Meaning of ds^2 according to Carroll

[guitarphysics]

Hi all, I need some help- I was reading Carroll's GR book, and on pages 71-71 he discusses the metric in curved spacetime. I have a few questions regarding this section:

(1) He says
In our discussion of path lengths in special relativity we (somewhat handwavingly) introduced the line element as $ds^2=\eta_{\mu\nu}dx^{\mu}dx^{\nu}$, which was used to get the length of the path. Of course now that we know that $\text{d} x^{\mu}$ is really a basis dual vector, it becomes natural to use the terms "metric" and "line element" interchangeably, and write $$ds^2=g_{\mu\nu}\text{d}x^{\mu}\text{d}x^{\nu}$$
My question here is: isn't what he's doing right now just as handwavy as before? It's very hard to see a concrete reason justifying what he's doing, unless he's defining $ds^2$ by the above expression, which I'd have no problem with. But just come out and say it man!

(2) Why does he talk about $ds^2$ in terms of basis dual vectors, instead of just differentials? What's the point of doing this? I realize that this is not just Carroll's idea; that's the way it is in GR. But in treating $ds^2$ as a (0,2) tensor, instead of the square of a differential length, don't we lose the physical meaning inherent in $ds^2$? How do we go back to interpreting $ds^2$ as a length so quickly when we changed its meaning so drastically?

Thanks in advance!

 

[PeterDonis]

isn't what he's doing right now just as handwavy as before?

It is in the sense that he isn't giving any more justification than he did before. But that doesn't mean there isn't more rigorous justification--it only means that he's not going to that level of rigor (which requires some fairly intensive differential geometry, more than is usually done in a physics course).

Also, the expression using coordinate dual basis vectors is correct mathematically in a way that the expression using differentials is not. See below.

Why does he talk about $ds^2$ in terms of basis dual vectors, instead of just differentials?

Because the metric $g_{\mu \nu}$ is an (0, 2) tensor, and to get a scalar, $ds^2$, from an (0, 2) tensor, you need to contract it with two vectors. See below.

in treating $ds^2$ as a (0,2) tensor

It isn't. $ds^2$ is a scalar. The (0, 2) tensor is $g_{\mu \nu}$, as noted above. So what the equation is saying is that you take the (0, 2) metric tensor and contract it with two coordinate dual basis vectors to get a scalar, $ds^2$. You can't do that with differentials; the idea of contracting an (0, 2) tensor with differentials makes no sense.

To be more precise, you would put back the integral signs in the equation, to obtain:

$$
s^2 = \int_A^B ds^2 = \int_A^B g_{\mu \nu} \text{d} x^\mu \text{d} x^\nu
$$

In other words, to get the squared length $s^2$ along a curve between two given events, A and B, you integrate the contraction of the metric tensor, an (0, 2) tensor, with the coordinate dual basis vectors, along the curve between those two events.

 

[Orodruin]

So what the equation is saying is that you take the (0, 2) metric tensor and contract it with two coordinate dual basis vectors to get a scalar, ds^2.

I find this confusing. By definition, (0,2) tensor should be contracted with two tangent vectors to give a scalar, not with two dual vectors. Of course, with $g_{\mu\nu}$ being the components of the metric, $g_{\mu\nu} dx^\mu \otimes dx^\nu$ is the metric tensor.

To me, the meaning of the line element does not become clear until you consider the squared length of the tangent vector of a curve $\gamma$, ie, $\dot\gamma = (dx^\mu/dt)\partial_\mu$ and obtain $g(\dot\gamma,\dot\gamma) = g_{\mu\nu}\dot x^\mu \dot x^\nu$, where you are dealing with actual coordinate displacements $\dot x^\mu$.

 

[guitarphysics]

Because the metric $g_{\mu \nu}$ is an (0, 2) tensor, and to get a scalar, $ds^2$, from an (0, 2) tensor, you need to contract it with two vectors. See below.

It isn't. $ds^2$ is a scalar. The (0, 2) tensor is $g_{\mu \nu}$, as noted above. So what the equation is saying is that you take the (0, 2) metric tensor and contract it with two coordinate dual basis vectors to get a scalar, $ds^2$. You can't do that with differentials; the idea of contracting an (0, 2) tensor with differentials makes no sense.

