Angle of deviation general relativity, trigo

AI Thread Summary
The discussion centers on understanding the derivation of the angle of deviation of a light beam in a gravitational field, specifically the expression d∅=gocos³(α)/c² *dz. The user seeks clarification on how to arrive at this formula, mentioning that their teacher provided it without detailed explanation. Key concepts include gravitational acceleration, the geometry of light paths, and the relationship between the reference frame and the gravitational field of the sun. Participants emphasize the need for clearer definitions of symbols and the context of the problem to facilitate understanding. The conversation highlights the complexity of relating gravitational effects to light behavior in general relativity.
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Homework Statement


Hello, I have a little problem with this figure, this is not an assignment or something else I just don't understand how we can find that d∅=gocos3(α)/c² *dz

Homework Equations


go=G*M/R² with R the sun's radius

The Attempt at a Solution


I wrote all what I can deduce
ge=gr*cos(α)
r=R/cos(α)
dt=dz/c
tg(∅)≈∅= R/dz
tg(α)=z/R

Please I really need to understand this :frown:
 

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What is this all about, exactly?
 
In the course the teacher advised us to do it and just gave us the expression, so I'm trying to, my exam is Tuesday and I want to understand how we find dphi because it's from this equation, by integration that we find that phi=2*G*M/(c²*R)
Thanks
 
To do "it" - what is "it"? Describe what the drawing represents and what you are supposed to find.
 
"It" is to prove that d∅=go*cos3(α)/c² *dz
It's the angle of deviation of a light beam in the gravitational field of the sun.
What I'm trying to find is this expression d∅=go*cos3(α)/c² *dz from the drawing
Thanks
 
Are you going to feed us one spoon at a time? Unless you explain what all those symbols mean - except ##R## and ##g_0## which you did explain - you won't get much help here.
 
dz is the width of the reference frame
r is the distance between the reference frame and the center of sun
gr gravitational interaction
ge is the effective gravitational acceleration
 
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None of this is making sense. There cannot be a distance to a "reference frame" because a reference frame fills up the entire space, and there is no width of it, for the same reason.

In the context of curving a light beam by the Sun, I would expect a distance source (star) and a receiver (observer) on the Earth. I cannot see any of that on your diagram.
 
In the example in the class the teacher explained that the little squares were local inertial reference frame (as elevators) and as the photon of a star were crossing them they fall in the direction of the sun.
I'm sorry for the lack of informations, I'm trying to remember because my notes are not complete..
 
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