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Given the following action $$S=\frac{1}{16\pi G}\int d^4x \sqrt {-g}(R+aR^2+bR_{\mu\nu}R^{\mu\nu}+cR_{\mu\nu\lambda\sigma}R^{\mu\nu\lambda\sigma}),$$ which is in 4D.

  1. How to we generalise this action into $d$ spacetime dimension? By which I will need the specific form of Scalar curvature, Ricci tensor, and Riemann tensor.

  2. Also I want to know if the proportionality factor $\frac{1}{16\pi G}$ changes?


Edited to add: One way I can think of answering myself is to write the Riemann Tensor in terms of Weyl Tensor as $$R_{\mu\nu\rho\sigma}=C_{\mu\nu\rho\sigma}+\frac{1}{D-2}(g_{\mu\rho}R_{\nu\sigma}+R_{\mu\rho}g_{\nu\sigma}-g_{\nu\rho}R_{\mu\sigma}-R_{\nu\rho}g_{\mu\sigma})-\frac{1}{(D-1)(D-2)}R(g_{\mu\rho}g_{\nu\sigma}-g_{\mu\sigma}g_{\nu\rho})$$ where one can use contraction to derive $R_{\mu\nu}$ and $R$. But is there any simpler way?

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    $\begingroup$ Change the 4 to a d. The only thing specific to four dimensions is the measure. $\endgroup$ – Jerry Schirmer Mar 17 '15 at 13:43
  • $\begingroup$ Yes, but then value for scalar curvatures, Ricci tensor, and Riemann tensor in d dimension? also what would be the change for $\frac{1}{16\pi G}$ $\endgroup$ – user67742 Mar 17 '15 at 13:46
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    $\begingroup$ As a hint, what are the dimensions (units) of: (1) the action [should be spacetime dimension independent], (2) the measure, (3) the curvature, and lastl but definitely not least in this question, (4) $G$? $\endgroup$ – Andrew Mar 17 '15 at 13:54
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    $\begingroup$ The expressions for the curvatures in terms of a generic metric will never change if you change the dimension, $d$. However, they may admit alternate, more convenient equivalent expressions, such as in the case $d=2$. $\endgroup$ – JamalS Mar 17 '15 at 14:08
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    $\begingroup$ In $4$ dimensions $R^2 - 4 R_{\mu \nu} R^{\mu \nu} + R_{\alpha \beta \gamma \delta} R^{\alpha \beta \gamma \delta}$ forms the Gauss-Bonnet invariant and integrates to zero trivially. Your action in $4$ dimensions is equivalent to the one you have written with $c=0$, while in $d > 4$, the dynamics will be different when $c \neq 0$. $\endgroup$ – Arthur Suvorov Mar 18 '15 at 5:14
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In terms of a metric $g_{ab}$, the Riemann curvature tensor is given by,

$$R^a_{bcd} = \partial_c \Gamma^a_{db} - \partial_d \Gamma^a_{cb} + \Gamma^a_{c e} \Gamma^e_{d b} - \Gamma^a_{de} \Gamma^e_{c b}$$

and consequently changing $d$ does not change the formula, though of course the actual numerical measures of curvature may change. Nevertheless, there are cases where there are additional, equivalent expressions, because things may simplify in certain dimensions. For the case $d=2$, we have,

$$ R_{abcd} = \frac{1}{2}R \left(g_{ac}g_{bd} - g_{ad}g_{bc} \right)$$

a convenient expression due to the symmetries of the tensors. It shows that in two dimensions, the fact that $R=0$ is sufficient to imply $R_{abcd} = 0$, i.e. full Riemann flatness. This is not generally the case for an arbitrary dimension, $d$, and is not the case in $d=4$ (where we usually do general relativity).

Hence, your expression does not require any adjustments for general $d$, except $4 \to d$. That being said, if you choose to change from $4$ to say, $2$ dimensions, some terms can be simplified since, e.g.$^\dagger$

$$\frac{1}{4\pi} \int_M d^2x \, \sqrt{g} \, R = \chi(M)$$

is a topological invariant, the Euler characteristic of the manifold. That being said, one should be mindful that changing the dimension $d$ of an action may have important phenomenological and physical implications, since for example renormalizability is dependent on $d$.


$\dagger$ This is a consequence of the Chern-Gauss-Bonnet theorem which in turn can be shown to follow from the Atiyah-Singer index theorem, a more general result. Note it applies iff $M$ is compact.

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  • $\begingroup$ Thanks @JamalS, yes as you mentioned the general form of Riemann tensor is given by your 1st expression. What I am asking is what you did in 2-$d$ but in n-$d$. what would be the form of measure? what would be the form of Chrisoffel Symbolds to begin with and etc... $\endgroup$ – user67742 Mar 17 '15 at 20:47
  • $\begingroup$ @Ilia Everything is the same. The measure in $d$ dimensions is just $d^d x \, \sqrt{|g|}.$ $\endgroup$ – JamalS Mar 17 '15 at 21:02
  • $\begingroup$ Ok, let me get this correctly. Riemann Tensor is defined as $$R_{\rho\sigma\mu\nu}=\frac{1}{2}(\partial_{\mu}\partial_{\sigma}g_{\rho\sigma}-\partial_{\mu}\partial_{\rho}g_{\nu\sigma}-\partial_{\nu}\partial_{\sigma}g_{\rho\mu}+\partial_{\nu}\partial_{\rho}g_{\mu\sigma})$$ is this definition the same in d-dimension? since for instance in d=2 it is simplified. $\endgroup$ – user67742 Mar 17 '15 at 21:13
  • $\begingroup$ @Ilia That's not valid for general $d$ dimensions. $\endgroup$ – JamalS Mar 17 '15 at 21:34
  • $\begingroup$ that's exactly my original question @JamalS what would be the form of it in d-dimension? also what would be the form of Ricci tensor and scalar curvature in d-dimension? $\endgroup$ – user67742 Mar 17 '15 at 21:35

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