MHB Coupled non-linear system of ODEs

[topsquark]

If you really want to know where this comes from I am solving the GR equations for a rectilinearly isotropic metric. In other words, I can express the metric as [math]d \tau ^2 = -T(x) dt^2 + X(x) dx^2 + dy^2 + dz^2[/math]. (It may be simpler to use a cylindrical coordinate system, but the equations come out easier this way. Cylindrical coordinates are in the batter's box, so to speak.)

Anyway no differential geometry is involved here...I simply need help in integrating the results:
[math]0 = -\frac{1}{2} \frac{d}{dx} \left ( \frac{T'}{T} \right ) + \frac{1}{4} \frac{T'^2}{XT} - \frac{1}{4} \frac{X'T'}{X^2}[/math]

[math]0 = \frac{1}{2} \frac{d}{dx} \left ( \frac{T'}{T} \right ) + \frac{1}{4} \frac{T'^2}{T^2} - \frac{1}{4} \frac{X'T'}{XT}[/math]

(The prime denotes a derivative with respect to x.)

The system of equations above is underconstrained. I am actually solving the GR equations backward...I know the motion I want and I need to find the metric based on that, so at this stage of the game I have no initial conditions on T or X.

Coupled DEQs are not a comfortable problem for me. Any help would be appreciated.

-Dan

 

[Ackbach]

Well, I've made some progress, I hope. Rewriting:
\begin{align*}
0&=-2\frac{d}{dx}\left(\frac{T'}{T}\right)+\frac{T'}{X}\left[\frac{T'}{T}-\frac{X'}{X}\right] \\
0&=2\frac{d}{dx}\left(\frac{T'}{T}\right)+\frac{T'}{T}\left[\frac{T'}{T}-\frac{X'}{X}\right].
\end{align*}
Let $\tau=T'/T$ and $\xi=X'/X$. The equations become
\begin{align*}
0&=-2\tau'+\frac{T'}{X}[\tau-\xi] \\
0&=2\tau'+\tau[\tau-\xi] \\
&\text{or} \\
2\tau'&=\frac{T'}{X}[\tau-\xi] \\
2\tau'&=-\tau[\tau-\xi].
\end{align*}
If we set the two RHS's equal, we obtain
$$\frac{\tau T}{X}[\tau-\xi]=-\tau[\tau-\xi],$$
where I've used $T'=\tau T$. Throwing everything on the same side yields
$$\left(\frac{T}{X}+1\right)\tau[\tau-\xi]=0.$$
There are three solutions to this equation:
\begin{align*}
T&=-X \\
\tau&=0 \\
\tau&=\xi.
\end{align*}
$\tau=0$ would probably lead to the trivial solution ($T$ and $X$ are both nonzero constants - this is a solution, by the way). I haven't projected out what $T=-X$ means, or what $\tau=\xi$ means, but they are definitely worth checking out.

 

[Ackbach]

According to my calculations, if we take the $T=-X$ solution, then $T'=-X'$, and hence
$$\frac{T'}{T}=\frac{X'}{X},$$
which is identical to the $\tau=\xi$ condition. That is, this is a double-root, essentially. Plugging this into the second DE yields
$$0=2\frac{d}{dx}\left(\frac{T'}{T}\right)+\left(\frac{T'}{T}\right)^{\!2}-
\left(\frac{T'}{T}\right)^{\!2}\quad\implies\quad \frac{T'}{T}=C.$$
The solution here is $T=T_{0}e^{Cx}$, and hence $X=-T_{0}e^{Cx}$. This is a non-trivial solution to the system of ODE's.

The first DE in the system will not, under these conditions, give you any other information.

 

[topsquark]

According to my calculations, if we take the $T=-X$ solution, then $T'=-X'$, and hence
$$\frac{T'}{T}=\frac{X'}{X},$$
which is identical to the $\tau=\xi$ condition. That is, this is a double-root, essentially. Plugging this into the second DE yields
$$0=2\frac{d}{dx}\left(\frac{T'}{T}\right)+\left(\frac{T'}{T}\right)^{\!2}-
\left(\frac{T'}{T}\right)^{\!2}\quad\implies\quad \frac{T'}{T}=C.$$
The solution here is $T=T_{0}e^{Cx}$, and hence $X=-T_{0}e^{Cx}$. This is a non-trivial solution to the system of ODE's.

The first DE in the system will not, under these conditions, give you any other information.

Very interesting. Yesterday I tweeked out the solution for T'/T = constant as an assumption. I had two reasons to do that, one to simplify the ODEs and one to simplify the motion equations, which I didn't post here. I got (almost) the same solutions you did so maybe the system isn't underconstrained after all.