I'm sorry, but that directly contradicts a lot of things that I'm pretty sure are true. Among them, and this one being directly from Carroll's book:
In fact our notation "$ds^2$" does not refer to the differential of anything, or the square of anything; it's just conventional shorthand for the metric tensor, a multilinear map from two vectors to the real numbers. Thus, we have a set of equivalent expressions for the inner product of two vectors $V^{\mu}$ and $W^{\nu}$: $$g_{\mu\nu}V^{\mu}W^{\nu}=g(V,W)=ds^2(V,W)$$

You also seem to have misunderstood the equation in the beginning of my post, as Orodruin pointed out. I think the reason for misunderstanding here is that $g_{\mu\nu}$ isn't the metric tensor, as you interpreted it; they're the components of the metric tensor, and the $\text{d}x^{\mu} \text{d}x^{\nu}$ next to it are basis dual vectors; not vectors. Thus resulting in a (0,2) tensor, $g$, or $ds^2$.

Still, thank you for trying to help- I initially got mixed up with exactly the same thing as you.

Orodruin, thank you too for your post, that helped a lot :)

 

[PeterDonis]

By definition, (0,2) tensor should be contracted with two tangent vectors to give a scalar, not with two dual vectors.

Hm, yes, good point.

with $g_{\mu\nu}$ being the components of the metric, $g_{\mu\nu} dx^\mu \otimes dx^\nu$ is the metric tensor.

Yes, this is a more precise way of expressing the tensor. Are you interpreting the RHS of Carroll's equation $ds^2 = g_{\mu \nu} \text{d} x^\mu \text{d} x^\nu$ as just a different way of writing this tensor (i.e., leaving out the $\otimes$ symbol for brevity)? That doesn't seem right either. Or are you interpreting Carroll's equation as a hand-wavy way of writing $ds^2 = g_{\mu \nu} \dot{x}^\mu \dot{x}^\nu$?

 

[guitarphysics]

Hm, yes, good point.Are you interpreting the RHS of Carroll's equation $ds^2 = g_{\mu \nu} \text{d} x^\mu \text{d} x^\nu$ as just a different way of writing this tensor (i.e., leaving out the $\otimes$ symbol for brevity)? That doesn't seem right either.

That is indeed what Carroll meant, I think. It's unfortunate, but apparently the tensor product is often omitted.

 

[PeterDonis]

Thus resulting in a (0,2) tensor, $g$, or $ds^2$.

You're right that the RHS of Carroll's equation can be interpreted as a tensor, as Orodruin says. But $ds^2$ is a scalar, no matter how you interpret the rest of the equation. See my response to Orodruin.

 

[Orodruin]

As mentioned by the OP, I think Carrol uses it as a "short-hand" (actually, "long-hand") for the metric.

 

[guitarphysics]

As mentioned by the OP, I think Carrol uses it as a "short-hand" (actually, "long-hand") for the metric.

Wait, so this (unfortunate) notation isn't universal in GR speak?

 

[Orodruin]

I would not say $ds^2$ is given by $g_{\mu\nu}\dot x^\mu \dot x^\nu$ but rather add a $dt^2$ to that, where $dt$ is a differential increment of the curve parameter. In this way, you will have a direct relation between the differential increment in the parameter and the corresponding distance on the manifold.

 

[guitarphysics]

Hmm what do you mean by add a $dt^2$ to that?

 

[guitarphysics]

Never mind, I'm guessing you meant that $ds^2 dt^2$, when acting on two tangent vectors, would give you the classic product of the metric tensor components with differential length elements.

 

[PeterDonis]

I would not say $ds^2$ is given by $g_{\mu\nu}\dot x^\mu \dot x^\nu$ but rather add a $dt^2$ to that, where $dt$ is a differential increment of the curve parameter.

