Still learning about tensors

[grzz]

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

 

[WannabeNewton]

If the torsion free condition is applied to the affine connection on the tangent bundle then [itex]\triangledown _{\mu }g_{\alpha \beta } = 0[/itex]. Remember that [itex]g_{p}:T_{p}(M)\times T_{p}(M) \mapsto \mathbb{R}[/itex] 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 [itex]\triangledown _{\mu }g_{\alpha \beta } = 0[/itex] is a statement of how the lengths of vectors (but not necessarily angles) are preserved under parallel transport.

 

[Phrak]

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

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.

 

[JDoolin]

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

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?

[tex]\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}[/tex]

The columns of this thing are:

[itex] A_r=\begin{pmatrix}
\frac{\partial x}{\partial r}\\
\frac{\partial y}{\partial r}\\
\frac{\partial z}{\partial r}
\end{pmatrix} [/itex], [itex]A_\theta=\begin{pmatrix} \frac{\partial x}{\partial \theta}\\ \frac{\partial x}{\partial \theta}\\ \frac{\partial x}{\partial \theta} \end{pmatrix}[/itex], and [itex]A_\varphi=\begin{pmatrix} \frac{\partial x}{\partial \varphi}\\ \frac{\partial x}{\partial \varphi}\\ \frac{\partial x}{\partial \varphi} \end{pmatrix}[/itex]

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, A_θ 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.

[tex]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}[/tex]

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.

 

[WannabeNewton]

You don't need to explore it for different metric tensors to see it. Its pretty straightforward to show in general: [tex]\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}[/tex]
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 [itex]g_{p}:T_{p}(M)\times T_{p}(M) \mapsto \mathbb{R}[/itex], the same ordered pair of vectors that is an element of [itex]T_{p}(M)\times T_{p}(M)[/itex] will have a different associated real value at different p on M.

 

[Fredrik]

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: [tex]\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}[/tex]
You can edit your post for 11 hours and 40 minutes.

 

[WannabeNewton]

Please keep the lines short when you use LaTeX. It's annoying to have to scroll to the right to read the text.

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.

 

[JDoolin]

Thank you, WannabeNewton.

I have five questions regarding your proof:

[tex]\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}[/tex]

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.

 

[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.

 

[Fredrik]

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.

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.

Thank you, WannabeNewton.

I have five questions regarding your proof:

1. Definition of covariant derivative of a (0,2) tensor field.
2. Definition of [itex]\Gamma_{\alpha\beta\gamma}[/itex] (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.

 

[WannabeNewton]

Thank you, WannabeNewton.

I have five questions regarding your proof:

[tex]\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}[/tex]

The first "why" is the coordinate representation of the co-variant derivative with a torsion free connection. In general, you can write it as [tex]\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 }[/tex] The second "why" is that you are contracting the christoffel symbol with the metric i.e. the [itex]\sigma [/itex] index is being summed over and since it involves the metric you can contract it with the christoffel symbol - [itex]g_{\sigma \beta }\Gamma ^{\sigma }_{\mu \alpha } = \Gamma _{\beta \mu \alpha } [/itex]. 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: [tex]\Gamma _{\alpha \beta \gamma } = \frac{1}{2}(\partial _{\beta }g_{\alpha \gamma } + \partial _{\alpha }g_{\beta \gamma } - \partial _{\gamma }g_{\alpha \beta })[/tex] and the last "why" is simply that the metric is completely symmetric so [itex]g_{\alpha \beta } = g_{\beta \alpha }[/itex].

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

 

[Fredrik]

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

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.

 

[JDoolin]

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

(Post 4)
You can generate the metric tensor by taking dot products of these column-vectors.

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

[tex]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}[/tex]

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

[tex]\triangledown _{\mu }g_{\alpha \beta }[/tex]

of this thing?

(I'm not sure how to do it, because I think of [itex]\triangledown _{\mu }[/itex] 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.

