We know the following:

  1. Two masses are attracted to one another, as represented by Newtonian gravity

    $F = \frac{GMm}{R^{2}}$

  2. Light is massless and bends in the curvature of space-time which can be created by a mass, where the deflection is calculated with Einstein's general relativity and is twice that calculated by Newtonian gravity, https://en.wikipedia.org/wiki/Eddington_experiment.

  3. $E = mc^{2}$, where m is the relativistic mass.

Is there something special about mass energy that produces a curvature of spacetime? If that rest mass energy were converted to light energy (say, by annihilation of an equally large amount of matter and anti-matter, a planet sized "anti-matter bomb"), would the spacetime curvature resulting from the originating mass essentially instantaneously disappear?

The inverse process, say through a "pair-production bomb," could instantaneously create space-time curvature. Of course, "instantaneous" here is nearly achieved anyway since the light source is traveling at the speed of light toward the about-to-be-born mass.

Of course, this also answers the question of whether light attracts itself (through "gravity").


The source of gravity is energy-momentum, not just mass. Light has energy-momentum so it is a source of gravity.

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    $\begingroup$ This answer needs citations, and preferable a couple equations posted to justify the statements made. $\endgroup$ – Carl Witthoft Jul 31 '20 at 11:04
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    $\begingroup$ @CarlWitthoft: No it doesn't. It is informative and self-contained. $\endgroup$ – TonyK Jul 31 '20 at 11:07
  • $\begingroup$ What is meant by energy-momentum? $\endgroup$ – Tfovid Jul 31 '20 at 12:00
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    $\begingroup$ Just want to add that light bends spacetime much less than matter, due to m=E/c^2. The amount of energy in light is "diluted" by c^2, so the equivalent amount of spacetime bending matter is very very small. $\endgroup$ – ZenFox42 Jul 31 '20 at 14:39
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    $\begingroup$ @ZenFox42 A given energy density bends spacetime by the same amount regardless of whether the energy density is for matter or for light. There is no sense in which light bends it “less” than matter does. $\endgroup$ – G. Smith Jul 31 '20 at 16:20

Yes. You can have, for example, a Friedmann universe filled with nothing but electromagnetic radiation. The light’s gravity, due to its energy-momentum tensor, is the only gravity determining the dynamics of that cosmology.

Our own universe, which we believe is homogeneous and isotropic on the largest scales and thus can be modeled using the Friedmann metric, had an early radiation-dominated phase. The gravity of electromagnetic radiation was very important in determining the rate of expansion during that phase. The “radiation” then also included ultrarelativistic particles with mass, so it wasn’t 100% “light”. But the form of its energy-momentum tensor, and thus its gravity, was almost the same as if it were, because everything was traveling either at or very near the speed of light.

  • $\begingroup$ @safesphere A Friedmann universe filled with nothing but a free electromagnetic field has no interactions that make it opaque. But it has an energy-momentum tensor, and it has curvature. This is perhaps the purest example of what the OP asked. I have no interest in discussing the interacting case, and whether your understanding of it is correct or incorrect. I do not believe that either MTW or Weinberg’s textbooks on GR take your point of view when discussing the radiation-dominated phase. $\endgroup$ – G. Smith Aug 2 '20 at 23:51

Archibald Wheeler consider the simpler case than a kugelblitz, of a collection of photons dense enough to be gravitationally bound, which he called a geon. He thought they were extremely unlikely to be stable.

Photon-photon scattering can occur where combined energy is above that need for electron-positron pair production.


We all know that gravity according to Albert Einstein is the curvature of space-time: $$\underbrace{G_{\mu\nu}}_{\text{curvature of space-time}}=\underbrace{T_{\mu\nu}}_{\text{stress-energy }}\qquad{\mathrm{}} $$ I'll say simply: the EM field is a source of stress-energy which determines curvature in general relativity.


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