I Parallel transport general relativity

[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.

 

[PeterDonis]

If we state $e_\theta=(1, 0)$ and $e_\phi =(0, 1)$

If $e_\phi$ is supposed to be a unit vector in the $\phi$ direction, then it is not $(0, 1)$, it is $(0, 1 / \sin \theta)$.

 

[Jufa]

If $e_\phi$ is supposed to be a unit vector in the $\phi$ direction, then it is not $(0, 1)$, it is $(0, 1 / \sin \theta)$.

I don't think it is a needed that the vectors are unitary. Indeed the only effect of choosing a non Unit vector in the basis is a change in the metric.

 

[PeterDonis]

Actually we have $e_\theta=(1, 0)$ and $e_\phi =(0, 1)$ and the problem you point out is solved but the inconsestency remains.

You might be confusing yourself by switching equations too many times. Here's how I would work this problem using Equation 4.17 in the reference you gave earlier.

Equation 4.17 reads

$$
\frac{d A^\mu}{d \lambda} + \Gamma^\mu{}_{\nu \alpha} A^\alpha \frac{d x^\nu}{d \lambda} = 0
$$

Here the vector $A^\mu$ is $e_\phi = (0, 1 / \sin \theta)$. Since our curve is the equator, we have $\theta = \theta_0 = \pi / 2$ all along the curve, and therefore $d A^\mu / d \lambda = 0$. So we expect to find $\Gamma^\mu{}_{\nu \alpha} A^\alpha \frac{d x^\nu}{d \lambda} = 0$ for all possible index combinations.

Since $A^1 = 0$ (we have $1$ as the $\theta$ component index and $2$ as the $\phi$ component index), only $\alpha = 2$ needs to be considered. So the only connection coefficients that need to be considered are $\Gamma^1{}_{22} = - \sin \theta \cos \theta$ and $\Gamma^2{}_{12} = \cot \theta$. Both of these vanish for $\theta = \theta_0 = \pi / 2$. So Equation 4.17 is indeed satisfied for transporting $e_\phi$ along the equator.

Note that for other values of $\theta$ besides $\pi / 2$, Equation 4.17 will not be satisfied for transporting $e_\phi$ along a curve of constant $\theta$. That is because such curves are not geodesics for $\theta \neq \pi / 2$.

 

[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?

Oh. So Peter I think the result here is correct. I just needed to remember that $\theta_0 =\pi/2$. Many thanks!