# Still learning about tensors

1. Aug 25, 2011

### grzz

Can someone explain why the covariant derivative of g$_{\alpha\beta}$ with respect to x$^{\lambda}$ is always zero?
I am asking for a physical reason why it must be so.

2. Aug 25, 2011

### WannabeNewton

If the torsion free condition is applied to the affine connection on the tangent bundle then $\triangledown _{\mu }g_{\alpha \beta } = 0$. Remember that $g_{p}:T_{p}(M)\times T_{p}(M) \mapsto \mathbb{R}$ at each point p on the manifold M and the affine connection allows you to compare vectors from tangent space to tangent space via parallel transport. So $\triangledown _{\mu }g_{\alpha \beta } = 0$ is a statement of how the lengths of vectors (but not necessarily angles) are preserved under parallel transport.

3. Aug 25, 2011

### Phrak

In setting-up the machinery of general relativity it is defined to be zero. It doesn't have to be zero to define the geometry. It makes the definition of the connection (the Chistoffel connection in the case of general relativity) easier to manipulate.

It would not be incorrect to have the covariant derivative ranging, but the action of the resulting convariant derivative would be different--and harder to deal with.

You can define any number of connections on a manifold you wish. None is more correct than the next.

4. Aug 25, 2011

### JDoolin

I don't know why it must be so, but maybe with a bit of help on the vocabulary, I might be able to discover it for myself, and then we could explain why.

First of all, what is this equation called, and how is it expressed in tensor notation?

$$\begin{pmatrix} dx\\ dy\\ dz \end{pmatrix}=\begin{pmatrix} \frac{\partial x}{\partial r} &\frac{\partial y}{\partial \theta} & \frac{\partial z}{\partial \varphi} \\ \frac{\partial x}{\partial r} &\frac{\partial y}{\partial \theta} & \frac{\partial z}{\partial \varphi}\\ \frac{\partial x}{\partial r} &\frac{\partial y}{\partial \theta} & \frac{\partial z}{\partial \varphi} \end{pmatrix} \begin{pmatrix} dr\\ d\theta\\ d\phi \end{pmatrix}$$

The columns of this thing are:

$A_r=\begin{pmatrix} \frac{\partial x}{\partial r}\\ \frac{\partial y}{\partial r}\\ \frac{\partial z}{\partial r} \end{pmatrix}$, $A_\theta=\begin{pmatrix} \frac{\partial x}{\partial \theta}\\ \frac{\partial x}{\partial \theta}\\ \frac{\partial x}{\partial \theta} \end{pmatrix}$, and $A_\varphi=\begin{pmatrix} \frac{\partial x}{\partial \varphi}\\ \frac{\partial x}{\partial \varphi}\\ \frac{\partial x}{\partial \varphi} \end{pmatrix}$

These are covariant vectors of some name or other. (For vocabulary help, what are the column vectors called? I know the adjective is "covariant" but what's the noun?) I think maybe they're called "co" variant because they tell you how much (x,y,z) vary if you only vary r, or only vary θ, or only vary φ. For instance, Aθ tells you how much (x,y,z) co-vary, if you let θ vary.

You can generate the metric tensor by taking dot products of these column-vectors.

$$g=\begin{pmatrix} \left (\vec A_{r} \cdot \vec A_{r} \right ) & \left (\vec A_{r} \cdot \vec A_{\theta} \right ) & \left (\vec A_{r} \cdot \vec A_{\varphi} \right ) \\ \left (\vec A_{\theta} \cdot \vec A_{r} \right ) & \left (\vec A_{\theta} \cdot \vec A_{\theta} \right ) & \left (\vec A_{\theta} \cdot \vec A_{\varphi} \right ) \\ \left (\vec A_{\varphi} \cdot \vec A_{r} \right ) & \left (\vec A_{\varphi} \cdot \vec A_{\theta} \right ) & \left (\vec A_{\varphi} \cdot \vec A_{\varphi} \right )\end{pmatrix}$$

If I've got this definition right, then we could at least explore the covariant derivative of a few metric tensors, and see whether every component came out to be zero. Maybe then a physical reason would become obvious.

