Normalization Constant in Einstein-Hilbert Action

[Haorong Wu]

TL;DR

Why the normalization constant in Einstein-Hilbert action is chosen to be $1/16\pi G$?

Hello, there. Looking at the Einstein-Hilbert action $$S=\frac 1 {16\pi G}\int R \sqrt{-g}d^4 x,$$ I am wondering why the normalization constant is $1/16\pi G$. In the textbook by Carroll, he mentions that the action is so normalized to get the right answer. I think this is related to Einstein field equation, i.e., $$R_{\mu\nu}-\frac 1 2 R g_{\mu \nu} =8\pi G T_{\mu\nu}.$$

Particularly, if I write a matter action in a 3-dimensional space, such as $$S_m=\int d^3x \mathcal L(x^1,x^2,x^3, \dot x^1,\dot x^2,\dot x^3 )$$ where the time coordinate has been eliminated. Should the Einstein-Hilbert action reduce to 3-dimension as well? Will the normalization constant change?

I looked in Lectures in (2+1)-Dimensional Gravity. In it, Eq. (2.27) tells me that the constant is unaltered (the units are chosen to let $16\pi G=1$).

So could I infer that the constant is independent of the dimension of space?

 

[vanhees71]

The overall normalization is chosen such that the gravitational constant is defined as by Newton, i.e., such that the Newtonian limit of the field equations comes out right,
$$\Delta \Phi=4 \pi G \rho,$$
where $\Phi$ is the Newtonian gravitational potential. Indeed this leads to the solution for a point mass ($\rho=m \delta(\vec{x})$),
$$\Phi(\vec{x})=-\frac{G m}{|\vec{x}|}$$
as it should be.

 

[HomogenousCow]

You have to include a "matter" term in the lagrangian to see how the normalization comes out, without one the overall normalization is irrelevant. i.e. $$S= \int (\frac{1}{16\pi G}R + \mathcal {L}_M) \sqrt{-g}d^4 x.$$

 

[PeterDonis]

You have to include a "matter" term in the lagrangian to see how the normalization comes out, without one the overall normalization is irrelevant

This is not correct. Even in a vacuum solution (no matter present), matching the known Newtonian formula for the gravitational potential in the appropriate limit requires a specific normalization in the Lagrangian. While this normalization can in part be viewed as a choice of units, that is also true when matter is present. But in any case the normalization is only in part a choice of units; the fact that there is a coupling constant present on the curvature term that is not dimensionless is independent of any unit choice, and also is true whether or not matter is present.

 

[vanhees71]

But an overall factor in front of the action doesn't change the equations of motion. So that's $1/(16 \pi G)$ in front of the contribution with the Ricci tensor is determined by the coupling of the gravitational field to the sources, i.e., the energy-momentum-stress tensor of the "matter", which is determined by the variation of the matter Lagrangian wrt. $g_{\mu \nu}$, and then the relative factor between the matter Lagrangian and the gravitational-field Lagrangian $\propto R$ is fixed by defining $G$ to be Newton's gravitational constant, which describes the coupling strength between the matter energy-momentum tensor (source) and the gravitational field.

 

[PeterDonis]

an overall factor in front of the action doesn't change the equations of motion.

Without a factor with appropriate units, the $R$ term has the wrong units. So, as I said, there needs to be a coupling constant in front of the $R$ term. The exact numerical value of that coupling constant will depend on your choice of units, but you can't just eliminate it, even if there is no matter present.

 

[vanhees71]

Well, yes, if you insist that the action has the usual dimension then that's of course right. The physics is however at the end in the equations of motion, and the coupling constants and masses are then to be determined empirically (they are "manifestations of our ignorance" as my QFT&Standard Model professor used to say :-)).

 

[PeterDonis]

if you insist that the action has the usual dimension

If I want the equation of motion, which is derived by varying the action, to have the correct dimensions, then the action has to have the correct dimensions. I can't just stick things into the equation of motion that don't come from the action.

 

[vanhees71]

If I multiply $\mathcal{L}$ with an arbitrary non-zero constant, you don't change the equations of motion at all, because they are defined by the stationarity of the action, but I don't think that this is a very important point anyway.

