# Predicting angle of deflection of light

1. May 2, 2015

### William Henley

Hello,
I was recently reading relativity the general and special theory by Albert Einstein. When I came across the equation to calculate the angle of deflection for light as it approaches a massive body. I was just wondering if someone could explain why this works and also what "1.7 seconds of arc" means, as it is part of the equation.

2. May 2, 2015

### Staff: Mentor

The theory was made to match observations. What exactly do you mean with "this"?

Small angles are described with minutes and seconds of arc. 1 degree = 60 minutes = 3600 seconds.

3. May 2, 2015

### William Henley

Oh what I mean is what do the parts of the equation mean.

4. May 2, 2015

### William Henley

Oh yeah thanks, so is this what used to predict the angle of deflection?

5. May 2, 2015

### Staff: Mentor

That depends on the equations, there are different ways to derive that result and I don't know which set of equations you are looking at. I also don't know how much you know about general relativity.
This is used to quote the result. It is just a unit, like meters for length, seconds for time or degrees for angles.
Instead of 1.7 seconds of arc, you can also say "an angle of 1.7/3600 degrees".

6. May 2, 2015

### William Henley

I'm talking about a=1.7 seconds of arc over delta

7. May 2, 2015

### Janus

Staff Emeritus
The formula you are talking about comes from the formula

$$\theta = \frac{4GM}{rc^2}$$

Where r is the distance from The center of M that the light passes. The answer will be in radians

Basically if you use the mass of the Sun for M, set delta equal to 1 solar radius for r, and convert it to give an answer in second of arc rather than radians, you get

$$\theta = \frac{1.7}{\Delta}$$ seconds of arc.

For the angle of deflection of light passing at a distance of delta from the center of the Sun where delta is measured in sun radii.

8. May 3, 2015

### William Henley

Ok yeah that makes sense, thanks for the answer.