Find Geodesics in Dynamic Ellis Orbits Metric

[Onyx]

Well, I wouldn't bother with the square root, but that's up to you. Set $\theta=\pi/2$ for simplicity's sake, which makes $\partial\mathcal{L}/\partial\theta=\partial\mathcal{L}/\partial\dot\theta=0$. What do the three remaining Euler-Lagrange equations give you?

$\partial\mathcal{L}/\partial\dot\phi=r^2d\phi$
$\partial\mathcal{L}/\partial\dot t=\frac{r_s-r}{r}dt$
$\partial\mathcal{L}/\partial\dot r=\frac{r}{r-r_s}dr$

 

[Ibix]

No. $\mathcal{L}=\left(1-\frac{r_S}{r}\right)\dot{t}^2-\left(1-\frac{r_S}{r}\right)^{-1}\dot{r}^2-r^2\dot{\phi}^2$. Where are the differentials coming from on your right hand side?

 

[Onyx]

No. $\mathcal{L}=\left(1-\frac{r_S}{r}\right)\dot{t}^2-\left(1-\frac{r_S}{r}\right)^{-1}\dot{r}^2-r^2\dot{\phi}^2$. Where are the differentials coming from on your right hand side?

$\partial\mathcal{L}/\partial\dot\phi=2r^2d\phi$
$\partial\mathcal{L}/\partial\dot t=2\frac{r_s-r}{r}dt$
$\partial\mathcal{L}/\partial\dot r=2\frac{r}{r-r_s}dr$

 

[Ibix]

Why are there differentials on your right hand sides?

 

[Onyx]

Why are there differentials on your right hand sides?

$\frac{dx^u}{d\tau}$

 

[Onyx]

$\frac{dx^u}{d\tau}$

Just the chain rule applied to the right side when differentiating wrt the dotted variables.

 

[Onyx]

Just the chain rule applied to the right side when differentiating wrt the dotted variables.

$d\phi = \dot\phi$

 

[Ibix]

$d\phi = \dot\phi$

No it doesn't. $\dot{\phi}=d\phi/d\tau$. What is the correct expression for $\partial\mathcal{L}/\partial \dot x^i$? Don't randomly replace symbols with unrelated ones.

 

[Onyx]

No it doesn't. $\dot{\phi}=d\phi/d\tau$. What is the correct expression for $\partial\mathcal{L}/\partial \dot x^i$? Don't randomly replace symbols with unrelated ones.

Just replace the differentials in post #38 with dotted variables.

 

[Onyx]

Just replace the differentials in post #38 with dotted variables.

If not that, then I don't know.

 

[PeterDonis]

Just replace the differentials in post #38 with dotted variables.

No, you replace the differentials with dotted variables. If you want help, you need to do that part of the work yourself. We're not going to auto-correct what you post for you.

 

[Ibix]

Just replace the differentials in post #38 with dotted variables.

I suspect that's correct, but you seem to have some odd ideas about differentials so it would be a good idea for you to write it out.

You will also need the three $\partial\mathcal{L}/\partial x^i$, which I don't think you've posted.

 

[Onyx]

I suspect that's correct, but you seem to have some odd ideas about differentials so it would be a good idea for you to write it out.

You will also need the three $\partial\mathcal{L}/\partial x^i$, which I don't think you've posted.

Oh, I see what you mean. In that case, $\partial\mathcal{L}/\partial \phi=\partial\mathcal{L}/\partial t=0$. Meanwhile, $\partial\mathcal{L}/\partial r$ is a function of $r$.

 

[Ibix]

So what do you now know about $dt/d\tau$ and $d\phi/d\tau$?

 

[Onyx]

Okay, so for the Ellis metric, I get from the EL equation that $\frac{\partial \dot t}{\partial \tau}=-4t\dot \phi^2$ and $\frac{\partial \dot p}{\partial \tau}=5p\dot \phi^2$. But I'm unsure how to proceed. Could I divide the equations to get an equation of $\frac{dp}{dt}$? My thinking is that would take care of the $\dot \phi^2.$

 

[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.