Curved Space-time and Relative Velocity

[Anamitra]

Is it meaningful to talk of relative velocity between two moving points at a distance in curved space-time? This interesting issue came up in the course of discussion of the thread "Curved Space-time and the Speed of Light".I remember Dalespam giving some formidable logic with a very good example[Thread :#25]The content of the thread #19 was also very interesting (and similar in some sense). I would like the to repeat the basic idea that could prevent relative velocity from becoming a valid concept in the framework of general relativity :
To calculate relative velocity we need to subtract one velocity vector from another at a distance.For this we have to bring the vectors to a common point by parallel transport. We could keep one vector fixed[let us call this the first vector]and parallel transport the other [the second vector]to the position of the first vector. Now parallel transport may be performed along several routes. If different routes lead to different directions of the second vector in the final position, relative velocity does not have a unique meaning and becomes mathematically unacceptable.Again I refer to the illustration in thread #25.

My queries: 1)If two points(or observers) are moving relative to each other in curved space-time how should the motion of one point appear to the other physically? Is such observation meaningless from the physical point of view if we are unable to interpret it in the existing framework of mathematics?
2)On Parallel transport: We start with a familiar example on parallel transport:
A vector e on a globe at point A on the equator is directed to the north along a line of longitude.We parallel transport the vector first along the line of longitude until we reach the north pole N and then (keeping it parallel to itself) drag it along another meridian to the equator.Then (keeping the direction there) subsequently transport it along the equator(moving the vector paralley) until we return to point A. Then we notice that the parallel-transported vector along a closed circuit does not return as the same vector; instead, it has another orientation. "

It is interesting to observe that the route followed in parallel transport [in the above example and the example cited by Dalespam in thread #25]involves sharp bends or joints where derivatives cannot be defined. Incidentally from the mathematical point of view the definition of parallel transport[Wald:page 34] involves derivatives[covariant derivatives to state accurately] and we know very well that derivatives do not exist at sharp junction.Can we entertain the above examples to refute the concept of relative velocity in curved space time, considering the fact that such examples use paths containing sharp junctions?
Interestingly one may perform the above examples by " rounding off" the north pole edge and the ambiguity will be removed.One may draw "smooth(closed) curves" on "curved surfaces" and carry out examples of parallel transport. The vectors will coincide in their initial and final positions![If one is to perform an experiment by drawing smooth curves on a basket ball he should take care to move the vector parallely without bothering about what angle it is making with the curve after the motion has been started.Angles should be noted only at the initial and the final points/stages.]

If the arguments in point (2) are correct I may conclude that
1)The concept of relative velocity is mathematically consistent in relation to the notion of curved space time.
2)The ideas portrayed in the thread "Curved Space-time and the Speed of Light" are correct.

 

[George Jones]

I'm not sure I follow you, so let's consider a simple example. Consider two static observers in Schwarzschild spacetime who both have the \theta and \phi values, but who hover at different values of r. Using the method of parallel transport, what is their relative velocity? I think that this is fairly straightforward to compute, at least for geodesic radial paths.

 

[bcrowell]

The thread referred to in #1 is https://www.physicsforums.com/showthread.php?t=408994 Post #25 by DaleSpam is here https://www.physicsforums.com/showpost.php?p=2758350&postcount=25

 

[bcrowell]

I'm not sure I follow you, so let's consider a simple example. Consider two static observers in Schwarzschild spacetime who both have the \theta and \phi values, but who hover at different values of r. Using the method of parallel transport, what is their relative velocity? I think that this is fairly straightforward to compute, at least for geodesic radial paths.

If I'm understanding you correctly, you're saying that this is an example that confirms, as claimed in DaleSpam's #25, that relative velocities of distant objects are not well defined. A free-falling observer moving along a radial geodesic passes static object A and records its velocity relative to him. He parallel-transports that velocity vector along with him as he continues on his geodesic, and when he passes static object B, he sees that B's velocity relative to him differs from A's velocity relative to him. This is different from the equality of velocities that most people would expect when initially presented with this example, and it's also different from the result you get with parallel transport along other paths. For example, if we go from A to B along an (approximately) elliptical orbit, we'll pick up a difference in direction due to the geodetic effect.

