# Homework Help: Light Deflected by Sun

1. Jan 20, 2014

### Chemp_93

1. The problem statement, all variables and given/known data
Estimate the deflection of starlight by sun using an elementary analysis.

gsun = 275 meters / sec2

Diameter of sun = 1.4 * 109 meters.

In the following, assume that the light just grazes the surface of Sun in passing.

A) Determine an "effective time of fall" from the diameter of Sun and speed of light. From this time of fall deduce the net velocity of fall toward Sun produced by the end of the whole period of gravitational interaction.

2. Relevant equations

C = 3*108 meters/ second

3. The attempt at a solution

This is an odd problem so I have the correct answer.
A) By dimensional analysis and by what is given this is what I did (and got the right answer, but why is this the right answer?)

Diametersun / C = effective time fall = 4.67 seconds (which is correct).

So I imagine the light travelling horizantally to the right (+x direction), then once it grazes the sun it bends slightly towards the -y direction.

But how is the time for this to occur related to the Diameter of the sun??

2. Jan 20, 2014

### Staff: Mentor

The length of the path where the gravitational acceleration is close to (or at least not far below) gsun depends on the diameter.

3. Jan 21, 2014

### Chemp_93

Can you elaborate more on this? Maybe some math will help

4. Jan 21, 2014

### BvU

Actually the diameter of the sun only comes in here because of the 'grazing' : in your x-y picture, the distance of closest approach (to the center of gravity of the sun) is a scale factor for the vertical component of the gravitational force. The rest is simple geometry, with a theta to integrate over from 0 to pi. So what you work out here is the maximum possible deflection case.

5. Jan 21, 2014

### Staff: Mentor

Does this help?

#### Attached Files:

• ###### dimanalysis.png
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6. Jan 21, 2014

### Chemp_93

Thanks I appreciate the picture.

Okay but shouldn't we be interested in using radius rather than diameter in order to calculate the effective time since the gravitational force depends on the distance from the center of mass (center of sun) to the edge of the sun (where particle grazes the sun).

7. Jan 21, 2014

### Staff: Mentor

It is a rough estimate, a factor of 2 does not matter. I guess the diameter is a bit better than the radius as you have deflection both before and after the point of closest approach, but that is beyond the accuracy needed for the problem. A proper calculation would need an integration.

8. Jan 22, 2014

### BvU

Ha! The OP is a longer exercise and we are looking at the first step.

At the point of grazing, the vertical acceleration is at a maximum gsun. Photon is exposed to that only a short time.

Look at the picture by Mentor: even far away the 'negligible force' has a vertical component.
Better to speak of acceleration. Something with a sine in it.
The acceleration itself is gsun R2/r2.
So happens that R/r is the same sine !

Angle θ runs from 0 to -π when the photon travels from far left to far right.

In (a), they first want to know an effective time only. The time a particle has to hang around at the point of grazing to experience the same amount of acceleration towards the center of the sun.

They want us to integrate dx / c from -∞ to +∞ with a weight sin3Θ

(Hint: x = R/tanθ ; change to dθ ) My guess is that the integral gives 2R/c.

Contrary to what Mentor claims: the 2 is still important. Spectacularly so in 1919 when Einstein was proved right...

9. Jan 23, 2014

### Staff: Mentor

My nick is mfb, not Mentor.

Theories cannot be proven, only disproven. An experiment can be in agreement with the theory prediction, but that does not mean the theory has to be right.

The factor of 2 is important if you want a proper calculation. It is irrelevant if you are interested in the order of magnitude (as in "do I have a chance to see the effect at all?").

10. Jan 23, 2014

### BvU

How is Chemp doing ?