1
$\begingroup$

Suppose we assume that the gravitational constant is varying with time: $$G=G_0 e^{-\frac{t}{t_0}}$$ where $t$ is the time from the Big Bang, and $G_0, t_0$ being some constants.

Assuming matter-only universe, which one of the expansion equation should we use to obtain the Hubble's parameter $H$?

Equation 1:

$$H^2=\left\{\frac{\dot R}{R}\right\}^2=\frac{8\pi G}{3}\rho_0\frac{R_0^3}{R^3}$$

Equation 2:

$$\frac{\partial (H^2R^2)}{\partial (R^2)}=\frac{\ddot R}{R}= -\frac{4\pi G}{3}\rho_0\frac{R_0^3}{R^3}$$

$\endgroup$
2
  • 1
    $\begingroup$ Making $G$ vary with time isn't as simple as modifying those equations of motion. You'll need to go all the way back to the Einstein-Hilbert action with the $G$ moved inside of the integral, then vary the action to derive new equations of motion. I'm unsure whether you should include varying the action with respect to $G$ to get an equation that determines how $G$ varies or not. $\endgroup$ – Sean Lake Oct 10 '16 at 5:20
  • $\begingroup$ Side question: If G = G(t) instead of G = G(x, y, z, t) as in this question, only one equation would not be the same, is that correct? Since varying the EH action in tensor notation is just doing 4 equations at once. $\endgroup$ – Doug Oct 27 '16 at 22:49

Your Answer

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

Browse other questions tagged or ask your own question.