# Schwarzschild Orbits in Cartesian coordinates

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

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Chris Hillman
Continuing (my post grew too long!):

the Dirac equation in a gravitational field is particularly simple in Painleve-Cartesian coordinates. This may not make a hill of beans to a GR specialist,

For other readers: I think Carl was being disingenuous (or sarcastic) here, but for the record, simplifying the Dirac equation on the Schwarzschild or Kerr vacuums, or other "favorite gtr models", is in fact a "minor industry" (meaning that these topics are of considerable interest to researchers working in the field of gravitation physics!)--- maybe that's the "industry" Carl had in mind when he mentioned this term in an earlier post.

But nevertheless, this is a subject of interest in physics at this time, and this is precisely the language that is used to describe it. The reason is sort of subtle and has to do with the fact that GR can be derived from a graviton theory on a flat space.

My first guess was that Carl was referring here to Deser's approach to obtaining a "local mimic" of gtr by starting with a naive quantization formulated on a flat spacetime background, then fixing up inconsistencies, noticing new inconsistencies, fixing those, and so on until, miraculously, one winds up with gtr. (This is discussed briefly in MTW, including a citation to the original paper.)

However,
An example of a recent paper in the literature using "Gullstrand-Painleve-Cartesian" coordinates is:
http://arxiv.org/abs/gr-qc/0411060
The above has 12 citations and gives easily understood reasons for looking at things this way that can be understood at a high-school level.
the Hamilton/Lisle eprint (I seem to have overlooked this--- thanks for bringing it to my attention!) seems, at a glance, to refer to the flat metric inherited by the constant Painleve time hyperslices. They write "picture space as flowing like a river into the Schwarzschild black hole", which suggests to me they are thinking of projecting $\vec{e}_0$ into $T=0$ to form a flow (in the sense of differential equations on manifolds).

(The world lines of the Doran-Lemaitre observers in the Kerr vacuum form an irrotational congruence, hence a hypersurface orthogonal congruence, but the orthogonal hyperslices no longer have vanishing three-dimensional Riemann tensor, so this presumably would not apply to the Kerr generalization.)

Continuing with Carl's program, his transformation yields the line element
$$ds^2 = -\left(1-\frac{2m}{\sqrt{x^2+y^2+z^2}} \right) \, dT^2 + 2 \, \frac{\sqrt{2m}}{(x^2+y^2+z^2)^{3/4}} \, \left( x \, dx + y \, dy + z \, dz \right) \; dT + dx^2 + dy^2 + dz^2,$$
$$-\infty < T, \, x, \, y, \, z < \infty, \; x^2+y^2+z^2 \neq 0$$
The Lemaitre frame field becomes:
$$\vec{e}_0 = \partial_T - \frac{\sqrt{2m}}{(x^2+y^2+z^2)^{3/4}} \; \left( x \, \partial_x + y \, \partial_y + z \, \partial_z \right)$$
$$\vec{e}_1 = \frac{1}{\sqrt{x^2+y^2+z^2}} \; \left( x \, \partial_x + y \, \partial_y + z \, \partial_z \right)$$
$$\vec{e}_2 = \frac{z}{\sqrt{x^2+y^2+z^2} \, \sqrt{x^2+y^2}} \; \left( x \, \partial_x + y \, \partial_y \right) - \frac{\sqrt{x^2+y^2}}{\sqrt{x^2+y^2+z^2}} \; \partial_z$$
$$\vec{e}_3 = \frac{1}{\sqrt{x^2+y^2}} \; \left( -y \partial_x + x \partial_y \right)$$
To see this, we simply write down the coordinate vectors of the old chart in terms of the new one and plug these into our previous expressions for the frame field. The easiest way to transform a coordinate vector is to simply apply it to the new coordinates, which are simply functions on our manifold! For example,
$$\partial_r z = \partial_r (r \cos(\theta)) = \cos(\theta) = \frac{z}{\sqrt{x^2+y^2+z^2}}$$
and so forth, so that
$$\partial_r = \frac{1}{\sqrt{x^2+y^2+z^2}} \; \left( x \, \partial_x + y \, \partial_y + z \, \partial_z \right)$$
and so on.

The geodesic equations turn out to be a bit messy. It's much easier to state what the four standard Killing vectors look like:
$$\partial_T, \; -y \partial_x + x \partial_y, \; -z \partial_y + y \partial_z, \; -x \partial_z + z \partial_x$$
We should in fact take this simple result as the defining characteristic of "the cartesian-type Painleve chart" for our static spherically symmetric vacuum solution-- maybe that was Carl's point in one of his remarks above.

