# Painleve orbits in Cartesian Coordinates

## Is Carl going to find the Painleve orbits in Cartesian DE form?

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1. Sep 9, 2006

### CarlB

With the success of my effort to write the orbits of the Schwarzschild metric in "Cartesian" coordinates, (see https://www.physicsforums.com/showthread.php?t=126996 ) it is now time to compute the orbits for Painleve coordinates. When I'm done, I will have an applet that allows the computation of orbits for not just massive and massless particles in the Newton, Einstein/Schwarzschild, and Einstein/Painleve theories, but also tachyons.

The Java applet now has the Newton and Schwarzschild differential equations:
http://www.gaugegravity.com/testapplet/SweetGravity.html

To write the Painleve orbits as a simple differential equation, I again begin with the Painleve metric in Cartesian coordinates. It turns out that there is a paper on the web that tells how to get the Kerr metric (i.e. rotating black hole) into Painleve form in a particularly elegant way, and this paper conveniently includes the Kerr metric in Painleve form in Cartesian coordinate form:

A New Form of the Kerr Solutions
Chirs Doran, Cambridge Geometry Group
A new form of the Kerr solution is presented. The solution involves a time coordinate which represents the local proper time for free-falling observers on a set of simple trajectories. Many physical phenomena are particularly clear when related to this time coordinate. The chosen coordinates also ensure that the solution is well behaved at the horizon. The solution is well suited to the tetrad formalism and a convenient null tetrad is presented. The Dirac Hamiltonian in a Kerr background is also given and, for one choice of tetrad, it takes on a simple, Hermitian form.
http://www.arxiv.org/abs/gr-qc/9910099

Equation (7) of the above paper is the Kerr metric in Painleve form. After removing the rotating part by setting a=0, and normalizing the mass by setting M=1, and eliminating the z coordinate is:

$$ds^2 = \left(1 -\frac{2}{r}\right)dt^2 -\frac{\sqrt{8}}{r^{1.5}}(xdx+ydy)dt -dx^2 -dy^2.$$

This is a particularly simple equation (!!!) and will give very simple equations of motion. Following the method of extremizing the integral of proper time over coordinate time, I need to compute "I" and its various partial derivatives. To make the calculations easier, it helps to factor them so that I can count powers of the radius in the denominator. Because the Painleve metric uses a sqrt(r), it makes sense to count powers in sqrt(r) instead of r, and so I write the equation in $$w=\sqrt{r} = (x^2+y^2)^{0.25}$$ rather than r. The result for I and its needed partial derivatives are simply:

$$\begin{array}{rcl} I &=& (w(w^2-2) - \sqrt{8}(x\dot{x}+y\dot{y})-w^3(\dot{x}^2+\dot{y}^2))/w^3\\ \left(\frac{\partial I}{\partial x}\right) &=& (2xw +\sqrt{18}x(x\dot{x}+y\dot{y})-\sqrt{8}\dot{x}w^4)/w^7\\ \left(\frac{\partial I}{\partial y}\right) &=& (2yw +\sqrt{18}y(x\dot{x}+y\dot{y})-\sqrt{8}\dot{y}w^4)/w^7\\ \left(\frac{\partial I}{\partial \dot{x}}\right) &=& (\sqrt{8}x-2w^3\dot{x})/w^3\\ \left(\frac{\partial I}{\partial \dot{y}}\right) &=& (\sqrt{8}y-2w^3\dot{y})/w^3\\ \left(\frac{\partial^2 I}{\partial \dot{x}^2}\right) &=& -2w^3/w^3\\ \left(\frac{\partial^2 I}{\partial \dot{y}^2}\right) &=& -2w^3/w^3\\ \left(\frac{\partial^2 I}{\partial \dot{x}\dot{y}}\right) &=& 0\\ \left(\frac{\partial^2 I}{\partial x\dot{x}}\right) &=& (x^2-\sqrt{8}y^2)/w^7\\ \left(\frac{\partial^2 I}{\partial y\dot{y}}\right) &=& (y^2-\sqrt{8}x^2)/w^7\\ \left(\frac{\partial^2 I}{\partial x\dot{y}}\right) &=& 0\\ \left(\frac{\partial^2 I}{\partial y\dot{x}}\right) &=& 0 \end{array}$$

In the above, I've left some factors of w in place so that I can count powers easier later on. No less than three of the derivatives are zero, and two more are constants. This is far simpler than the Schwarzschild partial derivatives of post #37:
https://www.physicsforums.com/showpost.php?p=1072836&postcount=37

I think this is going to be simpler than I expected. After I'm done, I may go back and redo the calculation with the full on Kerr metric.

Carl

Last edited: Sep 9, 2006
2. Sep 10, 2006

### CarlB

Counting powers of w in the denominators, I see that A, B, D and E each will have w^6 in their denominators, while C and F will have w^10. I can therefore ignore all the w's in the deonminators of the above partial derivative, and just add a w^4 = r into the denominator of the final result. With this modification, the partial derivatives are all in polynomial form and can be worked on by MAXIMA with ease. The resulting program is:

ii:w*(w^2-2)-sqrt(8)*(x*xx+y*yy)-w^3*(xx^2+yy^2);
didx:2*x*w+sqrt(18)*x*(x*xx+y*yy)-sqrt(8)*xx*w^4;
didy:2*y*w+sqrt(18)*y*(x*xx+y*yy)-sqrt(8)*yy*w^4;
didxx:sqrt(8)*x-2*w^3*xx;
didyy:sqrt(8)*y-2*w^3*yy;
d2idxx2:-2*w^3;
d2idyy2:-2*w^3;
d2idxxdyy:0;
d2idxxdx:x^2-sqrt(8)*y^2;
d2idyydy:y^2-sqrt(8)*x^2;
d2idxxdy:0;
d2idyydx:0;
aa:ii*d2idxx2-0.5*didxx^2;
bb:ii*d2idxxdyy-0.5*didxx*didyy;
cc:(0.5*didxx*didx-ii*d2idxxdx)*xx+(0.5*didxx*didy-ii*d2idxxdy)*yy+ii*didx;
dd:ii*d2idxxdyy-0.5*didyy*didxx;
ee:ii*d2idyy2-0.5*didyy^2;
ff:(0.5*didyy*didy-ii*d2idyydy)*yy+(0.5*didyy*didx-ii*d2idyydx)*xx+ii*didy;
xxxn:bb*ff-cc*ee;
yyyn:cc*dd-aa*ff;
denom:expand(bb*dd-aa*ee);

