B Find Geodesics in Dynamic Ellis Orbits Metric

[Onyx]

TL;DR

Orbits in dynamic Ellis Wormhole

Does anyone see a way I can find geodesics in the metric $ds^2=-dt^2+dp^2+(5p^2+4t^2)d\phi^2$ (ones with nonzero angular momentum)? I'm hoping it can be done analytically, but that may be wishful thinking. FYI, this is the metric listed at the bottom of the Wikipedia article about Ellis Wormholes.

 

[PeterDonis]

Is there a reason why you can't use the techniques already mentioned in the previous thread you had on geodesics?

 

[PeterDonis]

the Wikipedia article about Ellis Wormholes

Do you mean this article?

https://en.wikipedia.org/wiki/Ellis_wormhole

 

[Onyx]

Do you mean this article?

https://en.wikipedia.org/wiki/Ellis_wormhole

Yes.

 

[Onyx]

Is there a reason why you can't use the techniques already mentioned in the previous thread you had on geodesics?

The techniques mentioned in the FLRW metric paper?

 

[Onyx]

The techniques mentioned in the FLRW metric paper?

What techniques are you referring to?

 

[PeterDonis]

What techniques are you referring to?

You had a previous thread on geodesics in the FLRW metric that discussed techniques for solving for geodesics. See, for example, post #6 of that thread:

https://www.physicsforums.com/threa...geodesics-in-flrw-metric.1047783/post-6827495

 

[Ibix]

It may help to note that $\partial_\phi$ is manifestly a Killing field, so should yield a conserved quantity that you can probably use to eliminate derivatives of $\phi$.

 

[PeterDonis]

It may help to note that $\partial_\phi$ is manifestly a Killing field, so should yield a conserved quantity that you can probably use to eliminate derivatives of $\phi$.

The geodesic Lagrangian method that I referred to in the previous thread automatically does this kind of elimination.

 

[Onyx]

The geodesic Lagrangian method that I referred to in the previous thread automatically does this kind of elimination.

I have not tried this yet. But there are no $t$ or $p$ killing fields, so I don't know what I would do with the $t$ and $p$ even if I did.

 

[Onyx]

I have not tried this yet. But there are no $t$ or $p$ killing fields, so I don't know what I would do with the $t$ and $p$ even if I did.

Would doing the Lagrangian method produce equations that are more solvable than $ds^2=-dt^2+dp^2+\frac{L^2}{5p^2+4t^2}$?

 

[PeterDonis]

there are no $t$ or $p$ killing fields, so I don't know what I would do with the $t$ and $p$ even if I did.

The geodesic Lagrangian method will tell you, probably more easily for this metric than the brute force way of computing the geodesic equation. That's because your metric has only one function of the coordinates (the coefficient of $d\phi^2$ is a function of $t$ and $p$), so most of the Christoffel symbols vanish. The geodesic Lagrangian method automatically filters out those cases and focuses attention on the small number of cases where the Christoffel symbols don't vanish and the geodesic equation has some meaningful content.

 

[PeterDonis]

Would doing the Lagrangian method produce equations that are more solvable than $ds^2=-dt^2+dp^2+\frac{L^2}{5p^2+4t^2}$?

I don't know what you mean by "more solvable". The metric isn't an equation you "solve" in the first place. You certainly can't just "solve" the metric if you want to find the geodesics. You have to solve the geodesic equation.

Basically it seems like you are looking for some magical shortcut that will tell you the answer without you actually having to do the math. There isn't one.

 

[PeterDonis]

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

This is a different metric from the one you gave in the OP. Which one are you using?

 

[Onyx]

I don't know what you mean by "more solvable". The metric isn't an equation you "solve" in the first place. You certainly can't just "solve" the metric if you want to find the geodesics. You have to solve the geodesic equation.

Basically it seems like you are looking for some magical shortcut that will tell you the answer without you actually having to do the math. There isn't one

I should have said that I'm wondering if there is a way to integrate geodesics exactly instead of using Euler or Runge-Kutta.

This is a different metric from the one you gave in the OP. Which one are you using?

This is the metric:
$ds^2=-dt^2+dp^2+(5p^2+4t^2)d\phi^2$

.

 

[PeterDonis]

I should have said that I'm wondering if there is a way to integrate geodesics exactly instead of using Euler or Runge-Kutta.

Some differential equations can be integrated exactly and some can't. The only way to find out which kind you're dealing with is to look at the equations.

This is the metric:
$ds^2=-dt^2+dp^2+(5p^2+4t^2)d\phi^2$

Ok.

 

[Onyx]

I don't know what you mean by "more solvable". The metric isn't an equation you "solve" in the first place. You certainly can't just "solve" the metric if you want to find the geodesics. You have to solve the geodesic equation.

Basically it seems like you are looking for some magical shortcut that will tell you the answer without you actually having to do the math. There isn't one

I should have said that I'm wondering if

The geodesic Lagrangian method will tell you, probably more easily for this metric than the brute force way of computing the geodesic equation. That's because your metric has only one function of the coordinates (the coefficient of $d\phi^2$ is a function of $t$ and $p$), so most of the Christoffel symbols vanish. The geodesic Lagrangian method automatically filters out those cases and focuses attention on the small number of cases where the Christoffel symbols don't vanish and the geodesic equation has some meaningful content.