 

[bcrowell]

If the arguments in point (2) are correct I may conclude that
1)The concept of relative velocity is mathematically consistent in relation to the notion of curved space time.
2)The ideas portrayed in the thread "Curved Space-time and the Speed of Light" are correct.

Right, the presence of kinks in the curves doesn't affect the argument.

 

[Anamitra]

I have been ,for quite some time, trying to explore the possibility of breaking the speed barrier within the "confines of relativity"? Locally we cannot do it. The laws are very strong in this context.The only option would be to explore the matter in a "non-local" consideration. If I am standing at some point in curved space-time and a ray of light is coming from a distant point (close to some dense object)it would be my normal interest to know the speed of light at each and every point as it comes towards me[Of course I continue to stand at the same point]. With this idea in mind I wrote "Curved Space-time and the Speed of Light".

I am repeating the basic aspects of my considerations in the following calculations:
Let us consider two points A and B separated by a large distance with different values of the metric coefficients.Observers at A and B consider a light ray flashing past 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 from B ie, c ={Spatial separation at B}/{sqrt{g(00)} at B.dt}

[Noting that the speed of light is locally "c"]
Speed of light at B as observed from A= c sqrt{(g(00) at B)/(g(00) at A)}

The left side of the above relation exceeds the speed of light if

g(0,0) at B>g(0,0) at A
[spatial separation is the same for both the observers while the temporal separations[physical] are different---the clocks have different rates at the two points]

It is important to note that general relativity seems to avoid "non-local considerations" [Please do correct me if I am mistaken] and we can always take advantage of this fact to investigate the speed of light at one point as it is observed from another in case, we can find some interesting result.

This exercise does not contradict any law in Special or General Relativity.

 

[Anamitra]

George's reply seems to favor me if I am not incorrect.And I have liked his signature very much.

A relevant point:
If I am standing still at one point V=0. What do I get if I transport a null vector? The nature of the curve will not be a big factor.

 

[Anamitra]

For example, if we go from A to B along an (approximately) elliptical orbit, we'll pick up a difference in direction due to the geodetic effect.

Geodetic effect is a physical effect . It should not be confused with the concept of parallel transport which is an imaginary procedure related to the geometry of the transport.

 

[bcrowell]

Geodetic effect is a physical effect . It should not be confused with the concept of parallel transport which is an imaginary procedure related to the geometry of the transport.

The way you calculate the geodetic effect is by using parallel transport. See
http://www.lightandmatter.com/html_books/genrel/ch06/ch06.html#Section6.2 , subsection 6.2.5.

 

[George Jones]

If I am standing still at one point V=0. What do I get if I transport a null vector?

Careful; think some more about this.

 

[Anamitra]

The way you calculate the geodetic effect is by using parallel transport. See
http://www.lightandmatter.com/html_books/genrel/ch06/ch06.html#Section6.2 , subsection 6.2.5.

An interesting line to quote from the above site:

"The definition of a geodesic is that it parallel-transports its own tangent vector, so the
velocity vector has to stay constant."

If a particle moves solely under the influence of gravity it should follow a geodesic and the above definition(which is common to all texts) settles the issue.

Now we may think of forced motions where forces other than gravity are operating and the bodies are constrained to move along lines that are not geodesics. [We may think of an aircraft following a line of latitude instead of a great circle and another one which moves along a path which is neither a line of latitude or longitude or a great circle]In such cases one may follow the logic suggested in thread#1 ,Point 2, to make relative velocity unique.

