Some questions about the derivation steps in the Gravitational deflection of light section in Schutz

In summary, the conversation discusses the derivation of equation (11.53) from equation (11.49). The speaker initially thought that the added term of ##O(M^2u^2)## was obtained from neglecting ##2Mu^3## due to ##Mu\ll 1## and approximating ##y\approx u##. However, the correct derivation involves using the expression ##(1-2Mu)^{-1}## and the fact that ##y^2 = u^2(1-2Mu)##. This results in the final equation of ##\mathrm{11.53}##. The speaker acknowledges their mistake and thanks the other person for correcting them.
  • #1
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My question is referring to some derivation from pages 293-294 of Schutz's second edition (2009) of A First Course in GR.
In the screenshots below there are the equations (11.49) and (11.53).

I don't understand how did he derive equation (11.53) from Eq.(11.49)?
From (11.49) I get: ##d\phi/dy= d\phi/du du/dy = (1/b^2-u^2+2Mu^3)^{-1/2}(1+2My)##.
It seems he neglected the ##2Mu^3## since ##Mu\ll 1##, so ##y\approx u##, but how do we get the added term of ##O(M^2u^2)##?

schutz2.pngschutz1.png
schutz1.png
schutz2.png
 
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  • #2
I would have thought\begin{align*}
\dfrac{d\phi}{dy} = \dfrac{d\phi}{du} \left( \dfrac{dy}{du} \right)^{-1} &= \dfrac{( 1 - 2Mu)^{-1} }{\left( b^{-2} - u^2(1-2Mu) \right)^{1/2}}
\end{align*}then because ##y^2 = u^2(1-Mu)^2 \sim u^2(1-2Mu)## and also ##(1-2Mu)^{-1} \sim 1+2Mu + O(M^2u^2) \sim 1 + 2My + O(M^2u^2)## you get ##\mathrm{11.53}##...
 
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  • #3
ergospherical said:
I would have thought\begin{align*}
\dfrac{d\phi}{dy} = \dfrac{d\phi}{du} \left( \dfrac{dy}{du} \right)^{-1} &= \dfrac{( 1 - 2Mu)^{-1} }{\left( b^{-2} - u^2(1-2Mu) \right)^{1/2}}
\end{align*}then because ##y^2 = u^2(1-Mu)^2 \sim u^2(1-2Mu)## and also ##(1-2Mu)^{-1} \sim 1+2Mu + O(M^2u^2) \sim 1 + 2My + O(M^2u^2)## you get ##\mathrm{11.53}##...
It seems I got it wrong. Thanks for correcting me.
 

1. What is the concept of gravitational deflection of light?

The gravitational deflection of light refers to the bending of light rays as they pass through a region of strong gravitational field, such as near a massive object like a star or a black hole. This effect was first predicted by Albert Einstein's theory of general relativity.

2. How is the gravitational deflection of light calculated?

The gravitational deflection of light can be calculated using the equation Δθ = 4GM/rc², where G is the gravitational constant, M is the mass of the object causing the deflection, r is the distance of closest approach, and c is the speed of light. This equation is derived from the principles of general relativity.

3. What are the main steps in the derivation of gravitational deflection of light?

The main steps in the derivation of gravitational deflection of light involve using the principle of equivalence, which states that an accelerating frame of reference is equivalent to a gravitational field, and the principle of general covariance, which states that the laws of physics should be the same in all reference frames. These principles are used to derive the geodesic equation, which describes the path of a light ray in a curved space-time.

4. What is the significance of the gravitational deflection of light?

The gravitational deflection of light is a key prediction of general relativity and has been observed and confirmed through various experiments and observations. It also has important implications in astrophysics, as it allows us to study and understand the properties of massive objects, such as black holes, by observing the bending of light around them.

5. Are there any practical applications of the gravitational deflection of light?

Yes, the gravitational deflection of light has practical applications in fields such as astronomy and navigation. In astronomy, it allows us to study distant objects and their gravitational fields, while in navigation, it is used to correct for the bending of light in GPS systems to ensure accurate location tracking.

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