Curved Space-time and Relative Velocity

[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.

 

[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.