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Light deflection

  • Thread starter fikus
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1. Homework Statement

I'm trying to derive deflection angle of light in weak gravitational field.

2. Homework Equations

Metric tensor in weak field [tex] g_{00}=(1+\frac{2\Phi}{c^2}), ~ g_{ii}=-(1-\frac{2\Phi}{c^2})[/tex]

3. The Attempt at a Solution

From Lagrangian [tex] L=\sqrt{g_{\mu \nu} \dot{x}^{\mu}\dot{x}^{\nu}} [/tex], I obtain equations of geodesic

[tex] g_{\mu \nu} \ddot{x}^{\nu} + \frac{\partial{g_{\mu \nu}}}{\partial{x^{\lambda}}} \dot{x}^{\lambda} \dot{x}^{\nu} - \frac{1}{2}\frac{\partial{g_{\lambda \nu}}}{\partial{x^{\mu}}} \dot{x}^{\lambda} \dot{x}^{\nu}~=~0 [/tex]

Putting in metric tensor and writing only equations for spatial coordinates I get :

[tex] -\left(1-\frac{2\Phi}{c^2}\right)\ddot{x}^i + \frac{2}{c^2}\frac{\partial{\Phi}}{\partial{x^j}} \dot{x}^j \dot{x}^i - \frac{\partial{\Phi}}{\partial{x^i}}\dot{t}^2- \frac{1}{c^2}\frac{\partial{\Phi}}{\partial{x^i}}(\dot{x}^2+\dot{y}^2+\dot{z}^2)~= 0 [/tex]

and for time coordinate
[tex] \dot{t}c^2(1+\frac{2\Phi}{c^2})= p_{t} [/tex]
which is constant of motion.
Noting that [tex] \dot{t}^2=(\dot{x}^2+\dot{y}^2+\dot{z}^2) [/tex] (from ds = 0 for light, in zeroth order in [tex]\Phi/c^2[/tex] ), and writing

[tex] \dot{x}^i=\frac{dx^i}{dt}\dot{t} = \frac{dx^i}{dt} \frac{p_t}{c^2(1+2\Phi/c^2)}[/tex]

and equivalently

[tex] \ddot{x}^i=p_t^2\frac{d^2x^i}{dt^2} - 2p_t^2 \frac{dx^i}{dt} \frac{\partial{\Phi}}{\partial{x^k}}\frac{dx^k}{dt}[/tex]

Putting it all together, neglecting terms with phi/c^2 and writing in vector form I get:

[tex]\frac{d\vec{v}}{dt} ~ = ~ \frac{4}{c^2}(\vec{v} \cdot \vec{\nabla}\Phi) \vec{v} -2\vec{\nabla }\Phi = \frac{2}{c^2}(\vec{v} \cdot \vec{\nabla}\Phi) \vec{v} - 2\vec{\nabla}_{\perp}\Phi[/tex]

Where I noted that the first term is just component of gradient in direction of v, and then writing nabla with \perp, for component of gradient perpendicular to v. So my question now is how to interpret this first term on rhs. Does it change speed of light ? (is it maybe shapiro delay ?) And how to then obtain deflection angle ? If there would be only second term I get

[tex]\vec{\alpha} = \frac{2}{c^2} \int \vec{\nabla}_\perp \Phi~ dl [/tex]

which is correct result. I just don't know what to do with that first term.

Hope I wasn't tooo long and thanks in advance.
1. Homework Statement



2. Homework Equations



3. The Attempt at a Solution
 

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