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 $k.$

  • $\begingroup$ 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... $\endgroup$ – GRrocks Jul 13 '14 at 8:49

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.