Bending of light

We shall calculate the bending of light due to the presence of the sun. We shall use the fact that light takes a path with minimal time. We vary y to find the minimum time.

\($TextStyle = "Helvetica"; 

    locations = {xs -> \(-1.5\), xr -> 1, y0 -> .1, y -> .3}; 

    ListPlot[{{xs, y0}, {0, y}, {xr, y0}, {xs, y0}} /. locations, 
      PlotRange -> {{\(-1.55\), 1.45}, {\(-.1\), .4}}, PlotJoined -> True, 
      AspectRatio -> Automatic, Axes -> False\ , 
      Epilog -> {{Point[{0, 0}], PointSize[.05]}, 
          Text[{xs, y0}, {xs, y0} /. locations, {\(-1\), \(-1\)}], 
          Text[{xr, y0}, {xr, y0} /. locations, {\(-1\), 0}], 
          Text[{0, y}, {0, y} /. locations, {\(-1\), 0}], 
          Text["Sun", {0, 0}, {\(-1\), 0}]}]; \)

I define a simplifying function.

simpleSeries[f_, x_] := Normal[Series[f, x]]

The metric has 4 non-zero terms:

\(gtt = \((1 - 2  M/r)\); 
grr = \(-1\)/\((1 - 2  M/r)\); 

    gθθ = \(-r^2\); 
gϕϕ = \(-r^2\) Sin[θ]^2; 
    \)

Let us change radial coordinates:

substitute = 
      Solve[u == (r - M + Sqrt[r^2 - 2  M r])/2, r][[1]]

gtt = Simplify[(1 - 2  M/r) /. substitute]

guu = 
      Simplify[-1/(1 - 2  M/r) (D[r /. substitute, u])^2 /. 
          substitute]

gθθ = Simplify[-r^2 /. substitute]

Thus the metric is \(TraditionalForm
      \`\(ds\^2 = 
        \(\((\(1 - M\/\(2  u\)\)\/\(1 + M\/\(2  u\)\))\)\^2\) dt\^2 - 
          \(\((1 - M\/\(2  u\))\)\^4\) 
            \((du\^2 + 
                \(u\^2\)(dθ\^2 + \(sin\^2\) θ\ dϕ\^2))\)\ 
      \)\).
We can use a metric that looks Cartesian locally; \(TraditionalForm
      \`ds\^2 = 
        \(\((\(1 - M\/\(2  u\)\)\/\(1 + M\/\(2  u\)\))\)\^2\) dt\^2 - 
          \(\((1 - M\/\(2  u\))\)\^4\) \((dx\^2 + dy\^2 + dz\^2)\)\).
Since M/u is small, expand in terms of it.

gttSMR = simpleSeries[Sqrt[gtt], {M, 0, 1}]

gxxSMR = simpleSeries[Sqrt[-guu], {M, 0, 1}]

The time measured by an observer at infinity is \(TraditionalForm
      \`dt = \((1 + M\/u)\) \((dx\^2 + dy\^2 + dz\^2)\)/\((1 - M\/u)\)\), which to first order is \(TraditionalForm
      \`dt = \((1 + \(2  M\)\/u)\) \((dx\^2 + dy\^2 + dz\^2)\)\). The factor of 2 is important. One comes from the time term and one comes from the spatial, curvature term

The time to go from (x,y) to (0,y) is

t = Integrate[1 + 2  M/u /. u -> Sqrt[xx^2 + y^2], {xx, 0, x}]

Because the distance to the point mass is much larger than the distance of nearest approach,

tSmallY = 
      Part[t, 1] + Part[t, 2] + PowerExpand[Limit[Part[t, 3], y -> 0]]

The time has two parts: x, the distance in the absence of the point mass, and 2  M log[((2  x)/y)],  a correction that scales with the mass.

To find the actual path, consider paths that run from (xs,y0) to (0,y) and then to (xr,y0).  We then minimize the time over y.

deriv = D[tSmallY /. x -> Sqrt[xs^2 + (y - y0)^2], y]

PowerExpand[simpleSeries[deriv /. y -> y0 + δ, {δ, 0, 1}]]

Since δ is proportional to M, the terms in the parenthesis proportional to M are negligible.

%[[1]] + Limit[%[[2]], M -> 0]

Now consider the second part of the path (from (0,y) to (xr,y0)) and solve (dTime/d)δ = 0.

Solve[% + (% /. xs -> xr) == 0, δ][[1]]

Let the distance of the source, which we take to be a star, be infinite compared with that of the earth. The observed quantity is the deflection angle:

angle = Limit[δ/xr /. %, xs -> Infinity]

For the sun, the angle in arcsec for a path grazing its surface,

angle 180 3600/Pi /. {M -> 1.5  km, y0 -> 6.96*^5  km}

22 January 1998