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Gravitational lensing and light deviation is a well known subject in General Relativity. I'm wondering if the null geodesics equation could be recast into the classical lenses equation: \begin{equation} \frac{1}{p} + \frac{1}{q} = \frac{1}{f}. \tag{1} \end{equation} If it's possible, how should be defined the gravitational focal length $f$ ?

I never saw the geodesics equation written into the shape (1), even in the case of the Schwarzschild metric: \begin{equation}\tag{2} \frac{d k^{\lambda}}{d \sigma} + \Gamma_{\mu \nu}^{\lambda} \, k^{\mu} \, k^{\nu} = 0, \end{equation} where $k^{\mu} \propto \frac{d x^{\mu}}{d \sigma}$ is a null four-vector: $g_{\mu \nu} \, k^{\mu} k^{\nu} = 0$.

The weak field approximation is defined by the following metric: \begin{equation}\tag{3} ds^2 \approx (1 + 2 \phi) \, dt^2 - (1 - 2 \phi)(\, dx^2 + dy^2 + dz^2), \end{equation} where $\phi \ll 1$ is a gravitational potential (I'm using $c \equiv 1$ and metric-signature $(1, -1, -1, -1)$).

Could the geodesics equation (2) be reformulated into the shape of (1), assuming the special case of the Schwarzschild metric, or the weak field metric (3)?

EDIT: I'm thinking about this kind of rays, with their focal length dependant on the impact parameter: enter image description here

Cham
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    Duplicate: https://physics.stackexchange.com/q/92050/520 – dmckee --- ex-moderator kitten Jul 03 '19 at 16:26
  • @dmckee, this question isn't exactly the same. I'm asking about the geodesics equation reformulated as a lense equation. – Cham Jul 03 '19 at 17:12
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    The other questions answers this one; there is no focus so you can't have a equation describing the system that involves a focus. – dmckee --- ex-moderator kitten Jul 03 '19 at 17:17
  • @dmckee, I'm not sure this is true. The focal length $f$ may depend on the ray (like the chromatic and spherical aberrations on real lenses), so $f \equiv f(q, p)$. – Cham Jul 03 '19 at 17:51
  • In those aberrations there is almost a focus. In gravitational lens there is nothing remotely like a focus. – dmckee --- ex-moderator kitten Jul 03 '19 at 17:54
  • @dmckee, well, I don't understand this. A given light ray parallel to the optical axis (with some impact parameter $b$) will certainly converge to a point located on the optical axis, at a distance $f$ from the gravitational lense. – Cham Jul 03 '19 at 17:56
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    Cham, I am not an expert in gravitational lensing, but you can quickly see that there is a major difference with optics (convex lenses) : the larger the impact parameter is, the smaller the deviation is in grav. lensing. So no focus. Check also https://en.m.wikipedia.org/wiki/Gravitational_lensing_formalism – magma Jul 04 '19 at 01:54
  • @magma, thanks for the link. Helpfull. – Cham Jul 04 '19 at 02:03
  • You are welcome – magma Jul 04 '19 at 02:05

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