Britannica says this about the graviton:

graviton, postulated quantum that is thought to be the carrier of the gravitational field. It is analogous to the well-established photon of the electromagnetic field. Gravitons, like photons, would be massless, electrically uncharged particles traveling at the speed of light. Since gravitons would apparently be identical to their antiparticles, the notion of antigravity is questionable.

The graviton Wiki goes even further to state:

Antiparticle Self

My Question(s):

(1) Why graviton & anti-graviton being identical will make the notion of antigravity questionable?

(2) If graviton (G) & anti-graviton (AG) collide & annihilate each other to release radiation , then gravity also "disappears": If there was some some entity P having gravitation attraction before annihilation, that entity will now see a drop in gravitation attraction. Is this correct ? Diagrammatically:
P ----> (some gravitational attraction) ----> G + AG [[ before annihilation ]]
P -----------------------------------> radiation [[ annihilation ]]
P (no gravitational attraction) [[ after annihilation ]]

(3) If that is correct , then will P experience net 0 gravity , with the attraction of G + AG cancelling each other , even without annihilation ?

(4) In other words, when P is near only AG , will P experience negative gravity?

(5) If graviton & antigraviton are Identical, then will two such Particles get annihilated on interaction?

My Current thinking about this PARADOX is that graviton & anti-graviton are Exactly Identical & will not interact to annihilate. Thus it will always be attractive in nature, graviton never cancelling with anti-graviton.

Conclusion: graviton (& gravity) is INDESTRUCTIBLE!

Is that line of thought Correct?

  • $\begingroup$ There is no notion of particles and anti-particles for the graviton and the photon. There is no such thing as 'anti-graviton' since there is no notion of charge for them. A counter-example is the neutrinos whose electric charge is zero, so there is a notion of charge, even if the value is null. Mathematically all this is encoded in the objects we are using to describe the particles. For the electric charge, being a "complex version of another object" gives sense to the notion of charge. Photons and gravitons are no complex versions of vectors and tensors respectively. $\endgroup$ Commented Nov 17, 2022 at 12:04

1 Answer 1


There is a theorem (name and link to be remembered later) that in a quantum-mechanical interaction mediated by a boson, the directions of the force between like and unlike charges depends on whether the spin of the interaction field is even or odd. For the spin-one photon, like charges repel and unlike charges attract. For the spin-zero pion, it is like “charges” (but here, not electrical charges) which attract. This is why the pion holds the nucleus together. The graviton should have spin $2\hbar$, because gravity is a tensor field. That’s why masses all attract each other. Think of mass as “gravitational charge.”

You ask why these interaction bosons are (for the quantum ones) or should be (for the hypothetical graviton) unchanged under the mathematical transformation which transforms matters into antimatter. For the quantum particles, that’s an interesting question which is different from the one you asked. For the graviton, we don’t have a quantum theory at all, but there is no reason to think this detail will be different.

We do have evidence for anti-gravity: the expansion of the universe is speeding up instead of slowing down. But we don’t know what’s causing it. We refer to this unknown cause as “dark energy.” In general relativity, dark energy has “negative pressure”; the gravitational stress-energy tensor is more complicated than one might guess from pop-science descriptions.

Your bolded, quoted sentence suggests to me that Brittanica had trouble hiring a domain expert to write that article. Read skeptically.

  • $\begingroup$ +1 Thanks ! I wish you would remember the theorem & web-link to add to the Post ! $\endgroup$
    – Prem
    Commented Nov 18, 2022 at 5:41
  • $\begingroup$ Hmmmm. Apparently I also didn’t know this name in the past, when I was better-connected to the research literature than I am today. $\endgroup$
    – rob
    Commented Nov 18, 2022 at 11:10

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