Find Geodesics in Dynamic Ellis Orbits Metric

[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 $\dot \phi^2.$

 

[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.