# Whats with the 8 in Einstein's field equation in mathematical form?

So, I've been trying to learn a lot of physics and teach myself it. But recently I came across a problem. When I read through Einstein's field equation in mathematical form, I noticed there's an $8\pi$. I thought about it for a while and came up with a solution. Its because spacetime acts in 4 dimensions, we multiply $4 \cdot 2$ and get $8$. Which means $8 \pi$ is going to be pretty much $2 \pi$. But in 4 dimensions. Not sure if that makes sense, at all. But its a good thought (at least in my opinion). Can someone tell me if my thought was correct? And if it wasn't, can someone tell me what the $8 \pi$ means?

Here is the formula.

$$\large {} R_{\mu\nu}-\dfrac{1}{2}g_{\mu\nu}R+g_{\mu\nu}\Lambda=\dfrac{8\pi G}{c^4}T_{\mu\nu}$$

Thanks so much! All helps appreciated!

By the way, please explain the reason for the downvotes so I can ask a better question later.

Nevermind, thanks so much guys! I guess my thought was mostly right. Thank you so much for that comment

• – AccidentalFourierTransform Jan 17 '17 at 14:44
• The DV are because this point is covered in textbooks , although I did not d/v myself – user140606 Jan 17 '17 at 15:01

The $8\pi$ appears when you take the classical limit of the field equations.

Let's suppose we have derived the field equations through the variational principle $$\delta S = \delta \int_\Omega \mathcal{L} \sqrt{-g} \ d^4x = 0,$$ we can separate the lagrangian into a term that corresponds to matter-energy ($\mathcal{L}_M$) and another that corresponds to the gravitational field ($\mathcal{L}_G$), so that we write $$\mathcal{L} = \mathcal{L}_G - 2 \ \kappa \ \mathcal{L}_M.$$ $\mathcal{L}_G = R$ ($R$ is the curvature invariant), that gives us the the Einstein-Hilbert Action. $$\delta S = \delta \int_\Omega \left( R - 2 \kappa \mathcal{L}_M \right) \sqrt{-g} \ d^4x = 0$$ Subtituting the above equation in the action, you get: $$R_{\mu \nu} - \frac{1}{2} R g_{\mu \nu} = \kappa T_{\mu \nu},$$ where $T_{\mu \nu}$ is the Stress-Energy tensor, and depends on $\mathcal{L}_M$.

Since Newtonian mechanics is very well tested, it's expected that it's a limit case of this equation, so we assume a weak field approximation: the Minkowski metric ($\eta_{\mu \nu}$) is under a small perturbation ($h_{\mu \nu}$ and the the mass is perfect fuild ($T_{\mu \nu} \approx T_{00} = \rho c^2$).

$$g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu}.$$

If you work with this, you arrive at: $$\frac{\partial^2 h_{00}}{\partial x^k \partial x^k} =\kappa \rho c^2$$

Comparing with the Newtonian equation for the potencial (Poisson's equation): $$\frac{\partial ^2\Phi}{\partial x^k \partial x^k} = 4 \pi G \rho$$

Then, we can assume that $h_{00} = \frac{2}{c^2} \Phi$ (The $1/c^2$ term is necessary to adjust the dimensions) and, therefore, $\kappa = 8 \pi G/c^4$: $$R_{\mu \nu} - \frac{1}{2} R g_{\mu \nu} = \frac{8 \pi G}{c^4} T_{\mu \nu}.$$

Nothing to do with dimensions.

(Since you said you are studying GR, I am assuming you kind of know topics like curvature invariant and variational principle).