#### [SOLVED] Do photons bend spacetime or not?

where Safesphere says in a comment:

Actually, photons themselves don't bend spacetime. Intuitively, this is because photons can't emit gravitons, because, as any massless particles not experiencing time, photons can't decay by emitting anything. The latest theoretical results show that the gravitational field of a photon is not static, but a gravitational wave emanating from the events of the emission and absorption of the photon. Thus the spacetime is bent by the charged particles emitting or absorbing photons, but not by the photons themselves.

Is there experimental evidence that massless particles such as photons attract massive objects?

where John Rennie says:

as far as I know there has been no experimental evidence that light curves spacetime. We know that if GR is correct it must do, and all the experiments we've done have (so far) confirmed the predictions made by GR, so it seems very likely that light does indeed curve spacetime.

Now this cannot be right. One of them says photons do bend spacetime, since they do have stress-energy, but it is hard to measure it since the energy they carry is little compared to astronomical body's stress-energy. So they do bend spacetime, it is just that it is hard to measure it with our currently available devices.

Now the other one says that photons do not bend spacetime at all. It is only the emitting charge (fermion) that bends spacetime.

Which one is right? Do photons bend spacetime themselves because they do have stress-energy or do they not? #### @Cham 2019-06-07 16:26:22

Here's an indirect proof that "photons" do bend spacetime. Consider the Peres metric (I'm using $$c \equiv 1$$ and the $$(1, -1, -1, -1)$$ convention): $$\begin{equation}\tag{1} ds^2 = dt^2 - dx^2 - dy^2 - dz^2 + F(x, y, t - z)(dt - dz)^2, \end{equation}$$ where $$F(x, y, u)$$ is an arbitrary function of three independant variables ($$u = t - z$$). Substitute this metric into Einstein's equation. First: without any stress tensor (and no cosmological constant): $$\begin{equation}\tag{2} G_{\mu \nu} = 0. \end{equation}$$ After some algebra, you then get a constraint on $$F(x, y, u)$$: $$\begin{equation}\tag{3} \frac{\partial^2 F}{\partial x^2} + \frac{\partial^2 F}{\partial y^2} = 0. \end{equation}$$ Thus, $$F$$ must be an harmonic function in $$x$$ and $$y$$. The simplest non-trivial solution (with spacetime curvature) is a linear superposition of quadratic functions (there are two independant polarization states for the gravitational wave): $$\begin{equation} F(x, y, u) = \mathcal{A}(u)(\, x^2 - y^2) + \mathcal{B}(u) \, x \, y, \end{equation}$$ where $$\mathcal{A}(u)$$ and $$\mathcal{B}(u)$$ are abitrary functions of $$u = t - z$$. Metric (1) then describes a planar gravitational wave propagating in vacuum.