 

[PeterDonis]

If I multiply $\mathcal{L}$ with an arbitrary non-zero constant, you don't change the equations of motion at all

As long as the constant has the right units. If it doesn't, you've changed the units of the equations of motion as well, which should not happen.

 

[samalkhaiat]

Summary:: Why the normalization constant in Einstein-Hilbert action is chosen to be $1/16\pi G$?

Hello, there. Looking at the Einstein-Hilbert action $$S=\frac 1 {16\pi G}\int R \sqrt{-g}d^4 x,$$ I am wondering why the normalization constant is $1/16\pi G$. In the textbook by Carroll, he mentions that the action is so normalized to get the right answer. I think this is related to Einstein field equation, i.e., $$R_{\mu\nu}-\frac 1 2 R g_{\mu \nu} =8\pi G T_{\mu\nu}.$$

Particularly, if I write a matter action in a 3-dimensional space, such as $$S_m=\int d^3x \mathcal L(x^1,x^2,x^3, \dot x^1,\dot x^2,\dot x^3 )$$ where the time coordinate has been eliminated. Should the Einstein-Hilbert action reduce to 3-dimension as well? Will the normalization constant change?

I looked in Lectures in (2+1)-Dimensional Gravity. In it, Eq. (2.27) tells me that the constant is unaltered (the units are chosen to let $16\pi G=1$).

So could I infer that the constant is independent of the dimension of space?

When the dimension of the integrand is not that of energy density (i.e., $M (\frac{L}{T})^{2}L^{-3}$), one needs to multiply by a dimension-full constant in order to get the correct dimension for the action (i.e., $M(\frac{L}{T})^{2}T$ ). So, if one gives you $$S = - \frac{1}{2 \beta} \int d^{4}x \ \sqrt{-g} R , \ \ \ \ \ \ \ \ \ (1)$$ you work out the dimension of the constant $\beta$ as follows: substituting for the action the dimension $M(\frac{L}{T})^{2}T$, $[R] = L^{-2}$ and for $[d^{4}x] = [dt d^{3}\mathbf{x}] = TL^{3}$, we find $$[\beta] = (\frac{L}{T})^{-4} (M^{-1}T^{-2}L^{3}) = [\mbox{speed}]^{-4} [G_{N}] .$$ So, in order for the action (1) to describe a relativistic gravity, we must have $$\beta = a \ \frac{G_{N}}{c^{4}}, \ \ \ \ \ \ \ \ (2)$$ for some (dimension-less) number $a$. Below, we will show that $a = 8 \pi$.

If $\mathcal{L}$ is the Lagrangian density of matter (the source of “gravity”), then the (gravitational) field equation can be obtained by varying the full action with respect to $g^{\mu\nu}$ : $$\delta S = \int d^{4}x \ \sqrt{-g} \left( - \frac{1}{2 \beta} G_{\mu\nu} + \frac{1}{2} T_{\mu\nu} \right) \delta g^{\mu\nu} ,$$ where $$G_{\mu\nu} = R_{\mu\nu} - \frac{1}{2}g_{\mu\nu}R,$$$$T_{\mu\nu} = \frac{2}{\sqrt{-g}} \frac{\delta}{\delta g^{\mu\nu}} (\sqrt{-g}\mathcal{L}) .$$ Thus $\delta S = 0$ leads to $$R_{\mu\nu} - \frac{1}{2}g_{\mu\nu}R = \beta T_{\mu\nu} .$$ From this, it follows that $$R^{\mu}_{\nu} = \beta \left( T^{\mu}_{\nu} - \frac{1}{2}\delta^{\mu}_{\nu}T \right), \ \ \ \ \ (3)$$ where $$T = g^{\mu\nu}T_{\mu\nu} = - \frac{R}{\beta} .$$

