Predicting angle of deflection of light

[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.

 

[mfb]

I was just wondering if someone could explain why this works

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.

 

[William Henley]

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.

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

 

[William Henley]

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

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

 

[mfb]

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

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.

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

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".

 

[William Henley]

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".

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

 

[Janus]

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

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.

 

[William Henley]

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.

Ok yeah that makes sense, thanks for the answer.