I Parallel transport general relativity

[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.)

 

[PeterDonis]

For everyone's information, a side discussion about covariant derivative notation has been moved to a new thread:

https://www.physicsforums.com/threads/covariant-derivative-notation.1001698/

 

[Jufa]

I think I have encountered where my misconception comes from. I think it comes from the definition of the Christoffel symbols $\Gamma^k_{ij}$. These are definied in such a way that:

$$ \nabla_j \vec{e_i} = \Gamma^k_{ji}\vec{e_k} $$

What I need to know is what $\nabla_j \vec{e_i}$ actually is. Its definition involvs the Christoffel symbols but the Christoffel symbols' definition involves the definition of covariant derivative.

 

[Jufa]

I think the definition of the Christoffel symbols can be made in terms of the metric following this reasoning:

$$ \partial_j g_{mn} = \partial_j \big( \vec{e_m} \cdot \vec{e_n} \big) =\nabla_j \big( \vec{e_m} \cdot \vec{e_n} \big) =
\nabla_j \big( \vec{e_m} \big)\cdot \vec{e_n} + \vec{e_m} \cdot \nabla_j \big( \vec{e_n} \big) = \Gamma^k_{jm} \vec{e_k} \cdot \vec{e_n} + \vec{e_m} \cdot \Gamma^k_{jn}\vec{e_k} = $$

$$= \Gamma^k_{jm}g_{kn} + \Gamma^k_{jn} g_{km} $$

The only thing that I am struggling with is the third equality (the chain rule for the covariant derivative involving an inner product). Could someone please confirm me that this equality holds and prove it?

 

[Orodruin]

I think I have encountered where my misconception comes from. I think it comes from the definition of the Christoffel symbols $\Gamma^k_{ij}$. These are definied in such a way that:

$$ \nabla_j \vec{e_i} = \Gamma^k_{ji}\vec{e_k} $$

What I need to know is what $\nabla_j \vec{e_i}$ actually is. Its definition involvs the Christoffel symbols but the Christoffel symbols' definition involves the definition of covariant derivative.

I explained this at length in post 43. The Christoffel symbols are the connection coefficients of the Levi-Civita connection.

 

[Orodruin]

I think the definition of the Christoffel symbols can be made in terms of the metric following this reasoning:

$$ \partial_j g_{mn} = \partial_j \big( \vec{e_m} \cdot \vec{e_n} \big) =\nabla_j \big( \vec{e_m} \cdot \vec{e_n} \big) =
\nabla_j \big( \vec{e_m} \big)\cdot \vec{e_n} + \vec{e_m} \cdot \nabla_j \big( \vec{e_n} \big) = \Gamma^k_{jm} \vec{e_k} \cdot \vec{e_n} + \vec{e_m} \cdot \Gamma^k_{jn}\vec{e_k} = $$

$$= \Gamma^k_{jm}g_{kn} + \Gamma^k_{jn} g_{km} $$

The only thing that I am struggling with is the third equality (the chain rule for the covariant derivative involving an inner product). Could someone please confirm me that this equality holds and prove it?

It should be noted that there are three tensors involved in the equality you wish to prove, not just the two basis vectors, but also the metric tensor. This is somewhat obscured by the notation using the $\cdot$ for the inner product rather than writing out the metric tensor. In general
$$
\partial_j g_{mn} = \nabla_j g(e_m,e_n) = (\nabla_j g)(e_m, e_n) + g(\nabla_j e_m, e_n) + g(e_m, \nabla_j e_n),
$$
where we have applied the Leibniz rule for a general connection (note that the connection is defined to be satisfying the Leibniz rule (also, the Leibniz rule - or product rule for derivatives - is not the same thing as the chain rule)). This is then where the metric compatibility of the Levi-Civita connection comes into play and tells us that $\nabla_j g = 0$, which leaves us with the terms where $\nabla_j$ acts on the basis vectors.

Apart from that, to get the expression for the Christoffel symbols, you will also need to use that the Levi-Civita connection is torsion free - which amounts to $\Gamma^i_{jk} = \Gamma^i_{kj}$.

 

[romsofia]

What I need to know is what $\nabla_j \vec{e_i}$ actually is.

