23
$\begingroup$

In Feynman's Lectures on Physics (Volume 2, chapter 42) he states that Einstein's field equation is equivalent to the statement that in any local inertial coordinate system the scalar curvature of space (which he gives in terms of the radius excess of a small sphere) is proportional to the energy density at that point. Does anyone know of an elegant proof of this fact?


I have a half-elegant proof based on the following:

  1. Show that the $tt$-component of the stress-energy tensor, $T$, is equal to energy density in inertial coordinates (straightforward)
  2. Show that the $tt$-component of the Einstein curvature tensor, $G$, is proportional to the scalar curvature of space in inertial coordinates (horrific)
  3. Argue that since $T$ and $G$ are both symmetric 2-tensors, and symmetric 2-tensors form an irreducible representation of the lorentz group, if the $tt$-components are equal in every local inertial frame then the tensors themselves must be equal (straightforward)

The problem with the above is that the only way I can find to prove step 2. is to explicitly calculate the radius excess in terms of the metric in local inertial coordinates. This requires finding the geodesic paths of length $r$ to order $r^3$, expanding the metric to third order at the surface of the generated sphere, and calculating the surfacearea (or volume, I found area easier). The integral for the surface area ends up with about 40 terms to take care of (which do not seem to be easily reducible through e.g. symmetry considerations), and took me several pages to do. Since the relationship between scalar curvature and the curvature tensors is quite fundamental I expect there must be a more direct, abstract argument, but I cannot find it.

$\endgroup$
  • $\begingroup$ I'm interested in the following aside: How would one prove the converse: Scalar curvature $\sim \rho \implies$ EFE? Seems like a fascinating question: It's always worth checking Feynman's interesting claims! :) +1 $\endgroup$ – Danu Jan 9 '15 at 7:34
  • $\begingroup$ @Danu: I feel like I recently read that the EFE isn't unique from the point of view you're asking for. $\endgroup$ – DanielSank Jan 9 '15 at 7:46
  • $\begingroup$ @Danu (and DanielSank): The proof I outline above goes both ways (so to speak), so it does demonstrate equivalence. Thus the Einstein curvature tensor is the unique symmetric 2-tensor for which this result holds (unless I've made a mistake in the proof). $\endgroup$ – Mark A Jan 9 '15 at 8:09
  • $\begingroup$ @WetSavannaAnimal: The main reason I ask is because I am teaching a course on GR next semester (the fact I am the most qualified person to do this tells you something about the size of our faculty!). Previously I have used Schutz's book, which is very good, but I find Feynman's approach much more physically intuitive (and clearer to see how Newtonian gravity appears from GR) so I was looking for a smooth way to go from there to the tensor equation. $\endgroup$ – Mark A Jan 9 '15 at 8:14
  • $\begingroup$ @Danu It's also interesting that the EFE you get from Feynman's condition must have zero cosmological constant. $\endgroup$ – Mark A Jan 9 '15 at 8:16
14
$\begingroup$

The key to proving item 2 is to express the metric in Riemann normal coordinates, which is usually what is meant when you say you are working in a locally inertial coordinate system. In these coordinates, the metric is equal to the Minkowski metric at a point, the first derivatives of the metric at the point vanish, and the second derivatives of the metric are given by the Riemann tensor at that point. The explicit form of the metric components are then (see e.g. this document)

$$g_{\mu\nu} = \eta_{\mu\nu}-\frac13R_{\mu\alpha\nu\beta}x^\alpha x^\beta + ...$$

where the dots represent higher order corrections in the coordinate distance from the origin, $x^\alpha = 0$.

We need to compute the volume of a sphere of coordinate radius $r$. For this we need the spatial metric, which is $h_{\alpha \beta} \equiv g_{\alpha\beta}+u_\alpha u_\beta$, and $u^\alpha$ is tangent to the inertial observer, so points in the time direction. The spatial volume element comes from the determinant of $h_{ij}$ as a spatial tensor ($i,j$ are only spatial indices). We have

$$h_{ij} = \delta_{ij} -\frac13R_{i\mu j\nu}x^\mu x^\nu+...$$

and the first order correction to the determinant just adds the trace of this tensor,

$$\sqrt{h} = 1 + \frac12\left(-\frac13\delta^{kl}R_{kilj}x^i x^j\right). $$

It will be useful to work with spacetime indices in a moment, where the background spatial metric is given by $\delta_{\mu\nu} = \eta_{\mu\nu} +u_\mu u_\nu$, and its inverse is $\delta^{\mu\nu} = \eta^{\mu\nu}+u^\mu u^\nu$. Now the volume of the sphere is simply

$$V = \int d^3x \sqrt{h} = \int d^3x\left(1-\frac16 \delta^{\mu\nu}x^\alpha x^\beta R_{\mu\alpha\nu\beta}\right).$$

(the limit of integration is over a coordinate sphere centered at the origin).

