1
$\begingroup$

Question

I was recently wondering about semi-classical gravity :

$$ G_{\mu \nu} = \frac{8 \pi G}{c^4} \langle \hat T_{\mu \nu} \rangle_\psi$$

Does this obey the equivalence principle?

My intuition

Let's say I am in a lift and I want to measure the standard deviation in the acceleration. Then with Heisenberg's equation of motion I can do so ,however , the left hand side $ G_{\mu \nu}$ is a classical object and will predict there is none whereas using the Hamiltonian in quantum mechanics $\hat T_{00}$ will only have zero standard deviation if it is in a acceleration eigenstate. But this view would only work in quantum mechanics and I'm not sure how to argue it for fields. Does this kind of argument still hold in QFT?

$\endgroup$
  • $\begingroup$ Clarifying question to OP. Say EM was coupled to GR through the following term: $\alpha \int d^4 x \sqrt{|g|} R F_{\mu \nu} F^{\mu \nu}$. Would that violate the equivalence principle (as in theory it is possible to distinguish a local gravity field from inertia by fine electromagnetic measurements)? AFAIK the popular point of view is that this term is absent from the Einstein-Maxwell Lagrangian not because of equivalence principle, but simply because the coupling $\alpha$ is irrelevant in the IR. Hence, there’s no real reason for the equivalence principle to play a fundamental role. $\endgroup$ – Prof. Legolasov Jun 1 at 22:34
  • $\begingroup$ @SolenodonParadoxus I'm not sure what you exactly mean by "there’s no real reason for the equivalence principle to play a fundamental role." Do you mean one does not expect "the equivalence principle" to hold in theories of quantum gravity? $\endgroup$ – More Anonymous Jun 4 at 13:31
  • $\begingroup$ Which equivalence principle? Weak, strong, Einstein? $\endgroup$ – Void Jun 10 at 14:57
  • $\begingroup$ To me it is a physical principle in which given Einstein's thought experiment setup: One cannot distinguish if the lift is under the influence of gravity or acceleration using any experiment within the lift $\endgroup$ – More Anonymous Oct 29 at 18:43
3
+50
$\begingroup$

Semiclassical gravity does violate the equivalence principle.

For example, consider the effect of QED in a gravitational field on photon propagation, discovered by Drummond & Hathrell:

  • Drummond, I. T., & Hathrell, S. J. (1980). QED vacuum polarization in a background gravitational field and its effect on the velocity of photons. Physical Review D, 22(2), 343, doi:10.1103/PhysRevD.22.343.

The effect could be understood in qualitative terms as a vacuum polarization from virtual electron–positron pairs $e^-e^+$. These pairs would experience tidal effects from anisotropic curvature tensor with a characteristic length scale, the Compton wavelength of the electron. As a result electromagnetic field would now gain coupling to the curvature tensor. Therefore a photon propagating in a curved spacetime would have a velocity dependent on its direction and polarization and differing from the “normal” speed of light $c$. We now have gravitational birefringence and the possibility of superluminal photons (such FTL photons would not violate causality).

Of course, such effects in a real astrophysical situations would be immeasurably tiny. For example, in the gravitational field of a Schwarzschild black hole the characteristic relative difference between velocities of different polarizations $$ \epsilon \approx \frac{\alpha}{30 \pi } \frac{r_s ƛ^2 }{r^3} , $$ where $r_s$ is the Schwarzschild radius of a black hole, $ƛ$ is the (reduced) Compton wavelength of the electron, $\alpha$ is the fine structure constant. For a black hole of 1 solar mass this is about $10^{-36}$ near the horizon.

$\endgroup$
0
$\begingroup$

So it seems like your question and intuition are asking two separate questions. Your intuition is asking about the way measurements are made in the classical world and in the quantum world. Semi-classical theories are known to be incomplete and so some parts of the physical process won't actually stand up, ie matching standard deviations.

That being said, we know GR is the most accurate field theory of gravity and most theories of gravity retain the equivalence principle in some form or another. In your question the equivalence principle is buried deep within the $G_{\mu\nu}$ and it's derivation. So with gravity, you will likely always use the equivalence principle it will just be disguised within the mathematics of the problem. But it is always there.

$\endgroup$
  • $\begingroup$ I'm not sure what you mean as in: is the answer negative or affirmative? As you have statements that suggest both? (Maybe I'm misunderstanding) ... "Semi-classical theories are known to be incomplete and so some parts of the physical process won't actually stand up, ie matching standard deviations." => No does not obey ... "In your question the equivalence principle is buried deep within the $G_{\mu \nu}$ and it's derivation. " => Yes it will obey $\endgroup$ – More Anonymous Jun 4 at 13:39
  • $\begingroup$ Maybe we differ in how we think of the equivalence principle. To me it is a physical principle in which given Einstein's thought experiment setup: One cannot distinguish if the lift is under the influence of gravity or acceleration using any experiment within the lift . $\endgroup$ – More Anonymous Jun 4 at 13:44

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.