Last edited: Aug 25, 2011
5. Aug 25, 2011

### WannabeNewton

You don't need to explore it for different metric tensors to see it. Its pretty straightforward to show in general: \begin{align}\triangledown _{\mu }g_{\alpha \beta } &= \partial _{\mu }g_{\alpha \beta } - \Gamma ^{\sigma }_{\mu \alpha }g_{\sigma \beta } - \Gamma ^{\sigma }_{\mu \beta }g_{\alpha \sigma } = \partial _{\mu }g_{\alpha \beta } - \Gamma _{\beta \mu \alpha } - \Gamma _{\alpha \mu \beta }\\ &= \partial _{\mu }g_{\alpha \beta } - \frac{1}{2} (\partial _{\beta }g_{\mu \alpha } + \partial _{\mu }g_{\beta \alpha } - \partial _{\alpha }g_{\beta \mu }) -\frac{1}{2} (\partial _{\alpha }g_{\mu \beta } + \partial _{\mu }g_{\alpha \beta } - \partial _{\beta }g_{\alpha \mu })\\ & = \partial _{\mu }g_{\alpha \beta } - \frac{1}{2}\partial _{\mu }g_{ \beta\alpha } - \frac{1}{2}\partial _{\mu }g_{\alpha \beta } = 0\end{align}
The thing is that we are using the levi civita connection here that is why the covariant derivative of the metric vanishes. If we use an affine connection on the tangent bundle without imposing the torsion - free condition then, since $g_{p}:T_{p}(M)\times T_{p}(M) \mapsto \mathbb{R}$, the same ordered pair of vectors that is an element of $T_{p}(M)\times T_{p}(M)$ will have a different associated real value at different p on M.

Last edited: Aug 25, 2011
6. Aug 25, 2011

### Fredrik

Staff Emeritus
Please keep the lines short when you use LaTeX. It's annoying to have to scroll to the right to read the text. I recommend that you use the align environment like this: \begin{align}\triangledown _{\mu }g_{\alpha \beta } &= \partial _{\mu }g_{\alpha \beta } - \Gamma ^{\sigma }_{\mu \alpha }g_{\sigma \beta } - \Gamma ^{\sigma }_{\mu \beta }g_{\alpha \sigma } = \partial _{\mu }g_{\alpha \beta } - \Gamma _{\beta \mu \alpha } - \Gamma _{\alpha \mu \beta }\\ &= \partial _{\mu }g_{\alpha \beta } - \frac{1}{2} (\partial _{\beta }g_{\mu \alpha } + \partial _{\mu }g_{\beta \alpha } - \partial _{\alpha }g_{\beta \mu }) -\frac{1}{2} (\partial _{\alpha }g_{\mu \beta } + \partial _{\mu }g_{\alpha \beta } - \partial _{\beta }g_{\alpha \mu })\\ & = \partial _{\mu }g_{\alpha \beta } - \frac{1}{2}\partial _{\mu }g_{ \beta\alpha } - \frac{1}{2}\partial _{\mu }g_{\alpha \beta } = 0\end{align}
You can edit your post for 11 hours and 40 minutes.

7. Aug 25, 2011

### WannabeNewton

Sorry didn't know; it showed up completely without a scroll bar on my screen so I naively assumed it would show up as such on other screens as well. My apologies.

8. Aug 25, 2011

### JDoolin

Thank you, WannabeNewton.

