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March 5

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Gravitational lens

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Let a given photon, move perpendicular to a given gravitational field created by a given star. According to General relativity, the photon's trajectory will be deflected by the angle toward the star, whereas: denotes the photon's velocity (i.e. ), denotes the universal constant of gravitation, denotes the star's mass, and denotes the distance between the star and the photon.

Question: What will the angle be, if we replace the photon by a massive particle, its properties being the same as before (except its velocity which will be slower than of course). Will the angle be a half of the angle mentioned above? HOTmag (talk) 15:11, 5 March 2024 (UTC)[reply]

There's no distinction - energy and mass are equivalent. The deflection of the light is just a much greater version of the same effect as the prcession of Mercury. NadVolum (talk) 18:54, 5 March 2024 (UTC)[reply]
Thx.
Here you've written "there's no distinction", but you've also written "No big difference" in the edit summary (see the history page), so I wonder what's more exact.
Additionally, what do you mean by "a much greater version"? HOTmag (talk) 19:43, 5 March 2024 (UTC)[reply]
Massive particles are deflected by half that angle. Ruslik_Zero 19:56, 5 March 2024 (UTC)[reply]
Now I'm a bit confused, because of the contradiction between your reply and the previous reply above yours. Of course, if anyone of you could supply a source (or any argument analogous to a source), I would be much less confused (if at all). HOTmag (talk) 20:23, 5 March 2024 (UTC)[reply]
The general formula, for massless and massive particles, is:[1]
where is the velocity at a large distance, before any acceleration due to the gravitational attraction. When the factor in parenthesis equals When it approaches  --Lambiam 10:47, 6 March 2024 (UTC)[reply]
Thanks. HOTmag (talk) 10:37, 13 March 2024 (UTC)[reply]