I The deflection Φ of light around a massive body...

1. Aug 28, 2016

Clive Redwood

In deriving the expression for the deflection Φ of light around a massive body, invoking Huygens' principle, Einstein arrived at the equation:

dΦ/dx2 = dγ/dx1

where γ is the speed of light in units of c, and the xμ are space coordinates.

How is this equation arrived at?

2. Aug 28, 2016

Staff: Mentor

Do you have a reference for where this equation comes from?

3. Aug 28, 2016

Clive Redwood

Foundation of the General Theory of Relativity, Einstein

4. Aug 29, 2016

Staff: Mentor

Can you be more specific? Chapter, section, page?

5. Aug 29, 2016

Clive Redwood

The equation is:

B = ∫(∂γ/∂x1)dx2

where B is the deflection, γ is the speed of light in units of c, and the subscript notation is used to index tensors.
This is found on page 163 in Section 22 of The Principle of Relativity, Dover Publications, 1952.

6. Aug 29, 2016

pervect

Staff Emeritus
I found this online at http://hermes.ffn.ub.es/luisnavarro/nuevo_maletin/Einstein_GRelativity_1916.pdf, it's on pg 199 of that version (still section 22). EInstein's treatment isn't very modern, if you're seriously studying GR I'd suggest a more modern textbook.

A modern approach to this problem would emphasize solving the geodesic equations for the particular case of light. Sean Caroll has some lecture notes that take this approach, if there's some interest I'll post a link, though it'd be easy enough to find if you search for it.

Einstein's rather old approach is basically choosing a particular coordinate system, computing the coordinate speed of light in those particular coordinates (this is also direction-dependent for, say, the Schwarzschild coordinates, though Einstein doesn't emphasize the point), then invoking Hugyen's principle to say that the bending of light can be regarded as a local phenomenon, so we can treat space as locally being flat.

At least I think that's what he's doing - it's not the approach I'd use, and I don't think I've fully justified to myself why it gives the right answer. In particular the fact that space is curved disturbs me, the application of Huygen's principle to suggest that this doesn't matter over a short enough distance seems a bit suspect. Perhaps it turns out to be only approximately true.

7. Aug 29, 2016

Clive Redwood

Einstein actually arrives at his stellar game changing prediction by solving the null geodesic equation (eq. 73)

gμνxμxν = 0

under the conditions of Huygens' principle expressed as

dB = (∂ϒ/∂x1)dx2

Both equations are expressed in tensors. So I don't think a special coordinate system was implied.

At this level, the notational changes of modern GR is of little relevance it seems to me.

8. Aug 30, 2016

pervect

Staff Emeritus
I don't see how $\gamma$ can be a tensor - from the lack of indices, it'd have to be a scalar. Among other issues, in plain Schwarzschild coordinates, $\gamma$ is different in the radial and non-radial directions. (One need isotropic coordinates to make the speed of light the same in all directions). So I don't think $\gamma$ can possibly be a coordinate-independent scalar.

Thus I don't see how Einstein's approach (which I have to admit that 'm not familiar with) could be described as being "a tensor equation", if $\gamma$ isn't a tensor.

The modern approaches I to calculate the deflection of light that I am famimlar with involve either using the geodesic equation directly, Hamilton-Jacobi equations, or even one simple-minded approach that uses dimensional analysis. Writing (& solving) the geodesic equations is probably the most direct approach, though potentially rather tedious.

9. Aug 30, 2016

vanhees71

The idea is just to calculate the null geodesics of the Schwarzschild pseudometric. The problem is that you cannot use directly the variational principle to use it for symmetry analysis. The trick is to consider time-like geodesics and take the limit to light-like geodesics as soon as the EoM (including the conservation laws as derived from symmetries, of which we have plenty for the Schwarzschild pseudometric).

For time-like geodesics we can use the proper time, defined by $\mathrm{d}s^2=g_{\mu \nu} \mathrm{d}q^{\mu} \mathrm{d} q^{\nu}$ as the world-line parameter. In any case we then have for the Lagrangian
$$L= \sqrt{g_{\mu \nu} \dot{q}^{\mu} \dot{q}^{\nu}},$$
and we can use the modified action principle with $\tilde{L}=L^2/2$ as the Lagrangian. Then the action reads
$$A=\int \mathrm{d} \lambda \frac{1}{2} g_{\mu \nu} \dot{q}^{\mu} \dot{q}^{\nu}.$$
Since now $\tilde{L}$ is quadratic in the $\dot{q}$ and since it doesn't depend explicitly on $\lambda$, the "Hamiltonian"
$$\dot{q}^{\mu} p_{\mu}-\tilde{L}=\tilde{L}=\text{const}$$
is automatically fulfilled, and we can just work with the action priniple without taking the constraint into account explicitly. From now on I write $L$ instead of $\tilde{L}$ for simplicity.

