Curved Space-Time and the Speed of Light

[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.

 

[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.

 

[starthaus]

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.

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)

 

[JesseM]

Several things:

-your derivations are incorrect
-it is well known that coordinate speed of light is not constant in GR (it is in SR)

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).

 

[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.

 

[JesseM]

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

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?

 

[starthaus]

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

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

The above considerations do have a physical significance.These issues have been seriously discussed in the uploaded file.

Not if both the physics and the math are wrong.

 

[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 black hole(from beyond the Schwarzschild Radius). Let a person observe it in a direction away from the black hole. 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 black hole.

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.

 

[starthaus]

The fact remains that we can always break the speed barrier.Just think of a light ray shooting out from the vicinity of a black hole(from beyond the Schwarzschild Radius). Let a person observe it in a direction away from the black hole. 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 black hole.

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'.

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:

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

 

[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

 

[starthaus]

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

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 believe that you would try to understand my calculations now

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

 

[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 AIn 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"]

 

[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?

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.

 

[Anamitra]

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 space-time curvature and no gravity as well.

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]

 

[JesseM]

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.

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'.

For non-local points(in the primed frame) you may try using my "suggested method"[from my last thread]

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.

 

[starthaus]

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?

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

 

[Anamitra]

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.

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.

 

[JesseM]

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.

If you're using a local inertial frame, then do you have any response to my first comment? Namely:

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.

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?

Again Jesse has used some simple rules of transformation (in her mathematical formulations) which are correct only in the classical context.

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 ), 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.

 

[Dale]

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.

 

[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.

 

[JesseM]

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.

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.

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.

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)?

 

[Anamitra]

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'.

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.

 

[Dale]

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.
(Emphasis added)

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

 

[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.

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.

 

[Dale]

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.

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?

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

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).

 

[Anamitra]

Relative speed ,it appears from your condiderans,is a troublesome concept.The concept of parallel transport adds a serious attribute of non-uniqueness to the whole idea.

Even then if I am standing at some point in curved space-time and an object flies past some other point(we consider somthing which is not a light ray), I should have some observation of relative motion physically.And I hope that from the physical point of view this observation should be of a unique natue!Is there any procedure in General Relativity that allows me to make such predictions ? In case there is some procedure/method can we apply it to a light ray?

Now I am coming to a second issue.

We consider the same two points(labeling them A and B this time) and consider a light ray passing from B to A.

We calculate the time from the formula

t=Integral dl/c
the integration extending from B to A along the path of the ray and dl being the spatial separation(physical)

Now observers at B and A are taking note of the times of sending the signal and receiving it.Later on they compare the times and take the difference. This difference should not be equal to the value of the integral. Even if the observer at A transforms the time [calculations shown below]of sending the signal to get his clock reading when the signal is flashed from B, the difference of times[between the sending and receiving of the signals] is expected to be different from the integral.So the average speed of light speed of light is indeed different from "c" here.


Calculations
dt(B)=dt(A) -------(coordinate separations)
dt(p,A)/{sqrt g(00) at A} =dt(p,B)/{sqrt g(00) at B}
[dt(p,A) is the physical separation(temporal) at A,and dt(p,B) is the physical separation(temporal) at B]

By integration we may find the relationship between t(p,A) and t(p,B)

 

[Dale]

I should have some observation of relative motion physically.And I hope that from the physical point of view this observation should be of a unique natue!

We seem to be going in circles. I have already answered this several times and I am not inclined to do so yet again.

Perhaps there is some miscommunication. What do you mean by the word "physically"? I have been interpreting it to mean "coordinate independent".

 

[Anamitra]

On the Uniqueness of Parallel Transport

The concept of parallel concept can be made unique by choosing a curve between a pair of points such that the inner product of any pair of vectors get preserved along the curve.

[Mathematically such a curve is one for which the components of the covariant derivatives of g(mu,nu) are zero]

This is a commonly accepted practice in relation to methods of General Relativity and has been discussed quite elaborately [from the mathematical point of view]by Robert M. Wald in the text "General Relativity" in Chapter Three[Curvature],section 3.1[Derivative Operators and Parallel Transport]

In the introduction to this chapter(ie chapter three) the author remarks "We will show in section 3.1 that a given metric(of any signature), there is a unique definition of parallel transport which presreves the the inner product of all pairs of vectors. The existence of a metric gives rise to a unique notion of parallel transport and ,thus, to an intrinsic notion of the curvature of the manifold."
Such a definition of uniqueness in relation to parallel transport ensures the the uniqueness of derivative operator for a given system of metrics.

I find no problem in applying this idea of "unique parallel transport"to my concepts.

 

[yuiop]

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

Let us say we have two stationary observers in the Schwarzschild metric. One is low down at r1 and the other is higher up at r2. They take continuous radar measurements and agree their spatial separation does not change over time. Now observer 2 at r2 measures the radar distance (r2-r1) to be longer than the radar distance measured by observer 1. The radar distance assumes the speed of light in a vacuum is constant everywhere. If on the other hand we make the assumption that distance (r2-r1) is equal to -(r1-r2) then the oobservations can be logically explained by the speed of light slowing down deeper in the gravitational field. The local speed of light appears to remain constant because clocks lower down are running slower and local rulers are length contracted. The outcome of calibrating local clocks and rulers using the assumption that the local speed of light is c, is that the local speed of light is always measured to be c. This is an artifact of the calibration and clock synchronisation convention used. If the clocks lower down are synchronised so that they run at the same rate as the clocks higher up, it would be obvious locally that the speed of light lower down is slower than the speed of light higher up. As Jesse has pointed out, there are are valid coordinate systems in GR where the local speed of light is not c.

