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}.