No, this is the wrong way to think about it. You're already lost in the abstraction, stop trying to make "more" sense of it in its' abstract form. Sit down, do the computation in #40 THEN ask yourself the results of it.

Because, seriously, look at those two objects. You have what amounts to a derivative acting on a basis vector. Roughly put (and you KNOW this from your previous studies at this point): Derivative = change, basis vectors = describes geometry. So, put the two together and you get that it tells you how much your basis (i.e your space) changes from point to point. But you're SO lost in the abstraction that you overlook the obvious things under your nose!

 

[Jufa]

As Romsofia suggested I tried to picture a simple example to see what's going on and I encounter something that I don't understand. Put this example: The surface of a sphere considereing the basis vectors as $\vec{e_\theta} = \big(cos(\theta)cos(\phi), cos(\theta)sin(\phi), -sin(\theta)\big)$ and $\vec{e_\phi} = \big(-sin(\theta)sin(\phi), sin(\theta)cos(\phi), cos(\theta)\big)$.
The metric of this space is defined such that $g_{11} = 1$ and $g_{22}=sin^2(\theta)$ being zero the rest of the metric components and with the association $\theta \rightarrow 1$ and $\phi\rightarrow 2$.

Let as now parallel transport the vector $\vec{e_\phi}$ from one point in the equator to another point on the equator, following the geodesic $x^1=\theta_0$ and $x^2= \phi$, being $\theta_0$ constant. The parallel transported vector, which we call $\vec{v}$, must fulfill at every point the following:

$$\Big(\nabla_j \vec{v}\Big)^k \frac{dx^j}{d\phi} = 0 $$

We know that the solution is $\vec{v} = \vec{e_\phi}$ but if we insert this solution we get the following:

$$\Big(\nabla_j \vec{e_\phi}\Big)^k \frac{dx^j}{d\phi} = \Gamma^k_{j2} \frac{dx^j}{d\phi}= \Gamma^k_{22} $$

Where we have used that $\frac{dx^j}{d\phi}=\delta^{j}_2$.
If we take k=1 we get:

$$\Big(\nabla_j \vec{v}\Big)^1 \frac{dx^j}{d\phi} = \Gamma^1_{22} = -\frac{1}{tg(\theta_0)}$$
which in general is different from zero.

Where am I wrong?

 

[PeterDonis]

The surface of a sphere

Is a 2-dimensional curved manifold, so your basis vectors should each have two components. But yours have three. That means you are doing what you have already been told is the wrong thing to do, namely, treating the 2-sphere as embedded in a 3-dimensional Euclidean space, and using the components of vectors in that 3-dimensional space.

Where am I wrong?

See above.

 

[PeterDonis]

Where am I wrong?

To expand on my previous answer: you are trying to determine whether the vector $e_\phi$, which points along the equator of the sphere, gets parallel transported along the equator. You (correctly) expect the answer to be that it is, because the equator is a geodesic of the 2-sphere and $e_\phi$ is its tangent vector, and a geodesic parallel transports its vector along itself. Using the correct math, the math that describes parallel transport on the 2-sphere, treated as a 2-dimensional manifold in its own right, would give the answer zero for the parallel transport equation.

However, the math you are using to check the above is the math that describes parallel transport in 3-dimensional Euclidean space; and the equator of the 2-sphere is not a geodesic of 3-dimensional Euclidean space, so of course you are going to get a nonzero answer. To put it another way, the vector $e_\phi$ that is tangent to the equator to the 2-sphere does change direction in the 3-dimensional Euclidean space as you go around the sphere. It does not change direction on the 2-sphere itself, considered as a manifold in its own right; but you cannot use the math of 3-dimensional Euclidean space to check this.

 

[Jufa]

Is a 2-dimensional curved manifold, so your basis vectors should each have two components. But yours have three. That means you are doing what you have already been told is the wrong thing to do, namely, treating the 2-sphere as embedded in a 3-dimensional Euclidean space, and using the components of vectors in that 3-dimensional space.
See above.

It is just a way of giving a physical sense to these basis vectors. The three components are never used during during the derivacions. Actually we have $e_\theta=(1, 0)$ and $e_\phi =(0, 1)$ and the problem you point out is solved but the inconsestency remains.