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For the sake of simplicity, lets take a 2+1 dimensional spacetime. Lets take the metric

$$ds^2 = g_{tt}dt^2 + g_{xx}dx^2 + g_{yy}dy^2 + g_{tx}dtdx + g_{xy}dxdy$$

What is the visualization or physical interpretation of the $g_{tx}$ and $g_{xy}$ terms of the metric? Does $g_{tx}$ mean motion of space i.e. on object in this spacetime point will be moving w.r.t an observer at infinity? What would $g_{xy}$ mean?

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    $\begingroup$ I miss the signature for your line element? $\endgroup$ – Eli Jan 8 at 8:16
  • $\begingroup$ Locally it is always possible to put $g$ in the diagonal frame due to the equivalence principle. I suspect the off-diagonal components may tell you more about the coordinate system in use, not about space and time. $\endgroup$ – Prof. Legolasov Jan 8 at 21:31
  • $\begingroup$ @Eli: The signature would be implicit in the metric $g_{\mu\nu}$. Some of the terms would be negative. $\endgroup$ – Angela Jan 9 at 3:03
  • $\begingroup$ @SolenodonParadoxus: Motivated by your comment, I've thought of a solution/interpretation, which I've posted online as an answer. Let me know what you think. $\endgroup$ – Angela Jan 9 at 3:05
  • $\begingroup$ @SolenodonParadoxus True, though for any practical calculation we'll be dealing with some extended body. Just a guess, but I'd expect the 0i, i0 terms to manifest in the tidal tensor as some kind of torque on a finite sized body. $\endgroup$ – R. Rankin Jan 9 at 5:08
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I am thinking of this possible answer motivated by the comments of Solenodon Paradoxus. Solenodon had mentioned that locally its possible to diagonalize the metric. I am wondering if this can be generalized, leading to an interpretation of the off-diagonal elements. Since, the metric is a symmetric square matrix, it should be always be possible to diagonalize it (even globally) by determining the eigenvalues and eigenvectors. Let $\Lambda$ be the eigenvalue diagonal matrix and $V$ be the eigenvector matrix. Then

$g_{\mu\nu} = V^{\dagger}\Lambda V$,

where, V and $\Lambda$ may in general be complicated non-linear functions of $x$ and $t$ (depending upon the original metric $g_{\mu\nu}$).

Now, the above can be interpreted as $\Lambda$ being the diagonal metric tensor of the modified coordinate system where

$\begin{bmatrix} dt'\\dx' \end{bmatrix} = [V] \begin{bmatrix} dt\\dx \end{bmatrix}$

Since $V$ can be any function of $x$ and $t$, $dx'$ and $dt'$ and hence, $x'$ and $t'$ can also be in general non-linear functions of $t$ and $x$. i.e. $t' = f(x,t)$ and $x' = h(x,t)$.

${\bf Interpretion~ of~ the~ non-diagonal~elements~of~the~metric~tensor~g_{\mu\nu}}$:

We now come to the interpretation part. $x' = h(x,t)$, means that the co-ordinate axis $x'$ is moving/changing with time $t$. The co-ordinate axis are dynamic. A person sitting in a car at point $x'$ with his engine off, would still be moving w.r.t. and observer at infinity, based on the $t$ dependence of $x'$.

If however, $g_{xt}$ were to be zero, $x'$ would not be a function of time $t$. Meaning if $g_{xt}$ were zero, $x'$ would be curved but stationary/static.

Thus $g_{xt}$ leads to a co-ordinate axis which is changing dynamically.

Please let me know if anyone sees an issue with this solution, of if this solution is incorrect.

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  • $\begingroup$ I think the eigenvector matrix transformed the coordinates $\Delta t $ etc not the t‘ ?. So to get the coordinate t‘ you have to integrate your equations $\endgroup$ – Eli Jan 9 at 7:11
  • $\begingroup$ @Eli, yes I also think so. I've edited the text to replace x and t by dx and dt etc. $\endgroup$ – Angela Jan 9 at 14:16
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First of all, your guess seems well-founded -- the metric tensor is closely related to energy and momentum. In fact, the relation may be stated as \begin{equation} T^{\mu\nu}={\delta S\over \delta g_{\mu\nu}}, \end{equation} where $S$ is some action, and $T^{\mu\nu}$ is the stress-energy tensor. For brevity, let us consider a perfect fluid that permeates all of spacetime -- this describes some mass/energy density. This is tied to the amount of matter and (and maybe photons) you have in this physical system.

Component-wise, we may write the stress-energy tensor as \begin{equation} \left[T^{\mu\nu}\right]= \begin{pmatrix} E&p^1&p^2\\ p^1&P^1&\tau\\ p^2&\tau&P^2 \end{pmatrix}, \end{equation} where $E$ and $p^i$ are the energy and momentum densities respectively. $P^i$ describes stress along the $i$-th direction, and $\tau$ is the shear stress; these quantities are defined in analogy to components of the usual stress tensor seen in fluid mechanics.

So $g_{tx}=g_{01}$ and $g_{xt}=g_{10}$ are related to momentum in the $x$-direction in this way. Likewise, $g_{tt}=g_{00}$ is related to energy. $g_{xy}=g_{12}$ would then be tied to the shear stress of the perfect fluid we considered earlier.

For more information on the stress-energy tensor, you may look at this post.

I don't know enough GR to comment on how frames of reference come in though.

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I dont know much physics but mathematically off diagonals elements on the metric tensor implies that your local coordinate system is not orthonormal. So i guess that can be interpretted as how much the spacetime looks squished from your perspective

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Your intiuition isn't especially crazy, but one might be careful about taking it too literally. Despite this, here is a coordinate transform that gives you a nice version of this:

Take the flat Robertson-Walker metric in standard form

$$ds^{2} = -dt^{2} + a(t)^{2}\left(dr^{2} + r^{2} d\Omega^{2}\right)$$

Now, make the coordinate transformation, $R = a\,r$, which gives $dr = \frac{dR}{a} - \frac{R{\dot a}\,dt}{a^{2}}$

This turns the metric into the form:

$$ds^{2} = -\left(1 - H^{2}R^{2}\right)dt^{2} - R\,H\left(2dt\,dr\right) + dR^{2} + R^{2}d\Omega^{2}$$

Where $H = \frac{\dot a}{a}$ is the Hubble "constant". If you remember Hubble's law, it gives precisely that the recession velocity of a distant galaxy is given by $HR$, which is what $g_{tr}$ is in these coordinates (which also have the nice property of having constant-time spatial sections, so this is the coordinate system where your local rulers don't change with respect to time)

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  • $\begingroup$ Very nice example. But why do you caution against taking this too literally? Is there any counter-example? Or any theoretical reason? $\endgroup$ – Angela Jan 10 at 1:09

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