Find Geodesics in Dynamic Ellis Orbits Metric

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SUMMARY

The discussion focuses on finding geodesics in the dynamic Ellis Orbits metric defined by the equation ##ds^2=-dt^2+dp^2+(5p^2+4t^2)d\phi^2##, particularly for cases with nonzero angular momentum. Participants reference the geodesic Lagrangian method, which simplifies the process by eliminating certain derivatives due to the presence of a Killing field, ##\partial_\phi##. The conversation emphasizes that while some differential equations can be integrated exactly, the complexity of the metric requires careful analysis to determine solvability. The geodesic equation must be solved rather than the metric itself.

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  • #91
I'm sure you have written it down. But where did it enter your differential equations? What would be different about them if the LHS was not zero? Or to put it another way, if you had used ##\mathcal{L}'=\mathcal{L}\pm 1## instead?

I don't see why the substitution wouldn't work, as long as you understand that you've introduced a third angular coordinate that isn't an angle in spacetime in any meaningful sense. I just suspect that inserting you null/timelike/spacelike constraint will be messy.

I'd be very doubtful of anything passing through the origin, but that's an issue with the spacetime not the maths.
 
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  • #92
Ibix said:
I'm sure you have written it down. But where did it enter your differential equations? What would be different about them if the LHS was not zero? Or to put it another way, if you had used ##\mathcal{L}'=\mathcal{L}\pm 1## instead?

I don't see why the substitution wouldn't work, as long as you understand that you've introduced a third angular coordinate that isn't an angle in spacetime in any meaningful sense. I just suspect that inserting you null/timelike/spacelike constraint will be messy.

I'd be very doubtful of anything passing through the origin, but that's an issue with the spacetime not the maths.

Ibix said:
I'm sure you have written it down. But where did it enter your differential equations? What would be different about them if the LHS was not zero? Or to put it another way, if you had used ##\mathcal{L}'=\mathcal{L}\pm 1## instead?

I don't see why the substitution wouldn't work, as long as you understand that you've introduced a third angular coordinate that isn't an angle in spacetime in any meaningful sense. I just suspect that inserting you null/timelike/spacelike constraint will be messy.

I'd be very doubtful of anything passing through the origin, but that's an issue with the spacetime not the maths.
Actually, I was wrong about substituting the trig functions yielding the desired result. Honestly I feel like ##t^2+p^2=f(\lambda)## might be as far as I can get in terms of anything exact. But that is probably enough. By the way, what do you mean by LHS?
 
  • #93
Onyx said:
By the way, what do you mean by LHS?
Left Hand Side. It's a completely standard abbreviation.
 

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