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Question

[Question Context: Consider the motion of a test particle of (constant) mass $m$ inside the gravitational field produced by the Sun in the context of special relativity.

In addition, consider the equations of motion for the test particle, which can be written as $$\frac{d(m\gamma c)}{dt} = \frac{\vec{v}}{c} \cdot \vec{F}$$

OR

$$\frac{d(m\gamma \vec{v})}{dt} = \vec{F},$$

where $\vec{v}$ is the speed of the test particle, $c$ is the (constant) speed of light, and by definition, $$\gamma \equiv \frac{1}{\sqrt{1- \frac{\vec{v}^2}{c^2}}} .$$

In addition, the gravitational force is given by $$\vec{F} \equiv -\frac{GMm}{r^2} \hat{e}_r$$

where $\hat{e}_r$ is the unit vector in the direction between the Sun (of mass $M$) and the test particle (of mass $m$).]

The Question Itself

Solve the previously found differential equation $$\frac{d^2u}{d\theta^2} + u \bigg( 1- \frac{G^2 M^2}{\ell^2 c^2} \bigg) - \frac{GMd}{\ell^2} = 0$$ for the trajectory, i.e. find the solution for $u(θ)$ (for all $θ$). What kind of trajectories do you find?

Source: [NOT APPLICABLE]


Personal Comment

Perhaps it's just me, however, I can't seem to solve this differential equation in a clean manner. For some reason, I always get a ton of constants and I feel like I am doing something wrong. With that in mind, any assistance, hints, or comments to help me toward the right answer would be much appreciated. Thank you for reading!

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    $\begingroup$ What if you called $$u\left(1 - \frac{G^2 M^2}{\mathcal{l}^2 c^2}\right) - \frac{GMd}{\mathcal{l}^2} = z.$$ What would be the differential equation for $z$? $\endgroup$ – Philip Jan 25 at 18:58
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    $\begingroup$ This question appears to be very similar to your earlier question: physics.stackexchange.com/q/525169/123208 $\endgroup$ – PM 2Ring Jan 25 at 19:45
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    $\begingroup$ You should not get “a ton of constants” from a 2nd-order ODE. You should get two. $\endgroup$ – G. Smith Jan 25 at 20:20
  • $\begingroup$ @PM2Ring, Indeed, it's similar simply because this is considered as a "follow up" question. $\endgroup$ – Athenian Jan 26 at 9:00
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    $\begingroup$ Well, $G$ and $M$ are there too. I was talking about constants of integration, not constants in the ODE to begin with. $\endgroup$ – G. Smith Jan 26 at 18:21
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Re-write for legibility: $$u''(\theta)+\alpha u(\theta)-\beta=0$$ Make a substitution: $$y=\alpha u(\theta)-\beta$$ So: $$y'=\alpha u'$$ And: $$y''=\alpha u''$$ $$\Rightarrow u''=\frac{y''}{\alpha}$$ $$\Rightarrow \frac{y''}{\alpha}+y=0$$ $$y''(\theta)+\alpha y(\theta)=0$$

Which is the classic ODE of the SHM. Solve and back-substitute. Don't neglect the BCs!

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  • $\begingroup$ Out of curiosity, would taking the ansatz: $u = a(1+e cos(bt))$ for some $a$, $b$ and substitute it back to the differential work in acquiring the solution? $\endgroup$ – Athenian Jan 26 at 23:30
  • $\begingroup$ When doing so, a friend said I should get $u = \frac{GMd}{\ell^2} \bigg(1- \frac{G^2 M^2}{\ell^2 c^2}\bigg)^{-1} \bigg(1+ e cos\bigg( \big(1- \frac{G^2 M^2}{\ell^2 c^2} \big) \theta \bigg) \bigg)$ as the answer ... $\endgroup$ – Athenian Jan 26 at 23:32
  • $\begingroup$ Not sure on both counts. It takes a while to work those out! $\endgroup$ – Gert Jan 27 at 2:21

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