From what I learned in tensor calculus so far, coordinate transformations are supposed to preserve the metric of the space. (Here I used GR notation, but the metric doesn't have to be the spacetime metric.) $$\Lambda{^\rho}{_\mu}\Lambda{^\sigma}{_\nu}g{_\rho}{_\sigma}=g{_\mu}{_\nu}$$

So, to find all possible $\Lambda$'s, I thought that I just have to use the rule described above. That is to find all $\Lambda$'s that give back the exact same metric.

However, for the transformations between Cartesian and Polar coordinates in a 2-d plane, the metric looks very much different in different coordinates, and yet they are equivalent. Is it because going from Cartesian to Polar is not a linear transformation or something? And the set of transformations that I get from the method above does not contain the Cartesian to Polar transformations? If so, then what kind of transformations are they?


You seem a bit confused. A general coordinate transformation is just any differentiable, bijective and with differentiable inverse function (called a diffeomorphism) between open sets in $\mathbb{R}^n$. So you can't really list them all; any function that satisfies the above conditions will work. Under a transformation $x'^\mu = x'^\mu(x)$, the metric changes as

$$g'_{\mu\nu} = \frac{\partial x^\alpha}{\partial x'^\mu} \frac{\partial x^\beta}{\partial x'^\nu} g_{\alpha\beta},$$

where the matrices $g_{\mu\nu}$ and $g'_{\mu\nu}$ will not in general be the same. A coordinate transformation preserves the metric in the sense that the abstract tensor is coordinate-independent, but its components do depend on the coordinates.

There is a special class of diffeomorphisms, called isometries, that do leave the components of the metric invariant:

$$g_{\mu\nu} = \frac{\partial x^\alpha}{\partial x'^\mu} \frac{\partial x^\beta}{\partial x'^\nu} g_{\alpha\beta}.$$

(Pay attention to the primes!) We think of them as the symmetries of our space. In Euclidean space they are rotations and translations; in Minkowski spacetime some rotations are replaced by Lorentz boosts. In a more general situation you might have less isometries: a black hole only has rotation and time translation symmetry, but not space translation or boost. A space might not have any isometries at all.

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  • $\begingroup$ Thanks for the answer! Now I understand the metric more than before. I have to admit that I was very confused because I haven't yet come across these concepts of isometries being the symmetries of the space. Could you point me toward some good textbooks that include these ideas? Once again, thank you. $\endgroup$ – Bao Oct 15 '18 at 5:48
  • $\begingroup$ @Bao most relatively advanced and/or complete texts have something about isometries. Zee's book Einstein Gravity in a Nutshell is probably the best place to start: there is some discussion about Killing vectors, isometries, and maximally symmetric space(time)s. And if you want to go full math mode (and I do mean full math mode), O'Neill's Semi-Riemannian Geometry with Applications to Relativity has quite a bit of very rigorous material. $\endgroup$ – Javier Oct 18 '18 at 18:12

It is actually not the metric tensor --- if it is considered as the ensemble of its components --- which is supposed to be preserved under coordinate transformations, but the invariant line element $ds$ respectively the square of the line element:

$$ds^2 = g_{ik}dx^i dx^k$$

Cartesian coordinates and the polar counterparts are very useful to demonstrate that: The square of the invariant line element in 2-dimensional coordinates is:

$$ds^2 = dx^2 + dy^2$$

A look on this formula shows us that the only non-zero components of the metric tensor are $$g_{11}= 1 = g_{22}$$ where we have identified for simplicity $x\equiv x^1$ and $y\equiv x^2$. If we now go over to polar coordinates, i.e. use $(r,\phi)$ instead of $(x=r\cos\phi,y=r\sin\phi)$, the components of the metric tensor undergo the following transformation:

$$ g_{\bar{i}\bar{k}} = \frac{\partial x^j}{\partial x^\bar{i}}\frac{\partial x^m}{\partial x^\bar{k}}g_{jm} \equiv \Lambda^j_\bar{i} \Lambda^m_\bar{k} g_{jm} $$

$$ g_{\bar{1}\bar{1}} = \frac{\partial x^1}{\partial x^\bar{1}}\frac{\partial x^1}{\partial x^\bar{1}}g_{11} +2\frac{\partial x^1}{\partial x^\bar{1}}\frac{\partial x^2}{\partial x^\bar{1}}g_{12} + \frac{\partial x^2}{\partial x^\bar{1}}\frac{\partial x^2}{\partial x^\bar{1}}g_{22}= \cos^2\phi g_{11} + 2\frac{\partial x^1}{\partial x^\bar{1}}\frac{\partial x^2}{\partial x^\bar{1}} \cdot 0 + \sin^2\phi g_{22} = 1$$ and $$ g_{\bar{2}\bar{2}} = \frac{\partial x^1}{\partial x^\bar{2}}\frac{\partial x^1}{\partial x^\bar{2}}g_{11} +2\frac{\partial x^1}{\partial x^\bar{2}}\frac{\partial x^2}{\partial x^\bar{2}}g_{12} + \frac{\partial x^2}{\partial x^\bar{2}}\frac{\partial x^2}{\partial x^\bar{2}}g_{22}= \cos^2\phi g_{11} + 2\frac{\partial x^1}{\partial x^\bar{1}}\frac{\partial x^2}{\partial x^\bar{1}} \cdot 0 + \sin^2\phi g_{22}= r^2 sin^2\phi g_{11} + 2\frac{\partial x^1}{\partial x^\bar{2}}\frac{\partial x^2}{\partial x^\bar{2}} \cdot 0 + r^2 cos^2\phi g_{22}= r^2 $$

It can equally easily checked in the same way that the component $g_{\bar{1}\bar{2}}=0$.

Therefore in polar coordinates the square of the invariant line element are:

$$ds^2 = 1\cdot dr^2 + r^2 d\phi^2$$ and as the name of $ds^2$ suggests, it does not change under the coordinate transformation:

$$ds^2 = dx^2 + dy^2 = 1\cdot dr^2 + r^2 d\phi^2$$.

You can of course consider the metric tensor in a coordinate-independent way, i.e.:

$$g= g_{ik} e^i \otimes e^k$$

One is free to choose the coordinates, e.g. cartesian coordinates or polar coordinates or what you like, but the components of the metric tensor $g_{ik}$ depend on the chosen coordinates, in any holonome coordinate set one would get:

$$g = g_{ik} dx^i \otimes dx^k=ds^2$$

The coordinate independent definition of the metric tensor turns out to be equivalent to the square of the invariant line element. $g\equiv ds^2$ does not change, but the coefficients $g_{ik}$, -- the components of g -- change according to the formula given above.

EDIT: Of course for most of the spaces which are described by a metric there are coordinate transformations which keep the components of the metric tensor invariant: For the N-dim. euclidian space these are rotations belonging to the group O(N) and translations, for the Minkowski space-time these are the Lorentz transformations and corresponding translations. For other spaces the search for the invariance group of the metric tensor is a problem of differential geometry.

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  • $\begingroup$ Thanks for the detailed answer. It really helped me understand my confusion. $\endgroup$ – Bao Oct 15 '18 at 5:50

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