In perusing the linearized Einstein equation, it appears that even a classical electromagnetic plane wave would always have to be associated with a tensor perturbation to the background spacetime. For a wave in the "z" direction say:

$$\partial_{\alpha}\partial^{\alpha}h_{zz}=kT_{zz} $$

where the plane em wave has the $T_{zz}$ as the only non-zero component of the stress energy tensor and $h_{\mu\nu}$ is the perturbation to the background metric.

From a qualitative view, they could never separate as the wave would always generate such a perturbation about it as it traverses the vacuum. To be consistent effects of an expanding universe would have to "redshift" both waves precisely the same.

Is this the case? And how come I never see it mentioned? It seems strange or rather fundamental that electromagnetic propagation would always have to be associated with this.

  • 3
    $\begingroup$ For this one I would go with... "In theory there is no difference between theory and practice. In practice there is." -- YogiBerra. Yes, there should be some coupling... but it's completely unmeasurable. $\endgroup$
    – CuriousOne
    Commented Apr 7, 2016 at 4:34
  • 1
    $\begingroup$ sciencedirect.com/science/article/pii/0550321385905255 $\endgroup$ Commented Apr 7, 2016 at 4:37
  • $\begingroup$ The short answer is yes. If you have any nonzero T, you shouldn't really be working in flat space. That distinction between infinitesimal and 0 is where analysis lies. $\endgroup$
    – AHusain
    Commented Jul 30, 2016 at 8:00
  • $\begingroup$ @AHusain I suppose, though not explicitly stated, I was wondering if it must also go the other way, that is: Shouldn't a gravitational wave traversing the vacuum always be associated with a nonzero stress tensor ( the EM one being the only one that makes sense? $\endgroup$
    – R. Rankin
    Commented Jul 30, 2016 at 8:18
  • $\begingroup$ @R.Rankin I suspect that the fact that gravitational waves arise only from quadrupole and higher moments plays a role in suppressing that coupling. $\endgroup$ Commented Jul 3, 2017 at 8:43

2 Answers 2


The same is true for anything else moving in vacuum: it perturbs the local spacetime. Everything carries about with it such a perturbation to spacetime. And similar things arise in purely electromagnetic effects: each charged object perturbs the electromagnetic field where it is.

Contributions to the total dynamics arising from this kind of effect are generally gathered under the name "self-force". The name refers to a contribution to the net dynamics that arises from the changes created by the object under discussion. It is a second-order effect and therefore usually small. In the case of electromagnetic waves and gravity it is especially small because gravity is weak.

Now the question arises, is a beam of light really in free fall (and therefore following a null geodesic) or isn't it? The answer is that a weak beam of light is in free fall, but a sufficiently intense beam of light is not. However the gravitational effect of a beam of light on itself is such as to produce no net focusing or defocusing at first order in the metric perturbation---which is already quite an interesting observation.

Finally, the redshift associated with cosmic expansion is first and foremost a freefall effect, so it can be calculated as usual for a weak beam of light. For sufficiently intense beams some sort of correction would come in, I suspect, but I have not seen it calculated. Perhaps people working on gamma ray bursts have found that they need to carry out such a calculation.


I will tackle this and please, somebody with better understanding should correct me.

To start with, the title talks of "a photon traversing the vacuum" and the content is only about classical electromagnetic waves and gravitational waves.

In the particle framework, assuming an effective quantization of gravity, one should be considering "photon graviton" interactions. In flat space the photon will be running straight without interacting . At the particle level, if there exists a curvature, the question becomes " is there a photon graviton interaction" in curved space? From what I see in searching there does not exist a definitive proposal even for effective quantized gravity. There are people studying this (example ), where compton like scatterings are introduced, but in no way can it be considered a standard. Certainly due to the weakness of the gravitational coupling this will be a very very small effect except near horizons of black holes.

In the particle framework the answer is that in curved space there should be gravitons which will be exchanged with the photons , but they will be off mass shell virtual gravitons. If there is a compton type interaction with a graviton of the curvature then one graviton will go off and the photon will lose energy.

A classical electromagnetic wave is built up from zillions of photons, and it is supposed that also the gravitational wave is composed by zillions of gravitons. The difference in the couplings is of order 10^-37. So for each 10^37 photons there may be a graviton produced to add up in a complicated manner to a gravitational wave derived from the photon beam, in curved space.

The above leads me to state that only in curved space a photon beam might generate a correlated gravitational beam .

Let us forget photons and gravitons and take the Maxwell's equations in curved space time. Yes , there will be a stress energy tensor associated with a beam of light, but as is true generally, both for electromagnetism and general relativity, waves are generated only through accelerations. There is a review talk here.. A plane electromagnetic wave does not undergo acceleration in flat space just because it has a stress energy tensor.

  • $\begingroup$ Assuming that an electromagnetic wave is produced by an accelerating electron, then you also have an accelerating mass (the electron) and an (undetectable) gravitational wave. $\endgroup$
    – R.W. Bird
    Commented Jul 16, 2020 at 18:24
  • $\begingroup$ @R.W.Bird classical assumption. electrons are quantum mechanical entities and each individual electron does not radiate classical light, but individual photons. It will also radiate individual gravitons ( if they exist) with very small probability as stated in the answer. $\endgroup$
    – anna v
    Commented Jul 16, 2020 at 18:39

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