Then add a planar monochromatic electromagnetic wave, of energy-momentum $$\begin{equation}\tag{4} T_{\mu \nu} = \Phi(x, y, u) \, k_{\mu} \, k_{\nu}, \end{equation}$$ where $$k^{\mu} = (\omega, 0, 0, \omega)$$ is the wave number and $$\Phi(x, y, u)$$ is arbitrary. Einstein's equation then becomes $$\begin{equation}\tag{5} G_{\mu \nu} = -\, \kappa \, T_{\mu \nu}. \end{equation}$$ Of course $$\kappa \equiv 8 \pi G$$. A lot of algebra gives the following constraint: $$\begin{equation}\tag{6} \frac{\partial^2 F}{\partial x^2} + \frac{\partial^2 F}{\partial y^2} = 2 \kappa \omega^2 \, \Phi. \end{equation}$$ I'm considering a planar monochromatic electromagnetic wave propagating in spacetime, with a circular polarization (this is a classical field which is the closest thing to a "quantum photon" of angular frequency $$\omega$$): $$\begin{equation}\tag{7} A^{\mu}(x, y, u) = \varepsilon_1^{\mu} \, \mathcal{F}(u) + \varepsilon_2^{\mu} \, \mathcal{G}(u), \end{equation}$$ where $$\varepsilon_{1, \, 2}^{\mu}$$ are the space-like polarization four-vectors, orthogonal to $$k^{\mu}$$, and \begin{align}\tag{8} \mathcal{F}(u) &= a_0 \cos{(\omega \, u)}, & \mathcal{G}(u) &= a_0 \sin{(\omega \, u)}. \end{align} The amplitude $$a_0$$ is just a constant. It is easy to verify that (7) and (8) give (4) with $$\Phi(x, y, u) = \text{cste} \propto a_0^2$$. Then (6) can be solved to give a simple non-trivial solution (the Riemann curvature tensor isn't 0): $$\begin{equation}\tag{9} F(x, y, u) = \frac{\kappa \, a_0^2 \, \omega^2}{8 \pi \alpha} \, (\, x^2 + y^2). \end{equation}$$ ($$4 \pi \alpha$$ is the electromagnetic coupling constant that appears in the energy-momentum tensor. It depends on your favorite units for the field amplitude $$a_0$$. I use the fine-structure constant $$\alpha \approx \frac{1}{137}$$). Metric (1) with the function (9) then describes a circularly polarized EM wave (and its associated gravitational wave) propagating in spacetime. The Riemann curvature isn't 0 (its components are constants, in that case, since the wave energy-momentum is homogeneous).

So the non-localisable "photon" do curves spacetime in a non-trivial way. Because of the circular polarization, curvature is homogeneous (but non-isotropic since the wave propagation defines a priviliged orientation). #### @safesphere 2019-06-08 23:48:53

Thank you, your input is helpful. As you have correctly stated, your "proof" is indirect and applies to classical electromagnetism rather than to photons. One example of the difference is given by the Vaidya metric: "It is notable that, the Vaidya field is a pure radiation field rather than electromagnetic fields. The emitted particles or energy-matter flows have zero rest mass and thus are generally called "null dusts", typically such as photons and neutrinos, but cannot be electromagnetic waves because the Maxwell-NP equations are not satisfied." en.wikipedia.org/wiki/Vaidya_metric #### @descheleschilder 2019-05-22 22:10:11

Imagine an isolated, spherical, and homogeneous spherical body somewhere in outer space and with zero velocity (as seen from a local inertial frame). If we let a parallel bundle (for creating more energy) of a high number of high-energy continuous laser beams (every beam consisting of real photons coherent in space and time; see here) pass this mass on one side, this bundle will, because of the curvature of spacetime around the massive object, change its direction toward the object.

This means that the "outgoing" bundle isn't parallel to the "incoming" one. In other words, the momentum of the bundle (and the photons constituting it) has changed direction. This, in turn, means that the momentum of the massive object has changed also to compensate for the change in momentum of the laser bundle. The only way this massive object can acquire this momentum (the three basic forces are not involved here) is because of a curved spacetime produced by the bundle lasers which consists of real photons.

Without the bundle of photons, the curvature around the mass is spherically symmetric, like Peter A. Schneider wrote rightly in a comment below. The only way for the massive body to acquire momentum is when the curvature of spacetime "surrounding" it is asymmetrical. It's obvious that the laser bundle is responsible for this asymmetry. Which means photons do curve spacetime.

EDIT In the answer given below by Ben Crowell (someone who knows what he's talking about) I read:

It would also be extremely problematic if light rays didn't produce gravitational fields, because we have detailed studies confirming that gravitational lensing works as predicted by GR. If the gravitational field of matter affected the momentum of light rays, but not the other way around, then conservation of momentum would be violated. This sort of thing is discussed in section 4.1.1 of Will and is parametrized by γ in the PPN framework. A variety of experiments constrains γ to be equal to the GR value to about 10−4.

