B Find Geodesics in Dynamic Ellis Orbits Metric

[PeterDonis]

Okay, so for the Ellis metric, I get from the EL equation that $\frac{\partial \dot t}{\partial \tau}=-4t\dot \phi^2$ and $\frac{\partial \dot p}{\partial \tau}=5p\dot \phi^2$.

No, that's not quite what you get from the EL equations. Since you haven't shown your work I can't tell where you went wrong, but you went wrong somewhere. Note that you should get three equations, not two; even though the metric does not depend on $\phi$, the EL equation for $\dot{\phi}$ still has non-negligible content.

 

[Onyx]

$\mathcal{L}=-\dot t^2+\dot p^2+\frac{1}{5p^2+4t^2}$
$\frac{\partial\mathcal{L}}{\partial t}=\frac{d}{d\tau}\frac{\partial\mathcal{L}}{\partial\dot t}$
$\frac{\partial\mathcal{L}}{\partial t}=-2\frac{d}{d\tau}\dot t$
$-\frac{8t}{(5p^2+4t^2)^2}=-2\frac{d}{d\tau}\dot t$
$\frac{4t}{(5p^2+4t^2)^2}=\frac{d}{d\tau}\dot t$

 

[Onyx]

$\mathcal{L}=-\dot t^2+\dot p^2+\frac{1}{5p^2+4t^2}$
$\frac{\partial\mathcal{L}}{\partial t}=\frac{d}{d\tau}\frac{\partial\mathcal{L}}{\partial\dot t}$
$\frac{\partial\mathcal{L}}{\partial t}=-2\frac{d}{d\tau}\dot t$
$-\frac{8t}{(5p^2+4t^2)^2}=-2\frac{d}{d\tau}\dot t$
$\frac{4t}{(5p^2+4t^2)^2}=\frac{d}{d\tau}\dot t$

And so it follows that...
$-\frac{5p}{(5p^2+4t^2)^2}=\frac{d}{d\tau}\dot p$

 

[Ibix]

You are missing a $\dot\phi^2$ in the last term in your expression for $\mathcal{L}$, so your expressions for $d\dot p/d\tau$ and $d\dot t/d\tau$ are both incorrect by that factor.

As Peter says, you also need the $\phi$ Euler-Lagrange equation which will enable you to eliminate $\dot\phi$ from the other two equations if you use it.

 

[PeterDonis]

$\mathcal{L}=-\dot t^2+\dot p^2+\frac{1}{5p^2+4t^2}$
$\frac{\partial\mathcal{L}}{\partial t}=\frac{d}{d\tau}\frac{\partial\mathcal{L}}{\partial\dot t}$
$\frac{\partial\mathcal{L}}{\partial t}=-2\frac{d}{d\tau}\dot t$
$-\frac{8t}{(5p^2+4t^2)^2}=-2\frac{d}{d\tau}\dot t$
$\frac{4t}{(5p^2+4t^2)^2}=\frac{d}{d\tau}\dot t$

As @Ibix pointed out, the last term in your $\mathcal{L}$ needs a $\dot{\phi}^2$. Also I don't understand why $5p^2 + 4t^2$ is in the denominator.

Your second line does correctly state the Euler-Lagrange equation for one coordinate, namely $t$. There are two others, for $p$ and $\phi$.

Your third line does correctly evaluate $d / d\tau ( \partial \mathcal{L} / \partial \dot{t} )$.

Your fourth and fifth lines, however, do not correctly evaluate $\partial \mathcal{L} / \partial t$.

 

[Onyx]

And so it follows that...
$-\frac{5p}{(5p^2+4t^2)^2}=\frac{d}{d\tau}\dot p$

But if I treated the lagrangian

As @Ibix pointed out, the last term in your $\mathcal{L}$ needs a $\dot{\phi}^2$. Also I don't understand why $5p^2 + 4t^2$ is in the denominator.

Your second line does correctly state the Euler-Lagrange equation for one coordinate, namely $t$. There are two others, for $p$ and $\phi$.

Your third line does correctly evaluate $d / d\tau ( \partial \mathcal{L} / \partial \dot{t} )$.

Your fourth and fifth lines, however, do not correctly evaluate $\partial \mathcal{L} / \partial t$.

I forgot to mention that I made the angular momentum $1$, which is why $5p^2+4t^2$ is in the denominator.

 

[PeterDonis]

I forgot to mention that I made the angular momentum $1$

Which is wrong. The Lagrangian does not specify particular values for any conserved quantities.

Do you understand how to convert an expression for the metric into a Lagrangian?

 

[Ibix]

I forgot to mention that I made the angular momentum $1$, which is why $5p^2+4t^2$ is in the denominator.

So you are using the remaining Euler-Lagrange equation, $\frac d{d\tau}\frac{d\mathcal{L}}{d\dot\phi}=0$, to imply $2\dot\phi(5p^2+4t^2)=\mathrm{const}$, then setting the constant to 1 and substituting into your expression for $\mathcal{L}$? In that case, I think you actually set your angular momentum to 2. And you should definitely make this calculation explicit rather than leaving usto guess that you are treating a special case.

It would make more sense to use the $\phi$ Euler-Lagrange equation to get the conservation law (as I did in my last paragraph) and then substitute it in to your other expressions.

 

[PeterDonis]

So you are using the remaining Euler-Lagrange equation, $\frac d{d\tau}\frac{d\mathcal{L}}{d\dot\phi}=0$, to imply $2\dot\phi(5p^2+4t^2)=\mathrm{const}$, then setting the constant to 1 and substituting into your expression for $\mathcal{L}$?

A more pertinent question would be, is doing that correct? And the answer to that is no.

