Confirming Covariant Derivative in Hartle's Gravity

[PeroK]

In Hartle's Gravity we have the covariant derivative (first in an LIF) which is:

$\nabla_{\beta} v^{\alpha} = \frac{\partial v^{\alpha}}{\partial x^{\beta}}$

As the components of the tensor $\bf{ t = \nabla v}$. But, it's not clear which components they are!

My guess is that $t^{\alpha}_{\ \beta} = \nabla_{\beta} v^{\alpha} = \frac{\partial v^{\alpha}}{\partial x^{\beta}}$

Rather than: $t^{\ \ \alpha}_{\beta} = \nabla_{\beta} v^{\alpha} = \frac{\partial v^{\alpha}}{\partial x^{\beta}}$

I'm obliged to anyone who could confirm this one way or the other.

 

[Ibix]

I think it doesn't matter - $t^\alpha{}_\beta=\nabla_\beta x^\alpha$, so the index labelled $\alpha$ is associated with the vector and the one labelled $\beta$ is associated with the derivative, whichever way round you put it on your $t$. So $t^\alpha {}_\beta$ is just as good as $t_\beta {}^\alpha$.

If you just give me $t^\alpha{}_\beta$ without an explanation of where it comes from then it's just a tensor and its components have properties that I know. If you tell me the definition in terms of the derivative then the association is obvious because of how you labelled the indices (also applies to $t_{\alpha\beta}=\nabla_\beta x_\alpha$).

 

[PeroK]

I think it doesn't matter - $t^\alpha{}_\beta=\nabla_\beta x^\alpha$, so the index labelled $\alpha$ is associated with the vector and the one labelled $\beta$ is associated with the derivative, whichever way round you put it on your $t$. So $t^\alpha {}_\beta$ is just as good as $t_\beta {}^\alpha$.

If you just give me $t^\alpha{}_\beta$ without an explanation of where it comes from then it's just a tensor and its components have properties that I know. If you tell me the definition in terms of the derivative then the association is obvious because of how you labelled the indices (also applies to $t_{\alpha\beta}=\nabla_\beta x_\alpha$).

I don't quite follow why it doesn't matter. If you have the same components indexed differently then don't you have a different tensor with different properties?

 

[Ibix]

I don't quite follow why it doesn't matter. If you have the same components indexed differently then don't you have a different tensor with different properties?

First off, ask yourself if you're thinking in terms of matrices, where the order of indices does matter. Tensors aren't matrices, despite an awful lot of suerficial (and not so superficial) similarities. My go-to example is the inner product of two vectors. In tensor notation that's just $g_{ab}V^aU^b$ and I can write those three elements in any order. If I want to represent those vectors and tensors by matrices I need to order them correctly and transpose as necessary: $\vec V^T \mathbf g \vec U$.

With tensors, if I have an upper index then I can contract it with a lower index, or transform it into another coordinate system. For either of those, I follow the same procedure whether the upper index came first or second. Ditto the lower index. So why should it matter which comes first and which comes second?

I do think that there can be some loss of information in this notation, I must say. For example the Riemann tensor $R^a{}_{bcd}$ tells me the vector $V^a$ I get if I parallel transport some vector $V^b$ around an infinitesimal parallelogram defined by following first $dx^c$ then $dx^d$. Written out explicitly like that you know exactly what each index means. But if I just tell you that the $R^0 {}_{123}$ component is such-and-such without previously laying out all of the explicit this-index-means-that stuff, I don't think you have any way to work out which index is which.

That isn't as ambiguous as it seems. Remember that the physically interesting stuff is always a scalar, so I'll never give you tensor components without using them for something which will provide that context.

 

[PeroK]

With tensors, if I have an upper index then I can contract it with a lower index, or transform it into another coordinate system. For either of those, I follow the same procedure whether the upper index came first or second. Ditto the lower index. So why should it matter which comes first and which comes second?
.

I have assumed that a tensor in general acts on an ordered pair of vectors. In which case, the order of the indices does matter. The metric tensor is a special case, so the order does not matter on that case.

