is another interesting issue involved problem. In special relativity if two particles X and Y are moving relative to each other the clock which X carries does not change its rate. But in GR as X moves his own clock does change its rate as it moves through different points of curved space-time. We need some complicated calibration of the spatial and the temporal axes to get the world lines before we can think of calculating the angles between them for different pairs of events..
Let me put the matter in this way.Say, motion takes place in the X-Y plane and there are two particles A and B moving in this plane. The Z-axis is taken to be the time axis. Now for each particle we should have different calibrations for the spatial and the temporal axes to get the world lines[before we can calculate the angles between them for different event-pairs], given the fact that the two particles traverse different regions of curved space-time. I am feeling a bit confused here and I would like to have some assistance from you.

25. Jun 12, 2010

### Staff: Mentor

I understand that you think it should be unique, but it is not. Perhaps a concrete example will help.

Suppose we have a 2D curved spacetime which is a sphere when embedded in a flat 3D space. We have a coordinate system on this spacetime like latitude and longitude lines such that north is the direction of increasing time coordinate and west is the dorection of increasing space coordinate.

Two objects on the equator separated by 90 deg are each going due north (locally at rest). We would like to compare their velocities so we take their tangent vectors which are each locally pointing due north. We then parallel transport the first to the location of the second. If we transport along the equator then we find that the first vector points north. If we transport along longitude lines to the north or south pole and then to the location of the second vector we find that it is now pointing due east or west. In fact, by parallel transporting along different paths we can wind up with the first vector pointing any direction including south.
Does this help you understand the non-uniqueness of parallel transport and the reason that you cannot compare distant speeds in a curved spacetime?

This is correct. So when calculating the tangent vector in order to determine the four-velocity we always normalize it to a unit norm (c).

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