I have five questions regarding your proof:

\begin{align}\triangledown _{\mu }g_{\alpha \beta } &\overset {why?} = \partial _{\mu }g_{\alpha \beta } - \Gamma ^{\sigma }_{\mu \alpha }g_{\sigma \beta } - \Gamma ^{\sigma }_{\mu \beta }g_{\alpha \sigma } \overset {why?} = \partial _{\mu }g_{\alpha \beta } - \Gamma _{\beta \mu \alpha } - \Gamma _{\alpha \mu \beta }\\ &\overset {why?} = \partial _{\mu }g_{\alpha \beta } - \frac{1}{2} (\partial _{\beta }g_{\mu \alpha } + \partial _{\mu }g_{\beta \alpha } - \partial _{\alpha }g_{\beta \mu }) -\frac{1}{2} (\partial _{\alpha }g_{\mu \beta } + \partial _{\mu }g_{\alpha \beta } - \partial _{\beta }g_{\alpha \mu })\\ & \overset {why?} = \partial _{\mu }g_{\alpha \beta } - \frac{1}{2}\partial _{\mu }g_{ \beta\alpha } - \frac{1}{2}\partial _{\mu }g_{\alpha \beta } \overset {why?} = 0\end{align}

I don't doubt what you're saying. I just find there is a massive chasm between my educational background and standard tensor notation, and I have no exposure to some of the things that seem obvious to you .

Also, I dout if I will be convinced that I have learned anything without seeing at least one example.

Last edited: Aug 25, 2011
9. Aug 25, 2011

### grzz

Much thanks for all contributions!
I do not know much ... but I learned enough to apply the covariant derivative to the metric as WannabeNewton did.
But what I would like is a reason for what I asked from the PHYSICAL point of view.
Is it correct to say that since spacetime is smooth then locally spacetime is flat and so the Christoffel symbols are zero which means that the covariant derivative of the metric is zero?
Thanks for your help.

10. Aug 25, 2011

### Fredrik

Staff Emeritus
No problem. Thanks for fixing it.

If I had maximized the window, I wouldn't have needed to scoll right either, but I don't think that's a pleasant way to view web pages. I prefer to have a window with a width that makes the word "naively" in the quote above the last word on the first line.

1. Definition of covariant derivative of a (0,2) tensor field.
2. Definition of $\Gamma_{\alpha\beta\gamma}$ (Christoffel symbols with the first index "lowered").
3. Follows from the formula for the relationship between the Christoffel symbols and the components of the metric. (I assume that's what he used, but I didn't check that he did it right. By the way, an alternative is to use that formula instead of step 2, and then simplify the result).
4. Because for all real numbers x, we have x-x=0.
5. Because for all real numbers x, we have x-x/2-x/2=0.

Last edited: Aug 25, 2011
11. Aug 25, 2011

### WannabeNewton

The first "why" is the coordinate representation of the co-variant derivative with a torsion free connection. In general, you can write it as $$\bigtriangledown _{\mu }T^{\alpha_{1} ...\alpha_{n} }_{\beta_{1} ...\beta _{m}} = \partial _{\mu }T^{\alpha _{1} ...\alpha _{n}}_{\beta_{1} ...\beta_{m} } + \Gamma ^{\alpha_{1} }_{\mu \sigma }T^{\sigma ...\alpha _{n}}_{\beta_{1} ...\beta _{m}} + ...\Gamma ^{\alpha _{n}}_{\mu \sigma }T^{\alpha _{1}...\sigma }_{\beta _{1}...\beta _{m}} -\Gamma ^{\sigma }_{\mu \beta_{1} }T^{\alpha_{1} ...\alpha _{n}}_{\sigma ...\beta _{m}} -...\Gamma ^{\sigma }_{\mu \beta _{m}}T^{\alpha _{1}...\alpha _{n}}_{\beta _{1}..\sigma }$$ The second "why" is that you are contracting the christoffel symbol with the metric i.e. the $\sigma$ index is being summed over and since it involves the metric you can contract it with the christoffel symbol - $g_{\sigma \beta }\Gamma ^{\sigma }_{\mu \alpha } = \Gamma _{\beta \mu \alpha }$. The same goes for the other term. The third "why" can be answered by looking at the defintion of the christoffel symbols of the first kind in terms of the metric: $$\Gamma _{\alpha \beta \gamma } = \frac{1}{2}(\partial _{\beta }g_{\alpha \gamma } + \partial _{\alpha }g_{\beta \gamma } - \partial _{\gamma }g_{\alpha \beta })$$ and the last "why" is simply that the metric is completely symmetric so $g_{\alpha \beta } = g_{\beta \alpha }$.

