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I am trying to study Wald's book on General Relativity. I thought the first two chapters were ok, but I'm completely stuck in chapter 3, on page 32 where he gives an example of a "derivative operator". I would very much appreciate some assistance in understanding this please.

He defined a "derivative operator", also known as a "covariant derivative", denoted $\nabla$ as a map that takes a $(k, l)$ tensor field to a $(k, l + 1)$ tensor field, that satisfies 5 properties.

Then, on page 32, he gives an example of such an operator:

Our first important task is to show that derivative operators exist. Let $\psi$ be a coordinate system and let $\lbrace\partial/\partial x^\mu\rbrace$ and $\lbrace\mathrm{d} x^\mu\rbrace$ be the associated coordinate bases. Then in the region covered by these coordinates we may define a derivative operators, $\partial_a$, called an ordinary derivative, as follows. For any smooth tensor field $T^{a_1 \cdots a_k}{}_{b_1 \cdots b_l}$ we take its components $T^{\mu_1 \cdots \mu_k}{}_{\nu_1 \cdots \nu_l}$ in this coordinate basis and define $\partial_c T^{a_1 \cdots a_k}{}_{b_1 \cdots b_l}$ to be the tensor whose components in this coordinate basis are the partial derivatives $\partial\left(T^{\mu_1 \cdots \mu_k}{}_{\nu_1 \cdots \nu_l}\right)/\partial x^\sigma$. All five conditions follow immediately from the standard properties of partial derivatives. Indeed, by the equality of mixed partial derivatives, the fifth condition holds for all tensor fields, not just scalar fields. Thus, given a coordinate system $\psi$, we can construct an associated derivative operator $\partial_a$. Of course, a different choice of coordinate system $\psi'$ will yield a different derivative operator $\partial'_a$, that is, the components of the tensor $\partial_c T^{a_1 \cdots a_k}{}_{b_1 \cdots b_l}$ in the new (primed) coordinates will not be equal to the partial derivatives of the primed components of $T^{a_1 \cdots a_k}{}_{b_1 \cdots b_l}$ with respect to the primed coordinates. Thus, a given ordinary derivative operator is coordinate dependent, i.e., it is not naturally associated with the structure of the manifold.

I don't follow how this object he has defined is a tensor.

Suppose, for brevity that the $(k, l)$ tensor is just a $(1,1)$ tensor. He is thus defining an object $\partial_cT^a_{ \ \ \ \ b}$. This should be a $(1, 2)$ tensor.

We are to pick some coordinate system and define the tensor $\partial_cT^a_{ \ \ \ \ b}$ to be the tensor that in some particular coordinates has components $\frac{\partial}{\partial x^{\sigma}}T^{\mu}_{ \ \ \ \ \nu}$. In other words, the full tensor should be $\left( \frac{\partial}{\partial x^{\sigma}}T^{\mu}_{ \ \ \ \ \nu}\right) \frac{\partial}{\partial x^{\mu}} \otimes dx^{\nu}\otimes dx^{\sigma}$.

But I don't see how this is a tensor. He doesn't get to just define the object in some coordinate system and call it a tensor. And he can't just declare that its components, by definition, transform however they need to, so that the object is a tensor. This is because he has defined the object in terms of both the partial derivative and a legitimate tensor, both of which we already know the transformation properties of. He seems to basically acknowledge this point towards the end of the above paragraph.

So, given all this, in what sense is this object he is calling a tensor actually a tensor?

End of original question.

EDIT: Thank you for the answers so far, which I am in the process of looking through. I am editing this postt to include a reply to one of the below answers.

@Michael Seifert. I tried to write a comment to your answer, but my comment was too long..

Thank you, but I still don't understand.

I am fine with the two common definitions of tensors. One, as you said, maybe favored by Mathematicians, and used is by Wald, and also is the one I prefer, saying (for example a (1,1) tensor) is a multilinear map from the product of $V^* \times V$ into $\mathbb{R}$. This is clearly basis independent, since it does not refer to a basis. But we then may pick a basis. And then we may want to change to a different basis. In order for the tensor to be invariant, the particular basis transformation requires the tensor components to transform a particular way. This leads to the tensor component transformation law. This is also explained in Wald.

