I Parallel transport general relativity

[Jufa]

TL;DR Summary

I am having serious difficulties to understand the concept of parallel transport.

Suppose you have a tensor quantity called "B" referenced in a certain locally inertial frame (with four Minkowski components for instance). As far as I know, a parallel transportation of this quantity from a certain point "p" to another point "q" consists in expressing it in terms of the reference frame of "q" in such a way that the tensor quantity "B" remains invariant. If it is the case that the space is flat and a coordinate independent reference frame is choosen (such as the Minkowski's), then "p" and "q" share local reference frames and therefore the components of "B" coincide in both locations. This is not the case in curved spaces, i.e. where gravity shows up.
Here is my problem:
Once known the metric of the space, we can fix a single local reference frame at every location (event). Then I don't understand how it is possible that the parallel transport can depend on the path or, equivalently, how can a parallel transport on a closed path give a different set of tensor coordinates than the initial ones. If the initial and final points of the path coincide their reference frame should be the same as well and therefore different coordinates in initial and final tensors would result into actually modifying the tensor quantity, which is not what parallel transport does.
The only solution I encounter to this apparent paradox is that, given the fact that local inertial frames (free falling frames) are not unique, one can choose the frame of the initial and final points of the path (which are the same) to be different. Why does this happen? I mean theoretically we are free to choose the local inertial frame that most suits our needs. But why would not we choose that the same location (although after covering a whole closed path) has the same reference frame? I guess that we are not that free to choose the reference frame of that final point once covering an entire close path. If we were, the difference of the tensor components after a certain parallel transport would not be even well defined.
My question is then the following:
Once fixing the initial point "p", its locally inertial reference frame and the curve we want to follow to carry out the parallel transport, how does this curve affect the choice of the locally inertial reference frames of every location covered? Maybe an example with a certain curve it would make it easier for me to understand it.
Thanks in advance.

 

[Dale]

As far as I know, a parallel transportation of this quantity from a certain point "p" to another point "q" consists in expressing it in terms of the reference frame of "q" in such a way that the tensor quantity "B" remains invariant.

I don’t recognize that as an accurate description of parallel transport.

Parallel transport intuitively means that you move a vector (or other tensor) without turning it. If your path turns 10 degrees left then a vector that doesn’t turn with you will be 10 degrees to your right.

A good example is on the 2D surface of a sphere. Suppose you start at the equator with a vector pointing north, and walk due north to the North Pole, the vector will point forward that whole segment. Now, at the North Pole turn 90 deg to the right, so the vector is now pointing to your left. Walk due south back to the equator and the vector will point to your left the whole way. On the equator turn 90 deg to the right again, so the vector is pointing behind you. Walk back to your start and the vector will continue to point behind you. When you return to your starting point turn 90 deg to the right. Finally, you have returned to your original location and bearing, but now the vector is pointing to your right (east) instead of north.

 

[Nugatory]

Summary:: I am having serious difficulties to understand the concept of parallel transport.

Then I don't understand how it is possible that the parallel transport can depend on the path or, equivalently, how can a parallel transport on a closed path give a different set of tensor coordinates than the initial ones. If the initial and final points of the path coincide their reference frame should be the same as well and therefore different coordinates in initial and final tensors would result into actually modifying the tensor quantity, which is not what parallel transport does.

Here's an easily visualized example.

At a point on the Earth's equator we have a vector pointing due north; for definiteness let's choose that point to be at zero degrees of longitude. We parallel transport the vector to the north pole; it is now pointing south along the 180 degrees of longitude line (if this is not clear, imagine that we started at the equator holding a spear pointing due north, and then started walking north until we reached the pole. What direction does the spear point now?). Next, parallel transport the vector along the 90 degree west longitude line until we reach the equator; now it points west. Parallel transport it along the equator to the east to return to our starting point, and now the vector is pointing west; the round trip rotated it by 90 degrees.

Next, try the same round trip in the other direction, again starting with the vector pointing due north: first 90 degrees west, then 90 degrees north, then 90 degrees south on the zero longitude line. We find that the vector ends up pointing east instead of west so we see different results on different paths.

We didn't do anything with reference frames or coordinates here, we just took advantage of the fact that a geodesic (the equator and the latitude lines are all geodesics) parallel transports a vector onto itself.

 

[nucl34rgg]

You're transporting the tangent vectors from the tangent space at one point on the curve to a different tangent space at a different point so that the vectors remain parallel as you travel an infinitesimal amount along a curve from the first point to the second point. Does that make sense?

 

[Jufa]

I don’t recognize that as an accurate description of parallel transport.

Well, the curve followed by a parallel transportation is defined as these curve where the covariant derivative of the vector vanishes. This means that the tensor quantity to transported does not change throghout the whole path. Am I not right?

 

[DrGreg]

Suppose you start at the equator with a vector pointing north, and walk due north to the North Pole, the vector will point forward that whole segment. Now, at the North Pole turn 90 deg to the right, so the vector is now pointing to your left. Walk due south back to the equator and the vector will point to your left the whole way. On the equator turn 90 deg to the right again, so the vector is pointing behind you. Walk back to your start and the vector will continue to point behind you. When you return to your starting point turn 90 deg to the right. Finally, you have returned to your original location and bearing, but now the vector is pointing to your right (east) instead of north.

Illustration from Wikipedia:

Source: https://commons.wikimedia.org/wiki/File:Parallel_Transport.svg
Creator: Fred the Oyster
Licence: CC-BY-SA 4.0

 

[PeterDonis]

I am having serious difficulties to understand the concept of parallel transport.