 

[WannabeNewton]

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

[tex]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}[/tex]

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

[tex]\triangledown _{\mu }g_{\alpha \beta }[/tex]

of this thing?

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

 

[robphy]

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

From an earlier thread
https://www.physicsforums.com/showthread.php?t=199500 ...

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

 

[grzz]

thanks Robphy for the 'livingreviews'.
I did not know about this online journal.

 

[JDoolin]

We can do the [itex]g_{rr}[/itex] component. The only non - trivial component to take the covariant derivative with respect to is [itex]\mu = r[/itex]. So we have [tex]\triangledown _{r}g_{rr} = \partial _{r}g_{rr} - \Gamma ^{\sigma }_{rr}g_{\sigma r} - \Gamma ^{\sigma }_{rr}g_{r\sigma } = \partial _{r}g_{rr} - 2\Gamma ^{\sigma }_{rr}g_{\sigma r}[/tex] Now you can just contract over the summed index and get [tex]\triangledown _{r}g_{rr} = \partial _{r}g_{rr} - 2\Gamma _{rrr}[/tex] Work out this component of the christoffel symbol and you will get [itex]\Gamma _{rrr} = r[/itex] so [tex]\triangledown _{r}g_{rr} = 2r - 2r = 0[/tex] You can repeat this for the other non trivial components if you want and you will see that they are all zero.

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).

 

[WannabeNewton]

You may still be overestimating my education. I don't think I know how to work the trivial components either.

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 [itex]\triangledown _{r }g_{\theta \theta }[/itex] not [itex]\triangledown _{r }g_{rr}[/itex] but the result is still the same. I'll edit the post.

 

[DrGreg]

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

If you're trying to use vector/matrix notation instead of tensor notation, you should think of [itex]\nabla_{\mu }[/itex] 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 [itex]M_{ab}[/itex], [itex]{M_a}^b[/itex], [itex]{M^a}_b[/itex], and [itex]M^{ab}[/itex]. And, as I think you realize from post #17, in [itex]\nabla_{\mu}g_{\alpha\beta}[/itex] 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.

 

[JDoolin]

(Post 8)
I have five questions regarding your proof:

[tex]\begin{align}\triangledown _{\mu }g_{\alpha \beta } &\overset {why1?} = \partial _{\mu }g_{\alpha \beta } - \Gamma ^{\sigma }_{\mu \alpha }g_{\sigma \beta } - \Gamma ^{\sigma }_{\mu \beta }g_{\alpha \sigma } \overset {why2?} = \partial _{\mu }g_{\alpha \beta } - \Gamma _{\beta \mu \alpha } - \Gamma _{\alpha \mu \beta }\\ &\overset {why3?} = \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 {why4?} = \partial _{\mu }g_{\alpha \beta } - \frac{1}{2}\partial _{\mu }g_{ \beta\alpha } - \frac{1}{2}\partial _{\mu }g_{\alpha \beta } \overset {why5?} = 0\end{align}[/tex]

(Post 10)
1. Definition of covariant derivative of a (0,2) tensor field.
2. Definition of [itex]\Gamma_{\alpha\beta\gamma}[/itex] (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.

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:

[tex]\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}[/tex]

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:
[tex]\triangledown _{\mu }g_{\alpha \beta } = \partial _{\mu }g_{\alpha \beta } - \Gamma ^{\sigma }_{\mu \alpha }g_{\sigma \beta } - \Gamma ^{\sigma }_{\mu \beta }g_{\alpha \sigma }[/tex]

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.

[tex]\Gamma^\sigma_{\mu\nu} = \frac{g^{\sigma \rho} (\partial_\mu g_{\nu \rho}+ \partial_\nu g_{\rho \mu}- \partial_\rho g_{\mu \nu})}{2}[/tex]
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.

