Why Is the Einstein-Hilbert Action Formulated with L Proportional to R?

[latentcorpse]

My notes read:

For the gravitational field, we seek an action of the form
$S[g]= \int_M d^4x \sqrt{-g} L$
where $L$ is a scalar constructed from the metric. An obvious choice for the Lagrangian is $L \propto R$. This gives the Einstein-Hilbert action
$S_{EH}[g]=\frac{1}{16 \pi} \int_M d^4x \sqrt{-g} R$

Why is it an obvious choice to pick $L \propto R$. This is definitely NOT obvious to me!Secondly, if you look at teh notes attached in this thread:
https://www.physicsforums.com/showthread.php?t=457123
On page 107,
where does equation (352) come from? Why is $\Delta^{\mu \nu}=gg^{\mu \nu}$?
And given eqn (353), how do we get (354)? Did we just det $g \rightarrow -g$? Where did the $\frac{1}{2}$ come from?

Thanks.

 

[dextercioby]

I address the last part of your post first.

1. The formula $\Delta^{\mu\nu} = g\cdot g^{\mu\nu}$ is just a fancy notation one uses to describe a thing you should definitely have learned in HS: How one computes the inverse of a square (in GR 4by4) matrix. The element $\mu\nu$ of the inverse is equal to the ratio between the determinant of the cofactor matrix for the element $\mu\nu$ and the determinant of the matrix you wish to find its inverse.

2. The g is indeed set to -g, because of the convention that the metric has a negative determinant (the free field limit of g is the Minkowski metric which has the determinant of "-1").

3. The 1/2 comes from differentiating the sqrt of the determinant of g by the elementary rules of differentiation.

4. There's a thread in PF linked to in one of my blog articles in which Samalkhaiat gives an excellent explanation as to why $\mathcal{L} = R$ for GR. Read that first and then I can send you a PM with more literature on the various attempts to derive the HE action.

 

[latentcorpse]

I address the last part of your post first.

1. The formula $\Delta^{\mu\nu} = g\cdot g^{\mu\nu}$ is just a fancy notation one uses to describe a thing you should definitely have learned in HS: How one computes the inverse of a square (in GR 4by4) matrix. The element $\mu\nu$ of the inverse is equal to the ratio between the determinant of the cofactor matrix for the element $\mu\nu$ and the determinant of the matrix you wish to find its inverse.

2. The g is indeed set to -g, because of the convention that the metric has a negative determinant (the free field limit of g is the Minkowski metric which has the determinant of "-1").

3. The 1/2 comes from differentiating the sqrt of the determinant of g by the elementary rules of differentiation.

4. There's a thread in PF linked to in one of my blog articles in which Samalkhaiat gives an excellent explanation as to why $\mathcal{L} = R$ for GR. Read that first and then I can send you a PM with more literature on the various attempts to derive the HE action.

Thanks. Still struggling with the 1/2 factor though!

Using (353),

$\delta \sqrt{-g} = \frac{\partial \sqrt{-g}}{\partial g_{\mu \nu}} \delta g_{\mu \nu}$
But if I differentiate that I get a 1/2 but I also get a $-(-g)^{-\frac{1}{2}}$ and I just get massively confused.
Initially my attempt had just been to change all the $g$'s to $\sqrt{-g}$'s in (353) but then I was missing the 1/2.

 

[dextercioby]

You can't blindly change that, because $|g|=\sqrt{|g|}\cdot\sqrt{|g|}$.

 

[latentcorpse]

You can't blindly change that, because $|g|=\sqrt{|g|}\cdot\sqrt{|g|}$.

Ok. Then I really don't get where the 1/2 is from then...

 

[dextercioby]

Well, the notes and my reply above point you in the right direction:

$$\frac{\partial \sqrt{-g}}{\partial g_{\mu \nu}} = \frac{\partial \left(-g\right)^{\frac{1}{2}}}{\partial g_{\mu \nu}} = \frac{1}{2}\frac{1}{\left(-g\right)^{\frac{1}{2}}} \frac{\partial \left(-g\right)}{\partial g_{\mu\nu}} = ...$$

 

[inline]

Why is it an obvious choice to pick $L \propto R$. This is definitely NOT obvious to me!

Because it is the simplest choice for scalar constructed from $g_{\mu\nu}$ and $\Gamma^\mu_{\nu\lambda}$.