# Does the Higgs VEV depend on the gravitational potential?

If I take the non-relativistic limit of the Klein-Gordon equation in a curved space-time I will get a contribution from the Newtonian gravitational potential. Shouldn't this gravitational contribution enter the Higgs-potential, so that the vacuum expectation value, strictly speaking, becomes a function of coordinates?

I see from basic estimates that this contribution is normally tiny and negligible. But it may get large close to black hole horizons. Is this correct? Or if not, what's wrong with my expectation?

The equation of motion for the Higgs field in a curved spacetime would be $$\nabla^\mu \nabla_\mu \phi = \frac{\partial V}{\partial \phi},$$ from which it is obvious that $$\phi\equiv\phi_0$$ (where $$\phi_0$$ is the VEV of a flat spacetime) would be a solution. So the answer to the title question would be no. This could be seen as another manifestation of the equivalence principle.
Possibility for the shift in vacuum expectation value of the Higgs field due to gravity exists if the Higgs field if coupled to gravity non-minimally, for example due to quantum vacuum polarization effects, the effective action could have terms proportional to $$\phi^2 R_{\mu\nu\lambda\rho} R^{\mu\nu\lambda\rho}$$. So in principle, in the strong gravity regions (around black holes, neutron stars) masses of elementary particles could be slightly different then in a flat spacetime.