# A Christoffel symbol definition

1. Jun 4, 2016

### mertcan

hi, I have seen that christoffel symbol definition or logic is shown in different ways. For instance, in first attachment ( RED box) you can see a normal vector (n) next to the christoffel symbol, but in second image everything is same except that there is a normal vector. Is there a confusion??? Which is the right representation of christoffel symbol??? Why does these 2 different sources put a normal vector or not put a normal vector????

By the way first image is quoted from "wikipedia", other one is quoted from "Pavel Grinfeld Introduction to Tensor Analysis and the Calculus of Moving Surfaces".

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2. Jun 4, 2016

### Brage

They use the same christoffel symbol, in image1 they have employed two different definitions for the covariant derivative and used substitution, it is not a definition for the Christoffel symbol. That being said, if you look at how they contract the raised index with the christoffel symbol in image1 with a tangent vector you see the same term as in picture2.

Picture2 is a definition, you can consider it as saying that how one basis vector changes will be related to contributions from the current basis vectors. I.e you already span your n-dimensional space with your current basis vectors, so modelling how one of them changes wrt displacements in space is inevitably describable by a vector in you n-dimensional space which can always be described by a weighted sum of your current basis vectors as they span this space. I hope this makes sense ;)

3. Jun 4, 2016

### mertcan

I think it makes sense, but I looked at the images again and I can say that the difference of 2 images is just "normal vector", although notations are different, meaning of 2 images is same except the "normal vector". If you don't mind Could you be more explicit???? (Also I can not see any contraction in image 1 as "Brage" said)

4. Jun 5, 2016

### mertcan

I really tried to compare or find some differences, but confusion still exists for me a little bit. I also want to share a different image called "image 3". So, if you combine image 3 and image 2 we can exactly obtain the same thing in image 1 "red box", for instance "R" in "image 3" corresponds to \psi in "image 1". Of course, normal vector disrupts everything. Could you explain this weird situation????

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5. Jun 5, 2016

????????

6. Jun 5, 2016

### stevendaryl

Staff Emeritus
mertcan, I'm not sure I understand what your question is. Are you asking why there is a normal vector involved in one of the definitions, and not in the other?

The difference is that in the one case, you are talking about curvilinear coordinates in a Euclidean space, and in the second case, you are talking about embedding a curved space into Euclidean space.

Why is a normal vector involved in the second case? Well, think about it in terms of parallel-transport. Think of the Earth, and you are at the equator. You have a tangent vector that is pointing North. Let's embed the Earth in 3-D space, so that
• the z-axis runs from the South pole to the North pole,
• the x-axis runs from the point (latitude = 0, longitude = 0) to the point (latitude = 0, longitude = 180),
• the y-axis runs from the point (latitude = 0, longitude = 90 West) to the point (latitude = 0, longitude = 90 East).
Then at the equator, a vector pointing north is in the z-direction.

Now suppose from the equator, you go north a few thousand miles. If you kept that vector point in the same direction, then it would no longer be tangent to the surface of the Earth. This is shown in the attached drawing. You start at the equator (the horizontal line in the picture) with a tangent vector pointing straight North. You move along the surface of the Earth to the North. Now your parallel-transported vector is still pointing in the z-direction, but it is no longer a tangent vector, because it is no longer tangent to the surface of the Earth. But we can decompose the parallel-transported vector into a tangent vector and a normal vector. So parallel transport in the embedded space (the surface of the sphere) is the same as parallel transport in the embedding space (3D Euclidean space) followed by subtracting off the normal component of the transported vector.

7. Jun 5, 2016

### stevendaryl

Staff Emeritus
I'm assuming you know the relationship between parallel transport and Christoffel symbols?

8. Jun 6, 2016

### mertcan

After your posts, I would like to add and ask: In parallel transport, covariant derivative is 0. Also I saw that covariant derivative is the projection of the ordinary derivative to the surface ( it may means that covariant derivative measures how the vector changes on the tangent plane of surface?). Now, If we make a parallel transport, magnitude of vector changes by proportion of christoffel symbol. That is why your vector's magnitude changes due to parallel transport ( embedded space (the surface of the sphere)) in post "6". To sum up: we both say in parallel transport covariant derivative is 0 and covariant derivative is the projection of the ordinary derivative to the surface ( covariant derivative measures how the vector changes on the tangent plane of surface). I think magnitude changes also correspond to vector change. How this is possible???????????

9. Jun 6, 2016

### mertcan

Also is there a mathematical proof of your expression[ parallel transport in the embedded space (the surface of the sphere) is the same as parallel transport in the embedding space (3D Euclidean space) followed by subtracting off the normal component of the transported vector]????????????????

10. Jun 6, 2016

### stevendaryl

Staff Emeritus
That's a little misleading. An "ordinary" derivative acts on real-valued functions, not vectors. It's just that in flat Euclidean space, using Cartesian coordinates, we can define a covariant derivative so that the basis vectors all have covariant derivative zero. So in that case, the covariant derivative of a vector field becomes just the ordinary derivative applied to each component.

The business about projections is about a curved space that is embedded into flat Euclidean space. In that case, you can relate the covariant derivative in the curved space to the covariant derivative in the Euclidean space. Not every curved space needs to be thought of as embedded in Euclidean space, however. Our universe is curved due to gravity, but it isn't usually thought of as being embedded in any larger space.

