In general relativity, light is subject to gravitational pull. Does light generate gravitational pull, and do two beams of light attract each other?


5 Answers 5


The general answer is "it depends." Light has energy, momentum, and puts a pressure in the direction of motion, and these are all equal in magnitude (in units of c = 1). All of these things contribute to the stress-energy tensor, so by the Einstein field equation, it is unambiguous to say that light produces gravitational effects.

However, the relationship between energy, momentum, and pressure in the direction of propagation leads to some effects which might not otherwise be expected. The most famous is that the deflection of light by matter happens at exactly twice the amount predicted by a massive particle, at least in the sense that in linearized GTR, ignoring the pressure term halves the effect (one can also compare it a naive model of a massive particle at the speed of light in Newtonian gravity, and again the GTR result is exactly twice that).

Similarly, antiparallel (opposite direction) light beams attract each other by four times the naive (pressureless or Newtonian) expectation, while parallel (same direction) light beams do not attract each other at all. A good paper to start with is: Tolman R.C., Ehrenfest P., and Podolsky B., Phys. Rev. 37 (1931) 602. Something one might worry about is whether the result is true to higher orders as well, but the light beams would have to be extremely intense for them to matter. The first order (linearized) effect between light beams is already extremely small.

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    $\begingroup$ All responses are good, but I have chosen this one as acccepted because of reference to actual calculations $\endgroup$ Mar 2, 2011 at 18:59
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    $\begingroup$ If two photons were traveling in the same direction, side by side, then wouldn't communication between them be prohibited anyway due to special relativity? Could that be a valid interpretation of the fact that parallel beams of light don't gravitate to each other? $\endgroup$ Aug 12, 2011 at 14:47
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    $\begingroup$ From your answer it seems like 'Yes' is more correct than 'it depends'. Especially if you look at the point you make about parallel light being the only exception (which is only true for one case out of an infinite measure of cases in the weak limit). $\endgroup$ Mar 3, 2015 at 12:37

According to general relativity, yes, two beams of light would gravitationally attract each other. Einstein's equation says that

$$R^{\mu\nu} - \frac{1}{2}R g^{\mu\nu} = 8\pi T^{\mu\nu}$$

The terms on the left represent the distortion ("curvature") of spacetime, and the term on the right represents matter and energy, including light. As long as $T^{\mu\nu}$ is nonzero, there will have to be some sort of induced distortion a.k.a. gravity, since $R^{\mu\nu} - \frac{1}{2}R g^{\mu\nu} = 0$ in flat spacetime.

In case you're interested, the relevant equations are the definition of the stress-energy tensor for electromagnetism,

$$T^{\mu\nu} = -\frac{1}{\mu_0}\biggl(F^{\mu\rho}F_{\rho}^{\ \nu} + \frac{1}{4}g^{\mu\nu}F^{\rho\sigma}F_{\rho\sigma}\biggr)$$

where $F$ is the electromagnetic field tensor, and the electromagnetic wave equation

$$D_{\alpha}D^{\alpha}F^{\mu\nu} = 0$$

where $D_{\alpha}$ is the covariant derivative operator. In principle, to calculate the gravitational attraction between two beams of light, you would identify the functions $F^{\mu\nu}$ that correspond to your beams (they would have to satisfy the wave equation), and then plug them in to calculate $T^{\mu\nu}$. When you put that into Einstein's equation, it places a constraint on the possible values of the metric $g_{\mu\nu}$ and its derivatives, and you could use that constraint to determine the geodesic deviation between the two beams of light, which, in a sense, corresponds to their gravitational attraction.

