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There is no position vector in general relativity. I was wondering whether a quantity like

$$k_\mu x^\mu$$

where $k_\mu$ are covariant vector components is to be treated like a scalar i.e. invariant under coordinate transformations?

A came across that problem, studying gravitational waves. In one coordinate system $x$ the wave is given by

$$h_{\mu\nu}=e_{\mu\nu}\exp(i k_\alpha x^\alpha)$$

For another coordinate system $x'$ I found the Ansatz

$$h'_{\mu\nu}=e'_{\mu\nu}\exp(i k_\alpha x^\alpha)$$

in a text book. This only makes sense to me, if $k_\alpha x^\alpha$ is invariant under coordinate transformations. Otherwise it should be

$$h'_{\mu\nu}=e'_{\mu\nu}\exp(i k'_\alpha x'^\alpha)$$

But as $x^\alpha$ are not the components of a vector, I don't understand why that would be true.

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  • $\begingroup$ Could you elaborate on "There is no position vector in general relativity." ? $\endgroup$
    – ohneVal
    Oct 26, 2020 at 12:35
  • $\begingroup$ Is your source considering perturbations of a flat Minkowski space per chance? $\endgroup$
    – TimRias
    Oct 27, 2020 at 18:25

2 Answers 2

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It is a misnomer to denote the coordinate as $x^{\mu}$ in General Relativity. Coordinates don't transform as 4-vectors, so the upper index is misleading and incorrect.

This is not even specific to General Relativity, but is a mathematical fact from differential geometry (with is heavily used in General Relativity).

One way to see this is to consider compactified space-time. Let's say it is topologically a $d$-torus $T^d$ (just as a mathematical example). Then your Fourier integral with waves $e^{i k_{\mu} x^{\mu}}$ doesn't make any sense mathematically. Instead, $k$ take discrete values, and $x^{\mu}$ are coordinates within a single chart of the atlas, that don't have a global value.

It is, however, true, that the infinitesimal differentials $dx^{\mu}$ transform as $4$-vectors under general coordinate transformations, which partially justifies this abuse of notation. They are also atlas-independent – coordinate differentials follow the usual 4-vector transformation laws in the intersections of charts of the space time manifold's atlas.

Anticipating the question – yes, I think Spiridon's answer is incorrect and misleading.

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  • $\begingroup$ So, the only way to make a meaning of this expression is to consider : $e^{i k_\mu d x^{\mu}}$? Kinda a locally plane wave? $\endgroup$ Oct 26, 2020 at 16:18
  • $\begingroup$ @spiridon_the_sun_rotator there are no plane waves in the general case of the differential manifold. Think different topologies and boundary conditions etc. For a given manifold, an abstract Fourier transform can be defined (a sum for compact dimensions and an integral for noncompact), however, these waves will not be the harmonics of the dalambertian operator associated to the metric, unless specifically chosen so. But if they are, they must depend on the metric and not only on $x^{\mu}$. $\endgroup$ Oct 26, 2020 at 17:58
  • $\begingroup$ thanks for the anser. Would you have any idea how one could explain then the ansatz for the gravitational wave in the second coordinate system? $\endgroup$ Oct 27, 2020 at 8:07
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In the general relativity the combination $k_\mu x^{\mu}$ is still a scalar. Performing a general change of coordinates $x \rightarrow x^{'}$, you get: $$ k_{\mu} x^{\mu} = \frac{\partial x^{\alpha}}{\partial x^{\mu '}} \frac{\partial x^{\mu}}{\partial x^{'\beta}} k_\alpha^{'} x^{' \beta} = \delta_{\beta}^{\alpha} k_\alpha^{'} x^{' \beta} = k_\alpha^{'} x^{' \alpha} $$ In the second equality we used the fact, that the $k_\mu$ and $x^{\mu}$ transform with the mutually inverse matrices.

Because you are considering graviational part, the polarization tensor will also transform: $$ e_{\mu \nu}^{'} = \frac{\partial x^{\alpha}}{\partial x^{\beta}} \frac{\partial x^{\mu}}{\partial x^{\nu}} e_{\alpha \beta} $$ So in the expression for graviational wave one sees the same exponent, but a transformed expression in front of it.

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    $\begingroup$ What you are saying amounts to claiming that $x^\mu$ transforms as a vector. Are you sure this is true? In general I think $x^\mu \rightarrow x'^\mu(x)$, where $x'^\mu$ is just some general function, and this law doesn't involve the derivative matrix. For example, I could take a coordinate transformation like $x'^\mu = (1+x^2) x^\mu$, which is not the same as $[ \partial x'^\mu / \partial x^\nu ]x^\nu = (1+ 3 x^2) x^\mu$. $\endgroup$
    – Andrew
    Oct 26, 2020 at 12:49
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    $\begingroup$ @Andrew is correct. The transformation law you propose is valid for infinitesimal differences $dx^{\mu}$ (which are elements of the tangent space), but not for functions $x^{\mu}$ themselves. $\endgroup$ Oct 26, 2020 at 12:52
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    $\begingroup$ @Andrew thanks for correction, I've been sloppy $\endgroup$ Oct 26, 2020 at 13:05
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    $\begingroup$ @spiridon_the_sun_rotator No worries! "An expert is someone who has made every mistake" :) $\endgroup$
    – Andrew
    Oct 26, 2020 at 14:14

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