Confusions regarding the metric - part 2

Question

I have a few questions regarding the metric in GR, if you wish to answer some of them, I would appreciate a detailed answer.

(2) Under a coordinate transformation from x to x' the metric $g_{\mu\nu}(\mathbf{x})$ transforms as $$g_{\mu'\nu'}(\mathbf{x'})=\dfrac{\partial{x^\mu}}{\partial{x^\mu{'}}}\dfrac{\partial{x^\nu}}{\partial{x^\nu{'}}}g_{\mu\nu}(\mathbf{x}).$$ Is it an assumption or is there some way to prove it? If it's an assumption, I would like to know any justification.

Please keep in mind that Im a beginner in GR and still don't know differential geometry at a mathematician level. Thanks in advance!

Answer 1

By requiring that the line element is invariant to a coordinate transformation. $$ds'^2=ds^2$$ $$g_{\bar{\mu}\bar{\nu}} dx^{\bar{\mu}}dx^{\bar{\nu}} = g_{\mu\nu}dx^{\mu}dx^{\nu}$$

Noting that (via the chain rule for a total differential) $$dx^{\bar{\mu}} = J^{\bar{\mu}}_\mu dx^\mu$$ $$dx^{\bar{\nu}} = J^{\bar{\nu}}_\nu dx^\nu$$

Where $$J^{\bar{\mu}}_\mu = \frac{\partial x^{\bar{\mu}}}{\partial x^\mu}$$

(There is a sum in $\mu$ and so we sum up all of the rates of change of each variable to find the total change $dx^{\bar{\mu}}$)

We obtain

$$J^{\bar{\mu}}_\mu J^{\bar{\nu}}_{\nu} g_{\bar{\mu}\bar{\nu}} dx^{\mu}dx^{\nu} = g_{\mu\nu}dx^{\mu}dx^{\nu} $$

$$[J^{\bar{\mu}}_\mu J^{\bar{\nu}}_{\nu} g_{\bar{\mu}\bar{\nu}} - g_{\mu\nu}]dx^{\mu}dx^{\nu} = 0 $$

We require that this statement is true for all dx, and so the bracket is independently zero

$$g_{\mu\nu} = J^{\bar{\mu}}_\mu J^{\bar{\nu}}_{\nu} g_{\bar{\mu}\bar{\nu}} $$

Answer 2

The metric is a second order tensor and transforms in the standard way. For a rigorous discussion of this topic see.

Informally this can be seen as a consequence of invariance. The distance in spacetime is independent of the chosen coordinate system and is given by $ds^2=g_{\mu\nu}dx^{\mu}dx^{\nu}$. Equating the spacetime interval in two different coordinate systems gives: $g_{\mu\nu}(x)dx^{\mu}dx^{\nu} = g_{\rho\sigma}(x')dx'^{\rho}dx'^{\sigma}$ and "dividing" by the $dx'$s gives the result.

Answer 3

As a mathematician, I think that there are various ways to answer your question, and they’re all equally right.

For starters, @jensen-paull gave a great answer showing how if you interpret the metric as “the thing that measures the actual lengths”, and by “length” I mean, “the thing you’ve learned in middle school”. Under this perspective, your equation is provable.

However, I think that this latter approach can only bring you this far. If you wish to dive into other Theoretical Physics you might want to adopt a more powerful and abstract perspective.

One possibility is the Mathematician’s explanation. Mathematicians love to define objects axiomatically, and, possibly, work in a coordinate-free way. To a Mathematician, a metric is a smooth assignment of an inner product on each tangent space of a manifold (yes, I am purposely being a little sloppy here). Notice how we don’t need to fix a coordinate system $\mathbf{x}$ to define this; let alone switching between two coordinate systems $\mathbf{x}$ and $\mathbf{x}^\prime$!

Besides, by carefully expanding on this definition, you can indeed find a coordinate description of this abstract object you have axiomatically conjured, and you can prove that two different coordinate descriptions refer to the same object if and only if the equation you have written is satisfied. So, in this sense, yes, the formula you are talking about is provable. Also, as many have already pointed out, this is true not only for metrics, but for any (0,2) tensor (field), of which metrics in general relativity are just one particular case.

Nonetheless, I think it’s important to add a third perspective, which in my opinion is the most important to Physicists (funny, considering I’m a Mathematician). You will see how Physics textbooks, especially General Relativity, Gauge Theory and Loop Quantum Gravity, will not introduce a new object by telling you explicitly what it is, like Mathematicians would have done, axiomatically. Instead, they will just show you an arbitrary coordinate description and tell you how that description changes under a coordinate transformation.

Every Mathematician I have spoken with hates this thing Physicists do, and I used to hate it too. But in hindsight, I think this is because most Mathematicians (and Physicists) don't have a perfectly clear understanding of what they're doing. To explain why Mathematicians define tensors as (to put it in @paulina's words) "linear maps which take vectors and output a number" and Physicists say they are tensors "because they transform like tensors", I need to go on a little detour and talk about fibre bundles.

Consider a scalar field in Classical Mechanics, such as a temperature distribution over a square metal plate $S$. This can be represented by a function $\Theta \colon S \rightarrow \mathbb{R}$, that to each point of the plate assigns a real number, which we interpret as the temperature at that point. You can also imagine the graph of this function. You can put the plate "flat", like on a table; and then the graph of the function $\Theta$ looks like a bunch of "hills" rising above the plate. So far so good.

You can take this one step further. Take the plate, and onto each point, attach a copy of the real number line $\mathbb{R}$, lying perpendicular to the plate. Now, you have constructed the cartesian product $S \times \mathbb{R}$, which should look like a cube, having the plate $S$ as base, and $\mathbb{R}$ spanning the height. This space is a simple example of a fibre bundle: you have a base space $S$, and to each point of $S$ you have attached a copy of another space, in this case $\mathbb{R}$, called the fibre.

A general fibre bundle is constructing by attaching repeated copies of a space (called the fibre) to a fixed space, called the base. The end result is called the total space. Naturally, the dimensions of the total spaces tend to be very high.

The reason why fibre bundles are useful is because now you can see the graph of the function $\Theta$ as lying inside the fibre bundle $S \times \mathbb{R}$. You can imagine having the cube $S \times \mathbb{R}$ and slicing through it with a katana, not necessarily in a straight line: be creative! Every possible curvy slice defines a graph of $\Theta$. Keep in mind how slicing the total space implicitly assigns a real number over each point of $S$.

Now, you can repeat this for much more complicated settings, but the general idea doesn't change. Take $S$ to be your spacetime manifold (dimension 4); at each point of $S$ you attach a copy of the vector space of all bilinear forms on each tangent space (dimension 10). In total, you get a fibre bundle of dimension 14. Each section of this fibre bundle corresponds to one assignment of a bilinear form to each point: namely, a metric.

It is much better to define the appropriate fibre bundle first, and then call metrics (or spinor fields, or connections, or whatever mathematical-physical object you wish) the sections of the bundle. All of the mathematical properties of metrics and similar objects are already contained in the geometric structure of the fibre bundle; especially how they transform under a coordinate change.

What does this have to do with coordinate transformations? It is simple: the fibre bundle construction theorem says that you can construct a fibre bundle out of smaller spaces, assuming you know how to glue the pieces together. Translating this into the language of Physics, the "smaller spaces" are the local coordinate patches: these spaces are so small that you can even explicitly write down your object in them. Knowing how to "glue" the pieces together means that you know how to switch between them when they overlap; which is to say, you know how they transform under a coordinate change.