Yes, you're right, there needs to be a differential on the RHS. So we would have this (I am using $\lambda$ as the parameter instead of $t$ to avoid confusion with the $t$ coordinate):

$$
ds^2 = g_{\mu \nu} \frac{dx^\mu}{d\lambda} \frac{dx^\nu}{d\lambda} d\lambda^2
$$

 

[George Jones]

I believe that there is another possible source of confusion here with respect to omitted tensor product symbols and non-diagonal metrics. I think that $dx^\mu dx^\nu$ is notation for

$$dx^\mu dx^\nu = \frac{1}{2} \left( dx^\mu \otimes dx^\nu + dx^\nu \otimes dx^\mu \right).$$

Then, we still have

$$\begin{align}
g_{\mu \nu} dx^\mu dx^\nu &= g_{\mu \nu} \frac{1}{2} \left( dx^\mu \otimes dx^\nu + dx^\nu \otimes dx^\mu \right)\\
&= \frac{1}{2} g_{\mu \nu} dx^\mu \otimes dx^\nu + \frac{1}{2} g_{\mu \nu} dx^\nu \otimes dx^\mu\\
&= g_{\mu \nu} dx^\mu \otimes dx^\nu ,
\end{align}$$

where relabeling of indices and the symmetry of $g_{\mu \nu}$ has been used.

Now, suppose that the metric written explicitly has a term like $f dt d \phi$. Thinking that there is only a single omitted tensor product, it is tempting to write

$$f dt d \phi = f dt \otimes d \phi,$$

so that $g_{t \phi} = f$.

Actually,

$$
f dt d \phi = f \frac{1}{2} \left( dt \otimes d\phi + d\phi \otimes dt \right)\\,
$$

so that $g_{t \phi} = f/2$.

This pitfall does not appear in diagonal metrics like Schwarzschild like, since the non-zero terms have $\mu = \nu$, and then

$$ \frac{1}{2} \left( dx^\mu \otimes dx^\nu + dx^\nu \otimes dx^\mu \right) = dx^\mu \otimes dx^\nu .$$

 

[vanhees71]

[corrected because of mistake found by PeterDonis in#17]

This shows, how important it is to distinguish between tensors (INvariant objects) and its components with respect to a basis and the corresponding dual basis of the (tangent-)vector space (COvariant objects).

A 2nd"=rank tensor $g$ is invariant and maps two elements of the the vector space to a scalar, and this is an invariant number. Now you introduce bases. The most simple ones are holnomous bases. Having generalized coordinates $q^{\mu}$ for the space-time manifold you take at any point the tangent vectors to the coordinate lines, i.e., those lines, where only one of the $q^{\mu}$ is varied. The (holonomous) basis vectors defined as the tangent vectors on these coordinate lines are denoted by $\partial_{\mu}$, and dual-basis vectors of the dual space to this basis are denoted by $\mathrm{d} q^{\mu}$.

A 2nd-rank tensor (here the pseudometric tensor) can be decomposed as
$$g=g_{\mu \nu} \mathrm{d} q^{\mu} \otimes \mathrm{d} q^{\nu}.$$
The components are the mapping of the basis vectors $\partial_{\mu}$, i.e.,
$$g_{\mu \nu}=g(\partial_{\mu},\partial_{\nu}).$$
Now it is also simple to get the various transformation properties. Take another map with coordiantes $q^{\prime \mu}$. Then you have (the mnemotechnically very easy to remember) transformation properties for the holonomous bases and co-bases,
$$\mathrm{d} q^{\prime \mu}=\frac{\partial q^{\prime \mu}}{\partial q^{\nu}} \mathrm{d} q^{\nu}, \quad \partial_{\mu}'=\frac{\partial q^{\nu}}{\partial q^{\prime \mu}} \partial_{\nu},$$
i.e., the transformations are contragredient to each other. Now it's easy to see, how the components transform
$$g_{\mu \nu}'=g(\partial_{\mu}',\partial_{\nu}')=g \left (\frac{\partial q^{\rho}}{\partial q^{\prime \mu}} \partial_{\rho} ,
\frac{\partial q^{\sigma}}{\partial q^{\prime \nu}} \partial_{\sigma} \right)=
\frac{\partial q^{\rho}}{\partial q^{\prime \mu}} \frac{\partial q^{\sigma}}{\partial q^{\prime \nu}} g(\partial_{\rho},\partial_{\sigma}) = \frac{\partial q^{\rho}}{\partial q^{\prime \mu}} \frac{\partial q^{\sigma}}{\partial q^{\prime \nu}} g_{\rho \sigma}.$$
It's also easy to see that
$$g=g_{\mu \nu} \mathrm{d} q^{\mu} \otimes \mathrm{d} q^{\nu} = g_{\mu \nu}' \mathrm{d} q^{\prime \mu} \mathrm{d} q^{\prime \nu}$$
etc.

