Wikipedia:Reference desk/Archives/Science/2024 March 5

= March 5 =

Gravitational lens
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 $$\theta = \frac{4GM}{v^2 r}$$ toward the star, whereas: $$v$$ denotes the photon's velocity (i.e. $$v=c$$), $$G$$ denotes the universal constant of gravitation, $$M$$ denotes the star's mass, and $$r$$ 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 $$v$$ which will be slower than $$c$$ of course). Will the angle be a half of the angle mentioned above? HOTmag (talk) 15:11, 5 March 2024 (UTC)


 * 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)
 * 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)
 * Massive particles are deflected by half that angle. Ruslik_ Zero 19:56, 5 March 2024 (UTC)
 * 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)
 * The general formula, for massless and massive particles, is:
 * $$\theta=\frac{2GM}{v^2r}\left(1+\frac{v^2}{c^2}\right),$$
 * where $$v$$ is the velocity at a large distance, before any acceleration due to the gravitational attraction. When $$v=c,$$ the factor in parenthesis equals $$2.$$ When $$v\ll c,$$ it approaches $$1.$$ --Lambiam 10:47, 6 March 2024 (UTC)
 * Thanks. HOTmag (talk) 10:37, 13 March 2024 (UTC)
 * where $$v$$ is the velocity at a large distance, before any acceleration due to the gravitational attraction. When $$v=c,$$ the factor in parenthesis equals $$2.$$ When $$v\ll c,$$ it approaches $$1.$$ --Lambiam 10:47, 6 March 2024 (UTC)
 * Thanks. HOTmag (talk) 10:37, 13 March 2024 (UTC)