Find Geodesics in Dynamic Ellis Orbits Metric

[Onyx]

$\mathcal{L}=-\dot t^2+\dot p^2+(5p^2+4t^2)\dot\phi^2$
Equation for $t$: $\frac{\partial\mathcal{L}}{\partial t }-\frac{d}{d\tau}\frac{\partial\mathcal{L}}{\partial\dot t}=0$
$\frac{\partial\mathcal{L}}{\partial\dot t}=-2\dot t$
$\frac{\partial\mathcal{L}}{\partial t}=8t\dot\phi^2$
$\frac{d}{d\tau}(-2\dot t)=-2\ddot t$

Therefore $8t\dot\phi^2=-2\ddot t$

Equation for $\rho$: $\frac{\partial\mathcal{L}}{\partial\rho }-\frac{d}{d\tau}\frac{\partial\mathcal{L}}{\partial\dot\rho}=0$
$\frac{\partial\mathcal{L}}{\partial\dot\rho}=2\dot\rho$
$\frac{\partial\mathcal{L}}{\partial\rho}=10\rho\dot\phi^2$
$\frac{d}{d\tau}(2\dot\rho)=2\ddot\rho$

Therefore $10\rho\dot\phi^2=2\ddot\rho$

Equation for $\phi$: $\frac{\partial\mathcal{L}}{\partial\phi }-\frac{d}{d\tau}\frac{\partial\mathcal{L}}{\partial\dot\phi}=0$
$\frac{\partial\mathcal{L}}{\partial\dot\phi}=(10\rho^2+8t^2)\dot\phi$
$\frac{\partial\mathcal{L}}{\partial\phi}=0$
$\frac{d}{d\tau}((10\rho^2+8t^2)\dot\phi)=(20\rho\dot\rho+16t\dot t)\dot\phi+(10\rho^2+8t^2)\ddot\phi$

Therefore $(20\rho\dot\rho+16t\dot t)\dot\phi+(10\rho^2+8t^2)\ddot\phi=0$

 

[PeterDonis]

@Onyx, I think you mistyped a couple of the $\phi$ equations. The first one should have $\partial \mathcal{L} / \partial \dot{\phi}$ on the LHS. The last one should have $0$ on the RHS.

Also, you seem to be writing both $\rho$ and $p$ for the same coordinate.

 

[Onyx]

@Onyx, I think you mistyped a couple of the $\phi$ equations. The first one should have $\partial \mathcal{L} / \partial \dot{\phi}$ on the LHS. The last one should have $0$ on the RHS.

Also, you seem to be writing both $\rho$ and $p$ for the same coordinate.

Let me fix that real quick.

 

[Onyx]

Let me fix that real quick.

I think everthing is corrected now.

 

[PeterDonis]

I think everthing is corrected now.

Yes, looks that way.

 

[Onyx]

Yes, looks that way.

This is the same final answer you got?

 

[PeterDonis]

@Onyx, so in post #71 you now have the geodesic equations for all three coordinates. Since you have a constant of the motion involving $\dot{\phi}$, you could, if you wanted, reduce them to two equations, for $\ddot{t}$ and $\ddot{\rho}$, by substituting the value of $\dot{\phi}$ in terms of the constant of the motion in those equations as you have them.

 

[Onyx]

@Onyx, so in post #71 you now have the geodesic equations for all three coordinates. Since you have a constant of the motion involving $\dot{\phi}$, you could, if you wanted, reduce them to two equations, for $\ddot{t}$ and $\ddot{\rho}$, by substituting the value of $\dot{\phi}$ in terms of the constant of the motion in those equations as you have them.

That just seems strange, given that I would have equations of $\ddot t$, $t$, $\ddot\rho$, $\rho$, but without $\dot t$ or $\dot\rho$. I'm not sure how I would deal with that, but I'll give it a try.

 

[Onyx]

That just seems strange, given that I would have equations of $\ddot t$, $t$, $\ddot\rho$, $\rho$, but without $\dot t$ or $\dot\rho$. I'm not sure how I would deal with that, but I'll give it a try.

Is there is any way to merge those two equations together at this point? Because otherwise I would have to deal with $t$ in the $\ddot\rho$ and $\rho$ in the $\ddot t$. It doesn't look like there is a way, though.