The first two geodesic equations are
$$0 = \ddot{T} + \frac{\sqrt{2m^3}}{(x^2+y^2+z^2)^{5/4}} \, \dot{T}^2 + \frac{2 m \, \dot{T}}{(x^2+y^2+z^2)^{3/2}} \; \left( x \dot{x} + y \dot{y} + z \dot{z} \right)$$
$$\hspace{0.5in} -\frac{\sqrt{m/2}}{(x^2+y^2+z^2)^{7/4}} \; \left( (x^2-2\,(y^2+z^2)) \, \dot{x}^2 + (y^2-2\,(z^2+x^2)) \, \dot{y}^2 + (z^2-2\,(x^2+y^2)) \, \dot{z}^2 \right)$$
$$\hspace{0.5in} + \frac{3 \, \sqrt{2m}}{(x^2+y^2+z^2)^{7/4}} \; \left( x y \, \dot{x} \dot{y} + y z \, \dot{y} \dot{z} + z x \, \dot{z} \dot{x} \right)$$
and
$$0 = \ddot{x} - m x \, \frac{2m - \sqrt{x^2+y^2+z^2}}{(x^2+y^2+z^2)^2} \, \dot{T}^2 - \frac{2 \sqrt{2 m^3} \, x \, \dot{T}}{(x^2+y^2+z^2)^{9/4}} \; \left( x \dot{x} + y \dot{y} + z \dot{z} \right)$$
$$\hspace{0.5in} -\frac{m x}{(x^2+y^2+z^2)^{5/2}} \; \left( (x^2-2 \, (y^2+z^2)) \, \dot{x} + (y^2-2 \, (z^2+x^2)) \, \dot{y} + (z^2-2 \, (x^2+y^2)) \, \dot{z} \right)$$
$$\hspace{0.5in} - \frac{6 m x}{(x^2+y^2+z^2)^{5/2}} \; \left( x y \, \dot{x}\dot{y} + y z \, \dot{y}\dot{z} + z x \, \dot{z}\dot{x} \right)$$
(similarly for $\ddot{y}, \, \ddot{z}$), where the dots denote differentiation wrt the affine parameter. These expressions are not particularly simple, but they are at least symmetric between the coordinates x,y,z (as must be the case). As a check, note that in relativistic units, the right hand sides have the dimensions of inverse length.

If I understand correctly, Carl's intention is not simply to write down the geodesic equations and numerically integrate them and plot results in illuminating ways (I enthusiastically second his recommendation of Andrew Hamilton's website, BTW; see the link at RelWWW below), but to express affine parameterized null geodesics (or their "tracks" in a constant time slice) in closed form. Most textbooks show how to do this (using the Schwarzschild exterior chart) in the exterior region, but to my knowledge none discuss how to do it in the future interior region (using for example ingoing Eddington or Painleve charts).

As a check, hardy readers can compute the connection one-forms and curvature two-forms, read off the Riemann components with respect to the Lemaitre frame (don't forget that this is an anholonomic frame!), and compute the characteristic polynomial, which is of course an invariant. If you do this correctly, you will obtain
$$\lambda^6 - \frac{6 m^2}{(x^2+y^2+z^2)^3} \, \lambda^4 + \frac{4 m^3}{(x^2+y^2+z^2)^{9/2}} \, \lambda^3 + \frac{9 m^4}{(x^2+y^2+z^2)^6} \, \lambda^2$$
$$\hspace{0.5in} - \frac{12 m^5}{(x^2+y^2+z^2)^{15/2}} \, \lambda + \frac{4 m^6}{(x^2+y^2+z^2)^9}$$
Of course this should agree with the result of simplying plugging Carl's transformation into the characteristic computed in the original Schwarzschild chart, and it does! Similarly, the principle Lorentz invariants of the Riemann tensor are
$$R_{abcd} \, R^{abcd} = \frac{48 m^2}{(x^2+y^2+z^2)^3}, \; \; R_{abcd} \, (\star R)^{abcd} = 0$$
(no intrinsic gravitomagnetism in the sense of Ciufolini and Wheeler).

