Stumped by a Lagrangian in Einstein's 1916 paper

[Isaek_]

TL;DR

Struggling to match the E-L equations from an assumed Lagrangian

I decided recently that I would work through Einstein's original paper on the foundation of GR and I've recently hit a roadblock. It starts with Einstein's Lagrangian for an empty space-time,

$$ \begin{align} \mathcal{L}= g^{\mu \nu} \Gamma^{\alpha}_{\nu \beta} \Gamma^{\beta}_{\mu \alpha} \tag{47a} \end{align}, $$

which, Einstein shows, gives the two quantities:

$$ \begin{align*} \frac{\partial \mathcal{L}}{\partial g^{\mu\nu}} = - \Gamma^{\alpha}_{\mu \beta} \Gamma^{\beta}_{\nu \alpha}; \\
\frac{\partial \mathcal{L}}{\partial (\partial_\sigma g^{\mu\nu})} = \Gamma^{\sigma}_{\mu \nu}. \tag{48} \end{align*} $$

I can follow the maths as given in the paper to arrive at this result, but I am confused by the first step. Namely, $\mathcal{L}$ is a function of $(g^{\mu\nu}, \partial_\sigma g^{\mu\nu})$, yet it is differentiated wrt $(g^{\mu\nu}, \Gamma^{\beta}_{\nu\alpha}).$ But isn't $\Gamma$ a function of both $(g^{\mu\nu}, \partial_\sigma g^{\mu\nu})$? When I differentiate wrt to these, I get the following quantities:

$$ \begin{align*}
\mathcal{L}(g^{\mu \nu}, \partial_\rho g^{\mu \nu}) &= g^{\sigma \tau} \Gamma^{\alpha}_{\sigma \beta} \Gamma^{\beta}_{\tau \alpha}; \\
\delta \mathcal{L} &= \left[ \Gamma^{\alpha}_{\sigma \beta} \Gamma^{\beta}_{\tau \alpha} + 2g^{\sigma\tau}\frac{\partial \Gamma^\alpha_{\sigma\beta}}{\partial g^{\mu\nu}} \right]\delta(g^{\mu\nu}) + 2\left[g^{\sigma\tau}\Gamma^{\alpha}_{\sigma\beta}\frac{\partial \Gamma^{\beta}_{\tau\alpha}}{\partial(\partial_\rho g^{\mu\nu})} \right]\delta(\partial_\rho g^{\mu\nu});\\
\frac{\partial \mathcal{L}}{\partial g^{\mu\nu}} &= 3\Gamma^{\alpha}_{\mu \beta} \Gamma^{\beta}_{\nu \alpha}; \\
\frac{\partial \mathcal{L}}{\partial (\partial_\sigma g^{\mu\nu})} &= \Gamma^{\sigma}_{\mu \nu}.
\end{align*}
$$

I guess my main question is, why are we disregarding the $g^{\mu\nu}$ dependency in $\Gamma$? I trace the difference to the second term in the first square bracket: if its sign were different, I would obtain the same answer! Thanks.

(Here's a link to the paper for ease of access ^^: https://en.wikisource.org/wiki/The_Foundation_of_the_Generalised_Theory_of_Relativity#§_15._Hamiltonian_Function_for_the_Gravitation-field._Laws_of_Impulse_and_Energy.)

 

[JimWhoKnew]

Welcome to PF!

I didn't carry out the calculations, only looked at the paper.

Please separate the equations (no multiple equations inside a single begin-end block). This way it will be easier to refer and quote them when needed.

There appears to be a sign inconsistency between equation (48) and the equation above it. I checked it up in the translation by Perret & Jeffery (Dover Publication) and they agree with me (the sign of the second term in "the above" should be +).

I can follow the maths as given in the paper to arrive at this result, but I am confused by the first step. Namely, $\mathcal{L}$ is a function of $(g^{\mu\nu}, \partial_\sigma g^{\mu\nu})$, yet it is differentiated wrt $(g^{\mu\nu}, \Gamma^{\beta}_{\nu\alpha}).$ But isn't $\Gamma$ a function of both $(g^{\mu\nu}, \partial_\sigma g^{\mu\nu})$?

By "the first step", do you mean the second equation below (47a)? It is not a differentiation, only a usual way to express the difference. If you read this paper, you are expected to be familiar with it. It is the chain rule of differentiation, and should be interpreted as$$\delta\Gamma=\frac{\partial\Gamma}{\partial g}\delta g+\frac{\partial\Gamma}{\partial (\partial g)}\delta \partial g\quad.$$So $\Gamma$ is indeed a function of both $g$ and $~\partial g~$, and Einstein addresses it in the following equation (third below (47a)).

I guess my main question is, why are we disregarding the $g^{\mu\nu}$ dependency in $\Gamma$?

As I said above, we don't.

$$\delta \mathcal{L} = \left[ \Gamma^{\alpha}_{\sigma \beta} \Gamma^{\beta}_{\tau \alpha} + 2g^{\sigma\tau}\frac{\partial \Gamma^\alpha_{\sigma\beta}}{\partial g^{\mu\nu}} \right]\delta(g^{\mu\nu}) + 2\left[g^{\sigma\tau}\Gamma^{\alpha}_{\sigma\beta}\frac{\partial \Gamma^{\beta}_{\tau\alpha}}{\partial(\partial_\rho g^{\mu\nu})} \right]\delta(\partial_\rho g^{\mu\nu})$$

The indices in the first term in the left square brackets are inconsistent. The second term should contain an additional $\Gamma$. I can't tell whether it is a typo or a calculation error.

I trace the difference to the second term in the first square bracket: if its sign were different, I would obtain the same answer!

I suspect you weren't careful enough with the indices.
(But assume you remembered to use (32))
Note that you effectively replace the $~g^{\mu\nu}~$ by $~g^{\sigma\tau}~$, so the negative and positive $~\partial_\rho g_{\alpha\beta}~$ may get "shuffled".

If you want people to find errors in your calculations, you should post them in detail. I don't promise to debug your calculations myself.

 

[Isaek_]

I see, thank you very much for all of this! I will have to go through the maths again.

The indices in the first term in the left square brackets are inconsistent. The second term should contain an additional . I can't tell whether it is a typo or a calculation error.

It is a typo! Sorry! I’m too hardwired into pressing shift+enter whenever I want to confirm an edit, which then posted this in a very incomplete state and I had to scramble to get everything finished…. It should read

$$ \begin{align*} 2g^{\sigma\tau} \Gamma^\alpha_{\sigma\beta}\frac{\partial\Gamma^\beta_{\tau\alpha}}{\partial g^{\mu\nu}} \delta(g^{\mu\nu}). \end{align*} $$

If you want people to find errors in your calculations, you should post them in detail. I don't promise to debug your calculations myself.

Sorry yes this is poor of me. I hadn’t intended to put the final block of equations in initially. I am content now in going back and scrutinising my work!

Thanks again :)

 

[anuttarasammyak]

TL;DR: Struggling to match the E-L equations from an assumed Lagrangian

I can follow the maths as given in the paper to arrive at this result, but I am confused by the first step. Namely, L is a function of (gμν,∂σgμν), yet it is differentiated wrt (gμν,Γναβ).

Einsteins seems just applied the general rule of variation for product H=ABC, i.e.,
$$\delta H = \delta A\ B\ C+A\ \delta B\ C+A\ B\ \delta C $$