Deriving Light Deflection Angle in Weak Gravitational Field | Homework Help

[fikus]

Homework Statement

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

Homework Equations

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

The Attempt at a Solution

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

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

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

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

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

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

and equivalently

$$\ddot{x}^i=p_t^2\frac{d^2x^i}{dt^2} - 2p_t^2 \frac{dx^i}{dt} \frac{\partial{\Phi}}{\partial{x^k}}\frac{dx^k}{dt}$$

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

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

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

$$\vec{\alpha} = \frac{2}{c^2} \int \vec{\nabla}_\perp \Phi~ dl$$

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.

 

[Jhero]

Thank you for your question. The first term on the right-hand side of your equation does not change the speed of light, as it is simply a component of the gradient vector in the direction of the velocity vector. This term represents the change in the direction of the light's path due to the presence of the gravitational field.

To obtain the deflection angle, you can use the small angle approximation, where the angle of deflection is approximately equal to the magnitude of the perpendicular gradient divided by the speed of light squared. So, in your equation, the deflection angle would be given by:

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

This approximation is valid for small angles of deflection, which is typically the case in weak gravitational fields. I hope this helps clarify the interpretation of the first term and how to obtain the deflection angle. Good luck with your derivation!