Non-relativistic (stationary weak field) approximation: Let us assume that $\partial_{t}g_{\mu\nu} = 0$ and write $g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu}$ with $|h_{\mu\nu}|\ll 1$. If we keep only terms of first order in $h_{\mu\nu}$, then the connection coefficients are $$\Gamma^{\mu}_{\nu\rho} = \frac{1}{2}\eta^{\mu\lambda}(\partial_{\rho}h_{\lambda \nu} + \partial_{\nu}h_{\lambda \rho} - \partial_{\lambda}h_{\nu\rho}) .$$ Since $\partial_{t}h_{00} = 0$, for $\Gamma^{j}_{00}$, we find $$\Gamma^{j}_{00} = \frac{1}{2}\partial_{j}h_{00}. \ \ \ \ \ \ \ \ \ \ (4)$$

Expanding the proper time element $$c^{2}d\tau^{2} = (1 + h_{00})(dx^{0})^{2} + 2 h_{0j} dx^{0}dx^{j} + (\eta_{ij} + h_{ij})dx^{i}dx^{j} .$$ Now, time reversal invariance requires $h_{0j} = 0$ and, since $$\frac{dx^{i}}{d\tau} \ll c \frac{dt}{d\tau} = \frac{dx^{0}}{d\tau} ,$$ we have approximately $$\frac{dt}{d\tau} = 1 - \frac{1}{2}h_{00}.$$ Using $\partial_{t}h_{00} = 0$, we find (the time component of the geodesic equation) $$\frac{d^{2}t}{d\tau^{2}} = 0.$$ By the same token, the spatial component of the geodesic equation becomes $$\frac{d^{2}x^{i}}{d\tau^{2}} + c^{2}\Gamma^{i}_{00} \left( \frac{dt}{d\tau}\right)^{2} = 0 , \ \ \ \ (5)$$ Since $\frac{dt}{d\tau}$ is approximately a constant, we can replace the differentiation $\frac{d}{d\tau}$ in (5) by $\frac{d}{dt}$. Doing that and using (4), we obtain $$\frac{d^{2}x^{j}}{dt^{2}} + \frac{1}{2}c^{2} \frac{\partial}{\partial x^{j}} h_{00} = 0 .$$ This becomes the equation of motion in the Newtonian theory, if we identify $\frac{c^{2}}{2}h_{00}$ with the gravitational potential $V(\mathbf{x})$ $$V = \frac{1}{2}c^{2}h_{00} . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (6)$$

Next we consider the non-relativistic approximation of the field equations. Here the energy density dominates, so of all the components $T^{\mu}_{\nu}$, we only consider $T^{0}_{0} = c^{2}\rho$. This means that the trace $T$ is also equal to $c^{2}\rho$. Thus, the field equation (3) reduces to $$R^{0}_{0} = \frac{1}{2} \ \beta \ T^{0}_{0}. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (7)$$ To calculate $R^{0}_{0}$ from the general expression for $R_{\mu\nu}$, we first note that terms containing products of $\Gamma$’s are of order $h^{2}$, so we can neglect them. We can also neglect the terms which contain $\partial_{0}h_{\mu\nu} = \frac{1}{c}\partial_{t}h_{\mu\nu}$. If you carry out this approximation, you end up with $$R^{0}_{0} = R_{00} = \partial_{j} \Gamma^{j}_{00} . \ \ \ \ \ \ \ \ (8)$$ Now, if you substitute (4) and (6) in (8), you get $$R^{0}_{0} = \frac{1}{c^{2}} \nabla^{2} V (\mathbf{x}) . \ \ \ \ \ \ \ (9)$$ Thus, the field equation (7) becomes $$\nabla^{2} V = \frac{1}{2}c^{2} \beta \ T^{0}_{0}. \ \ \ \ (10)$$ Comparing (10) with the Newton-Poisson equation $$\nabla^{2} V (\mathbf{x}) = 4 \pi G_{N} \rho (\mathbf{x}),$$ we find that the energy density $T^{0}_{0} = c^{2} \rho$ if and only if $$\beta = 8 \pi \frac{G_{N}}{c^{4}} .$$ And the action now is correctly normalised $$S = \int d^{4}x \ \sqrt{-g} \left( - \frac{c^{4}}{16 \pi G_{N}} \ R + \mathcal{L} \right) .$$