Now I don't care too much about someone's reputation and pointing out: "But the famous Mr. X has said..." but in this case, I find it strange that nobody said that his argument is circular (of which I obviously think it's not). Off course he gave also a lot of other great information but nevertheless... #### @Chris 2019-05-25 01:03:39

Comments are not for extended discussion; this conversation has been moved to chat. #### @safesphere 2019-06-07 08:24:09

@Chris The conversation was about the logical flaw in this answer and proved without a doubt that this answer was incorrect. Now you have deleted the proof effectively endorsing the incorrect answer. Is this really the best policy on a site that is supposed to be about science? #### @Chris 2019-06-07 14:24:46

@safesphere This was right on the threshold of what the automated system considers "worth saving." I've undeleted the chat room. #### @descheleschilder 2019-06-07 14:39:34

@safespere Hi there! You write that massless particles, like photons, can't emit gravitons. But gravitons don't experience the passage of time either. Doesn't put that them on equal footing? And what about the gluons emitted in connection with the strong force? Don't they emit other gluons so the strong force gets stronger with distance? Of course, gluons are virtual and can't exist isolated but these emissions of other virtual gluons are the cause which makes the strong force growing stronger if the distance to their source increases. Can't a photon decay to a state with less energy? #### @safesphere 2019-06-07 18:43:43

@descheleschilder A free massless particle cannot experience any change between the emission and absorption, because its proper time is zero. Gluons are not free. They are connected to each other (or to quarks) by "color strings". While each gluon may be massless, a system of two two interacting gluons is massive. Even a system of two non-coaxial photons is massive, but they don't interact, so they are not a system that can change, unlike gluons. Correct, virtual particles play by different rules. For example, virtual photons (e.g. a static electric field) do add mass to a charged object. #### @safesphere 2019-06-07 18:59:34

@descheleschilder "Can't a photon decay to a state with less energy?" - No, a photon does not decay. In fact, photons cannot even "redshift". The idea that photons "lose energy in the redshift" is a huge misconception. Whenever we see a "redshifted" photon, it is always emitted already redshifted in our frame of reference and doesn't lose energy in flight. This changes, if space expands with acceleration, but even if so, the acceleration is tiny while all the redshift we actually see is due to the reference frame where photons are emitted already redshifted and don't lose energy in flight. #### @descheleschilder 2019-06-07 23:20:57

Does the same hold for the CMBR? #### @safesphere 2019-06-09 04:43:24

@descheleschilder Yes, as long as we neglect the acceleration in the space expansion. #### @Árpád Szendrei 2019-07-04 11:15:23

@safesphere can you please tell me what you mean by "Even a system of two non-coaxial photons is massive" #### @safesphere 2019-07-04 15:55:07

@ÁrpádSzendrei A single photon has no mass or center of mass, as it moves with the speed of light. Same for two parallel photons. However, if two photons move in the opposite directions, their total momentum is zero and the center of mass can be defined. If they move on an angle, we can make them move in the opposite direction by a coordinate change. So it works out that, except for photons moving in the same direction, a system of more than one photon has a rest mass (this is well known): arxiv.org/abs/0708.4289 #### @descheleschilder 2019-07-05 03:24:37

@safesphere You write: A free massless particle cannot experience any change between the emission and absorption, because its proper time is zero. But the massless graviton also has a zero proper time. I think because of this the photon can emit gravitons. #### @Rob 2019-05-25 09:04:05

One of them says photons do bend spacetime, since they do have stress-energy, but it is hard to measure it since the energy they carry is little compared to astronomical body's stress-energy. So they do bend spacetime, it is just that it is hard to measure it with our currently available devices.

Now the other one says that photons do not bend spacetime at all. It is only the emitting charge (fermion) that bends spacetime.

Which one is right? Do photons bend spacetime themselves because they do have stress-energy or do they not?

Yes, photons bend spacetime. This same question went to lengthy debate on Research Gate's Forum, that's one place to continue it and view links to failed theories.

The photon's contribution of mass to a system is understood, but not agreed on. Since photons contribute to the stress–energy tensor, they exert a gravitational attraction on other objects, according to the theory of general relativity. Simplified, it's a mass-energy equivalence.