 

[Anamitra]

It is important to note that the space-time around the Earth is approximately flat and the nature of this space-time should not be confused with the example of the two aircraft I have given or with the spherical shape of the earth. The aircraft example have been given to emphasize that we may have motion along a geodesic (when gravity is the only agent) and we may have a non-geodesic motion if some other force,I mean some inertial force, is operating. If the effect of the inertial force is taken to be similar/equivalent to gravity we may think of adjusting the original metric to take care of the inertial force. At this point I may refer to Thread #8 of the posting "On the Speed of Light Again!" where the equivalence/similarity of gravity and acceleration has been highlighted with reference to papers in the archives of the "Scientific American" and the "Physical Review" as cited by Robert Resnick.
[If the original metric is altered/adjusted the lines of geodesic should change ]

 

[Dale]

We may think of an aircraft following a line of latitude instead of a great circle and another one which moves along a path which is neither a line of latitude or longitude or a great circle]In such cases one may follow the logic suggested in thread#1 ,Point 2, to make relative velocity unique.

Even if you restrict parallel transport to only be defined along geodesics it is still not unique. Consider parallel transporting a vector from the north pole to the south pole. Each longitude line is a geodesic and each one will result in a different vector.

Many issues in curved geometry can be worked around or defined away, but the non-uniqueness of parallel transport is something we just have to live with. Distant vectors are in different tangent spaces and cannot be compared.

PS when referring to posts in other threads a link would be helpful.

 

[Anamitra]

Let us consider the definition of parallel transport as we know in general relativity:

A vector is parallely propagated along a curve if its covariant derivative along the curve vanishes at each point.
So if the velocity vector is parallely propagated along a curve --"IT CANNOT CHANGE". Simple as that.

Conclusions:
1) The notion of Relative velocity is consistent with the mathematics of curved spacetime.
2)My assertions in the posting "Curved Space-time and the Speed of Light" are correct.

 

[Anamitra]

Can a space-time surface be exactly spherical?
Let us see.
A particle at the south pole sees several geodesics connecting it to the north pole. Which direction is to follow? It will be in a state of indecision.[We are assuming the presence of gravity only]

We may have several geodesics emerging from the same point.But they should not terminate on the same and the identical point on the other side.

 

[yuiop]

Can a space-time surface be exactly spherical?
Let us see.
A particle at the south pole sees several geodesics connecting it to the north pole. Which direction is to follow? It will be in a state of indecision.[We are assuming the presence of gravity only]

We may have several geodesics emerging from the same point.But they should not terminate on the same and the identical point on the other side.

I guess it makes some sort of sense if we define two particles on opposite sides of the "ball" to be infinitely far apart. When two particles are opposite each other like this, there is no preferred direction to move in, so they do not move. This is what we would intuitively expect if the consider the "force of gravity" loosely speaking to be proportional to the inverse of the distance squared, so that the "force" tends to zero when the particles are infinitely far apart. I might well be wrong. I am only just starting to "study" non Euclidean geometry and embedded surfaces.

 

[Anamitra]

Well, the space-time structure referred to in thread #15 was never created by the particles themselves,I mean the particles between whom the relative velocity is to be calculated. The mechanism of creation of the spacetime surface has not been described[and possibly cannot be described] . I have simply assumed its existence to prove that it cannot exist. The particles have been used as "test particles" whose fields are of negligible strength.They should not disrupt the existing field or interact between themselves but they should respond to the existing gravitational field created by some "other means".

Now let us consider a pair of gravitating particles separated by a large distance. The lines of force between them are basically parallel lines and the spacetime structure is "not a sphere". The particles if released from a large distance will move along a straight line. The geodesic is simply a straight line and the space you have described is flat space-time .[the shortest distance being a straight line]. If you kept the two initial particles fixed and released smaller particles midway between them ,they should be moving along straight lines.Of course, the "smaller particles" must be "small enough" not to disrupt the existing field.This in fact would be a better interpretation of the situation.

Now in the first paragraph I have used the term "test particles". It is important that you understand them in relation to the study of gravitational fields/spacetime structure.

