# How do you explain stretched metrics in the graviton picture?

Say you had a metric of flat space in a coordinate system where we multiplied the $$x$$-coordinate by 10:

$$g^{\mu\nu}(x,y,z,t) = \begin{bmatrix} 10 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \end{bmatrix}$$

A beam of light would travel 10 units in the $$x$$-axis for every 1 unit in the $$t$$-axis. But now, if we look at this in the graviton picture where $$g=\eta+h$$. Then somehow in the background space, this corresponds to a light ray interacting with gravitons on a Minkowski background and going 10 times the speed of light. Now, no doubt the two pictures are equivalent so it would not appear like the beam was travelling at this speed since all measuring rods would be stretched also by graviton interactions.

But... doesn't this contradict quantum field theory where no signal can travel faster than the speed of light?

This seems to me like a simple argument for why quantum gravity cannot be a simple quantum field theory of spin-2 particles on flat space. Yet I have not seen this argument before. Therefor, is this argument correct or if not, why is it wrong?

Or does a quantum field theory with spin-2 fields allow this seemingly faster than light signals (compared to the flat space-time)?

I would assume that the error lies in the expansion $$g=\eta+h$$ being only valid for small $$h$$. But maybe a similar argument could be made where $$g$$ is a flat metric but in polar coordinates.

• Your error is to assume that this configuration corresponds to a graviton. Not all configurations of the gravitational field correspond to wave solutions. Commented Mar 18, 2020 at 16:56
• @Raskolnikov Well all electronmagnetic fields corresponds to collections of photons. The idea is the same is true for gravitons. Otherwise there is no quantum theory of gravity in terms of gravitons.
– user84158
Commented Mar 18, 2020 at 19:27
• Not all states of the electromagnetic field correspond to waves. Think of the electric field of a static charge. No waves there. The photons here are only virtual. And virtual photons do not have to obey the law about speed. Commented Mar 18, 2020 at 19:40
• OK, so the gravitional field here corresponds to virtual gravitons. But how does that address the real photon moving at 10 times the speed of light?
– user84158
Commented Mar 18, 2020 at 20:13

A beam of light would travel 10 units in the $$x$$-axis for every 1 unit in the $$t$$-axis.
One such example is a Drummond–Hathrell effect (see e.g. this paper for a discussion) where the QED vacuum polarization in a background gravitational field leads to an effective action for EM field violating the strong equivalence principle (SEC), which in turn modify the lightcones and also lead to the propagation velocity depending on polarization (birefringence): $$k^2−\frac{2b}{m^2}R_{μλ}k^μk^λ+\frac{8c}{m^2}R_{μνλρ}k^μk^λa^νa^ρ= 0,$$ where $$k^\mu$$ is a photon wave vector, $$b$$, $$c$$ are constants, $$m$$ is an electron mass and $$a^\mu$$ is photon polarization vector. We see that the effect depends on Ricci and Riemann curvature tensors (and thus would be absent in the metric suggested in OP). Superluminal propagation can occur in physically interesting situations such as near rotating (Kerr) black holes, but even when it happens it does not lead to a causality violations.