# Visualization of $dtdx$ and $dxdy$ term in metric tensor

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?

• I miss the signature for your line element? – Eli Jan 8 at 8:16
• 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. – Prof. Legolasov Jan 8 at 21:31
• @Eli: The signature would be implicit in the metric $g_{\mu\nu}$. Some of the terms would be negative. – Angela Jan 9 at 3:03
• @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. – Angela Jan 9 at 3:05
• @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. – R. Rankin Jan 9 at 5:08

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.

• 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 – Eli Jan 9 at 7:11
• @Eli, yes I also think so. I've edited the text to replace x and t by dx and dt etc. – Angela Jan 9 at 14:16

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.

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

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)

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