I would refer to the book "Gravity" by James B. Hartley , Chapter 8,"Geodesics" in understanding the concept of "Test Particles"

 

[Dale]

A vector is parallely propagated along a curve if its covariant derivative along the curve vanishes at each point.
So if the velocity vector is parallely propagated along a curve --"IT CANNOT CHANGE". Simple as that.

Correct. The parallel transport of a vector along a curve is unique, but there is not a single unique curve connecting two events.

Conclusions:
1) The notion of Relative velocity is consistent with the mathematics of curved spacetime.
2)My assertions in the posting "Curved Space-time and the Speed of Light" are correct.

Neither of these conclusions is correct.

 

[Anamitra]

Let's see:

We consider two points A and B being connected by a pair of curves L1 and L2 lying on the curved space-time surface.We start with a vector V from A and propagate it parallely to B along L1.
V remains constant because its covariant derivative vanishes at each point during the parallel propagation.

Again we start from A with the same vector V and propagate it parallely along L2 to B. V does not change because the covariant derivative of V(as we move along L2)= always zero

In each case we start with the same vector and finish with the same vector.

It is to be seriously noted that constant vectors are not defined by constant components except in rectangular Cartesian systems. Rather the constancy of vectors is defined by the concept of parallel transport[the mathematical concept of parallel transport to state accurately].

Q)If I transport a vector from one point on the curved space time surface to another point,how do I make sure that it has remained constant?

Ans)By the concept of parallel transport. A parallel transport of the vector has to be executed from the first point to the second point.

 

[Dale]

I already gave counterexamples. It is useless to make assertions to which there are known counterexamples. Regardless of how compelling you find the chain of logic you know that there is a flaw since the conclusion is demonstrably false.

Can you find the flaw? Specifically, what is the geometric meaning of your phrase "remains constant"?

 

[Anamitra]

The parallel transport of a vector along a curve is unique, but there is not a single unique curve connecting two events.
DaleSpam in thread #18
This statement is indeed correct

It does not in any way contradict the assertions in thread #19

Specifically, what is the geometric meaning of your phrase "remains constant"?
Dalespam in Thread #20

You will find the answer in thread #19
Nevertheless I am repeating:
Q)If I transport a vector from one point on the curved space time surface to another point,how do I make sure that it has remained constant?

Ans)By the concept of parallel transport. A parallel transport of the vector [as defined mathematically]has to be executed from the first point to the second point.

You may perform this action along several curves connecting them.I have talked of the curves L1 and L2 connecting the points A and B in thread #19[but these two curves should not be geodesics at the same time connecting the same pair of points,A and B]

One should not confuse the concept of geodesics and the parallel transport of a vector along a curve[or along number of curves connecting a pair of points]. All good texts give us a clear concept of these two important(but distinct concepts) -------- that is, "geodesics" and "parallel transport".

Conclusions:
1) The notion of Relative velocity is consistent with the mathematics of curved spacetime.
2)My assertions in the attached article in "Curved Space-time and the Speed of Light"[the essence of which may be viewed in thread #6] are correct.

 

[DrGreg]

Anamitra, you don't seem to have understood the following example:

Even if you restrict parallel transport to only be defined along geodesics it is still not unique. Consider parallel transporting a vector from the north pole to the south pole. Each longitude line is a geodesic and each one will result in a different vector.

In this case the two dimensional surface of the Earth is a Riemannian manifold. Consider a vector at the North Pole pointing along the 0° meridian.

If you parallel transport it along the 0° meridian (a geodesic) it will at all times make an angle of 0° with the meridian, pointing south, and when you get to the South Pole it will be pointing along the 180° meridian.

If you parallel transport it along the 90°W meridian (another geodesic) it will at all times make an angle of 90° with the meridian. When you get to the Equator it will be pointing east and when you get to the South Pole it will be pointing along the 0° meridian.