I think part of the issue is that you are trying to think about parallel transport in terms of local reference frames, and you appear to have some misconceptions in your understanding of local reference frames. Also you appear to be thinking about the relationship between parallel transport and local reference frames backwards; see the comments at the end of this post.

If it is the case that the space is flat and a coordinate independent reference frame is choosen (such as the Minkowski's), then "p" and "q" share local reference frames

No, they don't. They share (assuming they are at rest relative to each other, or more precisely that observers located at p and q are at rest relative to each other) the same global reference frame, since they are at rest relative to each other, but their local reference frames will not be the same since they will be centered on different points.

Once known the metric of the space, we can fix a single local reference frame at every location (event).

You can do this even without knowing the metric, since local reference frames are treated as flat and any effects due to spacetime curvature are ignored. But there won't be just one local reference frame at any given event in spacetime; as you recognize later on in your post, there will be an infinite number of them, in fact a six-parameter infinite group of them, since it takes six independent parameters to specify a 4-velocity vector at an event (corresponding to the six independent parameters that specify an arbitrary local Lorentz transformation).

Once fixing the initial point "p", its locally inertial reference frame and the curve we want to follow to carry out the parallel transport, how does this curve affect the choice of the locally inertial reference frames of every location covered?

It doesn't have to affect the choice of local reference frames at all. But if you want to take the basis vectors of a local reference frame at p, and parallel transport those vectors along some curve (which we'll here assume to be a geodesic, to keep things simple) to q in order to obtain a local reference frame at q, that is indeed a well-defined operation. And if you do that operation around a closed curve in a curved spacetime, you will find, in general, that the final local reference frame you end up with will not be the same as the one you started with.

 

[Jufa]

It doesn't have to affect the choice of local reference frames at all.

Well, you say that if you choose a certain closed curve it might be possible that the final basis does not coincide with the inicial. Nonetheless if you don't perform any curve (which Can be thought as a particular case of a closed curve) then the final basis is clearly the same as the initial. From this point of view it seems clear to me that fixing the curve implies fixing the reference basis of the locations in the curve and I wonder what is the relation.

 

[Dale]

Well, the curve followed by a parallel transportation is defined as these curve where the covariant derivative of the vector vanishes.

Parallel transport doesn’t define a curve, you can do parallel transport along any curve.

This means that the tensor quantity to transported does not change throghout the whole path. Am I not right?

That is correct. But I don’t know what it has to do with reference frames.

 

[PeterDonis]

you say that if you choose a certain closed curve it might be possible that the final basis does not coincide with the inicial

Not "might be possible" if the spacetime is curved; the final basis will not coincide with the initial one, definitely.

if you don't perform any curve

Meaning the "curve" is just one point? Yes, this obviously would not change anything. So what?

From this point of view it seems clear to me that fixing the curve implies fixing the reference basis of the locations in the curve

No, all it means is that a "curve" consisting of only a single point is not a useful concept for this discussion. You have to look at curves that have a continuum of points.

What fixing the curve does is determine what you will end up with if you parallel transport a vector, or a set of vectors such as a basis, from one endpoint of the curve, along the curve to the other endpoint. But that has nothing whatever to do with choosing a basis at any point along the curve. The basis at the initial point is just one way of obtaining a set of vectors to be transported--and of course it's not the only way. You can pick vectors to be transported however you want. Once you've chosen vectors, how they are transported is independent of any choice of reference frame; that's why I said you were thinking of this backwards.

 

[Orodruin]

What strikes me when reading through all of this is that OP’s entire argument is made without referring back to the mathematical definitions and deductions which means that it is heuristic at best and ultimately leads to the wrong place. When all else fails, I find there is often a point in trying to formalise the argument as this may make it easier to see where the heuristic fails.

 

[FactChecker]

What strikes me when reading through all of this is that OP’s entire argument is made without referring back to the mathematical definitions and deductions which means that it is heuristic at best and ultimately leads to the wrong place. When all else fails, I find there is often a point in trying to formalise the argument as this may make it easier to see where the heuristic fails.

I think that one conceptual error is the implication that a tensor must have the same coordinates and coordinate system when parallel transport along a cyclical path gets back to the starting point. The tensor can have different coordinates in the new coordinate system as long as the tensor coordinate transformation requirements are satisfied.

 

[Ibix]

To me it has to do that the components of the transported tensor will along the curve as long as the reference frame choice changes along the curve.

There seems to be verb missing here, and I'm not at all sure what you are trying to say.

The point about parallel transport is that it defines what we mean by "not changing" for the tangent and cotangent spaces. If you parallel transport the basis vectors of a local frame along some path, those parallel transported vectors will always be the basis vectors of the local frame of someone following that path. If the path is a closed loop, the parallel transported frame may be different on a subsequent visit to a point from on its previous visit.

 

[Jufa]

If you parallel transport the basis vectors of a local frame along some path, those parallel transported vectors will always be the basis vectors of the local frame of someone following that path.

I think this is the key point. First of all. What do you mean by transporting the basis vectors? As far as I know what changes after a certain transport are the coordinates, not the vectors them selves. This is all I don't Understand.

 

[Ibix]

I think this is the key point. First of all. What do you mean by transporting the basis vectors? As far as I know what changes after a certain transport are the coordinates, not the vectors them selves. This is all I don't Understand.

Vectors at different points are in separate tangent spaces, so after parallel transport along a non-closed path you most definitely do not have the same vectors. The whole point is that there isn't even a way to compare the vectors at one point to those at another, except to define some mapping of a vector from one space to the other. That's what parallel transport does. It says that a vector in the tangent space at one point is, in some sense, pointing in the same direction as another vector in another tangent space.