"The Particle Data Group (PDG) cites the upper limit m$$_\gamma$$ < 8.4 × 10$$^{−19}$$ eV c$$^{−2}$$ (= 1.5 × 10$$^{−54}$$ kg) obtained by modelling the magnetic field of the solar system. However, this limit relies on assumptions about the form of the magnetic field and does not discuss measurement accuracy and errors. Another limit (m$$\gamma$$ < 4 × 10$$^{−52}$$ kg) has been derived from atmospheric radio waves has been reported (in https://doi.org/10.1103/PhysRevLett.93.043901). A more conservative approach was followed in an analysis of Cluster data, leading to an upper limit between 7.9 × 10$$^{−14}$$ and 1.9 × 10$$^{-15}$$ eV c$$^{−2}$$ (1.4 × 10$$^{−49}$$ and 3.4 × 10$$^{-51}$$ kg). It is clearly desirable to explore more direct and robust astrophysical constraints on a possible photon mass. This was the motivation for a study we made (here) (see also) showing how data from fast radio bursts (FRBs) could be used to constrain m$$_\gamma$$.".

In "Review of Particle Physics" (Aug 17 2018), by M. Tanabashi et al. (Particle Data Group) Phys. Rev. D 98, 030001 they reannounced the availability of tables of physical constants, specifically: Currently the PDG lists the mass of a photon as:

"< 1×10$$^{−18}$$ from Ryutov 2007 by MHD of solar wind"

So there is some mass, in a pp-wave spacetime.

A somewhat easy read is: "Gravitational properties of light — the gravitational field of a laser pulse" (Jan 29 2016), by Dennis Rätzel, Martin Wilkens and Ralf Menzel:

"... It is shown that the gravitational field of a linearly polarized light pulse is modulated as the norm of the corresponding electric field strength, while no modulations arise for circular polarization. In general, the gravitational field is independent of the polarization direction. It is shown that all physical effects are confined to spherical shells expanding with the speed of light, and that these shells are imprints of the spacetime events representing emission and absorption of the pulse. ...".

I checked for criticism of the theory and endorsement of the authors.

They offer a video with a simple explanation and a couple of easy to understand graphics: "Figure 6. These plots show the double logarithm of the metric perturbation $${h}^{{\rm{p}}}={h}_{00}^{{\rm{p}}}={h}_{{zz}}^{{\rm{p}}}=-{h}_{0z}^{{\rm{p}}}=-{h}_{z0}^{{\rm{p}}}$$ for a linearly polarized pulse of length L and central wavelength $$\lambda =\frac{2\pi c}{\omega }=\frac{2}{3}L$$ in the x-y-plane at $$t=50000L/c$$, after its emission at z = 0. $${h}^{{\rm{p}}}$$ is normalized to units of $$\kappa =4{{GAu}}_{0}/{c}^{4}$$ and then the logarithm of the logarithm is taken. The metric perturbation can be interpreted as the potential to the gravitational field. The front stemming from the emission event of the pulse is seen between $$z=6L+499994L$$ and $$z=7L+499994L$$. It shows oscillations with wavelength $$\lambda /2$$ and approaches the form of a plane fronted wave. The right plot shows the same situation for circularly polarized light where no modulations appear." The plots show the metric perturbation $${h}^{{\rm{p}}}={h}_{00}^{{\rm{p}}}={h}_{{zz}}^{{\rm{p}}}=-{h}_{0z}^{{\rm{p}}}=-{h}_{z0}^{{\rm{p}}}$$ for a pulse of length L in the coordinates $$({ct},x,y,z)$$ in the (x, y)-plane for different times t. $${h}^{{\rm{p}}}$$ is. normalized to units of κ and then the logarithm of the logarithm is taken.