 

[Ibix]

A more pertinent question would be, is doing that correct? And the answer to that is no.

Doesn't it just constrain the Lagrangian to apply to the case where the angular momentum is as specified? My maths suggests it works out correctly (at least in this case), but I did it with Maxima on my phone so the possibility of arithmetic slips is non-zero.

 

[PeterDonis]

Doesn't it just constrain the Lagrangian to apply to the case where the angular momentum is as specified?

You could look at it that way, but that just begs the question of why you would want to do that, particularly if you are after general equations for geodesics, as the OP is asking. General equations means general, not restricted to some particular choice of values for conserved quantities.

Also note that calling the conserved quantity associated with $\dot{\phi}$ "angular momentum" is, at least IMO, somewhat strange in this case, since this so-called "conserved" quantity is time-dependent. There are discussions in the literature of this type of case, which is called an "explicitly time-dependent constant of the motion", which has always seemed to me to be an oxymoron.

 

[Ibix]

You could look at it that way, but that just begs the question of why you would want to do that,

No argument from me here.

Also note that calling the conserved quantity associated with $\dot{\phi}$ "angular momentum" is, at least IMO, somewhat strange in this case, since this so-called "conserved" quantity is time-dependent. There are discussions in the literature of this type of case, which is called an "explicitly time-dependent constant of the motion", which has always seemed to me to be an oxymoron.

I'm not sure I see what your point is here, now. I think we agree that $\dot\phi=\mathrm{const}/2(5p^2+4t^2)$ from the $\phi$ Euler-Lagrange equation. Surely the angular momentum is the constant, and the "time varying" bit is in the relationship between $\dot\phi$ and the constant? Is that fundamentally different from the angular momentum and angular velocity having a relationship that depends on the radial distance, like in Schwarzschild spacetime?

 

[Onyx]

I'm actually not completely sure how to get the correct lagrangian from the line element. It looks like it is sometimes treated as the square root of the line element, but other times not. But Ibix seems to think the square root isn't necessary.

 

[PeterDonis]

I think we agree that $\dot\phi=\mathrm{const}/2(5p^2+4t^2)$ from the $\phi$ Euler-Lagrange equation

Yes.

Surely the angular momentum is the constant, and the "time varying" bit is in the relationship between $\dot\phi$ and the constant?

And $p$, yes.

Is that fundamentally different from the angular momentum and angular velocity having a relationship that depends on the radial distance, like in Schwarzschild spacetime?

It's different in that the $t$ coordinate is timelike, whereas $r$ in Schwarzschild spacetime is spacelike (outside the horizon). Whether that counts as a "fundamental" difference is a different question.

It does suggest a possible comparison that might be instructive, though: look at the angular momentum in Schwarzschild spacetime inside the horizon in Schwarzschild coordinates, where $r$ is timelike. You still have the same formula for angular momentum, but now the coordinate it depends on is timelike.

 

[Onyx]

I don't know how significant this is, but the article calls $rho$ the "geodesic radius" and $r$ the "circumferential radius. $r=\displaystyle \pm \sqrt{5p^2+4t^2}$.

 

[Onyx]

Ibix, if I understand what you said about substitution in post #58, I can say that $\frac{4t(constant)}{5p^2+4t^2}=-2\frac{d}{d\tau}\dot t$ and $\frac{5p(constant)}{5p^2+4t^2}=-2\frac{d}{d\tau}\dot p$. Not sure if that's what you intended, though.

 

[Onyx]

Ibix, if I understand what you said about substitution in post #58, I can say that $\frac{4t(constant)}{5p^2+4t^2}=-2\frac{d}{d\tau}\dot t$ and $\frac{5p(constant)}{5p^2+4t^2}=-2\frac{d}{d\tau}\dot p$. Not sure if that's what you intended, though.

This is based on my post #50. I'm not sure it's exactly right, but it looks at least like the right idea, with only $t$ and $p$ and their second derivatives. Is this the final answer you guys got?

 

[Ibix]

It looks like it is sometimes treated as the square root of the line element, but other times not. But Ibix seems to think the square root isn't necessary.

The Euler-Lagrange equations find trajectories that extremise $\int\mathcal{L}d\tau$. If there are no sign changes in $\mathcal{L}$ along an acceptable path (and there can't be in this case because $\mathcal{L}=g_{ab}\dot x^a\dot x^b$ is either 0 or $\pm$1 for paths of interest) then extreme values of the integral of something must also be extreme values of the integral of the square root of that something. And (at least in this case) the maths is a lot less messy without the root.

This is based on my post #50. I'm not sure it's exactly right, but it looks at least like the right idea, with only $t$ and $p$ and their second derivatives. Is this the final answer you guys got?

It's difficult to help you because you have bits of your work scattered across different posts, some with implicit assumptions and some without. It would help if you would state any assumptions you are making, and state your Lagrangian, your "raw" Euler-Lagrange equations, and your final results all in one place. Otherwise you have to wait until I have a chance to fire up my laptop and make notes.

 

[PeterDonis]

It looks like it is sometimes treated as the square root of the line element, but other times not. But Ibix seems to think the square root isn't necessary.

It isn't for the purpose you are using it, to find the geodesics. Sometimes it's more convenient, other times (as in this case, as @Ibix says), it isn't.

state your Lagrangian, your "raw" Euler-Lagrange equations, and your final results all in one place

@Onyx, this is excellent advice.

 

[Onyx]

It isn't for the purpose you are using it, to find the geodesics. Sometimes it's more convenient, other times (as in this case, as @Ibix says), it isn't.@Onyx, this is excellent advice.

Okay, I'll put it all in my next post.

 

[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.