But what about closed paths? In that case, parallel transport is a map from a tangent space to itself. It isn't necessarily an identity map, so a vector parallel transported around a curve isn't necessarily itself. The transported vector will typically point in a different direction to its original. So how could it be the same vector?

The key point is that parallel transport is a map from one vector space to another (or itself). It doesn't actually take a vector out of one tangent space and carry it to another space (what would that even mean?), so it can only be a mathematical method for mapping between two vectors (possibly in different spaces).

I'm not sure what point you are trying to make when you talk about "the coordinates [changing], not the vectors them selves". Are you confusing this with a basis change? In that case, you decide that you don't like your basis vectors anymore and pick new ones, and then you must re-express all your quantities in terms of your new basis - i.e., change their components without changing the vectors/tensors themselves. But in the case of parallel transport, we're applying some maths to find out what carrying our basis vectors around a loop would do to the direction they point. You could then change to this new basis, but you aren't required to do so.

 

[stevendaryl]

I think this is the key point. First of all. What do you mean by transporting the basis vectors? As far as I know what changes after a certain transport are the coordinates, not the vectors them selves. This is all I don't Understand.

Yes, the basis vectors change as you move around.

Let's look at a very simple case that has connection coefficients which is the 2-dimensional plane, in polar coordinates versus cartesian coordinates.

In cartesian coordinates, you have coordinates $x$ and $y$ and basis vectors $e_x$, which is a unit vector in the x-direction, and $e_y$, which is a unit vector in the y-direction. Transforming to polar coordinates $r$ and $\phi$ where $x = r cos(\phi)$ and $x = r sin(\phi)$, we have:

$e_r = \dfrac{\partial x}{\partial r} e_x + \dfrac{\partial y}{\partial r} e_y = cos(\phi) e_x + sin(\phi) e_y$
$e_\phi = \dfrac{\partial x}{\partial \phi} e_x + \dfrac{\partial y}{\partial \phi} e_y = - r sin(\phi) e_x + r cos(\phi) e_y$

Obviously, if $e_x$ and $e_y$ are constant vectors (which they are), then $e_r$ and $e_\phi$ are not constant vectors. They are functions of $r$ and $\phi$.

As a matter of fact, one way to define the connection coefficient $\Gamma^i_{jk}$ in a coordinate basis is this way:

$\Gamma_{ijk} = e_i \cdot \nabla_j e_k$

(Then $\Gamma^m_{jk} = g^{im} \Gamma_{ijk}$)

If the basis vectors didn't change from point to point, then the $\Gamma$ coefficients would all be zero.

 

[DrGreg]

It's easier to consider an example of a curved 2D space instead of a curved 4D spacetime, so consider the 2D surface of a sphere. The advantage is that we can draw diagrams of the 2D space embedded in 3D Euclidean space. The third dimension is never used in the maths but it can help you visualise what is happening.

Below we have a blue 2D manifold. The yellow squares represent 2D tangent spaces at various points on the manifold and the red and green arrows represent a choice for the two basis vectors for each tangent space.

The left diagram shows basis vectors that have been parallel-transported from a point $p$ along two different paths.

The right diagram shows basis vectors that have been parallel-transported from another point $r$ along two different paths.

And what is clear is that both diagrams agree over the basis vectors at $p$ and $r$, but they disagree over the basis vectors at $q$.

It also shows that parallel transport direct from $p$ to $q$ is not the same as parallel transport from $p$ to $r$ to $q$.

 

[A.T.]

I am having serious difficulties to understand the concept of parallel transport.

Maybe this mechanical model can help with the intuition:

Imagine a tank, with the gun turret rotation inversely coupled to the tank steering: When the tank hull turns X degree relative to the local ground, the turret turns -X degree relative to the hull, so the gun keeps its orientation relative to the local ground. This is how you parallel transport a gun.

On the sphere, the only way to prevent the turret from rotating relative to the hull, is to move on great circles (geodesics). To move on a circle of constant latitude off the equator, the tracks of the tank must be running at different speeds, so the tank is turning, and the turret is rotating relative the hull. When the tank arrives at the starting position, the gun will have a different orientation, then it started with.

 

[PeterDonis]

What do you mean by transporting the basis vectors? As far as I know what changes after a certain transport are the coordinates, not the vectors them selves.

No. Again you have it backwards. If you are talking about local coordinates centered on two different points, and all you have is the coordinates themselves, you have no way of comparing them at all. Components in local coordinates centered on point p have no relationship whatsoever with components in local coordinates centered on point q, if the coordinates are all you have.

The only way to have any relationship between anything at point p and anything at point q is to transport geometric objects, like vectors (you can also transport tensors of higher rank), from point p to point q, along some specific curve. And geometric objects are independent of any choice of coordinates; you have to get the geometric objects and how they transport straight first, before even looking at how they can be used to relate coordinates at one point to coordinates at another.

 

[FactChecker]

I think this is the key point. First of all. What do you mean by transporting the basis vectors? As far as I know what changes after a certain transport are the coordinates, not the vectors them selves. This is all I don't Understand.

It is important to distinguish between a "vector" in the physics sense versus a "vector" in the pure mathematical sense. In physics, a vector often means something physical that can be represented in many different coordinate systems but has a fixed physical meaning independent of the coordinate system. In mathematics, a vector is more general and some may be defined that do not represent anything physical and are dependent on the coordinate system. Orthonormal basis vectors can be used to define a coordinate system and are clearly dependent on the coordinate system.

 

[Jufa]

I am sorry guys but I definitely don't get it. In the sphere example, which I have seen everywhere, I don't know which is the rule to choose the local reference frame at a certain point of the curve. In the beginning it seems that the basis vectors chosen are the ones corresponding to the azimuth angle and the polar angle. This makes sense to me, because these two vectors do change along the curve chosen in such a way that in order to keep some vector (in the abstract physical sense) constant their components must change at every point. Nevertheless, when arriving to the north pole this basis vectors are no longer well defined, since it is a singular point.

