Help on some equations in Einstein's original papers

[Pyter]

TL;DR Summary

Help on tensor equations in Einstein's original General Relativity papers

Studying Einstein's original Die Grundlage der allgemeinen Relativitätstheorie, published in 1916's Annalen Der Physik, I came across some equations which I couldn't verify after doing the computations hinted at.

The first are equations 47b) regarding the gravity contribution to the stress-energy-momentum-tensor through the Hamiltonian:
$$ \begin{align} \frac{\partial}{\partial x_{\alpha}} \left( \frac{\partial H}{\partial g^{\mu\nu}_{,\alpha}} \right ) - \frac{\partial H}{\partial g^{\mu\nu}} = 0 \tag{47b} \end{align} $$
According to the discussion in the following paragraphs, this should be equivalent to:
$$ \frac{\partial}{\partial x_{\alpha}} \left( g^{\mu\nu}_{,\sigma}\frac{\partial H}{\partial g^{\mu\nu}_{,\alpha}} \right )
- \frac{\partial H}{\partial x_{\sigma}} = 0 $$
but you can see that in the original equation there is a term:
$$ - \frac{\partial H}{\partial g^{\mu\nu}} $$
not appearing in the modified equation, which doesn't seem to be vanishing.
Can you explain why?

The second are Equations 66) and 66a) regarding the electromagnetic contribution to the stress-energy-momentum-tensor:
$$ \begin{align} x_{\sigma} = \frac{\partial T_{\sigma}^{\;\nu }}{\partial x_{\nu}} - \frac{1}{2}g^{\tau\mu} \frac{\partial g_{\mu \nu}}{\partial x_{\sigma}}\,T^{\;\nu}_{\tau} \tag{66} \end{align} $$
$$ \begin{align} T^{\; \nu}_{\sigma} = -F_{\sigma \alpha}F^{\nu \alpha} + \frac 14 \delta_{\sigma}^{\; \nu}\; F_{\alpha \beta}F^{\alpha \beta} \tag{66a} \end{align} $$
According to the discussion in the previous pages, by substituting 66a) into 66) you should get three terms, but even before doing the actual computation, it's plain to see that you also get a fourth term which doesn't seem to vanish:
$$ -\frac 18\;F_{\alpha\beta}F^{\alpha\beta}\;g^{\mu\nu}\;g_{\mu\nu,\sigma} $$
Am I missing something?

P.S.: I've linked the original papers in German to make sure that the equations are indeed the original ones.

 

[Dale]

Are you learning this material for the first time or do you already understand it from modern textbooks and you are just exploring the history?

 

[Pyter]

I've studied it on modern textbooks but I like going directly to the source: Einstein, Minkowski, Lorentz. Often they're clearer, more concise and straight to the point.

 

[anuttarasammyak]

Summary:: Help on tensor equations in Einstein's original General Relativity papers

Can you explain why?

Multiply $\frac{\partial g^{\mu\nu}}{\partial x_\sigma}$ to (47b) and take sum for $\mu$ and $\nu$ . We get the formula in the paper.

When you would calculate $\frac{\partial H}{\partial x_\sigma}$, please be reminded that H is function of not only $g^{\mu\nu}$ but $g^{\mu\nu}_{\ \ \alpha}$ first derivative.

$$\frac{\partial H}{\partial x_\sigma}=\frac{\partial H}{\partial g^{\mu\nu}}\frac{\partial g^{\mu\nu}}{\partial x_\sigma}+\frac{\partial H}{\partial g^{\mu\nu}_{\ \ ,\alpha}}\frac{\partial g^{\mu\nu}_{\ \ ,\alpha}}{\partial x_\sigma }$$
clearer copies
1 http://myweb.rz.uni-augsburg.de/~eckern/adp/history/einstein-papers/1916_49_769-822.pdf
2 https://echo.mpiwg-berlin.mpg.de/EC.../Einst_Grund_de_1916/index.meta&mode=texttool

 

[Pyter]

@anuttarasammyak multiplying what?

