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I realize that many questions about deriving quantum gravity have been asked multiple times before on this forum, but it hasn't been asked exactly like I am doing here. I would like to know what specifically I get with this derivation; quantum gravity, quantum mechanics on curved space, something else, nothing? Also, if there are problems with it, what are they exactly --- non-renormalizable, transformations of GR violate the equations?


If I define an action as

$$ \mathcal{S}=\int \bar{\psi} (i\hbar c \gamma^\mu D_\mu - m c^2)\psi-\frac{1}{4 \mu_0} F_{\mu\nu}F^{\mu\nu} $$

Then, this is QED.

What if I gauge the wavefunction with respect to a general linear transformation:

$$ \psi'=g\psi g^{-1} $$

Then, I get the following gauge

$$ D_\mu \psi = \partial_\mu \psi -[iqA_\mu, \psi] $$

but, since the gauge is general linear, the field is:

$$ R_{\mu\nu}=[D_\mu,D_\nu] $$

Consequently, if I write the following action:

$$ \mathcal{S}=\int \bar{\psi} (i\hbar c \gamma^\mu D_\mu - m c^2)\psi-\frac{1}{4} R_{\mu\nu}R^{\mu\nu} $$

Is it quantum gravity. What are the problems with it?

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    $\begingroup$ Why would it be gravity? Classical GR isn't a Yang-Mills theory, why should quantum gravity be one? $\endgroup$
    – ACuriousMind
    Nov 4 '21 at 19:17
  • $\begingroup$ @ACuriousMind What did I get then... a fermion evolving in curved space-times... is it at least that? (and actually correct at describing that?) $\endgroup$
    – Anon21
    Nov 4 '21 at 19:19
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    $\begingroup$ Compared to QED, you just changed the gauge group and nothing else. Why would you expect a particular choice of gauge group to magically turn a special relativistic gauge theory into a theory about curved spacetimes? $\endgroup$
    – ACuriousMind
    Nov 4 '21 at 19:21
  • $\begingroup$ @ACuriousMind The Riemann tensor is in there... unless it's equal to zero somehow, then its doing something. I don't expect anything... I am ASKING what it does?! $\endgroup$
    – Anon21
    Nov 4 '21 at 19:22
  • $\begingroup$ @Anon21 gravity action is linear in the Riemann tensor, yours is quadratic. Your lagrangian doesn't describe gravity, but a certain Yang-Mills theory in flat spacetime. It is also not quantum, I have no idea why you would conclude that it is. $\endgroup$ Nov 4 '21 at 19:24
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There are only two small problems: this approach has nothing to do with gravity, and it is not at all quantum :)

Firt, your action doesn't describe gravity; it describes Yang-Mills theory with the group $GL(n) = U(1) \times SL(n)$. Not gravity.

There's a formulation of gravity in the gauge theory language but it uses a different action:

$$ S[e, A] = \int d^4 x | \det e | e_a^{\mu} a_b^{\nu} F^{ab}_{\mu \nu} $$ with $A$ a $SO(3,1)$ connection, $F$ its curvature tensor, and $e$ the tetrad field that maps the tangent space to a point in spacetime to the abstract space $R^4$ and is invertible by definition.

You can pass to ordinary variables by $$ g_{\mu \nu} (x) = \eta_{a b} e^a_{\mu} (x) e^b_{\nu} (x). $$

To couple to fermions, replace $\partial_{\mu}$ by the covariant derivative that acts on objects in the spinor representation of $SO(3,1)$.

The second problem is that nothing about this is quantum. This is a completely classical theory, as classical as they come. What makes you think this is a quantum theory?

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  • $\begingroup$ Can you elaborate what you mean with your objection that it is not quantum. QED, a well known quantum theory is defined the same way. Do you object that $\mathcal{S}=\int \bar{\psi} (i\hbar c \gamma^\mu D_\mu - m c^2)\psi-\frac{1}{4 \mu_0} F_{\mu\nu}F^{\mu\nu}$ is not quantum, hence cannot be QED? I guess I am confused by the comment. $\endgroup$
    – Anon21
    Nov 4 '21 at 19:35
  • $\begingroup$ Also, does my Lagrangian describe anything of value, such as the movement of fermions in curved space-times... anything like that? $\endgroup$
    – Anon21
    Nov 4 '21 at 19:37
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    $\begingroup$ @Anon21 QED is an extremely nontrivial procedure that involves writing down Feynman diagrams, renormalizing them, and interpreting results as an asymptotic series that approximates matrix elements of quantum operators. The lagrangian is just the starting point of this long, tedious path. While it works out for QED in the end, it famously does not work out for gravity, which is (one of the) actual reasons quantum gravity is hard. You didn't address any of the real problems. $\endgroup$ Nov 4 '21 at 19:43
  • $\begingroup$ Ah ok, so technically we do not know yet if my Lagrangian is or isn't quantum gravity; we will only know so at the end of the process. (but we can probably safely assume it will fail because it famously does for most attempts of this type). Are we on the same page? Sorry this is very confusing for me. $\endgroup$
    – Anon21
    Nov 4 '21 at 19:53
  • $\begingroup$ @Anon21 no worries, I'm happy to help. Yeah, that would be mostly accurate, except for the other part which is that your lagrangian is Yang-Mills, not gravity. (This is actually a very important difference, these are very different forces that behave in different ways, e.g. gravity is universally attractive and Yang-Mills isn't). $\endgroup$ Nov 4 '21 at 20:13

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