Is the Lagrangian method only for timelike geodesics?

.

 

[PeterDonis]

Is the Lagrangian method only for timelike geodesics?

The Wikipedia article describes it that way in the section I referenced, but you can actually use it for any kind of geodesic (with some technical complications required to justify the method for null geodesics, not discussed in the article). The key point is that the quantity whose variation is set to zero is the same regardless of the kind of geodesic.

 

[Onyx]

The Wikipedia article describes it that way in the section I referenced, but you can actually use it for any kind of geodesic (with some technical complications required to justify the method for null geodesics, not discussed in the article). The key point is that the quantity whose variation is set to zero is the same regardless of the kind of geodesic.

And you would recommend this method over just looking at $dt^2=dp^2+(5p^2+4t^2)d\phi^2$ in the null case?

 

[PeterDonis]

And you would recommend this method over just looking at $dt^2=dp^2+(5p^2+4t^2)d\phi^2$ in the null case?

What would just looking at that equation do for you?

 

[PeterDonis]

@Onyx, it seems like you're spending an awful lot of time asking questions instead of actually trying anything. Why don't you just try some things? The metric you're talking about is simple enough that the difference in computational effort between different methods isn't that much, and none of them will take a huge amount of effort. You could just try them all and see what works best.

 

[Ibix]

The geodesic Lagrangian method that I referred to in the previous thread automatically does this kind of elimination.

It is, indeed, quite neat. I just tried it for the Schwarzschild metric; you set $\theta=\pi/2$ and it immediately give you that $g_{\phi\phi}\frac{d\phi}{d\tau}=\mathrm{const}$ and $g_{tt}\frac{dt}{d\tau}=\mathrm{const}$. The $r$ equation is messy but you can decline to solve it and just plug the above results into $g_{ab}\dot x^a\dot x^b$ to get $dr/d\tau$ in terms of the effective potential.

I guess it's messier for a spacetime that isn't so symmetric as Schwarzschild, or with a poorer choice of coordinates, but it's still nice.

 

[Onyx]

It is, indeed, quite neat. I just tried it for the Schwarzschild metric; you set $\theta=\pi/2$ and it immediately give you that $g_{\phi\phi}\frac{d\phi}{d\tau}=\mathrm{const}$ and $g_{tt}\frac{dt}{d\tau}=\mathrm{const}$. The $r$ equation is messy but you can decline to solve it and just plug the above results into $g_{ab}\dot x^a\dot x^b$ to get $dr/d\tau$ in terms of the effective potential.

I guess it's messier for a spacetime that isn't so symmetric as Schwarzschild, or with a poorer choice of coordinates, but it's still nice.

Excellent. Can you post a walkthrough of how you did it with Schwarzschild?

 

[Ibix]

Excellent. Can you post a walkthrough of how you did it with Schwarzschild?

Do you know what the Lagrangian is? (Hint: see post #7.) Do you know the Euler-Lagrange equation? (Hint: google is your friend.)

 

[PeterDonis]

It may help to note that $\partial_\phi$ is manifestly a Killing field, so should yield a conserved quantity that you can probably use to eliminate derivatives of $\phi$.

For the non-dynamic Ellis wormhole (i.e., no $t$ dependence in the metric), this works easily. For the dynamic case (the one in the OP, with $t$ dependence in the metric), it's more complicated, because the "conserved" quantity you get is explicitly time-dependent.

 

[Ibix]

the "conserved" quantity you get is explicitly time-dependent.

Presumably because the circumference of a circle at fixed $p$ (which I think is meant to be $\rho$, by the way) is varying with time and that affects the conservation of angular momentum?

 

[PeterDonis]

Presumably because the circumference of a circle at fixed $p$ (which I think is meant to be $\rho$, by the way) is varying with time and that affects the conservation of angular momentum?

I think that would be a reasonable physical interpretation, yes.

 

[Onyx]

Do you know what the Lagrangian is? (Hint: see post #7.) Do you know the Euler-Lagrange equation? (Hint: google is your friend.)

Do you mean the method of defining the Lagrangian and then differentiate it w.r.t the one-forms ($dt, d\rho, d\phi$)?

 

[Ibix]

Yes.

 

[PeterDonis]

Do you mean the method of defining the Lagrangian and then differentiate it w.r.t the one-forms ($dt, d\rho, d\phi$)?

You don't differentiate with respect to the one-forms. You differentiate with respect to the coordinates and their derivatives with respect to the affine parameter.

Yes.

I don't think so. See above.

 

[Onyx]

Their derivatives with respect to the affine parameter.

This is what I meant by one-forms

 

[PeterDonis]

This is what I meant by one-forms

Then you should be aware that that is not correct terminology.

 

[Ibix]

So, @Onyx, what is the relevant Lagrangian for GR, and for Schwarzschild spacetime specifically?

 

[Onyx]

So, @Onyx, what is the relevant Lagrangian for GR, and for Schwarzschild spacetime specifically?

So, @Onyx, what is the relevant Lagrangian for GR, and for Schwarzschild spacetime specifically?

$L=\sqrt{{g_{uv}}u'v'}$

 

[Ibix]

$L=\sqrt{{g_{uv}}u'v'}$

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?

 

[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 $\dot \phi^2.$