 

[Pyter]

For all those interested, an accurate English edition of the original papers is available at this link (ISBN-13: 978-0486600819, ISBN-10: 0486600815). I couldn't find a free version unfortunately.
The text is translated, but the equations are scanned directly from the original papers.

 

[Sagittarius A-Star]

For all those interested, a freely available English translation of the original Einstein paper is available starting on document page 204 /258 (=printed page number 184) at
https://archive.org/details/einstein_relativity/page/n203/mode/2up

You can find the mentioned formula (47b) on document page 232 /258 (=printed page number 211).

You can find an annotated manuscript of the mentioned formula (47b) on document page 118 /258 (=printed page number 98).

 

[Sagittarius A-Star]

Here is a link to another English translation of the original Einstein paper. You can find the mentioned formula (47b) in §15.

The Foundation of the Generalised Theory of Relativity

Source:
https://en.wikisource.org/wiki/The_Foundation_of_the_Generalised_Theory_of_Relativity

 

[Pyter]

Thanks @Sagittarius A-Star, nice addition.

 

[Pyter]

The annotated manuscript is very cool. On 47b) it doesn't seem to add extra info with respect to the printed edition though, the apparently missing term is still missing
And watch out for the second source, the 47b) has an error, the covariant sigma on the first term should be an alpha. I'd double check all the equations against the original.

 

[Pyter]

Multiply $\frac{\partial g^{\mu\nu}}{\partial x_\sigma}$ to (47b) and take sum for $\mu$ and $\nu$ . We get the formula in the paper.

When you would calculate $\frac{\partial H}{\partial x_\sigma}$, please be reminded that H is function of not only $g^{\mu\nu}$ but $g^{\mu\nu}_{\ \ \alpha}$ first derivative.

$$\frac{\partial H}{\partial x_\sigma}=\frac{\partial H}{\partial g^{\mu\nu}}\frac{\partial g^{\mu\nu}}{\partial x_\sigma}+\frac{\partial H}{\partial g^{\mu\nu}_{\ \ ,\alpha}}\frac{\partial g^{\mu\nu}_{\ \ ,\alpha}}{\partial x_\sigma }$$
clearer copies
1 http://myweb.rz.uni-augsburg.de/~eckern/adp/history/einstein-papers/1916_49_769-822.pdf
2 https://echo.mpiwg-berlin.mpg.de/EC.../Einst_Grund_de_1916/index.meta&mode=texttool

That's the problem, he multiplies only the first term of 47b) by $g^{\mu\nu}_{\sigma}$, then he equates it to 0, but this would only be legal for the whole 47b) multiplied by that factor.

 

[anuttarasammyak]

@Pyter Please investigate the attached calculation.

 

[Pyter]

@anuttarasammyak my calculations are attached, as per the original paper. It's pretty straightforward, there's no need to derive $\frac{\partial H}{\partial \sigma}$ .

 

[Pyter]

@anuttarasammyak I see your point. You get the $\frac{\partial H}{\partial \sigma}$ with the total derivative of H, but I get it by simply canceling out the $\partial g^{\mu \nu}_{\alpha}$ at the numerator and denominator.

 

[anuttarasammyak]

@Pyter

Formula of H in (47a) shows
$$H =g \Gamma \Gamma=H(g^{\mu\nu},g^{\mu\nu}_\alpha)$$
All the parameters should be considered for partial derivative calculation. See derivative formula in post #4. Both RHS terms are necessary to make $\frac{\partial H}{\partial x_\sigma}$ though you deals only the 2nd term. Einstein prepared the both.

 

[Pyter]

@anuttarasammyak absolutely right. Now I see my error: given $H = H \left( \xi^{\nu} \right)\text{, }\xi^{\nu} = \xi^{\nu} \left( x^{\sigma}\right)$, I calculated:
$$ \frac {\partial H}{\partial \xi_1} \frac {\partial \xi_1}{\partial x^\sigma} = \frac {\partial H}{\partial x^\sigma}, $$
while it's in fact:
$$ \frac {\partial H}{\partial \xi_\nu} \frac {\partial \xi_\nu}{\partial x^\sigma} = \frac {\partial H}{\partial x^\sigma} .$$
I guess tensor calculus tricked me again. Well spotted.
Do you happen to have any thought about the 66) and 66a) too?

