Haorong Wu said:
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) .