 

[Jufa]

I will try to make my question as simple as possible:
Let us consider the physical quantity $A^{\mu}(x^{\nu}_0) e^\alpha_{\mu}(x^\nu_0)$ which it is obtained by evaluating the vectorial field $A^{\mu}(x^{\nu})$ at $x^\nu_0$, considering a local reference frame defined by the basis vectors $e^\alpha_{\mu}(x^\nu_0)$. This basis vectors, in contrast with the coordinates $A^{\mu}(x^{\nu}_0)$, reffer to an actual physical quantity and, therefore, do not depend on the reference frame chosen.
If we perform a parallel transport of this vector from $x^{\nu}_0$ to $y^{\nu}_0$ we obtain $\hat{A}^{\mu}(y^{\nu}_0)$ which satisfies the following:

$$ \hat{A}^{\mu}(y^{\nu}_0) e^\alpha_{\mu}(y^\nu_0) = A^{\mu}(x^{\nu}_0) e^\alpha_{\mu}(x^\nu_0) $$

My only question is how do we know the relation between the basis vectors $e_{\mu}(y^\nu_0)$ and $e^\alpha_{\mu}(x^\nu_0)$. Because it seems that they don't just depend on the position considered but also on the path we have followed to reach that position. I am asking for the rule we follow to choose the basis vectors at every single point of the curve, which ultimately determines how do the components of the vector transported must vary.

Someone might say that there is no way of relating $e^\alpha_{\mu}(y^\nu_0)$ and $e^\alpha_{\mu}(x^\nu_0)$ since they are vectors referred to different locations and therefore different local frames. But I don't see this clear, since $e^\alpha_{\mu}(y^\nu_0)$ and $e^\alpha_{\mu}(x^\nu_0)$ are actual physical quantities i.e. they do not depend of the reference frame considered. In the sphere example, for instance, the basis vectors are all 3-D vectors which can be related considering theme as free vectors, independently of their position.

 

[A.T.]

I am sorry guys but I definitely don't get it. In the sphere example, which I have seen everywhere, I don't know which is the rule to choose the local reference frame at a certain point of the curve.

Does the mechanical analogy in post #18 make sense to you? it doesn't explicitly use specific local reference frames.

 

[Jufa]

Does the mechanical analogy in post #18 make sense to you? it doesn't explicitly use specific local reference frames.

I'm affraid I don't get it.

 

[Jufa]

a choice for the two basis vectors for each tangent space.

Here you state that the reference frame at every point is out of choice.

 

[Jufa]

basis vectors that have been parallel-transported

Nevertheless you say here that the new basis at each point is not obtained by a mere choice but by parallel transporting it.

 

[DrGreg]

Nevertheless you say here that the new basis at each point is not obtained by a mere choice but by parallel transporting it.

What I mean is I have made a choice, and my choice is to use parallel-transported basis vectors. I could have chosen a different basis, but I didn't.

 

[Jufa]

@Jufa, try reading through [the relevant sections of] this document: https://arxiv.org/pdf/1412.2393.pdf

Many thanks. I'll give it a try.

 

[Orodruin]

In the sphere example, for instance, the basis vectors are all 3-D vectors which can be related considering theme as free vectors, independently of their position.

This is where (one of) your misconception lies. There is generally no need to refer to the embedding of the space in a higher dimensional one. The description is purely two-dimensional and the tangent spaces are fundamentally different. If you do want to consider the embedding, the tangent spaces at two different points are still different as a vector that is tangent to the sphere at one point will generally not be tangent to it at another point. Remember, the description is two-dimensional - you are utterly forbidden to use vectors that are not tangent to the sphere as those would represent translations out of the manifold.

 

[Jufa]

What I mean is I have made a choice, and my choice is to use parallel-transported basis vectors. I could have chosen a different basis, but I didn't.

Okey. Then the rule I was asking for that we follow to chose the reference frame at each point of the curve is that we transport the initial basis vectors. The only thing I don't get is how we do transport a basis vector. A basis vector is a physical quantity (independent of the reference system) and therefore it can not vary due to a parallel transport (maybe you refer to transport the coordinates of the reference basis). Also, in order to transport some vector to a a certain position "q" we need to know in advance what the reference frame we will use at "q". So to me it sounds weird to say that in order to know which reference basis we choose in "q" we need to perform a parallel transport, which already requires of knowing the reference frame in "q".

 

[PeterDonis]

In the sphere example, which I have seen everywhere, I don't know which is the rule to choose the local reference frame at a certain point of the curve.

There is no such rule. You can choose a local reference frame any way you like. Trying to think about parallel transport in terms of "which local reference frame" will only confuse you--as, indeed, it has throughout this thread.

To understand parallel transport, you need to forget about reference frames entirely and think about what is happening, geometrically, to vectors as they are transported along curves. Only once you understand that, without making use of reference frames at all, can you then go back and apply that understanding to transporting particular vectors you are interested in, such as the set of vectors that happen to be basis vectors of some chosen reference frame at some particular point.

Let us consider the physical quantity $A^{\mu}(x^{\nu}_0) e^\alpha_{\mu}(x^\nu_0)$ which it is obtained by evaluating the vectorial field $A^{\mu}(x^{\nu})$ at $x^\nu_0$, considering a local reference frame defined by the basis vectors $e^\alpha_{\mu}(x^\nu_0)$. This basis vectors, in contrast with the coordinates $A^{\mu}(x^{\nu}_0)$, refer to an actual physical quantity and, therefore, do not depend on the reference frame chosen.