By the way, it is easy to rationalize the original Painleve chart. Set
$$\tau=T, \; \rho = \sqrt{r}, \; \eta = \sin(\theta)$$
Writing $M=\sqrt{2m}$, the line element becomes
$$ds^2 = -(1-M^2/\rho^2) \, d\tau^2 + 4 M \, d\tau d\rho + 4\rho^2 \, d\rho^2 + \rho^4 \; \left( \frac{d\eta^2}{1-\eta^2} + \eta^2 \, d\phi^2 \right),$$
$$-\infty < \tau < \infty, \; 0 < \rho < \infty, \; 0 < \eta < 1, \; -\pi < \phi < \pi$$
The geodesic equations become
$$\ddot{\tau} + \frac{2 M^2}{\rho^3} \, \dot{\tau} \dot{\rho} + \frac{M^3}{2 \rho^5} \, \dot{\tau}^2 + \frac{2M}{\rho} \, \dot{\rho}^2 - M \rho \; \left( \frac{\dot{\eta}^2}{1-\eta^2} + \eta^2 \dot{\phi}^2 \right) = 0$$
$$\ddot{\rho} - \frac{M^3}{\rho^5} \, \dot{\tau} \dot{\rho} + (\rho^2-M^2) \; \left( \frac{M^2 \, \dot{\tau}^2}{4 \rho^7} + \frac{\dot{\rho}^2}{\rho^3} + \frac{1}{2 \rho} \; \left( \frac{\dot{\eta}^2}{1-\eta^2} + \frac{\dot{\phi}^2}{\eta^2} \right) \right) = 0$$
$$\ddot{\eta} + \frac{4}{\rho} \, \dot{\rho} \dot{\eta} + \eta \; \left( \frac{\dot{\eta}^2}{1-\eta^2} - (1-\eta)^2 \, \dot{\phi}^2 \right) = 0$$
$$\ddot{\phi} + 2 \, \dot{\phi} \; \left( \frac{2 \dot{\rho}}{\rho} + \frac{\dot{\eta}}{\eta} \right) = 0$$
In the second equation, note that the horizon is located at $\rho=M$, so this equation simplifies greatly there!

In the equatorial plane $\eta=1$, the geodesic equations become
$$\ddot{\tau} + \frac{2 M^2}{\rho^3} \, \dot{\tau} \dot{\rho} + \frac{M^3}{2 \rho^5} \, \dot{\tau}^2 + \frac{2M}{\rho} \, \dot{\rho}^2 - M \rho \, \dot{\phi}^2 \right) = 0$$
$$\ddot{\rho} - \frac{M^3}{\rho^5} \, \dot{\tau} \dot{\rho} + (\rho^2-M^2) \; \left( \frac{M^2 \, \dot{\tau}^2}{4 \rho^7} + \frac{\dot{\rho}^2}{\rho^3} + \frac{\dot{\phi}^2}{2 \rho} \right) \right) = 0$$
$$\ddot{\phi} + \frac{4}{\rho} \, \dot{\rho} \dot{\phi} = 0$$
where the last immediately yields the first integral
$$\dot{\phi} = \frac{L}{\rho^4}$$
Now plug this into the first two equations. We obtain a coupled system which is first order in $\dot{\tau}, \dot{\rho}$ provided that we pretend to forget the relationship between $\rho$ and $\dot{\rho}$. Thus, near some fixed locus of constant "rootradius", we can apply a phase plane analysis.

As this shows, for analyzing or plotting geodesics, it can pay to use several coordinate charts, e.g. using this one to numerically integrate the geodesic equations, then transforming to another chart, such as the "cartesian-type Painleve chart", to plot the results.

Hope this helps others to follow along--- speak up, lurking students, if you want me to continue trying to provide background as above!

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Homework Helper
Now where was I? The equations of motion for Painleve-Cartesian coordinates. As usual, be warned I make LOTS of mistakes in calculation, and corrections are appreciated (and the reason I do this publicly).

$$\frac{1}{2\sqrt{I}} \frac{\partial I}{\partial x} =\frac{d}{d t} \left[ \frac{1}{2\sqrt{I}} \left( \frac{\partial I}{\partial \dot{x}} \right) \right],$$ or

$$\frac{I^{-0.5}}{2} \frac{\partial I}{\partial x} =$$
$$\frac{I^{-0.5}}{2}\left( \frac{\partial^2 I}{\partial\dot{x}\partial x}\dot{x}+ \frac{\partial^2 I}{\partial\dot{x}\partial y}\dot{y}+ \frac{\partial^2 I}{\partial\dot{x}^2}\ddot{x}+ \frac{\partial^2 I}{\partial\dot{x}\partial\dot{y}}\ddot{y}\right)$$
$$-\frac{I^{-1.5}}{4}\left( \frac{\partial I}{\partial x}\frac{\partial I}{\partial\dot{x}}\dot{x}+ \frac{\partial I}{\partial y}\frac{\partial I}{\partial\dot{x}}\dot{y}+ \frac{\partial I}{\partial \dot{x}}\frac{\partial I}{\partial\dot{x}}\ddot{x}+ \frac{\partial I}{\partial \dot{y}}\frac{\partial I}{\partial\dot{x}}\ddot{y} \right)$$

