$c^4$ in Einstein field equations

I have read many derivations of Einstein field equations (done one myself), but none of them explain why the constant term should have a $c^4$ in the denominator. the $8{\pi}G$ term can be obtained from Poisson's equation, but how does $c^4$ pop up. Most of the books say that in units where $c$ is not equal to 1, you get $\frac{8{\pi}G}{c^4}$. There is no need or mention of an explicit assumption that $c=1$.

It's simple dimensional analysis. The theory has two fundamental parameters, Newton's constant $G$, which determines the strength of gravitational attraction. it has units $\frac{N\cdot m^{2}}{kg^{2}} = \frac{kg\cdot m}{s^{2}}\left(\frac{m^{2}}{kg^{2}}\right) = \frac{m^{3}}{kg\cdot s^{2}}$. Secondly, you have the speed of light, which tells you how much time you get for how much space, and it obviously has units $m/s$.

Then, you have Einstein's equation. Curvature has units $m^{-2}$ just from the fundamental equations for it, and you have

$$(\mbox{curvature terms}) = (\mbox{const.})(\mbox{stress-energy tensor})$$

What units should the constant have? Well, the stress-energy tensor has units of pressure, by its definition. This translates to:

$$\frac{N}{m^{2}} = \frac{kg\cdot m}{s^{2}\cdot m^{2}} = \frac{kg}{s^{2}\cdot m}$$

Therefore, if our equation is going to make any sense, the constant, assembled only from $G$ and $c$ and a pure number, must be of the form $C\,G^{n}c^{k}$, and it must have units of $\frac{s^{2}}{m\cdot kg}$

We note that only $G$ has a factor of kilograms, so $n$ must be 1.

Putting it all together, we have:

\begin{align}\frac{s^{2}}{m\cdot kg} &= \frac{m^{3}}{kg\cdot s^{2}}\frac{m^{k}}{s^{k}}\\ \frac{s^{4}}{m^{4}} &= \frac{m^{k}}{s^{k}} \end{align}

Thererfore, $k = -4$, and we have Einstein's equation:

$$R_{ab} - \frac{1}{2}Rg_{ab} = C \frac{G}{c^{4}}T_{ab}$$

The value of C cannot be determined from first principles. Comparison with the predictions of Newton's law give us the value $8\pi$, which fixes $G$ to have the same value as the $G$ in Newton's law.

You know that in GR you need a locally Minkioski spacetime. This, in each point of your manifold you can change the coordinates so that the metric is diagonal, and the square of the infinitesimal displacement is $ds^2=\left(ct\right)^2-x^2-y^2-z^2.$ So here is where the $c$ come from.

Then, when you want to compute the coupling constant $k=\frac{8\pi G}{c^4}$, if start you taking into account that there is a $c$ in the metric, you will find the right power of $c$ in $k.$

• OK, I get it @Antonio Ragagnin, But can you not get this from poisson's, because most of the books use it to expand for k in RHS... – GRrocks Jul 13 '14 at 8:49