Suppose we have a field $\phi(x)$ and the metric field is $g_{\mu \nu}(x)$. The action is the functional $S[\phi (x) , g_{\mu \nu } (x)] $. We want to do the path integral:

$$\int d[\phi (x)] d[g_{\mu \nu} (x)] e^{iS[\phi(x), g_{\mu \nu} (x)]}$$

The above probably isn't well-defined because we don't have a theory of quantum gravity.

But what if, instead of varying $[g_{\mu \nu} (x)]$ like an independent variable in the path integral, we make it dependent on the Energy tensor of $[\phi(x)]$ according to Einstein's relation between curvature and energy? And then we only vary $[\phi(x)]$ independently. Does this make the integral well-defined? Thanks.

  • $\begingroup$ Then you basically have to insert $i\alpha^{\mu \nu}(G_{\mu \nu}-\kappa T_{\mu \nu})$ inside the exponent, and integrate on $\alpha_{\mu \nu}$. This seems (to me) more complicated than just 'doing' the path integral you are showing. $\endgroup$ Aug 6 at 15:28
  • $\begingroup$ I don't think you can introduce Einstein's equations in the path integral. Note that in path integral formalism, the action is more fundamental while classical equation of motion is derived from it. This classical equation of motion will only contributes partly to the entire transition amplitude. $\endgroup$
    – KP99
  • $\begingroup$ @KP99 I just wanted to experiment with this. I just think it's pretty natural to have the curvature of space consistent with the path, even in the path integral. The path can still go randomly. I just want the curvature to be consistent with it. This is because gravity is very different in nature from the other forces. $\endgroup$
    – Ryder Rude


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