For the Schwarzschild case you have
$$L=\frac{1}{2} \left [A(r) \dot{t}^2 -\frac{1}{A(r)} \dot{r}^2 - r^2 (\dot{\vartheta}^2+\sin^2 \vartheta \dot{\varphi}^2)\right],$$
where
$$A(r)=1-\frac{r_s}{r}, \quad r_s=\frac{2GM}{c^2}.$$
Since $t$ is cyclic, the corresponding generalized momentum is conserved, which implies
$$E=\frac{\partial L}{\partial \dot{t}}=A \dot{t}=\text{const}.$$
From rotational invariance it's clear that the "angular momentum" is conserved either. This implies that the orbit is a plane orbit and we can choose the coordinate system such that $\vartheta=\pi/2=\text{const}$. To verify this we calculate the EoM for $\vartheta$:
$$p_{\vartheta}=\frac{\partial L}{\partial \dot{\vartheta}}=-r^2 \dot{\vartheta},$$
from which
$$\dot{p}_{\vartheta}=-2r \dot{r} \dot{\vartheta}-r^2 \ddot{\vartheta}=\frac{\partial L}{\partial \vartheta}=-2 r^2 \sin \vartheta \cos \vartheta,$$
and indeed $\vartheta=\pi/2=\text{const}$ is a solution. The other apparent constant solutions $\vartheta=0,\pi$ are excluded, because the coordinates have a singularity there (the usual singularity of spherical coordinates along the polar axis).

Further $\varphi$ is cyclic and thus
$$p_{\varphi}=\frac{\partial L}{\partial \dot{\varphi}}=-r^2 \sin \vartheta \dot{\varphi}=-r^2 \dot{\varphi}=-\ell=\text{const}.$$
Now we have enough first integrals, in terms of which the Lagrangian reads
$$L=\frac{1}{2} \left (\frac{E-\dot{r}^2}{A}-\frac{l^2}{r^2} \right).$$
Now for a null geodesic $L=0=\text{const}$, and we get
$$\dot{r}^2=E-\frac{\ell^2 A}{r^2}.$$
The perihelion is determined by $\dot{r}=0$, i.e., $\ell^2/r_P^2=E/A_P$. For the trajectory we divide by $\dot{\varphi}^2=\ell^2/r^4$, which after some simple steps leads to
$$r^{\prime 2}=\frac{r^2}{r_P^2} (A_P r^2-A r_P^2).$$
The total change of the angle for the incoming light beam is
$$\Phi=2 \int_{r_P}^{\infty} \mathrm{d} r \frac{r_P}{r} \sqrt{\frac{1}{A_P r^2-A r_P^2}}.$$
This is an elliptic integral, but for the case of our sun, where along the trajectory of the light beam $r \geq r_P \geq r_s$ we can use $\epsilon=r_s/r_P$ as expansion parameter. We have, again after some algebra
$$A_P r^2-A r_P^2=(r^2-r_P^2) \left [1-\epsilon \left (\frac{r_P}{r}+\frac{r}{r+r_P} \right) \right].$$
Since the term $\propto \epsilon$ is $\ll 1$ for all $r \in [r_P,\infty)$ we can expand the integrand to first power in $\epsilon$ and then evaluate the approximate integral analytically
$$\Phi=2 \int_{r_P}^{\infty} \mathrm{d} r \frac{r_P}{r \sqrt{r^2-r_P^2}} \left [1+ \frac{\epsilon}{2} \left (\frac{r_P}{r}+\frac{r}{r+r_P} \right) \right]=\pi + 2 \epsilon + \mathcal{O}(\epsilon^2).$$
The deviation from a straight line $\Phi=\pi$ is
$$\Delta \Phi=\Phi-\pi=2 \epsilon=\frac{2r_s}{r_P},$$
which is Einstein's final result of 1916.