I GR: Clarifying Different Forms of the Metric for Self-Studiers

[Rollo]

I am self-studying GR, using principally Carroll’s textbook and Alex Maloney’s online lectures, and nice book by a guy called Herbert Roseman. I am a bit confused by alternative ways of expressing the metric and it would be most helpful if someone could clarify J

Basically,

• I am perplexed by people’s writing the metric in the form ds^2 = [coefficient]t^2 + [coefficient]x^2 …
• My initial thought on seeing this was that it was the line element not the metric. Carroll specifically notes this as a misunderstanding in his textbook but I don’t fully follow his account as to why it is
• The metric is a symmetric covariant rank (0,2) tensor, right?
• …But the right hand side of this expression doesn’t seem to me to result in such a tensor. The quantities t, x, y, z are vectors, and the square presumably means an inner product, so the whole expression seems to me to be a scalar – which is consistent with it being the square of a length, aka the line element, but not with its being the metric.

Can someone enlighten me? I’m clearly making a schoolboy error but I am not sure what.

Thanks

Rollo

 

[phyzguy]

The interval ds^2 is just: ds^2 = g_{ij} dx^i dx^j. You're right that g_{ij} is a rank 2 tensor and ds^2 is a scalar. Basically, this is just a way of writing the metric. You can "pick off" the components of the metric tensor by looking at the expression for ds^2.

 

[Ibix]

The metric is a tensor and the line element a scalar, true. However, you can deduce the elements of the tensor from the scalar, since the coefficient of $dx^\mu dx^\nu$ is $g_{\mu\nu}$ when $\mu=\nu$ and $(g_{\mu\nu}+g_{\nu\mu})$ otherwise (since $dx^\mu dx^\nu=dx^\nu dx^\mu$). People are often sloppy and say that "the metric is..." and give the line element, because the same information is conveyed. And it keeps dentists in work from all the mathematicians grinding their teeth when they see it.

 

[Orodruin]

I am perplexed by people’s writing the metric in the form ds^2 = [coefficient]t^2 + [coefficient]x^2 …

I would be perplexed too. It should be $dt^2$ and $dx^2$, not $t^2$ and $x^2$.

My initial thought on seeing this was that it was the line element not the metric. Carroll specifically notes this as a misunderstanding in his textbook but I don’t fully follow his account as to why it is

If you want to be strict about it, it is the line element, but you can easily identify the metric components from it. If the line element is $ds^2 = g_{ij} dx^i dx^j$ (where the $dx^i$ are infinitesimal changes in the coordinates), then the metric tensor is $g = g_{ij} dx^i \otimes dx^j$ (where the $dx^i$ are the holonomic basis vectors for the cotangent space).

The metric is a symmetric covariant rank (0,2) tensor, right?

The "covariant" is superfluous here. It is a symmetric (0,2) tensor, meaning that its components in the holonomic basis transform covariantly.

…But the right hand side of this expression doesn’t seem to me to result in such a tensor. The quantities t, x, y, z are vectors

The quantities t, x, y, z are coordinates, not vectors. Also, note the difference between the coordinates $x^i$ themselves and the corresponding one-forms (or differentials) $dx^i$.

 

[Rollo]

Thanks all for speedy and helpful responses!

Ps annoyed with myself for leaving out the critical d before t, x, etc. Even I knew that! :)

 

[pervect]

I am self-studying GR, using principally Carroll’s textbook and Alex Maloney’s online lectures

Notation does vary, but typically

$$ds^2 = a \,dt^2 + b \, dx^2 + 2 c \,dx\,dt$$

would formally be considered to be a line element, and not a metric, because ds, dt, and dx are just numbers. Sometimes people are sloppy about making the distinction though, including yours truly.

Now, if we let dt be a dual vector, also called a one-form, which is a rank 1 tensor, we can write the rank 2 tensor g in terms of the rank 1 tensors using the tensor product $\otimes$ in index free notation as

$$\textbf{g} = a \, \textbf{d}t \otimes \textbf{d}t + b \, \textbf{d}x \otimes \textbf{d} x + c \,\textbf{d}x \otimes \textbf{d}t + c\,\textbf{d}t \otimes \textbf{d}x$$

The rank 1 tensor $\textbf{d}t$ is a map from a vector to a scalar, so it's an operator whose domain is vectors and whose range is a scalar. So, dt doesn't operate on anything, it's just a scalar number, while dt operates on a vector, and returns a scalar number, the time component of the vector it operates on.

The tensor product of two rank 1 tensor is a rank 2 tensor, so g has the correct rank.

You might see some variations on this notation - MTW uses the boldface notation as I've written it to distinguish between the number dt and the one-form dt, but not all textbooks or papers use the boldface notation. I can't say what your textbooks use, for instance.

Also not that the tensor product is not commutative, so that $\textbf{d}x \otimes \textbf{d}t$ is not equal to $\textbf{d}t \otimes \textbf{d}x$.

It's easier to write the line element because multiplication commutes. The above example should illustrate how we ensure the proper symmetry of g knowing the line element.

 

[vanhees71]

I am self-studying GR, using principally Carroll’s textbook and Alex Maloney’s online lectures, and nice book by a guy called Herbert Roseman. I am a bit confused by alternative ways of expressing the metric and it would be most helpful if someone could clarify J

Basically,

• I am perplexed by people’s writing the metric in the form ds^2 = [coefficient]t^2 + [coefficient]x^2 …
• My initial thought on seeing this was that it was the line element not the metric. Carroll specifically notes this as a misunderstanding in his textbook but I don’t fully follow his account as to why it is
• The metric is a symmetric covariant rank (0,2) tensor, right?
• …But the right hand side of this expression doesn’t seem to me to result in such a tensor. The quantities t, x, y, z are vectors, and the square presumably means an inner product, so the whole expression seems to me to be a scalar – which is consistent with it being the square of a length, aka the line element, but not with its being the metric.

Can someone enlighten me? I’m clearly making a schoolboy error but I am not sure what.

Thanks

Rollo

It's a very economic way to write down the metric components with respect to a given holonomous (coordinate) basis,
$$\mathrm{d} s^2=g_{\mu \nu} \mathrm{d} q^{\mu} \mathrm{d} q^{\nu}.$$
The metric itself is a 2nd-rank symmetric tensor
$$\boldsymbol{g}=g_{\mu \nu} \mathrm{d} q^{\mu} \otimes \mathrm{d} q^{\nu}.$$