The basis vectors, as vectors, are geometric objects independent of coordinates, yes. But that is not the same as saying that if you take those vectors, which are basis vectors at point p, and transport them along some curve to point q, they will still be basis vectors at point q. Whether or not that is the case is a choice that you can make either way--because it is equivalent to choosing local coordinates at q, which is a choice independent of your choice of local coordinates at p. The geometry of the situation, including the geometric properties of parallel transport, tells you nothing about such coordinate choices, which is why I keep telling you to forget about coordinates and reference frames and concentrate first on understanding the geometric properties involved.

What you should be looking at here is the inner product between the two vectors $A$ and $e^\alpha$, which is what the expression you wrote down represents. If we assume that $e^\alpha$ is the tangent vector to the curve you want to transport along, then the inner product between $A$ and $e^\alpha$ tells you the angle of $A$ relative to the direction along the curve. And parallel transport will keep that angle the same as both vectors get transported along the curve (assuming the curve is a geodesic). That is a key geometric property that you need to understand.

Now, consider the sphere example again. Suppose we start out at point p on the equator, with the vector $A$ pointing towards the North Pole, which is along the first geodesic segment we want to parallel transport along. Then $A \cdot e^\alpha = 1$ at p (assuming both vectors are unit vectors), and that property will be preserved by parallel transport, so when we reach the North Pole, we will have $\tilde{A} \cdot \tilde{e}^\alpha = 1$, i.e., the parallel transported vectors will still point in the same direction.

Now we switch curves, which may be where some of your confusion is happening. Our second curve segment is at right angles to the first, so it has a different tangent vector, $e^\beta$, which makes a different angle with our chosen vector $A$ than $e^\alpha$ did. In fact, since the curve is at right angles to our original one, we will have $\tilde{A} \cdot e^\beta = 0$. Notice that this statement, just like the statement that $A \cdot e^\alpha = 1$, has nothing to do with any choice of basis at the North Pole. The key point is not that any basis has changed, it's that we've changed curves, so the tangent vector has changed, so the angle between the tangent vector and $A$ has changed.

Now we transport along our second curve back down to the equator; call the point we reach on the equator point q. Parallel transport will preserve $\tilde{A} \cdot e^\beta = 0$, so that will still be true at q. At q, we switch curves once more, our third curve being the equator, so we have a third new tangent vector, $e^\gamma$, at right angles to $e^\beta$, and therefore we have $\tilde{A} \cdot e^\gamma = -1$. We then transport along the equator back to point p. And we find that the transported vector $A$ is now at right angles to the original vector $e^\alpha$: we have $\tilde{A} \cdot e^\alpha = 0$. Or, since the original vector $A$ pointed in the same direction as $e^\alpha$, we can say that parallel transporting around the closed curve has changed the vector $A$ by a right angle; the inner product of the transported vector with the original vector vanishes: $tilde{A} \cdot A = 0$. And note, once again, that I reached this conclusion without talking at all about any coordinates or reference frames or basis vectors. I only talked about invariant geometric properties: tangent vectors to curves and inner products of vectors.

 

[Orodruin]

Also, in order to transport some vector to a a certain position "q" we need to know in advance what the reference frame we will use at "q".

No you don’t. You can express the parallel transport equation in an arbitrary coordinate system using whatever basis vectors at each point that you like - or even express it in a coordinate independent way.

It seems to me that you need to take a step back in order to cement the basic concepts of what a tangent vector to a manifold is and why tangent spaces at different points are generally fundamentally different before you can move on to understanding parallel transport properly.

 

[Jufa]

There is no such rule. You can choose a local reference frame any way you like. Trying to think about parallel transport in terms of "which local reference frame" will only confuse you--as, indeed, it has throughout this thread.

To understand parallel transport, you need to forget about reference frames entirely and think about what is happening, geometrically, to vectors as they are transported along curves. Only once you understand that, without making use of reference frames at all, can you then go back and apply that understanding to transporting particular vectors you are interested in, such as the set of vectors that happen to be basis vectors of some chosen reference frame at some particular point.
The basis vectors, as vectors, are geometric objects independent of coordinates, yes. But that is not the same as saying that if you take those vectors, which are basis vectors at point p, and transport them along some curve to point q, they will still be basis vectors at point q. Whether or not that is the case is a choice that you can make either way--because it is equivalent to choosing local coordinates at q, which is a choice independent of your choice of local coordinates at p. The geometry of the situation, including the geometric properties of parallel transport, tells you nothing about such coordinate choices, which is why I keep telling you to forget about coordinates and reference frames and concentrate first on understanding the geometric properties involved.

What you should be looking at here is the inner product between the two vectors $A$ and $e^\alpha$, which is what the expression you wrote down represents. If we assume that $e^\alpha$ is the tangent vector to the curve you want to transport along, then the inner product between $A$ and $e^\alpha$ tells you the angle of $A$ relative to the direction along the curve. And parallel transport will keep that angle the same as both vectors get transported along the curve (assuming the curve is a geodesic). That is a key geometric property that you need to understand.

Now, consider the sphere example again. Suppose we start out at point p on the equator, with the vector $A$ pointing towards the North Pole, which is along the first geodesic segment we want to parallel transport along. Then $A \cdot e^\alpha = 1$ at p (assuming both vectors are unit vectors), and that property will be preserved by parallel transport, so when we reach the North Pole, we will have $\tilde{A} \cdot \tilde{e}^\alpha = 1$, i.e., the parallel transported vectors will still point in the same direction.