I want to get rid of the sqrt(I) and the 4s, and rearrange the terms because we need to solve for the 2nd derivatives in terms of the 1st derivatives and constants. Moving the 2nd derivatives to the right hand side we get:

$$2I\frac{\partial I}{\partial x} + \left(\frac{\partial I}{\partial x}\frac{\partial I}{\partial\dot{x}}- 2I\frac{\partial^2 I}{\partial\dot{x}\partial x}\right)\dot{x} + \left(\frac{\partial I}{\partial y}\frac{\partial I}{\partial\dot{x}}- 2I\frac{\partial^2 I}{\partial\dot{x}\partial y}\right)\dot{y} =$$

$$\left(2I\frac{\partial^2 I}{\partial\dot{x}^2}- \frac{\partial I}{\partial \dot{x}}\frac{\partial I}{\partial\dot{x}} \right)\ddot{x}+ \left(2I\frac{\partial^2 I}{\partial\dot{x}\partial\dot{y}}- \frac{\partial I}{\partial \dot{y}}\frac{\partial I}{\partial\dot{x}} \right)\ddot{y}$$

Similarly, the Euler Lagrange equation in y gives:

$$2I\frac{\partial I}{\partial y} + \left(\frac{\partial I}{\partial y}\frac{\partial I}{\partial\dot{y}}- 2I\frac{\partial^2 I}{\partial\dot{y}\partial y}\right)\dot{y} + \left(\frac{\partial I}{\partial x}\frac{\partial I}{\partial\dot{y}}- 2I\frac{\partial^2 I}{\partial\dot{y}\partial x}\right)\dot{x} =$$

$$\left(2I\frac{\partial^2 I}{\partial\dot{y}^2}- \frac{\partial I}{\partial \dot{y}}\frac{\partial I}{\partial\dot{y}} \right)\ddot{y}+ \left(2I\frac{\partial^2 I}{\partial\dot{y}\partial\dot{x}}- \frac{\partial I}{\partial \dot{x}}\frac{\partial I}{\partial\dot{y}} \right)\ddot{x}$$

The above equations mix the 2nd deriviatives. To use them, we need to solve for $$\ddot{x}$$ and $$\ddot{y}$$. The equations are linear so they can be written in 2x2 matrix form. To solve the equations requires that we invert the matrix, which can be done provided its determinant is not zero. The determinant is:

$$\left(2I\frac{\partial^2 I}{\partial\dot{x}^2}-\frac{\partial I}{\partial \dot{x}}\frac{\partial I}{\partial\dot{x}}\right)\left(2I\frac{\partial^2 I}{\partial\dot{y}^2}-\frac{\partial I}{\partial \dot{y}}\frac{\partial I}{\partial\dot{y}}\right) -\left(2I\frac{\partial^2 I}{\partial\dot{x}\partial\dot{y}}-\frac{\partial I}{\partial \dot{y}}\frac{\partial I}{\partial\dot{x}}\right)\left(2I\frac{\partial^2 I}{\partial\dot{y}\partial\dot{x}}-\frac{\partial I}{\partial \dot{x}}\frac{\partial I}{\partial\dot{y}}\right)$$

Next to calculate the various derivatives for the Painleve coordinates.