 

[Passionflower]

Let us say we have two stationary observers in the Schwarzschild metric. One is low down at r1 and the other is higher up at r2. They take continuous radar measurements and agree their spatial separation does not change over time. Now observer 2 at r2 measures the radar distance (r2-r1) to be longer than than than the radar distance measured by observer 1. The radar distance assumes the speed of light in a vacuum is constant everywhere. If on the other hand we make the assumption that distance (r2-r1) is equal to -(r1-r2) then the oobservations can be logically explained by the speed of light slowing down deeper in the gravitational field. The local speed of light appears to remain constant because clocks lower down are running slower and local rulers are length contracted. The outcome of calibrating local clocks and rulers using the assumption that the local speed of light is c, is that the local speed of light is always measured to be c. This is an artifact of the calibration and clock synchronisation convention used. If the clocks lower down are synchronised so that they run at the same rate as the clocks higher up, it would be obvious locally that the speed of light lower down is slower than the speed of light higher up.

Indeed. In that respect a quite a parallel view is expressed in the Lorentz ether theory for flat spacetime.

However, and here is a key difference, in the extreme case, passed the event horizon, the interpretation under GR must be different because of:

• The assumption that an unaccelerated observer close to the event horizon would pass the event horizon and reach a singularity in finite proper time.
• The assumption that inside the horizon imaginary proper time is equivalent to proper space.

Note that it was not always so 'apparently natural' that an interior solution must exist. For historical reasons it might be interesting to pinpoint the physicists who introduced and 'commonized' this, now standard, interpretation in GR.

 

[Dale]

I find no problem in applying this idea of "unique parallel transport"to my concepts.

Excellent. Then please do so rigorously, i.e. In a manifestly covariant way. Perhaps when you do so you will find a problem.

 

[nismaratwork]

On the Uniqueness of Parallel Transport

The concept of parallel concept can be made unique by choosing a curve between a pair of points such that the inner product of any pair of vectors get preserved along the curve.

[Mathematically such a curve is one for which the components of the covariant derivatives of g(mu,nu) are zero]

This is a commonly accepted practice in relation to methods of General Relativity and has been discussed quite elaborately [from the mathematical point of view]by Robert M. Wald in the text "General Relativity" in Chapter Three[Curvature],section 3.1[Derivative Operators and Parallel Transport]

In the introduction to this chapter(ie chapter three) the author remarks "We will show in section 3.1 that a given metric(of any signature), there is a unique definition of parallel transport which presreves the the inner product of all pairs of vectors. The existence of a metric gives rise to a unique notion of parallel transport and ,thus, to an intrinsic notion of the curvature of the manifold."
Such a definition of uniqueness in relation to parallel transport ensures the the uniqueness of derivative operator for a given system of metrics.

I find no problem in applying this idea of "unique parallel transport"to my concepts.

I say this with all due respect, but you last sentence sounds a bit mad to me. If you can do that, then do it. I've read this thread from the beginning, and you seem to be saying the same thing in a hundred different ways. I think the keyword here is "[YOUR] concepts" which sounds fun, but don't reflect physics as I understand them, or as anyone else here does it seems.

You're saying outright that DaleSpam is wrong, and that in a curved space you find something meaningful by comparing distant speeds? I'm sorry, but that makes no sense at all to me, can you show some work to support this?!

 

[yuiop]

... then the observations can be logically explained by the speed of light slowing down deeper in the gravitational field. ...

Indeed. In that respect a quite a parallel view is expressed in the Lorentz ether theory for flat spacetime.

However, and here is a key difference, in the extreme case, passed the event horizon, the interpretation under GR must be different because of...

Do you agree that if we dismiss the light slowing down notion, that the only alternative is to accept that a single stationary rod extending from r1 to r2, really does have two separate physical lengths at the same time?

 

[starthaus]

Do you agree that if we dismiss the light slowing down notion, that the only alternative is to accept that a single stationary rod extending from r1 to r2, really does have two separate physical lengths at the same time?

1. Light doesn't "slow down", the speed of light measured in a small vicinity is always c.
2. The rod does not "have two separate physical lengths", it is just that the local observer measures a different length from the length measured by a distant observer.

1. and 2. above can be proven really easily in three lines of simple computations.

 

[Passionflower]

Do you agree that if we dismiss the light slowing down notion, that the only alternative is to accept that a single stationary rod extending from r1 to r2, really does have two separate physical lengths at the same time?

Well, you can't have your cake and eat it too. If you assume that light slows down in a gravitational field then, at the event horizon, it must completely stop. Then how fast do you think it goes passed the event horizon? I know you can do fancy math tricks there but we are talking physics right?

Could you explain the case where you think you have two separate physical lengths at the same time?

1. Light doesn't "slow down", the speed of light measured in a small vicinity is always c.

Well obviously, as the gravitational effect in a small vicinity is always zero and the clock is running slower so it will not be able to detect the assumed slowing down of light.