Algebra/differentiation/trig expansions/ small approximation

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The forum discussion revolves around solving the equation ##\frac{\partial u}{\partial \phi}=-\frac{Me}{J^{2}}sin\phi + \frac{M^{3}}{J^{4}}(-\frac{1}{e}sin\phi+e^{2}sin2\phi+3e\phi cos \phi)## under the assumption that it equals zero at ##\phi=\pi+\epsilon##. The correct solution for ##\epsilon## is established as ##\epsilon=\frac{3M^{2}}{J^{2}}\pi##. Participants emphasize the importance of using small angle approximations, specifically ##\sin(\epsilon) \approx \epsilon## and ##\cos(\epsilon) \approx 1##, to simplify the calculations. Misinterpretations of the equation and the role of the variable ##e## are also discussed, highlighting common pitfalls in mathematical derivation.

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  • #31
binbagsss said:
In fact, the book does a similar exercise a few pages on (for timelike geodesics instead of null) and so the equations differ very slightly.
Is this problem considering a null geodesic? I figured it would be a timelike geodesic since you're comparing it to a Newtonian orbit.

I take it M=mass, J=angular momentum, e=eccentricity, and u=1/r. Is that right?

For what it's worth, I don't see how you're supposed to get the expected result without some other assumption that allows you to neglect more terms.
 
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  • #32
vela said:
Is this problem considering a null geodesic? I figured it would be a timelike geodesic since you're comparing it to a Newtonian orbit.

I take it M=mass, J=angular momentum, e=eccentricity, and u=1/r. Is that right?

For what it's worth, I don't see how you're supposed to get the expected result without some other assumption that allows you to neglect more terms.

Apologies yes it's time-like, I said this the wrong way around.
Yes, with ##e## as given by #27.

In justifying neglecting the ##u^{3}## the book gives an idea of figures for the case of the Earth orbitting the Sun, these are:
##2Mu## ~ ##10^{-8}##
##J^{2}u^{2}##~##10^{-8}##
##2MJ^{2}u^{3}##~##10^{-16}##
 

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