Curved Space-time and Relative Velocity

[DrGreg]

So if we parallel transport a vector along different paths starting from the same point the components do not change,when referred to the local inertial frames.

This is really a circular argument. At any event in spacetime there isn't a single "locally inertial" coordinate system. You have a choice: for a start, you can rotate the spatial axes and get an equally good locally inertial coordinate system. Also any coordinate system moving at a constant velocity to a locally inertial system is locally inertial.

Given that choice, how would you choose which family of locally inertial coordinate systems to use along a worldline? The answer is you'd use parallel transport!

If you have two different geodesics (or any other worldline) joining a pair of events, if you parallel-transport a locally inertial coordinate system from one event to the other, you could get two different locally inertial coordinate systems at the other end!

 

[Anamitra]

"Again, this is demonstrably false; DrGreg provided a very detailed counter example.

https://www.physicsforums.com/showpos...2&postcount=22

Starting from the same point the components of the final parallel-transported vector are in fact changed depending on the path. This is fundamental to understanding the very basic concept of curvature, and instead of trying to learn it you are just going around in circles making the same false assertion over and over."
Dalespam, Thread 35

This "demonstrably false" notion arises out of the fact that the singularities north and south poles have been chosen simultaneously.I have demonstrated my reason as to why they should not be chosen simultaneously in thread #15 ,https://www.physicsforums.com/showpost.php?p=2847718&postcount=15

If you chop off either the north or the south pole from the sphere the demonstration provided by Dr Greg fails

 

[Anamitra]

Remember geodesics in spacetime do not connect points in space, they connect events in spacetime. The decision as to which spacetime geodesic the particle will follow is determined entirely by the particle's velocity (or to be more precise and coordinate independent, by its 4-velocity tangent 4-vector). The point is, given an event, there are lots of particles that can pass through that event on a geodesic, and no way to single out one of those geodesics as being the "correct" one (unless you believe in aether).

Dr Greg has considered here several geodesics emanating from the same point. This is possible. If all these geodesics terminated on the same point we encounter a serious discrepency as shown in--->https://www.physicsforums.com/showpost.php?p=2847718&postcount=15

Thanks for the great explanations DrGreg

The Schwarzschild solution has all of the symmetry of a 2-sphere plus a lot of symmetries that the sphere does not have. These types of problems are actually more of an issue in 4D, not less.

In his example Dr Greg has used spatial geodesics!

 

[JesseM]

Dr Greg has considered here several geodesics emanating from the same point. This is possible. If all these geodesics terminated on the same point we encounter a serious discrepency as shown in--->https://www.physicsforums.com/showpost.php?p=2847718&postcount=15

Anamitra, look at page 118 from Relativity on Closed Manifolds--the diagram clearly shows two geodesics which intersect at two different points, and the text even gives a name for this phenomenon:

When two neighboring geodesics intersect twice, the points of intersection are termed conjugate

Similar diagrams and discussions can be found in this book and this one.

 

[Anamitra]

A "Seriously Heavy Point"

It is very important to consider the addition/subtraction of three velocities. If I am standing at Point A and I see a light ray flashing past past another B I would be interested in the three velocity of the light ray[my own three velocity being the null vector]. This is relevant to the issue in thread#6 --->https://www.physicsforums.com/showpost.php?p=2846710&postcount=6

In threads #19 and #25 [https://www.physicsforums.com/showpost.php?p=2756767&postcount=19 , https://www.physicsforums.com/showpost.php?p=2758350&postcount=25]of "Curved Space-time and the Speed of Light" DaleSpam has tried to counter the concept of Relative Velocity in Curved Space-time in by giving examples in relation to 4-D space,I mean by referring to four vectors.[ Of course these examples have failed in their mission]

He has kept silent on the issue of the addition/subtraction of 3-velocities!

The issue of subtraction of three velocities at a distance,especially when one is null, is extremely relevant to the discussion in thread #6

I am saying all this not to give any extra fortification to what ever I have said in relation to 4D considerations but because these points are seriously heavy. My assertions in relation to 4D concepts are strong enough to stand on their own feet.

 

[Dale]

This "demonstrably false" notion arises out of the fact that the singularities north and south poles have been chosen simultaneously.

There are no singularities on a sphere, the curvature is everywhere finite.

If you chop off either the north or the south pole from the sphere the demonstration provided by Dr Greg fails

No. Because of the symmetry you can do this from any point on the sphere, it is just easier to describe verbally from the poles.