No, christoffel symbols are not about the magnitude changing, but about the components changing.

Let's take the simplest example: flat 2-D space using polar coordinates $r$ and $\theta$. Then:

$g_{\theta \theta} = r^2$
$g_{r r} = 1$
(the rest of the components of $g$ are zero).

$\Gamma^r_{\theta \theta} = -r$
$\Gamma^\theta_{r \theta} = \frac{1}{r}$
$\Gamma^\theta_{\theta r} = \frac{1}{r}$

(all the other christoffel symbols are zero). Then let $\vec{V}$ be defined by: $\vec{V} = e_\theta$, the basis vector in the $\theta$ direction. Now, let's parallel-transport $\vec{V}$ around the path $\mathcal{P}(s)$ defined by letting $(\mathcal{P}(s))^r=1, (\mathcal{P}(s))^\theta = s$.

Then $\frac{d\mathcal{P}^\nu}{ds} \nabla_\nu V^\mu = 0$ for parallel transport. So:

$\frac{d\mathcal{P}^\nu}{ds} (\partial_\nu V^\mu + \Gamma^\mu_{\nu \lambda} V^\lambda) = 0$

Since $\mathcal{P}(s)$ takes us just in the $\theta$ direction, we have $\frac{d\mathcal{P}^\theta}{ds} = 1$ and $\frac{d\mathcal{P}^r}{ds} = 0$. So we have:

$(\partial_\theta V^\mu + \Gamma^\mu_{\theta \lambda} V^\lambda) = 0$

There are only two nonzero values of $\Gamma^\mu_{\theta \lambda}$: $\Gamma^r_{\theta \theta} = -r$ and $\Gamma^\theta_{\theta r} = \frac{1}{r}$. So we have:

$(\partial_\theta V^r - r V^\theta) = 0$
$(\partial_\theta V^\theta + \frac{1}{r} V^r) = 0$

So $V^r = -r \partial_\theta V^\theta$. Plugging this into the top equation gives:

$\partial_\theta (r \partial_\theta V^\theta) + r V^\theta = 0$

So

$(\partial_\theta)^2 V^\theta + V^\theta = 0$

The general solution is: $V^\theta = A cos(\theta) + B sin(\theta)$

Since $V^\theta = 1$ at $\theta = 0$, we have:

$V^\theta = cos(\theta)$

Then we can use our equation for $V^r$:
$V^r = -r \partial_\theta V^\theta = r sin(\theta)$.

The magnitude squared of $V$ is $g_{\mu \nu} V^\mu V^\nu = g_{rr} V^r V^r + g_{\theta \theta} V^\theta V^\theta = r^2 sin^2(\theta) + r^2 cos^2(theta) = r^2$. So since $r=1$, the magnitude of $V$ doesn't change.

The magnitude is not (usually) changed by parallel transport, if there is a metric and the covariant derivative is compatible with that metric.

11. Jun 6, 2016

### stevendaryl

Staff Emeritus
That's what your source is claiming. It seems plausible to me, but I don't know a proof.

12. Jun 6, 2016

### mertcan

thanks for your nice explanation "stevendarly"

13. Jun 6, 2016

### mertcan

also, I found proof for length conservation of vector during parallel transport "stevendarly". You can see attachment

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14. Jun 6, 2016

### mertcan

By the way, now we know length are preserved, also christoffel symbol are about the components change. But How do we know vector remains parallel to itself although the components change??????????

15. Jun 6, 2016

### stevendaryl

Staff Emeritus
In a sense, the covariant derivative DEFINES what it means for two vectors to be parallel.

Suppose you have a parametrized path $\mathcal{P}(s)$, and at each point along the path, you have a corresponding vector $V(s)$. To say that the vector is "the same" along the path is just to say that

$\frac{dV}{ds} = 0$

If we write it in terms of components, $V$ is a combination of basis vectors:

$V = V^\mu e_\mu$

where $e_\mu$ is the basis vector for coordinate $x^\mu$ (I'm assuming a coordinate basis). So the fact that $V$ doesn't change along the path is captured mathematically by:

$\frac{dV}{ds} = \frac{d}{ds} (V^\mu e_\mu) = \frac{dV^\mu}{ds} e_\mu + V^\mu \frac{d e_\mu}{ds} = 0$

or

$\frac{dV^\mu}{ds} e_\mu = - V^\mu \frac{d e_\mu}{ds}$

The basis vectors are functions of the coordinates $x^\mu$, so we can write:

$\frac{d e_\mu}{ds} = \frac{\partial e_\mu}{\partial x^\nu} \frac{dx^\nu}{ds}$

By definition, $\frac{\partial e_\mu}{\partial x^\nu} = \Gamma^\lambda_{\nu \mu} e_\lambda$, so the requirement that $V$ remains constant becomes:

$\frac{dV^\mu}{ds} e_\mu = - \Gamma^\lambda_{\nu \mu} V^\mu \frac{dx^\nu}{ds} e_\lambda$
$= -\Gamma^\mu_{\nu \lambda} V^\lambda \frac{dx^\nu}{ds} e_\mu$

So

$\frac{dV^\mu}{ds} = - \Gamma^\mu_{\nu \lambda} V^\lambda \frac{dx^\nu}{ds}$