 

[WannabeNewton]

I believe Caroll gives a good motivation for the covariant derivative no? It comes from the need to correct the fact that the derivative with respect to a coordinate of, say, a vector in a general coordinate system also takes into account the rate of change of the basis vectors with respect to that coordinate: [tex]\partial _{\alpha }\mathbf{\overrightarrow{V}} = \overrightarrow{e_{\beta }}\partial _{\alpha }V^{\beta } + V^{\beta }\partial _{\alpha }\overrightarrow{e_{\beta }} [/tex] You can see that if you were dealing with an orthonormal basis, like that of Cartesian coordinates, this would reduce to the usual form of the directional derivative. Basically, you let [itex]\partial _{\alpha }\overrightarrow{e_{\beta }} = \Gamma ^{\mu }_{\beta \alpha }\overrightarrow{e_{\mu }} [/itex] and, after relabeling some dummy indeces, you get that the above derivative of a vector is the covariant derivative times the basis vector term. So it basically has to do with how the basis vectors themselves change relative to the coordinates in a non - orthonormal coordinate system.

 

[grzz]

thanks Robphy for the 'livingreviews'.
I did not know about this online journal.

... above journal is MUCH TOO advanced for me.

 

[robphy]

... above journal is MUCH TOO advanced for me.

For your original question (concerning physical interpretation)...

in http://relativity.livingreviews.org/open?pubNo=lrr-2004-2&page=articlesu3.html
skim over the mathematical calculations and
focus on the logical development and physical and mathematical] interpretation
of the various levels of geometrical structure.

Although "Living Reviews in Relativity" is fairly advanced reading,
you can still skim through it.


Other possible places to look for answers:
http://www.google.com/search?q=non-metricity

 

[JDoolin]

I believe Caroll gives a good motivation for the covariant derivative no? It comes from the need to correct the fact that the derivative with respect to a coordinate of, say, a vector in a general coordinate system also takes into account the rate of change of the basis vectors with respect to that coordinate: [tex]\partial _{\alpha }\mathbf{\overrightarrow{V}} = \overrightarrow{e_{\beta }}\partial _{\alpha }V^{\beta } + V^{\beta }\partial _{\alpha }\overrightarrow{e_{\beta }} [/tex] You can see that if you were dealing with an orthonormal basis, like that of Cartesian coordinates, this would reduce to the usual form of the directional derivative. Basically, you let [itex]\partial _{\alpha }\overrightarrow{e_{\beta }} = \Gamma ^{\mu }_{\beta \alpha }\overrightarrow{e_{\mu }} [/itex] and, after relabeling some dummy indeces, you get that the above derivative of a vector is the covariant derivative times the basis vector term. So it basically has to do with how the basis vectors themselves change relative to the coordinates in a non - orthonormal coordinate system.

I think there may be a typo; the first derivative should be a total derivative. (Otherwise the term on the left-hand-side, and the first term on the right are identical.)
(1)[tex]d _{\alpha }\mathbf{\overrightarrow{V}} = \overrightarrow{e_{\beta }}\partial _{\alpha }V^{\beta } + V^{\beta }\partial _{\alpha }\overrightarrow{e_{\beta }} [/tex]

After that, I think I agree based on the use of the chain-rule. But I remain a little bit confused about whether the chain rule should be used here.

In general, if we have a function of two variables: f(x,y) then the total derivative is[tex]d(f(x,y))=\frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy[/tex],

So I may need to convince myself that a function of spherical coordinates [tex]f(r,\theta,\phi)[/tex] must actually be expressed as a function of the coordinates AND the unit vectors: [tex]f(r,\theta,\phi,\hat e_r, \hat e_\theta, \hat e_\phi)[/tex].​

For now I will take that as given.