This is an example in a 2D Riemannian manifold (space); something similar could happen in a 4D pseudo-Riemannian manifold (spacetime).

 

[Anamitra]

You are getting these results because you are considering a spherical space-time surface which should not exist in practice. I have tried to explain this in thread #15

 

[DrGreg]

You are getting these results because you are considering a spherical space-time surface which should not exist in practice. I have tried to explain this in thread #15

Can a space-time surface be exactly spherical?
Let us see.
A particle at the south pole sees several geodesics connecting it to the north pole. Which direction is to follow? It will be in a state of indecision.[We are assuming the presence of gravity only]

We may have several geodesics emerging from the same point.But they should not terminate on the same and the identical point on the other side.

The spherical space-only manifold is just an easy-to-understand example. It can happen in other manifolds too.

In spacetime, the geodesic that a particle follows depends on its velocity. There are an infinite number of different geodesics through any single event, corresponding to particles with different velocities. In flat spacetime there is only one geodesic joining a given pair of events, but in curved spacetime there can be more than one. In fact, a good example is lots of different satellites all at the same height following geodesics orbiting a planet. There are lots of ways of orbiting a planet to reach an antipodal point.

 

[Dale]

Thanks for the great explanations DrGreg

You are getting these results because you are considering a spherical space-time surface which should not exist in practice. I have tried to explain this in thread #15

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.

 

[Anamitra]

"In flat spacetime there is only one geodesic joining a given pair of events, but in curved spacetime there can be more than one. In fact, a good example is lots of different satellites all at the same height following geodesics orbiting a planet."
DrGreg in Thread #24

The geodesics in the neighborhood of the earth(or any planet for that matter) are simply radial lines,considering the flat(or nearly flat nature) of the space-time surrounding it.If I drop a body from somewhere (from some height above the Earth's surface ) it will simply fall down. It will not move along a great circle. I am ready to give you a assurance on that issue. If I am flying an air craft in a great circle around the Earth I am "using fuel" to drive the aircraft. The motion is not "under the action of gravity alone" .
But in case of satellites matters are somewhat complicated.In the parking orbits they move under the action of the "centripetal force" which simply an effect of space-time curvature.Motion is under the action of gravity alone. But far away from the Earth's surface the geodesics are again straight lines. If I want to go from London to New York over the Earth's surface the shortest route would be the great circle connecting London and New York. But if one considers a pair of point very high above the eart's surface, the shortest route would be again a straight line[remembering the essentialy flat nature of spacetime there,I mean at the large height considered] and not a line parallel to a great circle on the Earth's surface.Near the Earth's surface the "geodesics" are great circles possibly due to the impenetrability of the Earth and not due the"strong" curvature of space -time .

 

[Dale]

The geodesics in the neighborhood of the earth(or any planet for that matter) are simply radial lines

This is incorrect, purely radial geodesics are not the only geodesics. There are also geodesics whose projections are circles, ellipses, and hyperbolas. All of these types of geodesics can exhibit the polar symmetry that DrGreg and I described, and often other symmetries as well. The counterexample is directly (and obviously) applicable to the Schwarzschild spacetime.

Also, (this is a minor point) remember that we are talking about spacetime not just space. Even if the projection onto a 3D hyperslice is a straight radial line the actual 4D worldline is curved.

 

[Anamitra]

"This is incorrect, purely radial geodesics are not the only geodesics. "
DaleSpam in Thread #27

This is true.

"There are also geodesics whose projections are circles, ellipses, and hyperbolas. All of these types of geodesics can exhibit the polar symmetry that DrGreg and I described, and often other symmetries as well."
DaleSpam in Thread #27
This again true .

In fact if one considers a portion of a "great circle" at a large height it is indeed a straight line consistent with the "flatness" or the "near flatness" of space-time.Of course the "nearly flat" nature of space-time is responsible for the curving in the salellite orbits[and this curving is consistent or nearly consistent with the action of the "classical" centripetal force].