The result doesn't reduce without hand cranking. I got:
$$denom = -4w^3(\sqrt(2)(x\dot{x}+y\dot{y})(8w^4 + 4w^6(\dot{x}^2+y^2))+w^9(\dot{x}^2+\dot{y}^2+1) + 16w^3(x\dot{x}+y\dot{y})^2))$$

I'm still working on the numerator (i.e. xxxn)...

Carl

Last edited: Sep 10, 2006
3. Sep 13, 2006

### CarlB

I couldn't get the equations to factor, and when I simulated it, it was awful. Knowing there had to be an error, I quickly found that the second derivatives were wrong. The correct values appear to be:

$$\begin{array}{rcl} I &=& (w(w^2-2) - \sqrt{8}(x\dot{x}+y\dot{y})-w^3(\dot{x}^2+\dot{y}^2))/w^3\\ \left(\frac{\partial I}{\partial x}\right) &=& (2xw +\sqrt{18}x(x\dot{x}+y\dot{y})-\sqrt{8}\dot{x}w^4)/w^7\\ \left(\frac{\partial I}{\partial y}\right) &=& (2yw +\sqrt{18}y(x\dot{x}+y\dot{y})-\sqrt{8}\dot{y}w^4)/w^7\\ \left(\frac{\partial I}{\partial \dot{x}}\right) &=& (\sqrt{8}x-2w^3\dot{x})/w^3\\ \left(\frac{\partial I}{\partial \dot{y}}\right) &=& (\sqrt{8}y-2w^3\dot{y})/w^3\\ \left(\frac{\partial^2 I}{\partial \dot{x}^2}\right) &=& -2w^3/w^3\\ \left(\frac{\partial^2 I}{\partial \dot{y}^2}\right) &=& -2w^3/w^3\\ \left(\frac{\partial^2 I}{\partial \dot{x}\dot{y}}\right) &=& 0\\ \left(\frac{\partial^2 I}{\partial x\dot{x}}\right) &=& \sqrt{2}(x^2-2y^2)/w^7\\ \left(\frac{\partial^2 I}{\partial y\dot{y}}\right) &=& \sqrt{2}(y^2-2x^2)/w^7\\ \left(\frac{\partial^2 I}{\partial x\dot{y}}\right) &=& \sqrt{18}xy/w^7\\ \left(\frac{\partial^2 I}{\partial y\dot{x}}\right) &=& \sqrt{18}xy/w^7 \end{array}$$

These numbers simulate to give [more] reasonable orbits. (I have not yet released the updated Java as I expect MAXIMA to succeed in factoring the equations and simplifying the DE to get it into rational form.) But I think that there is yet another error. My suspicion is in the form of the Painleve metric; I will check that next.

And I realized that I can't expect the orbits to match relativity exactly as the initial velocities need to be adjusted. I think I should include this.

Carl

Last edited: Sep 13, 2006
4. Sep 13, 2006

### pervect

Staff Emeritus
Chris Hilbert (CH) has taken an interest in this thread, and has posted some comments on it on my Wikipedia talk page. It's still unclear if CH will be joining PF, though I'm hoping he will - he apparently has some privacy concerns.

In the meantime, his (CH) comments are available at http://en.wikipedia.org/wiki/User_talk:Pervect

5. Sep 14, 2006

### CarlB

pervect,

I can't get the above equations to make decent orbits. They look vaguely relativistic but they don't conserve energy (gasp). They tend to spiral into the center, even when they have way more than enough angular momentum to avoid that fate.

Furthemore, the equations don't reduce like I expect them to. The conversion from polar to Cartesian should be something like this:

$$\begin{array}{rcl} \ddot{x} &=& d(d(r\cos(\theta))/dt)/dt \\ &=& d(\cos(\theta)\dot{r}-r\sin(\theta)\dot{\theta})/dt\\ &=& \cos(\theta)\ddot{r}-r\sin(\theta)\ddot{\theta} -2\sin(\theta)\dot{r}\dot{\theta}-\cos(\theta)(\dot{\theta})^2\\ &=& \frac{x\ddot{r}}{r}-y\ddot{\theta} -2\frac{y\dot{r}\dot{\theta}}{r}-\frac{x(\dot{\theta})^2}{r}\\ &=&(x\ddot{r} -yr\ddot{\theta}-2y\dot{r}\dot{\theta}-x(\dot{\theta})^2)/r \end{array}$$

and

$$\begin{array}{rcl} \dot{x} &=& \cos(\theta)\dot{r} -r\sin(\theta)\dot{\theta})\\ &=&(x\dot{r}-yr\dot{\theta})/r \end{array}$$

and one sees that the Cartesian version of the equations of motion should have a few more power of r in the denominator but be otherwise no more complicated than the Polar coordinate version.

Now MAXIMA tells me that the forms I have for numerator and denominator do not divide like this. And I'm fairly sure that this is the case. But on the other hand, some of the orbits that the computer produces look a lot like a division by zero.

Carl

Last edited: Sep 14, 2006