In general, I thought that

$t(\bf{a,b}) \ne t(\bf{b,a})$

Otherwise, why not simply write $t^{\alpha}_{\beta}$ for both cases?

 

[Orodruin]

I have assumed that a tensor in general acts on an ordered pair of vectors. In which case, the order of the indices does matter. The metric tensor is a special case, so the order does not matter on that case.

In general, I thought that

$t(\bf{a,b}) \ne t(\bf{b,a})$

This does not even make sense if you have a type (1,1) tensor as one of the arguments is reserved for a tangent vector and the other for a dual vector.

Otherwise, why not simply write $t^{\alpha}_{\beta}$ for both cases?

If you are dealing with a type (1,1) tensor and there is no direct connection to a (0,2) or (2,0) tensor, e.g., through the metric, you can just as well write $t^\alpha_\beta$, there would be no possible confusion. However, you still need to be conscious of where you put the arguments if you want to write it as $t(a,b)$.

 

[PeroK]

This does not even make sense if you have a type (1,1) tensor as one of the arguments is reserved for a tangent vector and the other for a dual vector.

I wonder if Hartle is using a nonstandard approach. He doesn't seem to distinguish between vectors and dual vectors or different types of tensor.

A rank 2 tensor has four possible sets of components, depending on whether the two indices are upper or lower. It's the same with vectors.

Interesting!

 

[Orodruin]

As long as you have a non-degenerate 2-form (such as the metric or a symplectic form) you have a natural map from the tangent space to the dual space. When that is the case, you also have a natural map between different types of rank 2 tensors. You could then make the identification as "the same" tensor to some extent, but it does obscure what is actually going on and there are some tensors (vectors even) that are more naturally seen as having a particular type. For example, curve tangents (such as the 4-velocity) are natural tangent vectors whereas differentials are natural dual vectors (which is why you have a natural map to scalars being the derivative of the function with respect to the curve parameter).

 

[PeroK]

Thanks to both of you. Everything is working out with my assumptions so far! Let me see how I get on.

 

[Orodruin]

As for your original question, you could define it either way you want. Of course, $t$ will generally not be a symmetric tensor (in fact, if $v$ is a Killing field, it will be anti-symmetric). It really is up to you where you want to put the indices, but your definition should make it clear which argument goes with which index. As to Hartle has chosen to do it, I will leave unsaid because it really should be clear from his definition of $\nabla {\bf v}$. Personally, I would let the first index be the one belonging to the derivative.

 

[Ibix]

I have assumed that a tensor in general acts on an ordered pair of vectors. In which case, the order of the indices does matter. The metric tensor is a special case, so the order does not matter on that case.

In general, I thought that

$t(\bf{a,b}) \ne t(\bf{b,a})$

As Orodruin says, if $t$ is a (1,1) tensor then this doesn't work. You can write $t^\alpha{}_\beta a_\alpha b^\beta$, but switching a and b makes no sense: $t^\alpha{}_\beta b^\alpha a_\beta$. But that isn't what switching the order of the indices does as long as you don't switch them on the vector and one-form. So there's nothing wrong with writing your original tensor as $t_\alpha{}^\beta$ or $T^\beta{}_\alpha$ (using t and T to be clear that I haven't just raised and lowered indices - I've swapped the meanings of the first and second indices), because in either case when you contract it with a vector and a one-form you get the same thing: $t_\alpha{}^\beta U^\alpha V_\beta=T^\beta{}_\alpha{} U^\alpha V_\beta$.

Otherwise, why not simply write $t^{\alpha}_{\beta}$ for both cases?

Because you can raise and lower indices. You can go from a (2,0) tensor to a (1,1) tensor and to a (0,2) tensor, and you need to maintain the order of the indices to keep track of which one is which. Your original example was the covariant derivative of a vector. When you define $t$ you are free to choose whether you associate the first index with the derivative or the vector, but you need to be consistent, and that's what the index ordering does.

Edit: Note that when you can't raise and lower indices (e.g. the Christoffel symbols) or when it makes no difference (e.g. $\delta^\alpha_\beta$) you can be sloppy about index ordering