Now we switch curves, which may be where some of your confusion is happening. Our second curve segment is at right angles to the first, so it has a different tangent vector, $e^\beta$, which makes a different angle with our chosen vector $A$ than $e^\alpha$ did. In fact, since the curve is at right angles to our original one, we will have $\tilde{A} \cdot e^\beta = 0$. Notice that this statement, just like the statement that $A \cdot e^\alpha = 1$, has nothing to do with any choice of basis at the North Pole. The key point is not that any basis has changed, it's that we've changed curves, so the tangent vector has changed, so the angle between the tangent vector and $A$ has changed.

Now we transport along our second curve back down to the equator; call the point we reach on the equator point q. Parallel transport will preserve $\tilde{A} \cdot e^\beta = 0$, so that will still be true at q. At q, we switch curves once more, our third curve being the equator, so we have a third new tangent vector, $e^\gamma$, at right angles to $e^\beta$, and therefore we have $\tilde{A} \cdot e^\gamma = -1$. We then transport along the equator back to point p. And we find that the transported vector $A$ is now at right angles to the original vector $e^\alpha$: we have $\tilde{A} \cdot e^\alpha = 0$. Or, since the original vector $A$ pointed in the same direction as $e^\alpha$, we can say that parallel transporting around the closed curve has changed the vector $A$ by a right angle; the inner product of the transported vector with the original vector vanishes: $tilde{A} \cdot A = 0$. And note, once again, that I reached this conclusion without talking at all about any coordinates or reference frames or basis vectors. I only talked about invariant geometric properties: tangent vectors to curves and inner products of vectors.

So you are telling me that parallel transport of a vector $A^\mu$ along a curve $x^\mu(\lambda)$ fulfills the following

$$ \frac{d}{d\lambda}\Big( \frac{d x^\nu(\lambda)}{d\lambda}\tilde{A}(x(\lambda))_\nu\Big) = \frac{d}{d\lambda} \Big(\tilde{A}(\lambda) \cdot t(\lambda) \Big) = 0 $$
i.e. the inner product of the vector tangent to the curve $t^\nu =\frac{d x^\nu(\lambda)}{d\lambda}$ and the transported vector $\tilde{A}^\nu$ remains constant along the curve.

Nevertheless, the condition I found that a parallel transported vector $\tilde{A}^\mu$ must fulfill is:

$$ \frac{D}{d\lambda}\tilde{A}(x(\lambda))^\mu = \big(\nabla_\nu \tilde{A}(x(\lambda))^\mu\big)\frac{dx^\nu}{d\lambda} = 0 $$

Where $\nabla_\nu \tilde{A}^\mu$ is the $\mu$-th component of the covariant derivative of $\tilde{A}^\mu$ w.r.t. $x^\nu$.
I don't think both conditions are equivalent. Am I using a bad definition of parallel transport? This would explain a lot of things

 

[PeterDonis]

the inner product of the vector tangent to the curve $t^\nu =\frac{d x^\nu(\lambda)}{d\lambda}$ and the transported vector $\tilde{A}^\nu$ remains constant along the curve.

Yes.

Am I using a bad definition of parallel transport?

Apparently yes, as you were told all the way back in post #2 of this thread.

 

[PeterDonis]

the condition I found

Where did you find this condition?

 

[Nugatory]

I don't know which is the rule to choose the local reference frame at a certain point of the curve.

You don't need any rule to choose a reference frame, because you don't need a reference frame, nor coordinates.

Go back to @A.T. example of the tank with a turret-mounted gun and treads. Changes of direction are found in a completely coordinate-independent way just by comparing the speed of the tracks on each side - equal means following a geodesic across the surface, not equal means deviating from the geodesic towards one side or the other. We don't need any coordinates or reference frames for this, we just need a rev counter on the drive shafts on each side. Now we turn the turret back and forth according to the difference between the two track speeds; again we don't need any coordinates or reference frames, we just need to turn the turret to the left when the left-hand track is running faster and to the right when the right-hand track is running faster. The exact ratio of turret motion to track deviation will depend on the curvature of the surface we're moving over - but once we know that quantity everything else can be calculated from the directly observable and local properties of the machinery on the tank.

Parallel transport is about calculating where the gun will point after following various paths across the surface.

 

[Jufa]

Where did you find this condition?

In the notes I was given on General relativity at my University. Here I attach both the condition and the definition of covariant derivative I was also given.

 

[PeterDonis]

In the notes I was given on General relativity at my University. Here I attach both the condition and the definition of covariant derivative I was also given.

Equation 4.17 in what you gave looks like a correct equation for parallel transport in a global coordinate chart--i.e., a chart that covers the entire curve you are going to parallel transport along. It is not an equation that is useful in a local coordinate chart centered on just one point. I suspect you have been trying to think of it in the latter fashion; that doesn't work. In any case, as I and others have pointed out repeatedly now, it's best to not even think about coordinates at all until you first understand parallel transport in a coordinate-independent way.

(In a local coordinate chart, at least if it is constructed in the standard way, i.e., Riemann normal coordinates, the connection coefficients all vanish so equation 4.17 just reduces to the partial derivatives of the components all being zero. But that only works within the small patch surrounding the chosen point, and in that small patch, all effects of curvature are being ignored, so there's no way of even dealing with the kinds of scenarios we've been discussing in this thread, such as on a sphere.)

 

[PeterDonis]

I don't think both conditions are equivalent.

