Do gravitons interact with photons?

[johne1618]

Does string theory predict that gravitons interact with photons?

Can a particle traveling linearly at the speed of light curve space-time around it?

I guess a photon could instantaneously have rest mass it if were energetic enough to convert to an electron-positron pair and back again.

John

 

[Kevin_Axion]

I thought gravitons interact with the space-time causing curvature. Thus the concentration of energy and mass in a certain volume indicates the amount of curvature and also the amount of gravitons interacting with space-time. In this sense photons interact with the space-time as all things follow geodesics in curved surfaces.

 

[crackjack]

Photons are described by massless open-string state, while gravitons by massless closed-string states. So, the amplitude could be calculated by inserting a graviton vertices in the bulk of a disk and photon vertices on its boundary.

 

[chrispb]

You can write down a (nonrenormalizable) EFT with gravity and photons. Gravity does indeed couple to photons; the term in the action would look like int d^4x sqrt(-g)(-1/4) F^2.

As far as the one-loop diagram that looks like photon -> e+ e- -> photon goes, computing the correction to the self-energy results in no shift in the photon mass, but an infinite wavefunction renormalization. The question of the infinity and wavefunction renormalization is a separate one; let's focus on the mass. We discover that the correction to the mass is actually zero, as previously stated. This is good because mass terms are not gauge-invariant. So, the unbroken local gauge symmetry protects the photon's mass. Doing loops involving gravitons in the EFT stated above should result in the same conclusion, despite the math being significantly more awful. Hope this helps.

 

[CHIKO-2010]

Does string theory predict that gravitons interact with photons?

Yes it does! the low energy limit of string theory reproduces GR + electrodynamics + other stuff.

Can a particle traveling linearly at the speed of light curve space-time around it?

I guess a photon could instantaneously have rest mass it if were energetic enough to convert to an electron-positron pair and back again.

If you are asking whether photons couple to gravity in GR then answer is again yeas! The reason is that gravitons are spin-2 particles and they can directly couple to spin-1 photons without violating neither gauge invariance (charge conservation), nor diffeomorphism invariance (recall, in GR the energy-momentum tensor couples to the graviton, not just a mass).

Would the graviton be described by spin-0 particle, there would not be coupling of gravitons with photons (at least in the classical approximation), because the spin-0 particles can interact with the trace of energy-momentum tensor, which is 0 for the electromagnetic field. This crucial difference has been verified experimentally by Sir Arthur Stanley Eddington in 1919 (bending of starlight by Sun's gravitational field)

 

[QuantumClue]

You can write down a (nonrenormalizable) EFT with gravity and photons. Gravity does indeed couple to photons; the term in the action would look like int d^4x sqrt(-g)(-1/4) F^2.

As far as the one-loop diagram that looks like photon -> e+ e- -> photon goes, computing the correction to the self-energy results in no shift in the photon mass, but an infinite wavefunction renormalization. The question of the infinity and wavefunction renormalization is a separate one; let's focus on the mass. We discover that the correction to the mass is actually zero, as previously stated. This is good because mass terms are not gauge-invariant. So, the unbroken local gauge symmetry protects the photon's mass. Doing loops involving gravitons in the EFT stated above should result in the same conclusion, despite the math being significantly more awful. Hope this helps.

I just want to rewrite this in terms of latex. I hate reading non-latexed stuff. I find myself leaving many pages because it is riddled with all the latex talk, but no latex to show :)

\int \sqrt{-g} -\frac{1}{4} F_{\mu \nu} F^{\mu \nu} d^4 x

 

[CHIKO-2010]

\int \sqrt{-g} -\frac{1}{4} F_{\mu \nu} F^{\mu \nu} d^4 x

In more details the above equation is:

\int \sqrt{-g} -\frac{1}{4}g^{\mu \rho}g^{\nu \sigma} F_{\mu \nu} F_{\rho \sigma} d^4 x