The first term will give the flat space volume of the sphere, so we need to compute the second term to get what Feynman is calling the spatial curvature of space. Remember that the Riemann tensor is taken to be constant since it is evaluated at the origin. Also, when integrated over a spherical region, only the trace of $x^\alpha x^\beta$ contributes, the other parts canceling out, so we can replace $x^\alpha x^\beta \rightarrow \frac13 r^2 \delta^{\alpha\beta}$. So the integral we are computing becomes

$$\Delta V = -\frac16\frac{4\pi}{3}\delta^{\mu\nu}\delta^{\alpha\beta}R_{\mu\alpha\nu\beta}\int_0^{r_s} r^4 dr = -\frac{2\pi}{45} r^5\delta^{\mu\nu}\delta^{\alpha\beta}R_{\mu\alpha\nu\beta}.$$

The numerical coefficient is not important, we only care about the dependence on the Riemann tensor. Re-writing the $\delta$'s in terms of the background metric $\eta^{\mu\nu}$ and $u^\alpha$, we get

$$\delta^{\mu\nu}\delta^{\alpha\beta}R_{\mu\alpha\nu\beta}=(\eta^{\mu\nu}+u^\mu u^\nu)(\eta^{\alpha\beta}+u^\alpha u^\beta)R_{\mu\alpha\nu\beta} = R + 2 R_{\mu\nu}u^\mu u^\nu,$$

where $R$ is the Ricci scalar at the origin. Now we can easily check that this is proportional to the $uu$-component of the Einstein tensor (using $u^\mu u^\nu g_{\mu\nu} = -1$),

$$G_{\mu\nu}u^\mu u^\nu = \frac12(2 R_{\mu\nu}u^\mu u^\nu + R) \checkmark$$

Then from the rest of your arguments, we arrive at Feynman's conclusion: the energy density is proportional to the spatial curvature in all locally inertial frames.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Something went wrong in the second equation; I'm pretty sure the $\mu$ and $\nu$ are contracted, while the $i,j$ should be free. I'm editing that in, you can roll it back if you think it's a mistake on my part. Great answer otherwise, +1 $\endgroup$ – Danu Jan 10 '15 at 10:29
  • $\begingroup$ Thanks, this is much simpler than the way I did it. I just have one comment before I accept the answer: radius excess is defined in terms of a geodesic ball, whereas in your answer you used the coordinate ball. In local inertial coordinates the deformation is of order $\sim r^3$, so in principle this can also contribute a term of order $\sim r^5$ to the volume. I included this deformation in my calculation (which made it much more difficult) but the fact that my answer is the same as yours implies that it must come out as zero, but I can't see any simple reason why that should be the case. $\endgroup$ – Mark A Jan 12 '15 at 3:36
  • 1
    $\begingroup$ Oh, hold on... we can just use geodesic coordinates from the start. Then in these coordinates there is no deformation and since geodesic coordinates are also local inertial coordinates the rest of your argument above is unchanged. :-) $\endgroup$ – Mark A Jan 12 '15 at 4:05
  • $\begingroup$ Okay good that sounds right. The Riemann normal coordinates are defined so that geodesic distance corresponds with coordinate distance. $\endgroup$ – asperanz Jan 12 '15 at 4:21
7
$\begingroup$

Here is the easiest way to generalize that statement that I know of, which leverages the ADM formulation of relativity. I'm glossing over a lot below, and I might be off by some factors of 2 and some signs, but the argument below is generally correct.

In the neighborhood of an arbitrary point, pick an arbitrary timelike direction $t^{a}$ to be our time coordinate and a local spacelike surface $\Sigma$, which will have unit timelike normal $n^{a}$. In general, we have:

$$t^{a} = \alpha n^{a} + \beta^{a}$$

where $\alpha$ is called the lapse function, and $\beta$ is called the shift function. (and $\beta$ is tangent to $\Sigma$). Now, if we do this, it should be easy to show that $\gamma_{a}{}^{b} = \delta_{a}{}^{b} + n_{a}n^{b}$ is a projection operator onto $\Sigma$ and that $\gamma_{ab} = \gamma_{a}{}^{c}\gamma_{b}{}^{d}g_{cd}$, so we call $\gamma_{ab}$ the metric of $\Sigma$

Finally, we note that the surface has an Extrinsic curvature, which tells us how $\Sigma$ bends inside of our 4-d spaceitme (a cylinder in $\mathbb{R}^{3}$ is no intrinsic curvature, but it does have extrinsic curvature). It turns out that the extrinsic curvature is defined by $K_{ab} = -\gamma_{a}{}^{c}\nabla_{c}n_{b}$