EDIT: I did this before seeing Fredrik's post so sorry for the redundancy but he explains everything.

12. Aug 25, 2011

### Fredrik

Staff Emeritus
I don't think there's a physical reason why it must be so. It's just the simplest way to define the connection and the covariant derivative, and it happens to give us a theory of physics that makes very accurate predictions.

13. Aug 25, 2011

### JDoolin

I thank you, WannabeNewton and Fredrik for your explanations, but to understand it, I stil want to work through an example.

I have worked out the metric tensor for cartesian to spherical coordinates, and I believe it agrees with what everybody else gets:

$$g=\begin{pmatrix} 1 & 0 & 0\\ 0 & r^2 & 0\\ 0& 0 & r^2 \sin^2(\theta) \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0\\ 0 & (x^2+y^2+z^2) &0 \\ 0& 0 & (x^2+y^2) \end{pmatrix}$$

One thing I'm not sure of, is exactly how to take the covariant derivative

$$\triangledown _{\mu }g_{\alpha \beta }$$

of this thing?

(I'm not sure how to do it, because I think of $\triangledown _{\mu }$ as a vertical vector, and that can't be multiplied from the left by a 3X3 matrix.)

It will also be helpful, I'm sure, to calculate the connection coefficients, and even to visualize why they must be taken into account.

Last edited: Aug 25, 2011
14. Aug 25, 2011

### WannabeNewton

We can do the $g_{\theta \theta}$ component. Let's do the $\mu = r$. So we have $$\triangledown _{r}g_{\theta \theta} = \partial _{r}g_{\theta \theta} - \Gamma ^{\sigma }_{r \theta}g_{\sigma \theta} - \Gamma ^{\sigma }_{r \theta}g_{\theta \sigma } = \partial _{r}g_{\theta \theta} - 2\Gamma ^{\sigma }_{r \theta}g_{\sigma \theta}$$ Now you can just contract over the summed index and get $$\triangledown _{r}g_{\theta \theta} = \partial _{r}g_{\theta \theta} - 2\Gamma _{\theta r \theta}$$ Work out this component of the christoffel symbol and you will get $\Gamma _{\theta r \theta} = r$ so $$\triangledown _{r}g_{\theta \theta} = 2r - 2r = 0$$ You can repeat this for the other non trivial components if you want and you will see that they are all zero.

Last edited: Aug 25, 2011
15. Aug 25, 2011

### robphy

From an earlier thread

One view is that
"[torsion-free] metric compatibility" means that the metric tensor field carries all of the information of the geometry of spacetime....
Physically, this means that the metric tensor field determines the motion of free particles (the geodesic structure) and the propagation of light (the conformal structure..and causal structure).

This might be a useful resource:
http://relativity.livingreviews.org/open?pubNo=lrr-2004-2&page=articlesu3.html [Broken]

Last edited by a moderator: May 5, 2017
16. Aug 25, 2011

### grzz

thanks Robphy for the 'livingreviews'.

17. Aug 25, 2011

### JDoolin

Okay, now it's making more sense. Taking the vector (covariant derivative) times each of nine scalars (components of the metric tensor) yields a set of nine vectors, so it will take 27 times altogether (to evaluate that each component is zero).

Last edited: Aug 25, 2011
18. Aug 25, 2011

### WannabeNewton

When I say trivial I just mean the components that simply vanish but I did make a huge mistake with the notation. I should have done $\triangledown _{r }g_{\theta \theta }$ not $\triangledown _{r }g_{rr}$ but the result is still the same. I'll edit the post.