 

[PeterDonis]

That just seems strange, given that I would have equations of $\ddot t$, $t$, $\ddot\rho$, $\rho$, but without $\dot t$ or $\dot\rho$. I'm not sure how I would deal with that, but I'll give it a try.

Equations of the form $\ddot{q} = F(q)$, where $F(q)$ is some function of $q$, are fairly common in physics; there's nothing very strange about "skipping" the $\dot{q}$ term.

Is there is any way to merge those two equations together at this point?

I would not expect there to be, because fundamentally the metric depends on two coordinates, $t$ and $\rho$, and there's no way to eliminate that. (In Schwarzschild spacetime, the reason everything can be reduced to a single effective potential equation is that, taking into account the spherical symmetry, the only meaningful coordinate dependence in the metric is on $r$.)

 

[PeterDonis]

This is the same final answer you got?

Yes, except that I didn't bother actually evaluating $d / d\tau ( \partial \mathcal{L} / \partial \dot{\phi} )$. I just left the expression for $\partial \mathcal{L} / \partial \dot{\phi}$ as a constant of the motion.

 

[Onyx]

Yes, except that I didn't bother actually evaluating $d / d\tau ( \partial \mathcal{L} / \partial \dot{\phi} )$. I just left the expression for $\partial \mathcal{L} / \partial \dot{\phi}$ as a constant of the motion.

Okay, so I guess it's a set of coupled second-order differential equations.

 

[Ibix]

Yes. And it's only the spherical symmetry that lets you have as few as two coupled ones, I'm afraid.

 

[Onyx]

I think I may have something. I can use the method described here where $E=\frac{1}{2}[\dot x^2+\dot y^2]+V(x,y)$, and then as he says express everything in terms of $r$ and $\phi$. In the description of variable separation for spherical coordinates on Wikipedia, the final result is in term of $r$ and $S_r$ only, but I still don't understand what $S$ or $S_r$ represent in this case. I think I can assume that the final equation can be independent of $\phi$, though.

 

[Ibix]

Handle with care. $\rho$ and $t$ are not Cartesian coordinates, so I'm not at all sure that you can merrily convert to polars without consequence. Note also that the $\phi$ in the StackExchange reply is not the same as the $\phi$ in your metric, and solutions will most definitely depend on it.

 

[Onyx]

Is there anything wrong with defining a coordinate $r^2=t^2+\rho^2$ and $\theta=arctan(\frac{y}{x})$? After all, both $t$ and $p$ can be negative, so I would think that one could visualize them as perpendicular axes. When substituting that into the line element, (including $\frac{d(rcos\theta)}{d\lambda}$ and $\frac{d(rsin\theta)}{d\lambda}$), and considering that I already have an equation of $r$ in terms of $\lambda$, and seeing as how there are trig identities that cancel out when expanding, I am left with an equation of just $\theta$ and $\lambda$. I hope this is meaningful, but as you said, they don't have a Cartesian relationship, so not sure.

 

[Ibix]

At some point you still need to apply a constraint that $g_{ij}\dot x^i\dot x^j$ is 1 (or 0). Given that the space you are working in is locally Minkowski not locally Euclidean, those trigonometric functions may induce nasty maths down the line.

 

[Onyx]

At some point you still need to apply a constraint that $g_{ij}\dot x^i\dot x^j$ is 1 (or 0). Given that the space you are working in is locally Minkowski not locally Euclidean, those trigonometric functions may induce nasty maths down the line.

I was already setting the line element to $0$, and after chain-ruling and expanding I subtracted the $\theta$ stuff to the other side.

 

[Ibix]

I was already setting the line element to $0$

I don't see where you did that.

 

[Onyx]

$0=-dt^2+dp^2+\frac{L^2}{5p^2+4t^2}$

Sorry, I didn't write that down recently, but I thought I posted that near the beginning of the thread. But anyway it sounds like you are saying that because $t$ and $\rho$ have a non-Euclidean relationship, what I'm suggesting wouldn't work. Is that what you're saying?

 

[Ibix]

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.

 

[Onyx]

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.

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?

 

[Ibix]

By the way, what do you mean by LHS?

Left Hand Side. It's a completely standard abbreviation.