We now consider a the intersection of a pair of great circles at a large height. We raise a satellite to this intersection point. Of course for that we need " extra forces" , I mean forces other than gravity!Then a gain to get the satellite moving along a particular great circle we need the "extra forces" once more----- again gravity is not the only force in action.If we took the satellite to the "point of singularity/intersection of the great circles" and left it there it would have simply fallen down along a "radial line ". That should explain the issue.

 

[JesseM]

The geodesics in the neighborhood of the earth(or any planet for that matter) are simply radial lines,considering the flat(or nearly flat nature) of the space-time surrounding it.If I drop a body from somewhere (from some height above the Earth's surface ) it will simply fall down.

But if you give it some initial horizontal velocity--tossing it sideways--it will follow a curved arc down towards the surface. If the object had no electromagnetic interaction with the Earth so it could pass right through solid matter like a neutrino, that arc would be part of a highly eccentric orbit that would pass very close to the center of the Earth.

It will not move along a great circle. I am ready to give you a assurance on that issue. If I am flying an air craft in a great circle around the Earth I am "using fuel" to drive the aircraft. The motion is not "under the action of gravity alone" .

That's only because air crafts don't travel fast enough to be in an orbit above the surface (and it would be impossible in practice for anything to orbit at that height because of atmospheric drag--an object could potentially orbit at the height of an aircraft around a body without an atmosphere like the moon, though).

But in case of satellites matters are somewhat complicated.In the parking orbits they move under the action of the "centripetal force" which simply an effect of space-time curvature.Motion is under the action of gravity alone. But far away from the Earth's surface the geodesics are again straight lines.

Why do you say "the geodesics are again straight lines"? It's possible to have an orbit arbitrarily far from the Earth (assuming no other bodies to disrupt it), and all orbits are geodesics. I think (based also on your comment about great circles) that you may be confusing geodesics in space with geodesics in spacetime--general relativity says objects in free-fall with no non-gravitational forces acting on them follow geodesics in spacetime, not geodesics in space. Geodesics in spacetime are not the paths with the shortest spatial distance, rather they are the paths with the greatest proper time (in curved spacetime, the proper time is only 'greatest' when compared with other 'nearby' worldlines--worldlines that only deviate infinitesimally from the geodesic worldline--since as DrGreg mentioned you can have multiple geodesics passing through the same pair of points). Think for example of the twin paradox, where the inertial twin (following a geodesic in flat spacetime) ages more than the non-inertial twin (experiences more proper time), regardless of the precise path taken by the non-inertial twin.

If I want to go from London to New York over the Earth's surface the shortest route would be the great circle connecting London and New York.

That would be a spatial geodesic on the curved 2D surface of the Earth, it wouldn't be a geodesic in spacetime unless you were traveling at orbital speed, nor would it be a geodesic in 3D space (the path through 3D space with the shortest spatial distance would travel underneath the surface of the Earth).

Near the Earth's surface the "geodesics" are great circles possibly due to the impenetrability of the Earth and not due the"strong" curvature of space -time .

No, the fact that spatial geodesics on the curved 2D surface of the Earth are great circles has nothing to do with spacetime curvature, it's just a geometric issue. Again, general relativity only says that free-falling objects follow geodesics in spacetime, not geodesics in space, and a great circle along the surface is not a geodesic in spacetime unless you were traveling along the surface at orbital speed. And the "impenetrability of the earth" is also irrelevant to determining geodesics in spacetime--a spacetime geodesic is the hypothetical path that would be taken by an object that wasn't affected by any non-gravitational forces, so it'd be able to travel straight through solid matter like a neutrino, which is completely unaffected by the electromagnetic force (although the weak nuclear force has some effect on neutrinos, so even neutrinos wouldn't quite follow geodesics).