Homework Helper
The first order partial derivatives of I are as follows:
$$\frac{\partial I}{\partial x} = -2r^{-3}x - \sqrt{8}r^{-1.5}\dot{x} +\sqrt{18}r^{-3.5}x(x\dot{x}+y\dot{y}),$$
$$\frac{\partial I}{\partial y} = -2r^{-3}y - \sqrt{8}r^{-1.5}\dot{y} +\sqrt{18}r^{-3.5}y(x\dot{x}+y\dot{y}),$$
$$\frac{\partial I}{\partial \dot{x}} = 2\dot{x}-\sqrt{8}r^{-1.5}x,$$
$$\frac{\partial I}{\partial \dot{y}} = 2\dot{y}-\sqrt{8}r^{-1.5}y$$

The required second order partial derivatives are:

$$\frac{\partial^2 I}{\partial x\partial\dot{x}} = -\sqrt{8}r^{-1.5} + \sqrt{18}r^{-3.5}x^2$$
$$\frac{\partial^2 I}{\partial x\partial\dot{y}} = \sqrt{18}r^{-3.5}xy$$
$$\frac{\partial^2 I}{\partial y\partial\dot{x}} = \sqrt{18}r^{-3.5}xy$$
$$\frac{\partial^2 I}{\partial y\partial\dot{y}} = -\sqrt{8}r^{-1.5} + \sqrt{18}r^{-3.5}y^2$$

$$\frac{\partial^2 I}{\partial \dot{x}^2} = 2$$
$$\frac{\partial^2 I}{\partial \dot{y}^2} = 2$$
$$\frac{\partial^2 I}{\partial \dot{x}\partial\dot{y}} = 0$$

And the value of I was:
$$I =\frac{2}{r}-1 + \dot{x}^2 + \dot{y}^2-\sqrt{8}r^{-1.5}(x\dot{x}+y\dot{y})$$

The 2nd partials are fairly simple. Substituting them into the formula for the determinant gives:

$$\left(4I-\frac{\partial I}{\partial \dot{x}}\frac{\partial I}{\partial\dot{x}}\right)\left(4I-\frac{\partial I}{\partial \dot{y}}\frac{\partial I}{\partial\dot{y}}\right) - \left(\frac{\partial I}{\partial \dot{y}}\frac{\partial I}{\partial\dot{x}}\right)\left(\frac{\partial I}{\partial \dot{x}}\frac{\partial I}{\partial\dot{y}}\right)$$

Dividing by 4, this factors to:

$$I\left(4I - \left(\frac{\partial I}{\partial \dot{x}}\frac{\partial I}{\partial \dot{y}}\right)^2\right)$$

which is zero when I=0 or when

$$I = \frac{1}{4}\left(\frac{\partial I}{\partial \dot{x}}\frac{\partial I}{\partial \dot{y}}\right)^2$$

This is nice and simple, and I suspect it's correct. I should mention that when I last messed around with this, I ended up with very ugly Painleve calculations. MAXIMA barfed on it. This time I'm doing it by hand and it seems to be going better. Also, I was fixated at breaking out the singularity at the origin and I don't think that helped the calculation.

Putting (x,y) = (2,0) so that r=2 to see what happens on the event horizon eventually seems to lead to a result that I only end up with a zero determinant there for a light ray exiting the event horizon. I hope that I can divide these zeroes out of the equations of motion so that I'll have equations of motion that work for all particles, massless or massive (and perhaps also tachyonic), everywhere.

But before doing that, I'll go ahead and code up the above equations into the Schwarzschild applet to see if I can get motion that looks correct.

What I would really like would be motion that follows the Schwarzschild orbits but with different t so that the particles can penetrate the event horizon. However, to do this, I have to adjust the initial conditions between Painleve and Schwarzschild. I believe that initial positions are compatible with this hope, but that initial velocities depend on the difference in time. As usual, velocity here means dx/dt where t is parameter time, not proper time.

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Homework Helper
As usual, be warned I make LOTS of mistakes in calculation, and corrections are appreciated (and the reason I do this publicly).

As usual, I find at least one mistake in the above equations. The square root in the Painleve coordinates had the wrong sign. The correct sign for the Painleve coordinates is:

$$ds^2 =(\frac{2}{r}-1)dt^2 + dx^2 + dy^2 + \sqrt{8}r^{-1.5}(x\;dx\;dt+y\;dy\;dt)$$

This changes the sign of the square root here:

$$I =\frac{2}{r}-1 + \dot{x}^2 + \dot{y}^2+\sqrt{8}r^{-1.5}(x\dot{x}+y\dot{y})$$

and everywhere else the sqrt(8) or sqrt(18) appears.

But on correcting this, and putting the equations into my Java differential equation integrator, the behavior is fairly crazy, so I suspect that there is still another problem.