It seems to me, there are three things that should be clearly distnguished in any notation; the path, the coordinates, and the unit vectors: Let's expand equation (1) above using the following substitutions:
The path:[tex]\vec V(\lambda) = V^r(\lambda)\hat e_r +V^\theta(\lambda)\hat e_\theta +V^\phi(\lambda)\hat e_\phi[/tex]
The Coordinates:[tex]\alpha,\beta \in \lbrace r, \theta, \phi \rbrace[/tex]
The unit vectors[tex]\vec e_\beta \in \lbrace \hat e_r, \hat e_\theta, \hat e_\phi \rbrace[/tex]

When I substitute in, exactly (except for changing the order of [itex]\hat e_\beta \partial_r V^\beta \: \mathrm{ to }\: \left (\partial_r V^\beta \right )\hat e_\beta[/itex]. Those things are not commutative in matrix multiplication), I get:[tex]\begin{align*} d_r \vec V &= \frac{d}{dr}\left ( V^r \hat e_r + V^\theta \hat e_\theta + V^\phi \hat e_\phi \right ) \\ \hat e_\beta \partial_r V^\beta +V^\beta \partial_r \hat e_\beta &= \left ( \partial_r (V^r,V^\theta,V^\phi) \right )\begin{pmatrix}\hat e_r\\\hat e_\theta\\ \hat e_\phi \end{pmatrix} + (V^r,V^\theta,V^\phi)\left (\partial_r \begin{pmatrix}\hat e_r\\\hat e_\theta\\ \hat e_\phi \end{pmatrix} \right )\end{align*}[/tex]

So, yes, I think I see, provided I can convince myself that [itex]\vec V[/itex] is not just a function of the coordinates [itex](r, \theta, \phi)[/itex], but a function of the unit vectors as well, I should be able to see clearly why this is true. But I need to have more clarity on the topic, so I will continue to work on the specifics of spherical coordinates.

 

[WannabeNewton]

Yes it was a typo sorry but one thing I should point out is that you aren't using the unit basis vectors there, just the coordinate basis (in a general curved space - time the basis doesn't have to be of unit norm or orthonormal and not all curved spaces can be given non - coordinate bases, which are constructed to be orthonormal). You can use a non - coordinate (non - holonomic) basis but the coordinate basis is more natural. Just thought you would like to know. In any case, doing it on [itex]S^{2}[/itex] is probably a good way of getting used to it.

 

[Ben Niehoff]

not all curved spaces can be given non - coordinate bases, which are constructed to be orthonormal).

I don't think this is true, where did you get it? One can always find an orthonormal basis on any given tangent space, and the choice can always be extended continuously to the tangent bundle, at least in a coordinate patch. I would agree that there is not always a global choice, because that would require the manifold be parallelizable. But given any coordinate patch U, one can always find an orthonormal frame over all of U, so for the purposes of computation, there is no issue.

The metric is a symmetric matrix so it can always be diagonalized. Call this diagonal matrix D. Then all you need to do is write

[tex]D = |D|^{1/2} \eta |D|^{1/2}[/tex]

where [itex]\eta = \mathrm{diag}(-1,1,1,1)[/itex] and [itex]|\cdot|[/itex] is the absolute value (not the determinant). This whole process is algebraic, so it doesn't matter that the metric happens to be a matrix of functions.

 

[Ben Niehoff]

If the torsion free condition is applied to the affine connection on the tangent bundle then [itex]\triangledown _{\mu }g_{\alpha \beta } = 0[/itex].

This is also false. The torsion-free condition and metric-compatibility condition are independent!

 

[WannabeNewton]

This is also false. The torsion-free condition and metric-compatibility condition are independent!

Well the two conditions are [itex]\Gamma ^{\alpha }_{\beta \gamma } = \Gamma ^{\alpha }_{(\beta \gamma) } [/itex] and [itex]\triangledown _{\mu }g_{\alpha \beta } = 0[/itex] but if the torsion - free condition was taken off then the components of the affine connection in terms of the metric would have addition terms that would not result in a vanishing covariant derivative of the metric. Or is that circular since one is defined based on the other being true?