As you mention, some people would ignore the basis, and just define a tensor as an object that transforms according to this tensor transformation law. So, what the above person would call a tensor component, this person would call a tensor. I think I understand this issue and the distinction.

However, I don't see how Wald escapes here, since while he is using the abstract coordinate invariant way to define a tensor. In, in order for this to be a well-defined map, the components must satisfy the tensor component transformation law - and this example he gave doesn't.

No?

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    $\begingroup$ Fix a coordinate patch. Is $\partial_\mu \otimes dx^\nu \otimes dx^\sigma$ (for given values of the indices) a tensor? Obviously yes. Here, Wald is simply considering a linear combination of tensors. That is a tensor as well because tensors of definite kind define a vector space. $\endgroup$ Jun 11 at 13:19
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    $\begingroup$ What it is false is that the tensor definet that way has the same form (derivatives of components) in every coordinate system. $\endgroup$ Jun 11 at 13:25
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    $\begingroup$ Please consider this guidance about editing questions. Also, be cautious not to edit the question in a way which invalidates existing answers. Complex follow-up discussions might work better as separate questions; you can link the questions to each other to provide context. $\endgroup$
    – rob
    Jun 11 at 16:25
  • $\begingroup$ Please do not post images of texts you want to quote, but type it out instead so it is readable for all users and so that it can be indexed by search engines. For formulae, use MathJax instead. I edited the image into typed text. $\endgroup$ Jun 11 at 16:32

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He doesn't get to just define the object in some coordinate system and call it a tensor. And he can't just declare that its components, by definition, transform however they need to, so that the object is a tensor.

Actually, he can. And that's precisely what he does. The trick is: it is only a partial derivative in one specific choice of coordinates. In other coordinate systems, it will not be a partial derivative. This is very uncommon in GR textbooks and I don't recall seeing this convention anywhere else, but it is perfectly valid. It is similar to how we often prescribe tensors as their components in one coordinate system and just transform them appropriately when changing to other systems.

The thing is that instead of letting the partial derivative be a partial derivative in all coordinate systems, he ties it to a specific choice of coordinate system. It is the partial derivative in, e.g., Cartesian coordinates, but it won't be the partial derivative in spherical coordinates.

He also does a similar thing with Christoffel symbols, which, in his text, are tensors.

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An explicit example may help. Consider the manifold $\mathbb R^2$ and a vector field $V$ which, in a Cartesian coordinate system $(x,y)$, takes the form $$V = -y\frac{\partial}{\partial x} + x\frac{\partial}{\partial y}$$ In other words, the components of $V$ are $V^x= -y$ and $V^y = x$. Now I define the following object:

$$T = -\left(\mathrm dy \otimes \frac{\partial}{\partial x}\right)+\left(\mathrm dx\otimes\frac{\partial}{\partial y}\right)$$ which has components $T_x^{\ \ x}= 0,T_y^{\ \ x}=-1,T_x^{\ \ y}=1,T_y^{\ \ y}=0$. In more condensed notation, $T_i^{ \ \ j}= \partial_i V^j$.

The question now becomes, is $T$ a tensor field? The answer given by Wald is yes, of course, if we define it to be. It's entirely valid to choose a coordinate system, define any arbitrary set of components for a tensor in that coordinate system, then "lift" the result back up to the manifold level (after all, this is precisely what I've done to define $V$ in the first place). In this case, the set of components we've chosen are simply the partial derivatives of the components of $V$ with respect to the coordinates $(x,y)$.