All that is a long way to get to the fact that if you contract einstein's equation on $n^{a}n^{b}$, you get:

$$16\pi\rho = R - K^{ab}K_{ab} + K^{2}$$

, which is known as the Hamiltonian constraint. It turns out that there is always a choice of $\alpha$ and $\beta$ that can make $K_{ab}=0$ at an instant, and that since another contraction of Einstein's equation gives you

$${\dot \gamma_{ab}} = -2\alpha K_{ab} + \nabla_{(a}\beta_{b)}$$

this corresponds to choosing a locally Minkowski coordinate frame. In this frame, we then have:

$$16\pi \rho = R$$

which corresponds to Feynman's statement.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I must admit that I find this a little hard to follow... Is $R$ the scalar curvature of the submanifold $\Sigma$? Also, it looks like the meat of the calculation is getting to the result following the sentence 'All that is a long way to get to...' -- do you know of any reference where this calculation is shown explicitly? $\endgroup$ – Mark A Jan 12 '15 at 3:40
  • $\begingroup$ Wald has a nice treatment in Chapter 10 and Appendix E. There is also Eric Poisson's book on GR, which he has made available online here: physics.uoguelph.ca/poisson/research/notes.html under advanced GR. Chapter 4 covers the Hamiltonian formulation used above. $\endgroup$ – asperanz Jan 12 '15 at 4:26
  • $\begingroup$ @MarkA: yes, $R$ is the scalar curvature of $\Sigma$. And I did gloss over a whole chapter of most GR textbooks. $\endgroup$ – Jerry Schirmer Jan 12 '15 at 14:59
2
$\begingroup$

We know that Einstein tensor is given by

$G=R_{uv}-\frac{1}{2}g_{uv}R$

Now, assume an infinitesimally small spherical ball of uniform mass density at rest with respect to an observer at the center of the ball. We can define Minkowski's metric in this infinitesimal region of space such that the metric is given by:

$g_{uv}=g^{uv}=\begin{pmatrix}-1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix}$

Now, since $R=g^{uv}R_{uv}$

$\Rightarrow R=g^{aa}R_{aa}$ if we use Minkowski metric.

$\Rightarrow R=-R_{tt}+R_{xx}+R_{yy}+R_{zz}$

Hence, Einstein tensor becomes

$G=R_{uv}-\frac{1}{2}g_{uv}[-R_{tt}+R_{xx}+R_{yy}+R_{zz}]$

Now, getting back to EFE

$R_{tt}+\frac{1}{2}[-R_{tt}+R_{xx}+R_{yy}+R_{zz}]=kT_{tt}$

$\Rightarrow G_{tt}=\frac{1}{2}[R_{tt}+R_{xx}+R_{yy}+R_{zz}]$

Now, since, $R_{aba}^b=g_{aa}g^{bb}R_{bab}^{a}$

So, $R_{tat}^a=-R_{ata}^t$

Hence, $G_{tt}=\frac{1}{2}[R_{xnx}^n+R_{yny}^n+R_{znz}^n]$, where $n$ can vary only over spatial values of co-ordinate axis, namely $n=x,y,z$.

This proves that $G_{tt}$ is proportional to the spatial curvature, because $[R_{xnx}^n+R_{yny}^n+R_{znz}^n]$ is proportional to the deviation of the sphere drawn in space from its euclidean counterpart.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ @MarkA Thanks for pointing out that mistake! I actually did have that misconception about scalar curvature, which is now clear. Also, since the mistake made my answer fundamentally wrong, I have deleted it. But, your correction prompted me to write another answer which, I am afraid, is only a half proof of Feynman's statement. $\endgroup$ – Prem kumar Sep 30 '16 at 17:23
  • $\begingroup$ @MarkA your comment corrected yet another misconception of mine and solved the above problem, or so I think. I have corrected the answer for good! $\endgroup$ – Prem kumar Sep 30 '16 at 19:12
  • $\begingroup$ Now that is looking like a good answer! In summary the result follows just from the symmetries of the Riemann tensor. I knew there should be a simple answer to this, although we all failed to find it when I originally posted the question. Thank you! $\endgroup$ – Mark A Sep 30 '16 at 19:50
  • 1
    $\begingroup$ Two final quibbles: (i) you should be consistent with the indices on both sides of the equation, so for example "Hence, Einstein tensor becomes $G_{uv}$" and a couple of lines later should be $G_{tt}$ rather than just $G$; (ii) the assumption of "an infinitesimally small spherical ball..." is now not necessary as it is always possible to find coordinates where $g_{uv}$ locally takes the form you've assumed (regardless of the curvature of space-time). $\endgroup$ – Mark A Sep 30 '16 at 19:56

Not the answer you're looking for? Browse other questions tagged or ask your own question.