 

[WannabeNewton]

I don't think this is true, where did you get it? One can always find an orthonormal basis on any given tangent space, and the choice can always be extended continuously to the tangent bundle, at least in a coordinate patch. I would agree that there is not always a global choice, because that would require the manifold be parallelizable.

Yeah I was talking about globally because you can't just project a tensor onto the orthonormal basis everywhere [itex]t_{\hat{\alpha }\hat{\beta }} = (e_{\hat{\alpha }})^{\alpha }(e_{\hat{\beta }})^{\beta }t_{\alpha \beta }[/itex] when the [itex]{e_{\alpha }} [/itex] can only be extended to the [itex]T(M)[/itex] for a coordinate patch that is in the neighborhood of a point. Otherwise, its like saying you can convert to a global inertial frame on a manifold which isn't generally possible.

 

[Ben Niehoff]

To the OP and JDoolin:

First I think it would help if you read what I wrote about connections in another thread:

https://www.physicsforums.com/showpost.php?p=3436397&postcount=7

There are two different ways to think about the geometry of a given manifold. One is via a metric [itex]g_{\mu\nu}[/itex], which tells us how to measures lengths and angles. The other is via a connection [itex]\Gamma_{\mu\nu}^\lambda[/itex], which tells us whether two neighboring vectors are parallel. In principle, there is no reason these concepts should have anything to do with each other. To see why, think back to Euclid's postulates. The first 4 postulates are sufficient to tell us when triangles are congruent; that is, they tell us all the local information about lengths and angles. It was famously shown that the 5th postulate (on parallels) is independent of the first 4. So this is why the metric information and the parallelism need not depend on each other.

In order to define any sort of derivative, we need to quantify the notion of "change" of vectors between nearby points, and so naturally, we must have a notion of "equal" between nearby points. The only reasonable notion of "equal" comes from [itex]\Gamma_{\mu\nu}^\lambda[/itex], because it is the only object we have that relates vectors at nearby points. So we say a vector V at point x is "equal" to a vector W at point y if

[tex](y^\mu - x^\mu) \Gamma_{\mu\nu}^\lambda V^\nu= W^\lambda[/tex]
Of course, this only makes sense if the point y is infinitesimally close to the point x. So really we should write

[tex]\Delta x^\mu \Gamma_{\mu\nu}^\lambda V^\nu= W^\lambda[/tex]
where [itex]\Delta x^\mu[/itex] is some infinitesimal that points from x to y. It is important to note the asymmetry of the expression! The infinitesimal separation between points x and y is contracted into the first lower index of [itex]\Gamma_{\mu\nu}^\lambda[/itex].

Using this notion of "equal" between nearby points, we can derive the covariant derivative,

[tex]\nabla_\mu V^\lambda = \partial_\mu V^\lambda + \Gamma_{\mu\nu}^\lambda V^\nu[/tex]
Note again the asymmetry; the vector V is contracted into the second lower index.

Now we can ask what sort of properties [itex]\nabla[/itex] might have. Let's introduce a fancy notation as follows:

[tex]\begin{align}
(\nabla_X Y)^\lambda &= X^\mu \partial_\mu Y^\lambda + X^\mu \Gamma_{\mu\nu}^\lambda Y^\nu \\
(Xf)(x^\mu) &= X^\nu \partial_\nu f(x^\mu) \\
([X,Y])^\lambda &= (XY^\lambda - YX^\lambda) = X^\nu \partial_\nu Y^\lambda - Y^\nu \partial_\nu X^\lambda
\end{align}[/tex]
where X and Y are vector fields and f is a function. Then we can define two tensors, the torsion

[tex]T(X,Y) = \nabla_X Y - \nabla_Y X - [X,Y][/tex]
and the curvature

[tex]R(X,Y)Z = \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z - \nabla_{[X,Y]} Z[/tex]
In indices, the torsion comes out to be something nice and simple,

[tex]T^\lambda{}_{\mu\nu} = \Gamma_{\mu\nu}^\lambda - \Gamma_{\nu\mu}^\lambda[/tex]
The curvature, however, comes out to a longer expression that I don't feel like typing. It is also called the Riemann tensor, you can look it up.

