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Due to the mass-energy equivalence, both matter and EM radiation bend spacetime, and both are capable of forming singularities (black hole, white hole/kugelblitz). In light of this, why do photons traveling from the most distant reaches of the observable universe not lose energy due to the gravitational radiation they must emit? Furthermore, mustn't cosmic rays (e.g. protons) slow down or stop as they lose energy through the same mechanism?

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Photons or cosmic rays don't (normally) emit gravity waves.

Consider the comparison with radio waves. A moving electron doesn't emit radio waves. It has to be accelerating to emit EM radiation. Specifically radio waves are only emitted when there is a changing dipole moment.

So you wouldn't expect a particle moving at constant velocity (photon or otherwise) to emit gravity waves, and in fact unlike EM even an oscillating gravitational dipole won't emit gravity waves. For gravity wave emission you need an oscillating quadrapole moment. In principle a photon whose trajectory is bent by a gravitational potential could emit gravity waves, but in practice the intensity of the radiation emitted would be so small that you could never measure the energy loss.

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In light of this, why do photons traveling from the most distant reaches of the observable universe not lose energy due to the gravitational radiation they must emit?

There is a misconception here in "gravitational radiation they must emit" . There does not yet exist a unified theory of elementary particles and the three interactions well described by the standard model of physics to tell us if photons, which are elementary particles, radiate gravitons ( the putative elementary particle of quantized gravity) when they lose energy in an interaction with a gravitational field.

Edit after comments:

If one follows an effective field theory formulation and accepts the results of calculations as if a consistent unified model existed, one is talking of a bremsstrahlung type Feynman diagrams where the photon would interact with a virtual graviton from a gravitational source and emit a graviton.

Feynman diagrams are a shorthand for the calculations needed to predict the probability of the occurrence of an interaction. In these copied Feynman graphs

enter image description here

the solid red line would be the gravitational source, the squiggly the gravitons, and the blue the photon. The gravitational coupling is so weak in comparison with the other couplings, and it is needed twice for photon or proton gravitational bremsstrahlung that any effects for the high energy cosmic rays reaching us would be tiny, not measurable. (Thanks for the link to @MattReece , the conclusion is quite clear)

Furthermore, mustn't cosmic rays (e.g. protons) slow down or stop as they lose energy through the same mechanism?

An analogous diagram holds true for the corresponding calculation for protons and other massive cosmic rays.The interaction is so weak for individual particles that it cannot effect their trajectories measurably, let alone stop them.

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It's not true that we lack of a theory of how photons radiate gravitons. It's described by general relativity. –  Matt Reece Feb 17 at 15:24
    
@MattReece Gneral Relativity is not a quantum theory at unified level including photons etc. Photons as elementary particles are put in by hand if one hand waves a quantization of gravity.An interaction of photons and gravitons requires a unified consistent theory to be valid, a vertex in a Feynman type diagram, imo –  anna v Feb 17 at 18:23
    
No, it doesn't. General relativity coupled to the Standard Model is a perfectly good effective quantum field theory. The corrections we expect a complete theory of quantum gravity to compute would be tiny for questions like the radiation of gravitons from photons. –  Matt Reece Feb 17 at 18:37
    
@MattReece Can you give a readable link for these claims. ? –  anna v Feb 17 at 19:09
    
Maybe Cliff Burgess's review: relativity.livingreviews.org/Articles/lrr-2004-5 –  Matt Reece Feb 17 at 22:47
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