In his example Dr Greg has used spatial geodesics!

So what? Spacelike paths are perfectly acceptable paths for parallel transport and need not even be geodesic.


Anamitra, you don't seem to understand the very basics of parallel transport and intrinsic curvature. The most important example of parallel transport is to transport a vector around a closed loop back to its original position (NB a loop is generally a non-geodesic path). In a curved space the parallel transported vector will be rotated from the original vector by an amount which depends on the area enclosed by the loop as well as the direction of the loop. The Riemann curvature tensor describes exactly this property of curved space in the limit of infinitesimal loops.

In parallel transport the covariant derivative which is zero is the covariant derivative of the transported vector, the path along which it is transported need not have a zero covariant derivative, and need not even be smooth.

If you are so stuck on your preconcieved notions that you are not willing to learn these basic and fundamental geometric concepts then you may as well just stop even attempting to learn general relativity as it will be completely futile. I would recommend that you view Leonard Susskind's lectures on General Relativity which are available on YouTube.

 

[Anamitra]

The Sphere Again!

Let us consider the example of the spherical space-time surface in a mathematical way:

We calculate the distance[space-time separation] between the north pole and the south poles along the meridians and of course for a sphere we get the same value.One should use the relation ,space -time separation=integral ds along any meridian.Now if by some suitable trans formation we change the sphere to some other surface.The "meridians" will have different "lengths". The same pair of events[4D events] will have different separations. Since the sphere is full of antipodal points,better to leave aside the example of the sphere.

But these were 4-D considerations.Nevertheless I have a big interest in the 3-D issues I have specified in Thread #40, the seriously heavy points.

[NB: ds represents physical length on a space-time surface]

 

[bcrowell]

Anamitra, three people have been spending a lot of time trying to help you. In my opinion, all three know general relativity pretty well. You would be well advised to get out of this mode where you feel you have to defend a position you've staked out. It's not going to serve you well in learning general relativity.

 

[atyy]

I think you are asking about conjugate points. This is entertaining http://hawking.org.uk/old-site/pdf/time.pdf

 

[Anamitra]

I am very much interested in receiving replies in regard to Thread#40

Regarding Thread#42:

It is true that in many standard texts we have examples of several geodesics connecting a pair of space-time points. If they happen to be of unequal lengths is there going to be any problem, in the sense that space-time separation for the pair is no more unique?I am keen on receiving some answer to this issue so that I can improve my knowledge. This is just a request.

 

[Dale]

Anamitra, regarding post 40:

You cannot compare vectors unless they are in the same vector space. The tangent space at each point in a manifold is a different vector space, and it is only once you have mapped the vector in one tangent space to a vector in the other tangent space that you can make any comparisons. The process for doing this is called parallel transport. Parallel transport must come first, before any other vector operation is possible.

Now, please address post 41.

 

[Dale]

We calculate the distance[space-time separation] between the north pole and the south poles along the meridians and of course for a sphere we get the same value.One should use the relation ,space -time separation=integral ds along any meridian.Now if by some suitable trans formation we change the sphere to some other surface.The "meridians" will have different "lengths".

This is not true. ds is an invariant quantity (a tensor of rank 0), so it remains unchanged under any coordinate transformation.

 

[Anamitra]

Anamitra, you don't seem to understand the very basics of parallel transport and intrinsic curvature. The most important example of parallel transport is to transport a vector around a closed loop back to its original position (NB a loop is generally a non-geodesic path). In a curved space the parallel transported vector will be rotated from the original vector by an amount which depends on the area enclosed by the loop as well as the direction of the loop. The Riemann curvature tensor describes exactly this property of curved space in the limit of infinitesimal loops.

In parallel transport the covariant derivative which is zero is the covariant derivative of the transported vector, the path along which it is transported need not have a zero covariant derivative, and need not even be smooth.

We consider the definition of the covariant derivative:

covariant derivative= dA(mu)/dx(i) + affine connection part
Now How does one calculate dA(mu)/dx(i) on a sharp bend? I am quite confused

Now on to the aspect of the Rimannian curvature tensor.

May I refer to Wald: page 30[3.2 Curvature]
The diagram given[fig 3.3] we have a curve with sharp edges.The proof seems to be concerned with four separate parallel transports and not with a single transport at a stretch
If such a procedure defines the curvature of a surface in a proper manner it really does not contradict any thing.

But if one is interested in the parallel transporting a vector at a stretch along a curve it should not be one with sharp bends. In such an instance it cannot be called parallel transport in the totality of the operation.