 

[anuttarasammyak]

According to the discussion in the previous pages, by substituting 66a) into 66) you should get three terms, but even before doing the actual computation, it's plain to see that you also get a fourth term which doesn't seem to vanish:

Follow the calculations by your hand as Einstein taught and tell me what $\chi_\sigma$ you get.

 

[Pyter]

@anuttarasammyak using this link, as reference, starting from the (65a) I get the three terms mentioned right after in the paper, that is:
$$ x_\sigma = \frac{\partial}{\partial x_\nu} (F_{\sigma \mu}F^{\mu \nu})
+\frac 14 \frac{\partial}{\partial x_\sigma} (F^{\mu \nu} F_{\mu \nu})
+ \frac 12 F^{\mu \tau} F_{\mu \nu}g^{\nu \rho} \frac{\partial g_{\rho \tau}}{\partial x_\sigma}
$$
By substituting (66a) [the equation right after (66)] into the (66), I get the same three terms plus a fourth one which doesn't seem to vanish:
$$ -\frac 18\;F_{\alpha\beta}F^{\alpha\beta}\;g^{\mu\nu}\;g_{\mu\nu,\sigma} $$
and surely is not a tensor. Isn't $x_{\sigma}$ supposed to be a covariant vector?
I find this section really interesting because it gives the EMF components of the stress-energy-momentum tensor.

EDIT: watch out for the document I've linked, I've just noticed there's a misprint in the third term, there should be a rho instead of the sigma in the covariant index of the $\partial g$. I've just corrected it here.

 

[anuttarasammyak]

@Pyter
Thank you for showing calculation.
$$g^{\mu\nu}g_{\mu\nu,\sigma}=g^\mu_{\mu,\sigma}=0$$
because $g^\mu_\nu=1$ for $\mu=\nu$ otherwise 0, so constant anyway.

 

[Pyter]

@anuttarasammyak I thought so too at first, but apparently that contraction is not allowed.
You can verify by yourself with a random metric, i.e. $g_{\mu \nu} = diag[ 3x^1, 4x^2 ]$, that it doesn't vanish.
I guess it's because $g_{\mu \nu, \sigma }$ is not a tensor.

 

[kent davidge]

$g^{\mu \nu}g_{\mu \nu,\sigma} = (g^{\mu \nu}g_{\mu \nu})_{,\sigma} - g_{\mu \nu}g^{\mu \nu}{}_{,\sigma} = - g_{\mu \nu}g^{\mu \nu}{}_{,\sigma}$. Problem is that $g_{\mu \nu}g^{\mu \nu}{}_{,\sigma} \neq g^{\mu \nu}g_{\mu \nu,\sigma}$. As just shown one is minus the other. Therefore it doesn't vanish.

 

[anuttarasammyak]

@anuttarasammyak I thought so too at first, but apparently that contraction is not allowed.

Hum... Let us investigate more.
Following Einstein for $\chi_\sigma$, I got the same 1st and 2nd term with you that coincide with $T^{\ \ \ \nu}_{ \sigma,\nu}$ but the 3rd one is
$$-\frac{1}{4}F_{\alpha\beta}F_{\mu\nu}(g^{\mu\alpha}g^{\nu\beta})_{,\sigma}$$
that should coincide with
$$-\frac{1}{2}g^{\tau\mu}g_{\mu\nu,\sigma}T^{\ \ \nu}_\tau$$
accrording to (66).

Is it same with yours but different expression ?

 

[Pyter]

@anuttarasammyak if I follow the explanation up to 66) I get an expression with the 3 terms that are explicitly written in the paper (I did the math and they check out); if I substitute 66a) into 66) I get the same 3 terms, plus the one I wrote about that doesn't vanish.