However, if I change to a new chart, the components of $T$ will no longer be equal to the components of $V$ with respect the coordinates. For example, if I go to polar coordinates $(r,\theta)$, we would have $$\matrix{\frac{\partial}{\partial x} \rightarrow\cos(\theta)\frac{\partial}{\partial r} -\frac{\sin(\theta)}{r}\frac{\partial}{\partial \theta} & \frac{\partial}{\partial y} \rightarrow \sin(\theta)\frac{\partial}{\partial r} + \frac{\cos(\theta)}{r}\frac{\partial}{\partial \theta} \\ \mathrm dx \rightarrow \cos(\theta) \mathrm dr - r\sin(\theta) \mathrm d\theta & \mathrm dy \rightarrow \sin(\theta) \mathrm dr + r\cos(\theta) \mathrm d\theta}$$

and so our vector field becomes $V = \frac{\partial}{\partial \theta}$ while our tensor field becomes $$T = \frac{1}{r}\left(\mathrm dr\otimes \frac{\partial}{\partial \theta}\right) - r\left(\mathrm d\theta\otimes \frac{\partial}{\partial r}\right)$$

Obviously in polar coordinates, the components of $T$ are not the partial derivatives of the components of $V$ (which are constant). This is the point Wald is making - given any tensor field $V$, you may go to a coordinate chart, take the partial derivatives of its components in that chart, and then define the result to be the components of a new tensor $T$. However, under a change of chart this relationship between $T$ and $V$ will change; the components of $T$ are only the derivatives of the components of $V$ in the chart you used to define $T$ in the first place.

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Wald's definition of a tensor is not based on their coordinate transformation laws. Instead, it is akin to how mathematicians define a tensor. Earlier in Chapter 1 (I think), he defines a tensor to be a multi-linear map from some product of $V$ and $V^*$ to the real numbers. In this sense, $\partial_c T^{a} {}_b$ is a tensor: if we contract it with a dual vector $u_a$ and two vectors $v^b$ and $w^c$, then we get a real number, so this is a valid map from $V^* \otimes V \otimes V \to \mathbb{R}$. It's also not too hard to see that this map is linear. So under this definition, it's a tensor.

By defining a tensor this way, Wald avoids the seemingly vacuous notion that "a tensor is something that transforms like a tensor". Under this definition, it is perfectly possible to define a tensor in a way that depends on some underlying structure, such as a coordinate system.

It does lead to some statements that take some getting used to if you're used to the idea that a tensor's definition is based on its transformation properties, though. For example, under this approach, the Christoffel "symbols" are actually tensors, since they define a linear map from $V \otimes V \otimes V^* \to \mathbb{R}$. Wald discusses this briefly later in Chapter 3, saying that the Christoffels $\Gamma^c {}_{ab}$ are a tensor defining the "difference" between a particular coordinate derivative operator $\partial_a$ and the covariant derivative operator $\nabla_a$. If we change coordinates from the original coordinate system $\{x^a\}$ to a "primed" system $\{{x'}^a\}$, then the Christoffel tensor field ${\Gamma'}^c {}_{ab}$ associated with this coordinate system is different from ${\Gamma}^c {}_{ab}$. This explains why applying a coordinate transformation to the coordinate components of ${\Gamma}^c {}_{ab}$ in the $\{x^a\}$ does not yield the coordinate components of ${\Gamma'}^c {}_{ab}$ in the $\{{x'}^a\}$ system: ${\Gamma}^c {}_{ab}$ and ${\Gamma'}^c {}_{ab}$ are different tensor fields.

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  • $\begingroup$ I agree with your answer, but I would not say that the definition of tensor in terms of families of components with a certain transformation law is vacuous. It is actually equivalent to the definition in terms of multi linear maps you used in your answer. $\endgroup$ Jun 11 at 13:28
  • $\begingroup$ @ValterMoretti True. (I've added a word to soften the statement.) That said, the notion that "a tensor is a family of components with a certain transformation law" is a slippery notion to define when we consider tensors that themselves depend on coordinate systems (as we see here with partial derivatives of tensors or Christoffel tensors). At a later date I will try to come back to expand on that idea further. $\endgroup$ Jun 11 at 13:35
  • $\begingroup$ @Michael Seifert - I wrote an "answer" instead of a comment in reply to your answer. $\endgroup$
    – Gleeson
    Jun 11 at 14:16

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