# Deflection of light passing the Sun

• Office_Shredder
I think it is more likely that the measured deflection is double the computed value due to general relativity.In summary, the book suggests that the actual measured deflection from some experiments is about 9 times 10-6 radians, and this is double the value that a naive computation returns because of general relativity effects. So the goal is to compute a number that looks like half this. If suggests to do it by:Compute an effective distance traveled over the sun using the diameter of the sun, and multiply that by the maximum acceleration due to gravity the light experiences (i.e. at the surface).This comes to 6.7 times 10-6 radians, which is pretty close to half the total experimentalf

#### Office_Shredder

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Homework Statement
Paraphrasing 2-13 of spacetime physics, an attempt to compute how much light will be deflected by the sun

The sun has a diameter of ##1.4\times 10^9## meters, and the acceleration due to gravity at the surface of the sun is ##275\ m/s^2##. Compute how much light is deflected in radians if it just skims the surface of the sun while passing by
Relevant Equations
Gravitational acceleration of the sun is ##275(r_0/r)^2## when you are ##r## meters from the center, where ##r_0=0.70\times 10^9## is the radius of the sun.
The book says the actual measured deflection from some experiments is about ##9\times 10^{-6}## radians, and this is double the value that a naive computation returns because of general relativity effects. So I guess my goal is to compute a number that looks like half that. If suggests to do it by:

Compute an effective distance traveled over the sun using the diameter of the sun, and multiply that by the maximum acceleration due to gravity the light experiences (i.e. at the surface).

It then says calculus can prove this result is equivalent to the full deflection experienced by traveling past the sun. I wasn't really sure what the effective distance was supposed to mean, so I decided to try the calculus route.

Since the deflection is small, I can treat the light as passing by the sun in an approximately straight line that at its closest is ##0.7\times 10^9=r0## meters from the center of the sun, moving at the speed of light the whole time. If we let 0 be the point on the line closest to the sun and noting that every meter is passed in ##1/(3\times 10^8)## seconds, the total cumulative downward change in velocity can be written as
$$\int_{-\infty}^{\infty} \frac{275 r_0^2}{r_0^2+x^2} \frac{dx}{3e8}$$
$$=\frac{275}{3e8}\int_{-\infty}^{\infty}\frac{1}{1+(x/r_0)^2} dx$$

Substitute ##u=x/r_0## to get
$$=\frac{275}{3e8}r_0 \int_{-\infty}^{\infty}\frac{1}{1+u^2}du = \frac{275}{3e8}r_0 \left( \arctan(\infty)-\arctan(-\infty)\right)$$

This comes to ##\frac{275}{3e8} \pi r_0##. If the light traveled a distance equal to one half the circumference of the sun experiencing ##275\ m/s^2## of downward acceleration the whole time, the downward change in velocity would also be ##275 \frac{\pi r_0}{3e8}## which I guess means that's the interpretation of "effective distance". Then the angle of deflection is going to be approximately the new downward velocity divided by the original speed of light (I realize this is assuming the speed of light is not constant, but I think that doesn't change the result much) which is
$$\frac{275}{9e16} \pi r_0 =6.7\times 10^{-6}$$

This is pretty close to half the total experimental value, but not super convincingly so. Did I do this right and the experimental measured deflections were not really supposed to be 2x what I computed? Or am I missing something here?

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• Delta2
You have input the full Newtonian gravitational acceleration at every point. However, the acceleration orthogonal to the light path takes this value only when the object is closest to the Sun.

• Delta2
Well shoot, I was so focused on constructing a path whose integral I could compute I forgot to figure out the right thing to integrate. At a position of x, the fraction of acceleration which is downwards is going to be ##\frac{r_0}{\sqrt{r_0^2+x_0^2}}##. So the integral becomes

$$\int_{-\infty}^{\infty} \frac{275 r_0^3}{(r_0^2+x^2)^{3/2}} \frac{dx}{3e8}$$

I'll spare you my attempt to type out the trig substitution on my phone but I get about ##4.3\times 10^{-6}## which matches very well with the other claim.

That's said, I'm now uncertain what obvious path we could have pretended the light traveled that would match this answer without having to do any calculus.

Compute an effective distance traveled over the sun using the diameter of the sun, and multiply that by the maximum acceleration due to gravity the light experiences (i.e. at the surface).
This is the problem’s suggestion to get a rough estimate. With the diameter being ##2r_0##, you would obtain ##2r_0 g_0/c## as the estimated transversal velocity change.

Obviously, there is a certain amount of handwavium involved here and it is a coincidence that the correct result is obtained.

• Office_Shredder