To understand $\mathrm{d} s^2$ think about a curve parametrized with a parameter $\lambda$. Then
$$\left (\frac{\mathrm{d} s}{\mathrm{d} \lambda} \right)^2=\dot{s}^2 =g (\dot{q}^{\mu} \partial_{\mu},\dot{q}^{\nu} \partial_{\nu}) = g_{\mu \nu} \dot{q}^{\mu} \dot{q}^{\nu}$$
is the pseudo-length (pseudo, because the metric is not positive definite in GR but of signature (1,3) or (3,1) depending on whether you use west- or east-coast convention). The dot means derivative with respect to $\lambda$.

 

[lavinia]

Infinitesimally a metric is an inner product on each tangent space. It takes pairs of vectors and computes a scalar. Since it acts on pairs of vectors at each tangent space it is a dual 2 tensor. Thus, with respect to any tangent frame field $s_{i}$ it can be expressed as $g_{ij}s^{*}_{i}s^{*}_{j}$ where $s^{*}_{i}$ is the dual basis. If the frame field is a coordinate frame then the dual basis is $dx^{μ}$, the differentials of the coordinate functions. (However, if the frame field is not a coordinate field then the dual basis is not a set of differentials.)

If the metric is positive definite then the integral of its square root along a curve gives the length of the curve. Because of this the square root is called the line element. In Relativity, the metric is not positive definite but $ds^2$ is still a sum of products of dual tangent vectors, $g_{uv}dx^{u}dx^{v}$. This gives a rigorous definition of $ds^2$ in the case of a non-positive definite metric. Without seeing that the $dx^{u}$ are dual tangent vectors, one is only writing an expression by analogy with a positive definite metric. This is what Carroll meant by hand waving.

In both the case of a positive definite and a semi-definite metric, one can compute its value in terms of infinitesimal increments in the coordinate functions.

 

[PeterDonis]

These basis vectors are denoted by $\mathrm{d} q^{\mu}$

The components are the mapping of the dual vectors $\partial_{\mu}$

This is backwards. The quantities $\text{d} q^\mu$ are covectors, and the quantities $\partial_\mu$ are tangent vectors. The placement of the indexes can be confusing, but consider: the quantity $g$ is a (0, 2) tensor; the expression $g = g_{\mu \nu} \text{d} q^\mu \otimes \text{d} q^\nu$ is saying that this (0, 2) tensor is a particular linear combination of tensor products of two (0, 1) tensors, i.e., covectors. If those quantities were vectors, that equation would make no sense; you can't make a (0, 2) tensor from linear combinations of (1, 0) tensors.

As for the quantities $\partial_\mu$, they are directional derivatives along the coordinate basis directions, and there is a one-to-one-correspondence between directional derivatives and tangent vectors, so they can also be considered as tangent vectors. The equation $g_{\mu \nu} = g \left( \partial_\mu, \partial_\nu \right)$ then just says that $g_{\mu \nu}$ is a scalar--a number--obtained by contracting the (0, 2) tensor $g$ with a pair of (1, 0) tensors, i.e., vectors. If those quantities were covectors, that equation would make no sense; you can't contract a (0, 2) tensor with two (0, 1) tensors.

 

[PeterDonis]

$ds^2$ is still a sum of products of dual tangent vectors, $g_{uv}dx^{u}dx^{v}$.

This doesn't look right. $ds^2$ is supposed to be a scalar, so it can't be the sum of products of dual vectors; that would be an (0, 2) tensor. The expression $g_{uv} dx^u dx^v$ can be interpreted as a linear combination of products of coordinate differentials, with the coefficients being the metric coefficients; that would give a scalar. Unless you are using $ds^2$ just as an alternate (and confusing) notation for the tensor $g$. But I don't think that was Carroll's intention; he appears to refer to $ds^2$ as a "line element", i.e., a scalar denoting the arc length along an infinitesimal segment of a path.

