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- Thread starter Freespader
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Drakkith

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Annihilation (like electron-positron) doesn't just make the energy disappear. Other particles come out of such collisions; for example, gamma rays (photons) are generated by electron-positron annihilation. These byproducts are where the energy goes.

Now you may ask: gamma rays are massless and don't cause the gravitational attraction. Surely energy is lost. mgh = 0gh = 0, so where did it go.

mgh is only correct for the non-relativistic limit. For relativistic particles, like the gamma rays (photons) generated by annihilation collisions, you need to consider relativity. In general relativity, gravitational "attraction" is not a "force" in the usual sense of the word, it is more like a fictitious force (such as centrifugal force) caused by the curvature of spacetime. Even massless particles cause such curvature. Curvature is related to the presence of what many textbooks call "mass-energy;" I'm sure you've seen the equation E=mc^2. I'm not exactly sure what kind of general relativistic effects happen during annihilation (surely they touch on quantum gravity). However it is observationally confirmed that (massless) photons' paths bend around gravitational bodies--they do feel the gravitational "fictitious forces" as they travel along geodesics in curved spacetime.

Now you may ask: gamma rays are massless and don't cause the gravitational attraction. Surely energy is lost. mgh = 0gh = 0, so where did it go.

mgh is only correct for the non-relativistic limit. For relativistic particles, like the gamma rays (photons) generated by annihilation collisions, you need to consider relativity. In general relativity, gravitational "attraction" is not a "force" in the usual sense of the word, it is more like a fictitious force (such as centrifugal force) caused by the curvature of spacetime. Even massless particles cause such curvature. Curvature is related to the presence of what many textbooks call "mass-energy;" I'm sure you've seen the equation E=mc^2. I'm not exactly sure what kind of general relativistic effects happen during annihilation (surely they touch on quantum gravity). However it is observationally confirmed that (massless) photons' paths bend around gravitational bodies--they do feel the gravitational "fictitious forces" as they travel along geodesics in curved spacetime.

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Thanks! I think that clears it up.

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F=Gm1m2/r^2 Substitutes into U=mgh

So you have U=h*G*m1*m2/(r^2)

Removing the constants gets U=h/(r^2) and h is equal to r so U=1/r

So where does the U go this time? My best thought, and I'm doubtful of it, is that as the particle "catches up" with the light, it shortens the wave length of the light, increasing it's energy. But something just seems off about that. Thanks.

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In general though, for a quantum particle, the gravitational forces involved are several orders of magnitude smaller than the other fundamental forces, and can normally be neglected in calculations.

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