 

[Anamitra]

"But if you give it some initial horizontal velocity--tossing it sideways-..."
Jesse in Thread #29
You are calling in forces other than gravity."Originally Posted by Anamitra

It will not move along a great circle. I am ready to give you a assurance on that issue. If I am flying an air craft in a great circle around the Earth I am "using fuel" to drive the aircraft. The motion is not "under the action of gravity alone" .

That's only because air crafts don't travel fast enough to be in an orbit above the surface (and it would be impossible in practice for anything to orbit at that height because of atmospheric drag--an object could potentially orbit at the height of an aircraft around a body without an atmosphere like the moon, though)."

Jesse in Thread #29

To achieve the "Great mission" of getting the object/aircraft into the parking orbit some "extra force" has to be arranged by Jesse. Gravity won't do the job for her.
"Your confusing geodesics in space with geodesics in spacetime--general relativity says objects in free-fall with no non-gravitational forces acting on them follow geodesics in spacetime, not geodesics in space. Geodesics in spacetime are not the paths with the shortest spatial distance, rather they are the paths with the greatest proper time (in curved spacetime, the proper time is only 'greatest' when compared with other 'nearby' worldlines--worldlines that only deviate"
Jesse --in thread # 29

This is meaning ful

It does not stand as a hindrance to my arguments.

Just think of fact:
If we have two geodesics connecting a pair of points A and B ,the particle at the point of intersection would be in a state of indecision[as to which spacetime curve it should follow].
To help it decide Jesse has to provide some "extra force". Gravity will not help her!

Everybody will watch Jesse trying to help the particle decide what to do!

 

[DrGreg]

Just think of fact:
If we have two geodesics connecting a pair of points A and B ,the particle at the point of intersection would be in a state of indecision[as to which spacetime curve it should follow].
To help it decide Jesse has to provide some "extra force". Gravity will not help her!

Everybody will watch Jesse trying to help the particle decide what to do!

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).

 

[Dale]

Anamitra, none of your recent comments are in the least bit relevant to the issue of parallel transport. The only relevant points are:
1) There are multiple paths between any two events
2) The result of parallel transport depends on the path

The discussion about geodesics is not relevant since parallel transport is not defined in terms of geodesics (in fact, geodesics are defined in terms of parallel transport so attempting to define parallel transport that way would be a circular definition).

The discussion about satellites being launched is not only completely irrelevant, but wierdly so. I cannot fathom the thought process that would lead you to think that it could have any import to the non-uniqueness of parallel transport.

 

[Anamitra]

"The result of parallel transport depends on the path"

DaleSpam, Thread 32
Let me investigate the above statement:

Parallel transport is defined by the fact that covariant derivative of the vector as it moves along the curve should be zero .

Now the covariant derivative of a tensor has two parts :
1)The ordinary derivative.
2)The part containing the "affine connection".
The affine connection is zero in all local inertial frames while it may be non zero in non-inertial frames.

Covariant derivative= ordinary derivative[ie,dA(mu)/dx(i)] +part containing the affine connection

For parallel transport the left side of the above relation is zero

Therefore,

ordinary derivative + affine connection term= zero

For local inertial frames,affine connection part=0
ordinary derivative part, dA(mu)/dx(i)=0

We connect the end points of the path of parallel transport by a chain of "local inertial points"

Along the path, we have
dA(mu)/dx(i) = 0
Therefore the components of the tensor remain constant as we move along the "chain of local inertial points"

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.

[NB: If the components of a tensor are zero in any frame they are also zero in all other frames.
=> If the covariant derivative=0 in any frame it should be zero in all other frames,inertial or non-inertial]

 

[JesseM]

"But if you give it some initial horizontal velocity--tossing it sideways-..."
Jesse in Thread #29
You are calling in forces other than gravity."Originally Posted by Anamitra

It will not move along a great circle. I am ready to give you a assurance on that issue. If I am flying an air craft in a great circle around the Earth I am "using fuel" to drive the aircraft. The motion is not "under the action of gravity alone" .