They are, it just takes a little work to show. In fact we can show something more general than what you said: we can show that parallel transport preserves the inner product of any two vectors. I.e., we can show that

$$
\frac{d}{d \lambda} \left( A^\mu B_\mu \right) = 0
$$

for any $A^\mu$, $B_\mu$. To prove it, all you need is the corresponding formula to Equation 4.17 for parallel transport when the index on the vector is "down" instead of "up":

$$
\frac{d B_\mu}{d \lambda} - \Gamma^\alpha_{\mu \nu} B_\alpha \frac{d x^\mu}{d \lambda} = 0
$$

Then you just use the chain rule on the LHS of the first formula above, and use Equation 4.17 and the second formula above to substitute for $d A^\mu / d \lambda$ and $d B_\mu / d \lambda$. You should find that all of the terms cancel.

 

[stevendaryl]

My suggestion is that you work out some simple examples in detail. For example, take a two-dimensional flat plane. Use polar coordinates $r$ and $\theta$, which are related to rectangular coordinates $x$ and $y$ through $x = r cos(\theta)$ and $y = r sin(\theta)$.

Start at the point $r=1, \theta = 0$ with the vector $V$ that has components $V^r = 1$, $V^\theta = 0$. Parallel transport it along the curve $r = 1$ (so only $\theta$ changes) to get to the point $r=1, \theta = \pi/2$. What are the components of $V$ now?

 

[Orodruin]

They are, it just takes a little work to show.

Careful here. Yes, with a metric compatible connection, inner products of parallel transported vectors remain constant. This does not mean that the vectors are parallel transported if their inner product remains constant. In other words, they are not equivalent as one implies the other (with the constraint of the curve being a geodesic) but not vice versa.

the inner product of the vector tangent to the curve tν=dxν(λ)dλ and the transported vector A~ν remains constant along the curve.

It is not generally true that the inner product of the tangent vector and parallel transported vector remains constant. The relevant statement is that the inner product between two parallel transported vectors remains constant (or, more generally, index contraction commutes with parallel transport). If the curve is a geodesic, then its tangent is parallel along itself, which therefore means the inner product between the tangent and another parallel transported vector remains constant.

 

[Jufa]

Careful here. Yes, with a metric compatible connection, inner products of parallel transported vectors remain constant. This does not mean that the vectors are parallel transported if their inner product remains constant. In other words, they are not equivalent as one implies the other (with the constraint of the curve being a geodesic) but not vice versa.It is not generally true that the inner product of the tangent vector and parallel transported vector remains constant. The relevant statement is that the inner product between two parallel transported vectors remains constant (or, more generally, index contraction commutes with parallel transport). If the curve is a geodesic, then its tangent is parallel along itself, which therefore means the inner product between the tangent and another parallel transported vector remains constant.

What is the condition for parallel transport then? The conditions "index contraction between two parallel transported vectors remains constant" and equation 4.17 are definitely not equivalent. 4.17 implies the first but not vice versa.
(I have already discarded the condition involving the vector tangent to the curve.)

 

[Orodruin]

Equation 4.17 is the parallel transport equation as expressed in any given coordinate system. The problem seems to be that you are not applying the parallel transport equation, but instead using general statements about the parallel transport equation that do not really hold.

In normal coordinates for a given point $p$, it holds that the Christoffel symbols are equal to zero at $p$ and therefore the parallel transport equation states that the directional derivative of the vector components in the direction of the curve are zero. However, this is only true at $p$ and not necessarily in other points of the chart. It is also not saying anything about what is going to happen when you change coordinates. The basis it refers to is implicitly the holonomic basis of the given coordinate system.

Before you can properly understand parallel transport (other than an overall intuitive picture such as that provided by others on the sphere), you will need to understand the necessity and purpose of introducing a connection on a manifold. The first step to understanding this is to realize that the tangent spaces $T_p M$ and $T_q M$ at the different points $p$ and $q$ are not the same and there is no a priori way to relate vectors in these different spaces to each other. Because of this, you cannot introduce directional derivatives of vectors by using the typical limit construction of $\partial_\nu V = \lim_{\epsilon \to 0}[(V(x+\epsilon \delta_\nu) - V(x))/\epsilon]$ simply because the difference of vectors at different points is not well defined due to them belonging to different tangent spaces.

This is where the connection comes into define a directional derivative operator $\nabla$ that have several of the properties we expect from a directional derivative. We denote the directional derivative of a vector field $V$ in the direction $X$ by $\nabla_X V$ and demand that it satisfies the following

1. For a scalar field $\phi$, $\nabla_X \phi = X\phi = X^\nu \partial_\nu \phi$.
2. Linearity in $X$: $\nabla_{\phi_1 X_1 + \phi_2 X_2} V = \phi_1 \nabla_{X_1} V + \phi_2 \nabla_{X_2} V$, where $\phi_i$ are scalars and $X_i$ are vectors.
3. Leibniz rule is satisfied: $\nabla_X(VU) = V \nabla_X U + (\nabla_X V) U$ with any contractions between $V$ and $U$ remaining the same.

Given a basis of vector fields $E_{(\mu)}$, a connection can be uniquely defined by identifying its connection coefficients with respect to that basis $\Gamma_{\mu\nu}^\lambda$ defined by $\nabla_{E_{(\mu)}} E_{(\nu)} = \Gamma_{\mu\nu}^\lambda E_{(\lambda)}$. It should be noted that the connection generally by no means is unique. There are generally several different choices that would satisfy these conditions. However, for many applications, including GR, we impose additional requirements on the connection that make it unique. Those are that the connection should be metric compatible (directional derivative of the metric tensor = 0) and torsionless ($\nabla_X Y - \nabla_Y X - [X,Y] = 0$) and this uniquely identifies the so-called Levi-Civita connection. The metric compatibility is what will guarantee that the inner product between parallel transported vectors will remain constant.