19. Aug 25, 2011

### DrGreg

If you're trying to use vector/matrix notation instead of tensor notation, you should think of $\nabla_{\mu }$ as a horizontal rather than vertical vector, because it is covariant rather than contravariant.

But vector/matrix notation is quite limited and is unable to represent many things that can be expressed in tensor notation. For example, matrix notation can't distinguish between $M_{ab}$, ${M_a}^b$, ${M^a}_b$, and $M^{ab}$. And, as I think you realise from post #17, in $\nabla_{\mu}g_{\alpha\beta}$ there is no repeated index to be summed over, so you can't represent it as a multiplication of a row-vector by a matrix; the answer would have to drawn as a 3×3×3 cube.

20. Aug 26, 2011

### JDoolin

Alright. It seems to me that step 1 and step 3 are the major steps here. Step 4 and 5 are relatively obvious, and step 2 is notational rather than conceptual in nature.

My recommended programme of self-study, (for both me, and the Original Poster (OP)) then, is to understand more fully (1) why the covariant derivative of the tensor is different from the partial derivative of the tensor:

$$\bigtriangledown_\mu g_{\alpha\beta} \neq \partial_\mu g_ {\alpha \beta}=\begin{pmatrix} (\partial_r g_{rr} ,\partial_\theta g_{rr},\partial_\phi g_{rr}) & (\partial_r g_{r \theta} ,\partial_\theta g_{r \theta},\partial\ _\phi g_{r \theta}) & (\partial_r g_{r \phi} ,\partial_\theta g_{r \phi},\partial_\phi g_{r \phi}) \\ (\partial_r ,\partial_\theta ,\partial_\phi) g_{\theta r} & (\partial_r ,\partial_\theta ,\partial_\phi) g_{\theta \theta} & (\partial_r ,\partial_\theta ,\partial_\phi) g_{\theta \phi}\\ (\partial_r ,\partial_\theta ,\partial_\phi) g_{\phi r} & (\partial_r ,\partial_\theta ,\partial_\phi) g_{\phi \theta} & (\partial_r ,\partial_\theta ,\partial_\phi) g_{\phi \phi} \end{pmatrix}$$

Note: The covariant derivative of the metric tensor is here shown as a 3x3 matrix of 1x3 row vectors. I think this is in the spirit of DrGreg's advice from Post 19. The top row looks different because I expanded out the first row, but left the second and third row in differential form. I'm aware that the use of an "=" sign here is not entirely appropriate, because the term on the left is a single element, while the form on the right is a three-dimensional array. If anyone knows of a better notation, please let me know!
Instead, to find the covariant derivative, you must modify the partial derivative in the following way:
$$\triangledown _{\mu }g_{\alpha \beta } = \partial _{\mu }g_{\alpha \beta } - \Gamma ^{\sigma }_{\mu \alpha }g_{\sigma \beta } - \Gamma ^{\sigma }_{\mu \beta }g_{\alpha \sigma }$$

Part 1 of understanding would come from understanding why that is true, and part 2 of understanding "physically" would be finding each of the connection coefficients for a simple example such as cartesian-to-spherical case.

$$\Gamma^\sigma_{\mu\nu} = \frac{g^{\sigma \rho} (\partial_\mu g_{\nu \rho}+ \partial_\nu g_{\rho \mu}- \partial_\rho g_{\mu \nu})}{2}$$
and see whether we can find a physical explanation for each one. (This is Equation 3.21 in the Carroll Lectures http://www.blatword.co.uk/space-time/Carrol_GR_lectures.pdf), an equation he recommends committing to memory, because it is "one of the most important equations in this subject."

For me, I think it may be easier to go with part 2, first, and try understanding part 1 next.

Okay, I'm finally done editing.

Last edited: Aug 26, 2011