Finally, there is a third tensor we can define, which doesn't have an official name, but I'll call it the non-metricity

[tex]N(X,Y,Z) = Z(g(X,Y)) - g(\nabla_Z X, Y) - g(X, \nabla_Z Y)[/tex]
where g is the metric tensor. In indices, this becomes

[tex]N_{\mu\nu\lambda} = \partial_\lambda g_{\mu\nu} - \Gamma_{\lambda\mu}^\rho g_{\rho\nu} - \Gamma_{\lambda\nu}^\rho g_{\mu\rho}[/tex]

Now, for reasons explained in my other post linked above, our usual intuitions of geometry are most nearly matched by choosing the conditions

[tex]\begin{align}
T^\lambda{}_{\mu\nu} &= 0 \\
N_{\mu\nu\lambda} &= 0
\end{align}[/tex]
In fact, it turns out that these two conditions are sufficient to solve for [itex]\Gamma_{\mu\nu}^\lambda[/itex] explicitly in terms of [itex]g_{\mu\nu}[/itex], giving us the Christoffel connection.

But it is important to remember, we could have chosen any specific T and N, which would give us some other solution for [itex]\Gamma_{\mu\nu}^\lambda[/itex].

 

[JDoolin]

In any case, doing it on [itex]S^{2}[/itex] is probably a good way of getting used to it.

You may have noticed that I have posted and deleted twice now. (and now I have edited again, for emphasis) I couldn't seem to get the notation right. But my main point was that if we define a parametric path on the surface of a sphere, [tex](\theta(\lambda),\phi(\lambda))[/tex]

and you are asked to calculate the length of a differential section of that path, [tex]\left \| \left (\frac{\mathrm{d \theta} }{\mathrm{d} \lambda},\frac{\mathrm{d} \phi }{\mathrm{d} \lambda} \right )| \right \| =\sqrt {r^2 \frac{d \theta}{d\lambda}+ r^2 \sin^2(\theta)\frac{d \phi}{d\lambda}}[/tex]
you need to take into account the unit vectors; (i.e. the scale factors) how long a differential change in theta or phi actually is in the three-dimensional space.

However, if you are just asked for the differential of that path

[tex]\frac{\mathrm{d} \vec s }{\mathrm{d} \lambda}=\left (\frac{\mathrm{d \theta} }{\mathrm{d} \lambda},\frac{\mathrm{d} \phi }{\mathrm{d} \lambda} \right )[/tex]

you don't need to take into account how long the unit vectors are in the theta direction or phi direction, because the coordinates of the path already take that into account.

In particular, [tex]\frac{\mathrm{d} \vec s}{\mathrm{d} r},\frac{\mathrm{d} \vec s}{\mathrm{d} \theta}, \, \mathrm{and}\, \frac{\mathrm{d} \vec s}{\mathrm{d} \phi}[/tex]
should be identical to [tex]\frac{\partial \vec s}{\partial r},\frac{\partial \vec s}{\partial \theta}, \, \mathrm{and}\, \frac{\partial \vec s}{\partial \phi}[/tex].

If there is no flaw in my reasoning, this seems to directly conflict with:
[tex]d _{\alpha }\mathbf{\overrightarrow{V}} = \left (\partial _{\alpha }V^{\beta } \right )\overrightarrow{e_{\beta }} + V^{\beta }\left (\partial _{\alpha }\overrightarrow{e_{\beta }} \right )[/tex]

This equation also appears in various forms as equations 1a, 1b, 2a, 2b, here: http://www.mathpages.com/rr/appendix/appendix.htm
surely there must be a simply explained flaw in either my reasoning, or my understanding of what is meant.