 

[anuttarasammyak]

I am looking at (66) . It should say about covariant derivatives, so instead of $\frac{\partial g_{\mu\nu}}{\partial x^\sigma}$, Christoffel symbol $\Gamma_{\mu\nu\sigma}$ be there. So two more derivatives might be included. I do not know they happen to be zero here. This suspected difference might have caused your concern.

 

[Pyter]

@anuttarasammyak are you saying that there is a misprint in the equations of the original papers in German?

 

[Pyter]

Found it. The extra term vanishes because of equation (29):
$$
-\frac12 g^{\mu \nu}g_{\mu \nu,\alpha} = \frac {1}{\sqrt{-g}} \frac{\partial \sqrt {-g}}{\partial x^{\alpha}}
$$
with the additional hypothesis that $\sqrt{-g} = 1$.
That's another thing I didn't get about the papers, how can the fundamental tensor determinant be constant in every chart. Maybe I'll open another thread about it.

 

[PeterDonis]

how can the fundamental tensor determinant be constant in every chart

It can't. The condition $\sqrt{-g} = 1$ is a restriction on what coordinate charts are allowed.

 

[Pyter]

@PeterDonis I mean if it's even possible to find a non-local coordinate chart where that condition is satisfied everywhere.
For instance, for the 2-sphere of radius r inheriting the $E^3$ ambient metric, we have $g_{\mu \nu} = diag(r^2 sin^2 \theta, r^2)$ which doesn't satisfy the condition. Does it mean that we can't use spherical coordinates as a chart? And what should we use?

 

[kent davidge]

@Pyter , you are right that in general $\sqrt{|g|} \neq 1$. In spherical coordinates $(ct,r,\theta,\phi)$ we have $\sqrt{|g|} = r^2\sin\theta$.

It can't. The condition $\sqrt{-g} = 1$ is a restriction on what coordinate charts are allowed.

He was telling you that if you insist on having $\sqrt{|g|} = 1$, there is only a limited family of coordinate systems which will satisfy that.

 

[Pyter]

@kent davidge it's not me who's insisting, it's Einstein that is making this assumption from equation (18a) on. He mentions that eq. (49) is only valid for the coordinate systems where $\sqrt{-g}=1$. And apparently also the (66)-(66a), as I've just realized.
A limited family could reduce to the empty set. What would you use on a $S^2$, for instance?

 

[PAllen]

This condition on metric determinant is typically called the unimodular coordinate condition. It can always be achieved, though you may need more coordinate patches to cover a manifold than with other coordinate choices. In his early years, Einstein very much liked to impose this condition.

 

[haushofer]

See also "The road to relativity " of Gutfreund and Renn, page 71. The unimodular restriction imposes a significant simplification in the field equations.

 

[haushofer]

@PeterDonis I mean if it's even possible to find a non-local coordinate chart where that condition is satisfied everywhere.
For instance, for the 2-sphere of radius r inheriting the $E^3$ ambient metric, we have $g_{\mu \nu} = diag(r^2 sin^2 \theta, r^2)$ which doesn't satisfy the condition. Does it mean that we can't use spherical coordinates as a chart? And what should we use?

Why not? You have four coordinates, and the unimodular restriction is just one condition. So it even gives you plenty of other freedom to pick your coordinates.

And no, spherical coordinates are not unimodular.

 

[Pyter]

Does this mean that all the equations in the original papers, including the EFE, are only valid for those special coordinate systems, and would be much more complicated in a really general case?
What if you solved the EFE and found $g_{\mu\nu}$ not satisfying the condition?
For instance the Schwarzschild metric uses spherical coordinates, I've seen it computed with the "simplified" EFE, but its $\sqrt {-g} \neq 1$.

 

[vanhees71]

It can't. The condition $\sqrt{-g} = 1$ is a restriction on what coordinate charts are allowed.

Can't one impose this as one gauge condition? I've somehow in mind that Einstein preferred this choice of a gauge condition for the choice of the coordinates, but I've not the time to look for the source. It even may be in the here discussed paper.