General Relativity - Deflection of light

[unscientific]

Homework Statement

Find the deflection of light given this metric, along null geodesics.

Homework Equations

The Attempt at a Solution

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Conserved quantities are:

$$e \equiv -\zeta \cdot u = \left( 1 - \frac{2GM}{c^2r} \right) c \frac{dt}{d\lambda}$$
$$l \equiv \eta \cdot u = r^2 \left( 1 - \frac{2GM}{c^2r} \right) \frac{d\phi}{d\lambda}$$For a null, "time-like" vector, $u \cdot u = 0$:
$$- \left( 1 - \frac{2GM}{c^2r} \right)^{-1} e^2 + \left( 1 - \frac{2GM}{c^2r} \right) \left( \frac{dr}{d\lambda} \right)^2 + \frac{l^2}{r^2} \left(1 - \frac{2GM}{c^2r} \right)^{-1} = 0$$

Letting $b = \frac{l}{e}$:

$$\frac{1}{b^2} = \frac{1}{l^2} \left( 1 -\frac{2GM}{c^2r} \right)^2 \left( \frac{dr}{d\lambda} \right)^2 + \frac{1}{r^2}$$

Does this mean that the effective potential $W_{eff} = \frac{1}{r^2}$?

$$\frac{1}{b^2} = \frac{1}{l^2} \left( 1 -\frac{2GM}{c^2r} \right)^2 \left( \frac{dr}{d\lambda} \right)^2 + W_{eff}$$

To find deflection, we find $\frac{d\phi}{dr}$:
$$\frac{d\phi}{dr} = \frac{\frac{d\phi}{d\lambda}}{ \frac{dr}{d\lambda}} = \frac{1}{r^2 \sqrt{ \frac{1}{b^2} - W_{eff} }}$$

$$\Delta \phi = 2 \int_{r_1}^{\infty} \frac{1}{r^2 \sqrt{ \frac{1}{b^2} - W_{eff} }} dr$$

The turning point is when $W_eff(r_1) = \frac{1}{b^2}$, which means $b=r_1$.

$$= 2 \int_1^{0} \frac{w^2}{b^2} \left( \frac{1}{b^2} - \frac{w^2}{b^2} \right)^{-\frac{1}{2}} \cdot \frac{-b}{w^2} dw$$

$$= 2 \int_0^1 \left(1 - w^2\right)^{-\frac{1}{2}} dw$$

$$= 2 \left( \frac{\pi}{2} \right)$$

$$\Delta \phi = \pi$$

So this means no deflection, which is strange.

 

[TSny]

I'm not knowledgeable in this field, but I will just make a comment. The metric you are working with is an example of a "conformally flat" metric. In general, such a metric can be written in the form $g_{\mu \nu} = f \eta_{\mu \nu}$ where $\eta_{\mu \nu}$ is the Minkowski (flat) metric and $f$ is a positive function defined on the space-time manifold. It can be shown that there is no bending of light for a conformally flat metric. This is not too surprising if you note that any null direction for the flat metric $\eta_{\mu \nu}$ is automatically a null direction for the metric $g_{\mu \nu}$ and vice versa.