 

[vanhees71]

This is backwards. The quantities $\text{d} q^\mu$ are covectors, and the quantities $\partial_\mu$ are tangent vectors. The placement of the indexes can be confusing, but consider: the quantity $g$ is a (0, 2) tensor; the expression $g = g_{\mu \nu} \text{d} q^\mu \otimes \text{d} q^\nu$ is saying that this (0, 2) tensor is a particular linear combination of tensor products of two (0, 1) tensors, i.e., covectors. If those quantities were vectors, that equation would make no sense; you can't make a (0, 2) tensor from linear combinations of (1, 0) tensors.

As for the quantities $\partial_\mu$, they are directional derivatives along the coordinate basis directions, and there is a one-to-one-correspondence between directional derivatives and tangent vectors, so they can also be considered as tangent vectors. The equation $g_{\mu \nu} = g \left( \partial_\mu, \partial_\nu \right)$ then just says that $g_{\mu \nu}$ is a scalar--a number--obtained by contracting the (0, 2) tensor $g$ with a pair of (1, 0) tensors, i.e., vectors. If those quantities were covectors, that equation would make no sense; you can't contract a (0, 2) tensor with two (0, 1) tensors.

Ok,I mixed this up again. Indeed the tangent vectors on a curve given by $q^{\mu}(\lambda)$ in coordinate space are
$$t=\dot{q}^{\mu} \partial_{\mu}.$$
Thus indeed, the $\partial_{\mu}$ are the tangent vectors on the coordinate lines and $\mathrm{d} q^{\mu}$ the corresponding co-vectors (or dual vectors), i.e., linear mappings defined by $\mathrm{d} q^{\mu}(\partial_{\nu})={\delta^{\mu}}_{\nu}$.

I'll correct my previous posting.

 

[lavinia]

This doesn't look right. $ds^2$ is supposed to be a scalar, so it can't be the sum of products of dual vectors; that would be an (0, 2) tensor. The expression $g_{uv} dx^u dx^v$ can be interpreted as a linear combination of products of coordinate differentials, with the coefficients being the metric coefficients; that would give a scalar. Unless you are using $ds^2$ just as an alternate (and confusing) notation for the tensor $g$. But I don't think that was Carroll's intention; he appears to refer to $ds^2$ as a "line element", i.e., a scalar denoting the arc length along an infinitesimal segment of a path.

Yes it is correct. "products" means when evaluated on tangent vectors.

 

[PeterDonis]

"products" means when evaluated on tangent vetoers.

Which tangent vectors? There aren't any in the expression you wrote down.

 

[lavinia]

Which tangent vectors? There aren't any in the expression you wrote down.

not sure what you are getting at here. if you want to say more technically that it is not a sum of products until it is evaluated on a specific pair of tangent vectors, OK. But usual parlance doesn't require this level of formality. One just thinks of the coordinate differentials as infinitesimal increments - not yet evaluated - but whatever they are you get a weighted sum of products - which is a scalar.

This way of looking at the metric is classical and is implied in the notation $g_{ij}dx^{i}dx^{j}$. The more formal algebraic notation would be $g_{ij}dx^{i}⊗dx^{j}$ which then becomes a weighted product when evaluated.

 

[PeterDonis]

if you want to say more technically that it is not a sum of products until it is evaluated on a specific pair of tangent vectors, OK.

I would say it's not the sum of products of coordinate differentials, which is a scalar, until it's evaluated on a specific pair of tangent vectors. A sum of products of dual vectors that isn't evaluated is a (0, 2) tensor; evaluating that tensor on a specific pair of tangent vectors then gives a scalar.

 

[lavinia]

I would say it's not the sum of products of coordinate differentials, which is a scalar, until it's evaluated on a specific pair of tangent vectors. A sum of products of dual vectors that isn't evaluated is a (0, 2) tensor; evaluating that tensor on a specific pair of tangent vectors then gives a scalar.

Right. My language was technically sloppy - but not unusual.