In a later article: "Gravitational properties of light - The emission of counter-propagating laser pulses from an atom" (Oct 14 2016), by Dennis Rätzel, Martin Wilkens, Ralf Menzel they confirm their results:

"... the situation of two counter-propagating laser pulses emitted from a massive point particle was considered. The corresponding metric perturbation in the framework of linearized gravity and the corresponding curvature were derived. It was shown that the curvature is that of a massive point particle at all spacetime points lying in the causal future of the end of the emission process and in the causal past of the beginning of the emission process. It was concluded that the laser pulses only contribute to the curvature during their emission and their absorption. This is in agreement with the results presented in [our previous article], where only one pulse was considered and the gravitational effect of the emitter was neglected. In contrast to the model presented in the former article, in the model presented in this article, the emitter itself is taken into account, and the continuity equation of general relativity is fulfilled." #### @safesphere 2019-06-07 08:14:46

Thanks for supporting my point that photons in flight don't bend spacetime, as their gravitational field is not static, but a gravitational wave emitted by the processes of interaction of charged particles at the events of emission and absorption, exactly what your animation shows. From the article abstract: "It is shown that all physical effects are confined to spherical shells expanding with the speed of light, and that these shells are imprints of the spacetime events representing emission and absorption of the pulse." I would upvote your answer if you didn't contradict yourself. #### @Rob 2019-06-07 08:24:36 #### @safesphere 2019-06-07 08:56:27

My original quote in the question was based on the same article that your answer describes in details. The moderators deleted my link, but it is still there in the answer by knzhou. The article states that photons in flight do not bend spacetime. Thank you! #### @kpv 2019-05-25 02:54:59

Sun loses some mass (in the form of photons) every year and because of that earth's orbit gets bigger by 1.5 centimeters every year.

This means the space curvature by sun, decreases every year.

Where does that curvature go? Isn't it gone with the photons!

Photon's must curve space time, or they themselves must be that - tiny curves in space time depending upon how much energy they have. In both cases, it is evident they curve space time. It does not matter which (classical, or quantum) theory you apply, and what it indicates/predicts, the evidence is out there.

Therefore, the theory/prediction/reasoning that denies curving of space by photons must be wrong.

Because of minuscule size and enormous speed, experimentally proving it must be very challenging. #### @G. Smith 2019-05-22 02:48:06

Classical electromagnetic fields carry energy and momentum and therefore cause spacetime curvature. For example, the EM field around a charged black hole is taken into account when finding the Reissner-Nordstrom and Kerr-Newman metrics.

The question of whether photons cause spacetime curvature is a question about quantum gravity, and we have no accepted theory of quantum gravity. However, we have standard ways of quantizing linear perturbations to a metric, and reputable journals such as Physical Review D have published papers on graviton-mediated photon-photon scattering, such as this one from 2006. If such calculations are no longer mainstream, it is news to me. Given that photons have energy and momentum, it would surprise me if they do not induce curvature.

I also note that the expansion of the "radiation-dominated" early universe was caused by what is generally described as a photon gas and not as a classical electromagnetic field. So the idea that photons bend spacetime is part of mainstream cosmology, such as the standard Lambda-CDM model.

Finally, the idea of a kugelblitz makes no sense to me unless photons bend spacetime.

So in Rennie v. Safesphere, I am on the Rennie side, but I look forward to Safesphere defending his position in a competing answer.

Safesphere declined to answer; in a now-removed comment, he said that knzhou’s answer explains the disagreement. I don’t agree. I disagree with knzhou that “bends spacetime” is vague. It is commonly understood by most physicists to mean “contributes to the energy-momentum tensor on the right side of the Einstein field equations”. And most physicists believe that real photons do exactly this, for the reasons that Ben Crowell and I have stated. #### @Chris 2019-05-22 18:53:25

Comments are not for extended discussion; this conversation has been moved to chat. #### @tpg2114 2019-05-23 02:37:49

I won't clear the comments out (yet), but if it takes more than one on either side of the discussion, please continue it in the chat room for this answer. And if the answer needs clarified/updated based on the comments, please edit it in and then clean up the comments that aren't relevant. #### @G. Smith 2019-05-25 00:19:25 #### @knzhou 2019-05-22 10:59:09

Both posters are correct and their answers are not in contradiction. They're just talking about very different regimes. A classical electromagnetic field sources metric curvature, while a single isolated photon doesn't source real gravitons. Here safesphere is referring to this paper and transferring it to linearized gravity, which is a perfectly well-defined approach to quantum gravity at low energies.