 

[Anamitra]

This is not true. ds is an invariant quantity (a tensor of rank 0), so it remains unchanged under any coordinate transformation.

This is correct and I do not have any means to reject it.

But I will place certain questions to clarify my own concepts and not to contradict any body.

Well the length of any line connecting a pair of points and lying on the space time surface seems to be the space-time separation between them[s = integral ds along the said line]

Now if I take a line from the south to the north pole winding it several times on the body of the sphere is its length going to represent the space-time separation between the poles? Do we need geodesics to calculate space-time separations?

A subsidiary issue:

We may represent the space time sphere by the equation

x^2+y^2+t^2=a^2
I have taken the z-axis to be the time axis(t). Any motion perpendicular to the time axis represents infinitely fast motion forbidden by relativity.So the meridian perpendicular the time axis goes off.Keeping the axes fixed we may rotate the aforesaid plane[perp. to the time axis and going through the origin] about the x or y-axis and remove a huge number of meridians. Of course a huge number of meridians do . One may think of chopping off certain parts of the sphere using the light cone .Fact remains,I am confused.One may consider the meridian in the x-t plane.A particle moving round and round along it has "oscillatory time". It it were a human being his age would undergo periodical movement in the forward and backward directions of time.If the particle stays quiet at one point time would not flow. I am again confused.

[Lines of latitude perpendicular to the time-axis have to disappear to prevent infinitely fast motion. It seems ,that the sphere is in a certain amount of trouble]

 

[Passionflower]

Well the length of any line connecting a pair of points and lying on the space time surface seems to be the space-time separation between them[s = integral ds along the said line]

Now if I take a line from the south to the north pole winding it several times on the body of the sphere is its length going to represent the space-time separation between the poles?

If you mean by 'space-time separation' the metric distance then only a geodesic represents that.

 

[Dale]

May I refer to Wald: page 30[3.2 Curvature]
The diagram given[fig 3.3] we have a curve with sharp edges.The proof seems to be concerned with four separate parallel transports and not with a single transport at a stretch

I don't know how you reach that conclusion when the figure caption clearly reads "The parallel transport of a vector v around a small closed loop".

If you have this book then please examine carefully equation 3.1.19. Note that the tangent vector to the path appears in this equation, but not any derivatives of the tangent vector. So a sharp bend in the path does not cause any trouble. Note also the second sentence of section 3.2 where he explicitly states that the result is path-dependent.

 

[JesseM]

If you mean by 'space-time separation' the metric distance then only a geodesic represents that.

What do you mean? You can certainly calculate the integral of ds along a non-geodesic worldline, for a timelike path this would just be the proper time along that worldline (or i*c times the proper time, if you're using a definition of ds that is real-valued for spacelike paths). In GR I don't think physicists talk about the "space-time separation" between two events since there can be multiple geodesics between the same pair of events, the metric is understood to give you a notion of "distance" along particular worldlines.

 

[Dale]

Hi JesseM, I think I agree with Passionflower on this point. In flat spacetime the invariant interval ("spacetime separation") between two events is the integral of ds along a straight line from one event to the other. So in curved spacetime it should be the integral of ds along a geodesic. The invariant interval then may become ambiguous if there are multiple geodesics connecting the events, which is probably why, as you note, physicists do not speak of "space-time separation" in GR.

But, as it relates to this thread, parallel transport need not be restricted to geodesics.

 

[JesseM]

Hi JesseM, I think I agree with Passionflower on this point. In flat spacetime the invariant interval ("spacetime separation") between two events is the integral of ds along a straight line from one event to the other. So in curved spacetime it should be the integral of ds along a geodesic. The invariant interval then may become ambiguous if there are multiple geodesics connecting the events, which is probably why, as you note, physicists do not speak of "space-time separation" in GR.

Well, the fact that physicists don't really speak of "space-time separation" between events in GR was basically the point I was making to PassionFlower, with the additional point that it is still physically meaningful to integrate ds along non-geodesic wordlines.

 

[Dale]

I agree with both those points.

 

[Anamitra]

We consider a metric of the type shown below:

ds^2=g(00) dt^2-g(1,1) dx^2 - g(2,2) dy^2 - g(3,3) dz^2

ds^2= dT^2- dL^2 [dT--->Physical time,ds----> physical distance]

ds^2/dT^2 = 1- [dL/dT]^2

[ds/dT]^2 = 1-v^2

ds/dT = sqrt[1-v^2]

dT = gamma ds

[ c=1 here and ds is analogous to proper time]

This seems to be a counterpart of special relativity with a variable gamma . Here "v" corresponds to the notion of the three velocity.

