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CarlB

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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 [Broken]

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:

Chirs Doran, Cambridge Geometry Group

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:

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

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 [tex]w=\sqrt{r} = (x^2+y^2)^{0.25}[/tex] rather than r. The result for I and its needed partial derivatives are simply:

[tex]\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}[/tex]

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

The Java applet now has the Newton and Schwarzschild differential equations:

http://www.gaugegravity.com/testapplet/SweetGravity.html [Broken]

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:

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

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 [tex]w=\sqrt{r} = (x^2+y^2)^{0.25}[/tex] rather than r. The result for I and its needed partial derivatives are simply:

[tex]\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}[/tex]

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

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