Interestingly, Carroll could be saying that the idea of an infinitesimal increment is itself handwaving and that replacing it with the idea of a dual tangent vector makes it rigorous. In that case, the squared line element for a positive definite inner product is hand waving as well and the whole of classical physics and mathematics is formally erroneous.

 

[guitarphysics]

This doesn't look right. $ds^2$ is supposed to be a scalar, so it can't be the sum of products of dual vectors; that would be an (0, 2) tensor. The expression $g_{uv} dx^u dx^v$ can be interpreted as a linear combination of products of coordinate differentials, with the coefficients being the metric coefficients; that would give a scalar. Unless you are using $ds^2$ just as an alternate (and confusing) notation for the tensor $g$. But I don't think that was Carroll's intention; he appears to refer to $ds^2$ as a "line element", i.e., a scalar denoting the arc length along an infinitesimal segment of a path.

You seem to still be confused about this point. That was completely Carroll's intention, as he explicitly states in a quote I put above. In the context of his book (and maybe other parts in GR), the explicit meaning of $ds^2$ is not that of a scalar, as you keep saying, but rather that of a (0,2) tensor.

 

[PeterDonis]

In the context of his book (and maybe other parts in GR), the explicit meaning of $ds^2$ is not that of a scalar, as you keep saying, but rather that of a (0,2) tensor.

I don't think [or didn't--see edit below] that's what Carroll is saying. I think he is saying that, since there is an obvious correspondence between the scalar line element $ds^2$ and the (0, 2) tensor $g_{\mu \nu} \text{d} x^\mu \text{d} x^\nu$, we can for most purposes use those terms interchangeably. That doesn't mean they're the same thing; it just means that we can use the terms interchangeably for most purposes, since the correspondence between them is obvious.

Edit: I see that later on the same page Carroll does say that $ds^2$ is "conventional shorthand for the metric tensor". But I don't think that usage is very "conventional". ISTM that most GR references use $ds^2$ to refer to the scalar line element, not the metric tensor. And still later on the same page, Carroll says, referring to the line element on a 2-sphere, "This is completely consistent with the interpretation of $ds^2$ as an infinitesimal length". But an infinitesimal length is a scalar, so clearly Carroll is aware of the interpretation of $ds^2$ as a scalar.

 

[guitarphysics]

I agree that he's aware of the usual interpretation of $ds^2$ as a scalar, but I think he likes to have a more rigorous formalism behind this notion; see for example page 76- he's working with a metric given by $$ds^2=-\text{d}t^2+a^2(t)(\text{d}x^2+\text{d}y^2+\text{d}z^2)$$
Then later, on page 77, he insists on the interpretation of this $ds^2$ not as a scalar differential length element, but rather a rank 2 tensor (almost that entire page is a justification of a seemingly sloppy step relating to $ds^2$ being a tensor rather than a scalar).

(PS. I personally have not read many GR references. Carroll's is the most advanced level book I've looked into so far in GR, so I'm guessing since you have more experience here than me, this way of doing things is probably not done in many other places besides Carroll's book.)

 

[stevendaryl]

And still later on the same page, Carroll says, referring to the line element on a 2-sphere, "This is completely consistent with the interpretation of $ds^2$ as an infinitesimal length". But an infinitesimal length is a scalar, so clearly Carroll is aware of the interpretation of $ds^2$ as a scalar.

Some authors say that the notion of infinitesimals can be made rigorous in differential geometry by interpreting them as one-forms, rather than scalars. For example, people write things such as:

$d(sin(\phi)) = cos(\phi) d\phi$

which can be interpreted as a relationship between infinitesimals, but is also rigorously true as a relationship between one-forms.

 

[PeterDonis]

Some authors say that the notion of infinitesimals can be made rigorous in differential geometry by interpreting them as one-forms, rather than scalars.

Carroll explicitly says that he is not using $ds^2$ to denote a 1-form, or the square of one, or the square of anything; he says it's just a "conventional shorthand for the metric tensor".