On the other hand, here is my solution. Assume [math]\frac{T'}{T} = C[/math]. ie. [math]T = a e^{Cx}[/math]. Putting this into the first equation gives
[math]0 = -\frac{1}{2} \frac{d}{dx} (C) + \frac{1}{4} \frac{C^2 T^2}{X T} - \frac{1}{4} \frac{X'CT}{X^2}[/math]

This has the solution [math]X' = CX \implies X = b e^{Cx}[/math]. But this isn't your solution T = -X. (The second equation merely reduces to an identity so gives no more information.) I can clearly see where your negative sign comes from, but even after plugging this back into the metric and taking the limit I get b = a, not b = -a. I don't know how to explain why these are different.

-Dan

Edit: After some thought I see that both are actually solutions representing different behavior. Based on the equations both solutions are valid. Based on the dynamics I have to throw away the X = -T solution, but that is beyond the system of equations I posted. Why doesn't your solution method, which is far more general than mine, predict my solution??

 

[Ackbach]

Very interesting. Yesterday I tweeked out the solution for T'/T = constant as an assumption. I had two reasons to do that, one to simplify the ODEs and one to simplify the motion equations, which I didn't post here. I got (almost) the same solutions you did so maybe the system isn't underconstrained after all.

On the other hand, here is my solution. Assume [math]\frac{T'}{T} = C[/math]. ie. [math]T = a e^{Cx}[/math]. Putting this into the first equation gives
[math]0 = -\frac{1}{2} \frac{d}{dx} (C) + \frac{1}{4} \frac{C^2 T^2}{X T} - \frac{1}{4} \frac{X'CT}{X^2}[/math]

This has the solution [math]X' = CX \implies X = b e^{Cx}[/math]. But this isn't your solution T = -X. (The second equation merely reduces to an identity so gives no more information.) I can clearly see where your negative sign comes from, but even after plugging this back into the metric and taking the limit I get b = a, not b = -a. I don't know how to explain why these are different.

-Dan

Edit: After some thought I see that both are actually solutions representing different behavior. Based on the equations both solutions are valid. Based on the dynamics I have to throw away the X = -T solution, but that is beyond the system of equations I posted. Why doesn't your solution method, which is far more general than mine, predict my solution??

Well, a couple of things come to mind. First of all, this is a highly nonlinear system of ODE's. There are certainly no theorems of which I am aware that predict that it's even possible to come up with THE general solution.

Second of all, I think my method does predict your solution: it's in the $\tau=\xi$ part. So I think I may have spoken too soon when I said that the $T=-X$ solution is the same as the $\tau=\xi$ solution. Let's examine this more closely:
\begin{align*}
\tau&=\xi \\
\frac{T'}{T}&=\frac{X'}{X} \\
\ln|T|&=\ln|X|+C.
\end{align*}
Full stop. It's the magnitudes of $T$ and $X$ that we get here. So if $C=0$ here, then you could have either $T=X$ or $T=-X$. So this is where your solution is covered.

 

[topsquark]

Well, a couple of things come to mind. First of all, this is a highly nonlinear system of ODE's. There are certainly no theorems of which I am aware that predict that it's even possible to come up with THE general solution.

Second of all, I think my method does predict your solution: it's in the $\tau=\xi$ part. So I think I may have spoken too soon when I said that the $T=-X$ solution is the same as the $\tau=\xi$ solution. Let's examine this more closely:
\begin{align*}
\tau&=\xi \\
\frac{T'}{T}&=\frac{X'}{X} \\
\ln|T|&=\ln|X|+C.
\end{align*}
Full stop. It's the magnitudes of $T$ and $X$ that we get here. So if $C=0$ here, then you could have either $T=X$ or $T=-X$. So this is where your solution is covered.

Ah, I got stabbed with my own sword. I typically have to remind my students of the absolute value. :)

Okay, that makes a lot more sense. Just to give you the background: This wasn't my original problem, which was to find a mass distribution that would create a constant acceleration, but it got side-tracked into a problem I studied a few years ago and gave up on...the gravitational equivalent of the classic EM problem of the motion of a charge above an infinite plane of constant surface charge density. I know more Math now and I have discovered that the exponential in the metric is likely related to the existence of a continuous mass density reaching to infinity. GR doesn't play well with large gravitational fields over "small" spans of space-time and this problem has large gravitational fields over all space so the metric can be expected to be somewhat psychotic. (And if the coupled equations I posted here are non-linear you ought to see the equations of motion!)

I could see if there is a "Schwarzschild radius" for this problem but I likely won't bother except as an exercise...given the symmetry there shouldn't be a finite escape speed. My next project is to do this same problem using cylindrical coordinates. I can more easily remove the mass distribution at infinity by terminating it with a cut-off at r = R. Introducing a step function potential into the metric is the first goal of my overall problem, which is to add an additional constant acceleration (that of a rocket firing) while an object is falling into a black hole.

Thanks for the help!

-Dan