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

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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:^[1]

$\theta =\frac{2GM}{v^{2}r}(1+\frac{v^2}{c^2}),$

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)