 

[PeterDonis]

I dug out my copy of MTW to see how they handle this topic. On p. 77, Chapter 3 (which is part of the section on SR), in Box 3.2, "The Metric in Different Languages", part D. talks about the "connection to the elementary concept of line element". The last paragraph says:

Because the metric $\eta_{\mu \nu} \mathbf{d} x^\mu \otimes \mathbf{d} x^\nu$ and the line element $ds^2 = \eta_{\mu \nu} dx^\mu dx^\nu$ perform the same function of representing the squared length of an unspecified infinitesimal displacement, there is no conceptual distinction between them. One sometimes uses the symbols $\mathbf{ds}^2$ to denote the metric; one sometimes gets pressed and writes it as $\mathbf{ds}^2 = \eta_{\mu \nu} \mathbf{d} x^\mu \mathbf{d} x^\nu$, omitting the "$\otimes$"; and one sometimes even gets so pressed as to use nonbold characters, so that no notational distinction remains at all between metric and elementary line element:

$$
\mathbf{g} = \mathbf{ds}^2 = ds^2 = \eta_{\mu \nu} dx^\mu dx^\nu
$$

Note that I can't exactly reproduce MTW's typography here, what I am writing as a $\mathbf{d}$ is really a super-bold italicized "d", and similarly for the other bold characters.

Then, on p. 310, Chapter 13 (which is discussing Riemannian geometry), they say:

Just as modern differential geometry replaces the old style "differential" $df$ by the "differential form" $\mathbf{d}f$ (Box 2.3, page 63), so it also replaces the old-style "line element"

$$
ds^2 = g_{\mu \nu} dx^\mu dx^\nu = \text{("interval between } x^\alpha \text{ and } x^\alpha + dx^\alpha \text{")}
$$

by the bilinear machine ("metric tensor")

$$
\mathbf{g} = \mathbf{ds}^2 = g_{\mu \nu} \mathbf{d} x^\mu \otimes \mathbf{d} x^\nu
$$

So it looks like MTW recognize the distinction between "line element" (a scalar representing a squared infinitesimal displacement) and "metric" (a tensor which can be contracted twice with a tangent vector to obtain a scalar representing a squared infinitesimal displacement), but they also recognize that the notation $ds^2$ can refer to both. Most of the time, at least early on when they are laying the foundations of GR, they are careful and write $\mathbf{ds}^2$ to refer to the metric, and don't really mention "line element" at all (except in brief discussions like those quoted above, in which they explain their terminology and notation); but later in the text they seem much more likely to just write a line element, an equation with $ds^2$ (no boldface) on the LHS and an expression involving $dt^2$, $dr^2$, etc. (again no boldface) on the RHS, but then use the term "metric" interchangeably with "line element" to refer to it (see, for example, equation (23.27) on p. 607 and the accompanying text).

So while there is a recognized distinction between the scalar line element and the metric tensor, I would have to agree based on the above that the notation $ds^2$ is used interchangeably to refer to both. This doesn't cause any real difficulty for the physics, for the reasons already discussed.

 

[vanhees71]

Of course you have to distinguish the symbols. I'm an advocate of the good old Ricci calculus, because there it's much more intuitive. What I used, however in my previous postings in this thread is Cartan calculus on manifolds for the special case of holonomous coordinates. There the symbol $\mathrm{d} q^{\mu}$ is a tangent-space dual vector, i.e., a one-form, i.e., a linear mapping from tangent space at a point on the manifold, parametrized with real generalized coordinates $q^{\mu}$. The holonomous tangent-space basis at the point defined by the tangent vectors of the coordinate lines, i.e., keeping all $q^{\mu}$ constant except one. This basis is denoted by $\partial_{\mu}$. By definition the dual basis $\mathrm{d} q^{\mu}$ maps these holonomous basis vectors as follows $\mathrm{d} q^{\mu}(\partial_{\nu})={\delta^{\mu}}_{\nu}$, where the latter symbol is the usual Kronecker $\delta$.

In the Ricci calculus the $\mathrm{d} q^{\mu}$ are simply infinitesimal increments of the coordinates and thus
$$\mathrm{d} s^2=g_{\mu \nu} \mathrm{d} q^{\mu} \mathrm{d} q^{\nu}.$$
I've given the derivation of this in the Cartan calculus also in my previous posting.

 

[guitarphysics]

Thanks for that reference to MTW, Peter.. It becomes clearer to me as time goes on why that book is the bible of GR.