Integrating we have,
T2-T1=integral[ s1 to s2 along some path] gamma ds

Since the left side is path independent the right side is also path independent. While in special relativity we can take gamma outside the integral,here we cannot perform such an action.

 

[Anamitra]

This is in relation to what has been said in the Threads #40[https://www.physicsforums.com/showpost.php?p=2849247&postcount=40] and #46[https://www.physicsforums.com/showpost.php?p=2849745&postcount=46]


The space-time we live in is "nearly flat". The slight amount of curvature it has, is extremely important. For instance it keeps satellites/planets in their orbits. We do calculate the three velocities of these satellites from the ground---that is we calculate/assess experimentally the speed of objects at distance. And all this is done in disregard to whatever objections Dalespam has raised against the concept of relative speed in curved space-time.That should settle the issue.

 

[Dale]

Certainly. I never said otherwise.

In a flat space parallel transport is unique and path independent, so we can consider all vectors at any point in the manifold to be part of the same vector space and thus we can compare velocities of distant objects. In a "nearly flat" space, by definition, we can ignore the curvature and treat it as flat.

 

[JesseM]

We consider a metric of the type shown below:

ds^2=g(00) dt^2-g(1,1) dx^2 - g(2,2) dy^2 - g(3,3) dz^2

ds^2= dT^2- dL^2 [dT--->Physical time,ds----> physical distance]

Anamitra, can you respond to my question here about whether your notion of "physical time" and "physical distance" matches DrGreg's interpretation in post #18 of that thread? Do you indeed define them using two metrics which are different from the usual spacetime metric? If so, note that unlike the integral of ds which is always the same regardless of your choice of coordinate system, the integral of dT and dL along a particular path will be coordinate-dependent, since terms like g(00) will have different equations in different coordinate systems.

ds^2/dT^2 = 1- [dL/dT]^2

[ds/dT]^2 = 1-v^2

ds/dT = sqrt[1-v^2]

dT = gamma ds

Note that since you are defining gamma as a function of dL/dT, here gamma would be coordinate-dependent as well.

Integrating we have,
T2-T1=integral[ s1 to s2 along some path] gamma ds

Since the left side is path independent the right side is also path independent.

Why do you think the left side is path independent? If we consider a pair of events with two different worldlines that pass through the pair, the integral of dT (i.e. the integral of sqrt(g(00)) dt) on the first worldline may be different than the integral of dT along the second.

 

[Anamitra]

Certainly. I never said otherwise.

In a flat space parallel transport is unique and path independent, so we can consider all vectors at any point in the manifold to be part of the same vector space and thus we can compare velocities of distant objects. In a "nearly flat" space, by definition, we can ignore the curvature and treat it as flat.

We are not treating space-time as flat when we are considering the motion of planets or satellites.We are considering "with seriousness" the curvature of space-time---- the deviations from the flatness.

 

[Dale]

And when the deviations from flatness are significant then, by definition, you cannot compare distant velocities.

 

[TrickyDicky]

And when the deviations from flatness are significant then, by definition, you cannot compare distant velocities.

Would this be due to the fact that in GR there isn't even an unambiguous way to compare in general the ticking "rates" of inertial clocks, which leads us to not being able to compare velocities since for doing that you need an agreement about the proper time of the objects whose velocity you want to compare?
In the case of "nearly flat" scenarios with no significant curvature where you have the objects paths sharing two events, you can compare their velocities, right?
I'm just starting with the study of GR and wanted to check if I understand anything at all

 

[TrickyDicky]

Would this be due to the fact that in GR there isn't even an unambiguous way to compare in general the ticking "rates" of inertial clocks, which leads us to not being able to compare velocities since for doing that you need an agreement about the proper time of the objects whose velocity you want to compare?
In the case of "nearly flat" scenarios with no significant curvature where you have the objects paths sharing two events, you can compare their velocities, right?
I'm just starting with the study of GR and wanted to check if I understand anything at all

Anyone agrees?