That's only because air crafts don't travel fast enough to be in an orbit above the surface (and it would be impossible in practice for anything to orbit at that height because of atmospheric drag--an object could potentially orbit at the height of an aircraft around a body without an atmosphere like the moon, though)."

Jesse in Thread #29

To achieve the "Great mission" of getting the object/aircraft into the parking orbit some "extra force" has to be arranged by Jesse. Gravity won't do the job for her.

You're wrong, if there was no atmosphere to cause drag an object with a sufficiently high velocity could be orbiting at the height of an aircraft, there would be no need for me to arrange any extra force to make this orbit work (incidentally I am a he, not a she). For a perfectly circular orbit, the orbital velocity is easy to calculate in Newtonian physics (which is a good approximation to GR in the case of the weak spacetime curvature created by a planet)--just consider a rotating frame where the orbiting object is at rest, and in this case the "centrifugal" acceleration (equal and opposite to the centripetal acceleration) must have the same magnitude as the gravitational acceleration. The centripetal acceleration is just v^2/R, while the gravitational acceleration is GM/R^2 (where M is the mass of the planet), so setting them equal and multiplying both sides by R gives v^2 = GM/R, meaning the orbital speed would be \sqrt{GM/R}. This formula works just as well if R is at the radius of an aircraft as it would if R was at the radius of a satellite (and indeed the two radii only differ by a few km for a satellite in low Earth orbit)--for example, a craft 863 meters above the surface at the equator would have R=6379 km=6379000 meters, so with M = 5.9742 * 10^24 kg and G=6.67428 * 10^-11 meters^3 / (kg * s^2), this means GM/R = 6.25 * 10^7 meters^2/second^2, so if the orbital velocity is the square root of that, the necessary velocity would be about 7,900 meters/second, or 7.9 km/second, or about 28,000 km/hour. This is just slightly larger than the figure given here for the orbital velocity of a satellite at an altitude of 200 km, where the orbital velocity is quoted as 27,400 km/hour.

Of course, as I said it's not actually possible to orbit at a height like 863 meters above the Earth because the atmosphere would create too much drag, but if the Earth was an airless planet it would be quite possible. And likewise if we imagine a particle that is not affected by any non-gravitational forces so it can pass straight through solid matter unaffected (the neutrino is almost like this, although it is affected by the weak nuclear force so a tiny fraction of neutrinos won't pass straight through the Earth), it could orbit within the atmosphere, or even at a smaller R that is below the crust. Geodesic paths represent the paths that would be taken by such hypothetical particles which are only affected by gravity and not other forces.

"Your confusing geodesics in space with geodesics in spacetime--general relativity says objects in free-fall with no non-gravitational forces acting on them follow geodesics in spacetime, not geodesics in space. Geodesics in spacetime are not the paths with the shortest spatial distance, rather they are the paths with the greatest proper time (in curved spacetime, the proper time is only 'greatest' when compared with other 'nearby' worldlines--worldlines that only deviate"
Jesse --in thread # 29

This is meaning ful

It does not stand as a hindrance to my arguments.

Just think of fact:
If we have two geodesics connecting a pair of points A and B ,the particle at the point of intersection would be in a state of indecision[as to which spacetime curve it should follow].
To help it decide Jesse has to provide some "extra force". Gravity will not help her!

Everybody will watch Jesse trying to help the particle decide what to do!

Nope, no extra force is required. The "choice" of which geodesic path a particle follows from point A is totally determined by its instantaneous velocity at A (both direction and speed, as defined in some locally inertial frame at A)--particles with different velocities follow different geodesics. This is true in flat SR spacetime as well, where geodesics are just straight (inertial) worldlines--obviously you can have two straight lines going in different directions from a single point A, although unlike in GR, geodesics that cross at one point A can never cross again at a second point B in SR.

 

[Dale]

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.

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

https://www.physicsforums.com/showpost.php?p=2848002&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.

 

[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.