-5
$\begingroup$

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

$\endgroup$
3
$\begingroup$

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 \begin{equation} \delta S = \delta \int_\Omega \mathcal{L} \sqrt{-g} \ d^4x = 0, \end{equation} 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 \begin{equation} \mathcal{L} = \mathcal{L}_G - 2 \ \kappa \ \mathcal{L}_M. \end{equation} $\mathcal{L}_G = R$ ($R$ is the curvature invariant), that gives us the the Einstein-Hilbert Action. \begin{equation} \delta S = \delta \int_\Omega \left( R - 2 \kappa \mathcal{L}_M \right) \sqrt{-g} \ d^4x = 0 \end{equation} Subtituting the above equation in the action, you get: \begin{equation} R_{\mu \nu} - \frac{1}{2} R g_{\mu \nu} = \kappa T_{\mu \nu}, \end{equation} 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 ($T_{\mu \nu} \approx T_{00} = \rho c^2$).

\begin{equation} g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu}. \end{equation}

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

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

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, $\kappa = 8 \pi G/c^4$: \begin{equation} R_{\mu \nu} - \frac{1}{2} R g_{\mu \nu} = \frac{8 \pi G}{c^4} T_{\mu \nu}. \end{equation}

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.