Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

206 submissions , 164 unreviewed
5,103 questions , 2,249 unanswered
5,355 answers , 22,798 comments
1,470 users with positive rep
820 active unimported users
More ...

  Boundary Term in $f(R,T)$ Gravity

+ 1 like - 0 dislike
837 views

In standard $f(R)$ gravity we consider the Lagrangian of the form $L=\frac{1}{16\pi G}f(R)\epsilon$, where $\epsilon$ is the spacetime volume form and similarly, we consider the boundary term to be of the form $l=\frac{1}{8\pi G}f'(R)\epsilon_{\partial \mathcal{M}}$, where $\epsilon_{\partial \mathcal{M}}$ is the spacetime boundary form. Now, in $f(R,T)$ consider the following action
$$S = \int_{\mathcal{M}}L +\int_{\partial \mathcal{M}}l$$
where $L= \frac{1}{16\pi G}f(R,T)\epsilon$ and $l=\frac{1}{8\pi G}f'(R,T)\epsilon_{\partial \mathcal{M}}$, where $f'(R,T)=f_{R}\delta R+f_{T}\frac{\delta\left(g_{\alpha \beta}T^{\alpha \beta}\right)}{\delta g_{\mu \nu}}\delta g_{\mu \nu}$. Upon varying the Lagrangian we would obtain the following form

$$\delta L =\underbrace{\frac{1}{16 \pi G}\left(-R^{\mu\nu}f_{R}+f_{T}\frac{\delta\left(g_{\alpha \beta}T^{\alpha \beta}\right)}{\delta g_{\mu \nu}}\delta g_{\mu \nu} + \frac{1}{2}g^{\mu\nu}f\right)\epsilon\cdot \delta g_{\mu \nu}}_{=E^{\mu\nu}\delta g_{\mu \nu}} + d\Theta,$$

where

$$\theta^{\mu} = \frac{1}{16\pi G}\left(g^{\mu\nu}\nabla^{\nu}g_{\alpha \nu} - g^{\alpha\beta}\nabla^{\mu}\delta g_{\alpha \beta}\right)f_{R},$$

such that $\Theta = \theta\cdot \epsilon$. Now, the variation of the boundary term $l$ is quite messy and considering $\delta f'(R, T) = f_{RR}\delta R +f_{TT}\delta T +f_{RT}\left(\delta R + \delta T \right)$, terms with $\left(\delta g_{\mu\nu}\right)^{2}$ appear which cause problems in trying to fix the pullback of the variation of the metric tensor to the spatial slice after decomposing the boundary term. Computing $\Theta|_{\partial \mathcal{M}}+\delta l$ and imposing Dirichlet's boundary condition, i.e., fixing the pullback of $g_{\mu\nu}$ to $\Gamma$, the stationarity requirement $\left(\Theta +\delta l \right)|_{\Gamma} = dC$, where $C$ is a local $(d-2)$-form on $\Gamma$, is to be fixed. Thus, finally we have the variation of the action to be
$$\delta S = \int_{\mathcal{M}}E^{\mu\nu}\delta g_{\mu\nu} +\int_{\Sigma_{+}\Sigma_{-}}\left(\Theta +\delta l -dC\right), $$


where the following decomposition has been done: $\partial \mathcal{M} = \Gamma\cup\Sigma_{-}\cup\Sigma_{+}$. Firstly, is my boundary term correct, and how am I to fix the squared metric tensor variation term.

asked Aug 9, 2019 in Theoretical Physics by Naveen (85 points) [ no revision ]
retagged Aug 9, 2019 by Naveen

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsOverfl$\varnothing$w
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...