The main thing to understand however is that it is the connection that defines what it means for a vector field to "change" in a particular direction - not the expression of the components of a vector field in some particular basis. It may very well be that the components are constant in some basis but the vector field is still changing between points. One of the best examples of this is looking at the radial basis vector $\vec e_r$ from polar coordinates in the Euclidean plane. Its components relative to the basis $\vec e_r$, $\vec e_\theta$ are constant, but the vector $\vec e_r$ itself changes.

Generally, a field $V$ is said to be parallel if $\nabla_X V = 0$ for all $X$. However, the existence of parallel fields is not guaranteed (unless the connection is flat) as the system of differential equations is overdetermined. We can therefore introduce the concept of a field being parallel along a curve $\gamma$ if $\nabla_{\dot \gamma} V = 0$, where $\dot \gamma$ is the tangent vector of $\gamma$. This is a system of differential equations with as many unknowns as the number of equations and can therefore be solved given some initial condition $V(p)$ at the starting point $p$ of the curve. The resulting vectors along the curve are called the parallel transport of $V(p)$ along the curve. Note that, if you have two curves starting and ending at the same points $p$ and $q$, there is no guarantee that the end results at $q$ will be the same - in fact they typically will not be in a curved space.

Note that I really have not made any reference whatsoever to any coordinate systems throughout this discussion. The connection in itself is the geometrical object and as such is independent of whatever basis or coordinates that you choose to introduce, i.e., the connection comes first and is what defines parallel transport. If you do introduce a coordinate system with the corresponding holonomic basis, then the parallel transport equation takes the form given in 4.17. Hence, in order to fully understand parallel transport, you need to get it out of your mind that it somehow relates to some arbitrarily defined basis. You also need to understand the reasons for introducing the connection in the first place, which in turn requires understanding that tangent spaces at different points are, indeed, different.

 

[Jufa]

For a scalar field ϕ, ∇Xϕ=Xϕ=Xν∂νϕ.

I don't understand this line. Is there something missing?

 

[Dale]

The key point is that parallel transport is a map from one vector space to another (or itself). It doesn't actually take a vector out of one tangent space and carry it to another space (what would that even mean?), so it can only be a mathematical method for mapping between two vectors (possibly in different spaces).

Nice @Ibix, very well said.

 

[etotheipi]

I don't understand this line. Is there something missing?

Don't think so! A vector field $X$ is just a map from one $C^{\infty}$ function on the manifold to another $C^{\infty}$ function on the manifold, and for a scalar field $\varphi$ the quantity $\nabla_X \varphi$ is nothing but the directional derivative $X \varphi = X^{\mu} \partial_{\mu} \varphi$ along the vector field $X$. You can also view $X$ as assigning a tangent vector $X_p = X^{\mu} \partial_{\mu} |_{p}$ to any point $p \in M$ by $(X \varphi)(p) = X_p \varphi$.

 

[Dale]

In the sphere example, which I have seen everywhere, I don't know which is the rule to choose the local reference frame at a certain point of the curve.

Reference frames are not required. I would not even consider them at all.

The connection defines the concept of keeping a vector fixed in direction as you move. If you move in a straight line (geodesic) then the vector will keep its orientation with respect to you. If you turn then the vector will turn the opposite direction with respect to you.

Nevertheless, when arriving to the north pole this basis vectors are no longer well defined, since it is a singular point.

That is irrelevant. Every point on a sphere is equivalent. If you can imagine parallel transporting at any point on a sphere then you can imagine parallel transporting at the pole. Again, reference frames are not needed.

Nevertheless, when arriving to the north pole this basis vectors are no longer well defined, since it is a singular point.

I just realized that if this is a problem then we can simply start at the north pole. Start at the north pole with a vector pointing along the prime meridian. Walk south along the prime meridian to the equator. The vector will point straight ahead on this whole path. Turn 90 degrees to the right and the vector will be pointing to your left. Walk around the equator a quarter of the way around the globe and the vector will continue pointing to your left. and then turn 90 degrees right again. The vector will now be pointing behind you. Walk back to the pole and when you arrive your vector is still behind you. Turn 90 degrees to the right once more and you are now facing the same direction you were originally facing. But now your vector is to your right, which is 90 degrees away from where it started pointing straight ahead.

 

[stevendaryl]

I don't understand this line. Is there something missing?

This is a convention that takes a bit of getting used to (or at least, it did for me). For every vector $\vec{V}$ there is a corresponding operator, "the directional derivative in the direction of $\vec{V}$":
$\vec{V} \cdot \nabla$
which is an operator that takes a scalar field $\Phi$ and returns another scalar field. In a particular coordinate system, we can represent the effect of this operator in the following way:

$\vec{V} \cdot \nabla \Phi = \sum_j V^j \dfrac{\partial \Phi}{\partial x^j}$

Since every vector $\vec{V}$ corresponds to exactly one such operator, a modern approach is to just identify $\vec{V}$ with its directional derivative. So $V(\Phi)$ just means the same thing as $\vec{V} \cdot \nabla \Phi$

 

[stevendaryl]

Since every vector $\vec{V}$ corresponds to exactly one such operator, a modern approach is to just identify $\vec{V}$ with its directional derivative. So $V(\Phi)$ just means the same thing as $\vec{V} \cdot \nabla \Phi$

The conceptual advantage to this understanding of vectors is that you can make sense of vectors without introducing a particular basis or coordinate system.

 

[Orodruin]

I don't understand this line. Is there something missing?

Nothing is missing. It is saying that the directional derivative implied by the connection must coincide with the natural directional derivative of a scalar field when acting on a scalar field. (Note that this derivative is describable using the standard definition of a directional derivative since the difference in the definition is just a difference of numbers.)