 

[Passionflower]

Would this be due to the fact that in GR there isn't even an unambiguous way to compare in general the ticking "rates" of inertial clocks, which leads us to not being able to compare velocities since for doing that you need an agreement about the proper time of the objects whose velocity you want to compare?
In the case of "nearly flat" scenarios with no significant curvature where you have the objects paths sharing two events, you can compare their velocities, right?
I'm just starting with the study of GR and wanted to check if I understand anything at all

Take a simple example of a 1 meter rod free falling radially in the length in a Schwarzschild spacetime. Gravitation will cause an inertial acceleration between the front and the back of this rod (and of course everything in between). If it gets stretched do you consider the rod to be still one meter long? If the rod is very strong and accelerates everything inwards (proper acceleration!) to maintain its structure is the distance between the ends still one meter? How do you even want to define a meter in this situation?

You see that even in a simple scenario we could already get issues with distance.

 

[TrickyDicky]

Take a simple example of a 1 meter rod free falling radially in the length in a Schwarzschild spacetime. Gravitation will cause an inertial acceleration between the front and the back of this rod (and of course everything in between). If it gets stretched do you consider the rod to be still one meter long? If the rod is very strong and accelerates everything inwards (proper acceleration!) to maintain its structure is the distance between the ends still one meter? How do you even want to define a meter in this situation?

You see that even in a simple scenario we could already get issues with distance.

Yes, there is ambiguity also from the lengths part,as well as with the times (I assume what I say in my quote is basically correct, then), in the end, the final consequence is that as has been said here distant velocities can't be compared as opposed to what the OP was asking.

 

[Dale]

Sorry I missed this earlier. Essentially, yes. There is just no way to do this and arrive at a single unique answer. Meters, seconds, and directions are all affected by curvature.

 

[Anamitra]

"Additional Information"

In Thread #1, I have remarked that parallel transport of a vector in a round trip along a closed smooth on some surface (which in general may be curved)should not produce any change in the orientation of the vector. Let me investigate this in relation to an arbitrary vector without any reference to an inertial frame:

Covariant derivative along the curve=0 [the curve is parametrized by the variable t]

=> dA(mu)/dt + Appropriate Christoffel symbol * dA(lambda)/dt * A(nu) = 0 ---- (1)

[The second term on the left side is a summation term. In the aforesaid Christoffel tensor "mu" is the superscript while "nu" and "lambda" are subscripts]

When the vector comes back to the original location ,after the round trip,the value represented by the Christoffel tensor does not change. We have four identical differential equations at the beginning and the end of our journey as we perform the parallel transport.The equations should produce the same results, that is, the tensor components do not change------the vector does not change its orientation. But if we have sharp bends on our journey of parallel transport, derivatives become undefined on these "sharp bends" and the orientation of the vector is not expected to remain the same when it returns to its original location.
[Boundary conditions change at each sharp bend since the differential equations themselves become undefined on these sharp bends]
For a "smooth curve" which is "not closed" the vector remains parallel if we refer to a chain of inertial states .I have tried to show in Thread #33.

 

[Dale]

Anamitra, please post your equations (particularly tensor equations) using LaTeX. Also, please define your terms. What is t and what is A.

Hopefully when you do so you will note that the above equation is missing an impoortant variable.

 

[Anamitra]

I have uploaded the file which contains the formula in Latex.The formula may be referenced from "Gravitation and Cosmology" by Steven Weinberg, Chapter 4,Tensor Analysis,Section 9[Covariant Differentiation along a curve]

We must always keep in mind the boundary conditions when we are trying /considering to solve the four equations shown in the attachment[The equations have been represented by a single one]. The values of A(mu)/dt contribute to the boundary conditions. At the sharp bends we consider their limiting values. These values are unequal on either sides of any bend

 

[Dale]

OK, so looking at the equation you should be able to see that the curve along which you are parallel transporting is x, and A is the vector which is being parallel transported along x. Also, note that only first derivatives of x (the tangent vector) appear, so a sharp bend in x causes only a finite discontinuity in the tangent vector.

Note further that the equation is a first-order ordinary differential equation in A, and as such it has an analytical solution which involves integration. The integration is unaffected by any finite number of finite discontinuities in the tangent vector. This is why sharp bends are perfectly acceptable in parallel transport.

Furthermore, even if you restrict yourself to smooth curves you will find that your claim is completely false. When a vector comes back to its original location after a round trip (even with smooth curves) it will not match the original vector. This is essentially the DEFINITION of curvature. You are trying to flatten something that cannot be flattened, you need to let go of your Euclidean preconceptions.

I challenge you to take the formula that you have posted, use the metric on a sphere, and parallel transport a vector around the 45º lattitude. Work the problem out for yourself and see if the rest of us are wrong or not.