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    $\begingroup$ is attraction strength of two parallel beams of light traveling in opposite directions greater than if they traveling in one direction? $\endgroup$
    – voix
    Mar 2, 2011 at 6:24
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    $\begingroup$ @voix: I don't know offhand, you'd have to work through the math. My guess would be that the attraction (i.e. deflection) would be greater for beams traveling in the same direction since they would be close together for more than just a moment. $\endgroup$
    – David Z
    Mar 2, 2011 at 6:40
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    $\begingroup$ but for beams of relativistic neutrons? $\endgroup$
    – voix
    Mar 2, 2011 at 19:14
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    $\begingroup$ There is nothing wrong with the setup here, but note the paper Stan Liou cites shows that perfectly parallel beams happen to have no attraction. $\endgroup$
    – user10851
    Feb 4, 2014 at 2:49
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    $\begingroup$ Perfectly parallel beams in the small field limit only, which is not the OP question. $\endgroup$ Mar 3, 2015 at 12:40

Yes. The energy-momentum tensor (which is on the right hand side of Einstein's equation) is non-zero in the presence of any kind of energy density, including radiation. This means that the light beams will curve spacetime (measured by the left hand side of Einstein's equation) and hence affect the path that the light takes. But for typical light beams, this is very small and hence neglectable.

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    $\begingroup$ So, it's negligible for typical light beams. How about trillions of light beams , emitted from trillions of stars, all moving in various directions. Could they have an observable gravitational effect on galaxies? $\endgroup$
    – frodeborli
    Feb 20, 2015 at 19:45
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    $\begingroup$ Yes, photons have an effect on cosmological expansion precisely due to this. However, it is dwarfed by the contribution from dark energy and matter. Earlier in the universe, photons had the dominant effect though. $\endgroup$
    – dbrane
    Apr 17, 2017 at 4:39

In fact, one does not need the general relativity to show that two photons moving in the same direction do not attract each other. All that is needed is time foreshortening (from special relativity), the formula for energy of a moving particle, and the idea that gravitational effects of a photon are very similar to gravitational effects of an ultra-relativistic particle of the same energy.

Consider two particles of mass m at distance d. If they were at rest w.r.t. each other, then after t seconds the distance between them decreases by gm²t²/2d² (if t is small enough).

Now put them at distance L from a wall, and assume that they both move with velocity v towards the wall. If L is small, then when they hit the wall, the distance between them decreases by gm²L²/2d²v² (in Newtonian mechanics). Due to time foreshortening, the answer in special-relativity is (1−v²/c²)gm²L²/2d²v². In terms of energy E=mc²/√(1−v²/c²), this decrease of the distance is (1/v²−1/c²)²gE²L²/2d²c⁴.

Conclusion: when vc but the energy remains the same, the attraction between particles moving with the same velocity decreases to 0. (In the sense that the presence of a particle does not bend the path of the other particle.)

  • $\begingroup$ Conversely, if two photons are of different energy, there should be gravitational force? - "If they were at rest w.r.t. each other, ...": Rest mass is zero, hence gravitational force is zero? $\endgroup$ Sep 17, 2022 at 16:50
  • $\begingroup$ The notion of “rest mass” exists only in popular-science paraphrases of General Relativity. In GR, gravitational effects depend on the tensor of energy-momentum. Being a tensor, you cannot “express it” using a single number! (There are very few contexts where “rest mass” can be assigned some sense — as below — but still, it does not determine the gravitational effects!) $\endgroup$ Oct 19, 2022 at 11:30
  • $\begingroup$ Why this “conversely”?! If photons have different energies, just model them by ultra-relativistic particles of the same velocity but with different masses. $\endgroup$ Oct 19, 2022 at 11:32
  • $\begingroup$ In my preceding remark, it is better to replace “mass” by “momentum”. $\endgroup$ Oct 20, 2022 at 2:34

In the golden era of general relativity, A. Bonnor introduced Bonnor beams. Thin tubes of light in empty space with a momentum direction and energy density.

It turns out that if the beams have parallel momenta, they will stay as they are, while anti-parallel momenta will cause the beams to bend towards one another.

Intuitively this can be explained by frame dragging. As the beams have no mass they will not curve spacetime because of a mass factor and only a frame dragging component will be present. A mass initially at rest wrt to a faraway observer in an inertial frame will aquire a velocity towards the beam and a subsequent acceleration parallel to the momentum (frame-dragging is non-local).

So yes, beams of light exert gravity. It drags other light and drags as well as attracts massive particles.

  • $\begingroup$ Is it coherent to say (refering to your final statement): if earth exerts gravity on the photon, the photon exerts gravity on earth...? And, this does not imply, from this does not follow, that two photons exert gravity on each other (?). $\endgroup$ Sep 17, 2022 at 16:56

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