For a more familiar example, consider an isolated atom in its ground state. This system cannot source real photons, because it's already in the ground state. But at large distances one can measure an electric field, i.e. there is a dipole moment. The same thing is just going on here.

The only contradiction between the statements is a purely semantic point of what it means for "photons to bend spacetime", a phrase which was never perfectly defined anyway. Of course, electromagnetic fields can couple to gravity, as we know from experimental tests. #### @user4552 2019-05-22 05:42:28

In classical general relativity, electromagnetic fields do bend spacetime. They have a nonvanishing stress-energy tensor, and the Einstein field equations relate the stress-energy to the curvature.

We even have fairly direct experimental proof that electromagnetic fields interact gravitationally in this way, from Cavendish-like experiments. See Kreuzer, Phys. Rev. 169 (1968) 1007, which can be interpreted as confirming the correctness of the coupling of gravity to the pressure components of the stress-energy. For a discussion of Kreuzer and similar tests, including lunar laser ranging, see Will, “The Confrontation between General Relativity and Experiment,” The Kreuzer experiment is discussed in section 4.4.3.

We can also confirm that this holds for electromagnetic waves, not just static fields. One empirical confirmation of this comes from the fact that models of big bang nucleosynthesis (BBN) agree pretty well with observed data on things like the H/He ratio; during the BBN period, cosmological gravity was radiation-dominated.

It would also be extremely problematic if light rays didn't produce gravitational fields, because we have detailed studies confirming that gravitational lensing works as predicted by GR. If the gravitational field of matter affected the momentum of light rays, but not the other way around, then conservation of momentum would be violated. This sort of thing is discussed in section 4.1.1 of Will, and is parametrized by $$\gamma$$ in the PPN framework. A variety of experiments constrains $$\gamma$$ to be equal to the GR value to about $$10^{-4}$$.

There is no reason to think that the situation is any different when the electromagnetic field is quantized. By the correspondence principle, photons have to produce gravitational fields when the conditions are such that the classical theory is a good approximation (coherent states with lots of photons). In the case where the classical theory is invalid, and we really need to talk about photons, the best we can currently do, lacking a real theory of quantum gravity, is semiclassical gravity. Semiclassical gravity works by replacing the stress-energy tensor $$T$$ in the Einstein field equations with its expectation value $$\langle T \rangle$$. $$\langle T \rangle$$ can easily be nonzero. #### @safesphere 2019-05-22 06:20:01

"electromagnetic fields do bend spacetime" - One must be careful not to confuse a field with a wave. As already mentioned in another comment, there are no real photons flying around a permanent magnet or static charge. The quote by the OP refers to real photons, not fields or interactions. A massless particle cannot decay by emitting gravitons in flight. The quote doesn't state that photons are not associated with gravity, but that the gravitational field of the photon is a gravitational wave from the event of the emission and also absorption. See the link to the source in the comment above. #### @user4552 2019-05-22 16:21:12

@safesphere: What led to this question is that in a comment, you made the claim that "photons themselves don't bend spacetime." This is false. If you insist that it's true, please write an answer explaining in a coherent way why you think it's true. The purpose of comments isn't to give answers. Your comment above suggests that you want to change your mind or modify your previous, incorrect claim. If so, then please do that in an answer so we can tell what you actually want to claim. #### @safesphere 2019-05-22 16:40:55

@BenCrowell The original paper states that the gravitational field of a photon is produced by the event of emission, but not by the photon in flight. You are welcome to disagree, but it is just your opinion. #### @rob 2019-05-22 19:33:09

I've removed some discussion of the "move comments to chat" feature. Any user with 100 rep can make a chat room and put a link in a comment. To clarify when the diamond moderators do this on your behalf, please ask on Physics Meta.