I've received some PMs asking about the theoretical justification for this method of calculating geodesics. The calculation follows a problem from Kip Thorne's 2006-2007 class:

Exercise 24.5 Problem: Action principle for geodesic motion
Show, by introducing a specific but arbitrary coordinate system, that among all timelike world lines that a particle could take to get from event $$P_0$$ to $$P_1$$, the one or ones whose proper time lapse is stationary under small variations of path are the free-fall geodesics. In other words, an action principle for a timelike geodesic $$P(\lambda)$$ [i.e., $$x^\alpha(\lambda)$$ in any coordinate system $$x^\alpha$$ ] is

$$\delta\int_{P_0}^{P_1}d\tau = \int_0^1\left(g_{\alpha\beta}\frac{dx^\alpha}{d\lambda}\frac{dx^\beta}{d\lambda}\ \right)^{1/2} d\lambda = 0,$$

where $$\lambda$$ is an arbitrary parameter which, by definition, ranges from 0 at $$P_0$$ to 1 at $$P_1$$. [Note: unless, after the variation, you choose the arbitrary parameter $$\lambda$$ to be affine'' ($$\lambda = a\tau+b$$ where $$a$$ and $$b$$ are constants), your equation for $$d^2x^\alpha/d\lambda^2$$ will not look quite like (24.26).]
http://www.pma.caltech.edu/Courses/ph136/yr2004/0424.1.K.pdf
http://www.pma.caltech.edu/Courses/ph136/yr2006/

In my exercise, $$\lambda$$ is coordinate time, and the difference in time between the beginning and end of coordinate time is 1.

$$I = g_{\alpha\beta}\frac{dx^\alpha}{d\lambda}\frac{dx^\beta}{d\lambda} = \left(\frac{ds}{dt}\right)^2$$

Carl

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Chris Hillman
Just trying to help, as you requested...

Hi all, I am the PF member whom CarlB mentioned, although I don't think he understood my PMs very well.

As usual, I find at least one mistake in the above equations...wrong sign... putting the equations into my Java differential equation integrator, the behavior is fairly crazy, so I suspect that there is still another problem.

Well, you did ask in the thread for help checking your work, and it seems you could use some. Do you still want my input?

I feel that this series of threads has been unneccessarily confusing, due to the fact that Carl

1. has declined every opportunity to outline his motivation for seeming to do something simple in a peculiar and difficult way,

2. frequently uses standard terms to mean something different, without bothering to warn the readers whose help he requested,

3. at times appears to use two letters to mean the same quantity, and also to use one letter to mean two different quantities,

4. in various other ways writes with minimal clarity.

At this point, I am tiring of trying to help out, but for the benefit of other PF members, let me try to add a bit more clarification before I duck out.

Carl says he is trying to compute "the geodesic equations". Used without qualification, this phrase always means what all the textbooks say it means, indeed what Kip Thorne's notes which he just cited say it means. See (24.26) of Thorne's notes.

Carl also says he is trying to compute them "the Euler-Lagrange way". To those who have studied variational calculus, this can only mean "write down a Lagrangian and apply the Euler-Lagrange operator". This E-L operator has the schematic form
$$\frac{d}{d\lambda} \frac{\partial}{\partial \dot{q}} - \frac{\partial}{\partial q}$$
(See Peter J. Olver, Applications of Lie Groups to Differential Equations, 2nd ed. Springer-Verlag, 1993 for a discussion of the Euler-Lagrange operator in connection with the general Noether symmetry principle). To those who have studied gtr, the Lagrangian one expects is what MTW call in Box 13.3 the "dynamical Lagrangian", which is read right off the line element by replacing $dx$ by $\dot{x}$ and so on. Applying the E-L operator then yields a system of second order ODEs which can be written in the form
$$\ddot{x^i} + \left( \operatorname{quadratic} \, \operatorname{combination} \dot{x^j}, \dot{x^k} \right) = 0$$
This is very easy and efficient; see Box 14.4 for a worked example.

It is crucial to understand that in this standard approach, "dot" refers to differentiation with respect to an affine parameter for the unknown curve.

Unfortunately, it turns out that Carl is not seeking the geodesic equations at all. (Too bad, because I took the trouble to write them out for his "Cartesian-type" Painleve chart in the middle of post #51 in this thread.) Furthermore, he is not using the standard "E-L" way, nor is he using some alternative methods some of us guessed he might be working with. After fifty posts, one really ought to be to expect some sense of what he is actually trying to do, but unfortunately, I still am not very confident I understand what he is up to, still less whether he is likely to eventually shed a little light, after blowing so much smoke over what he called, in one of the posts above, his "secrets".

FWIW, it now comes to light that Carl is apparently seeking not "the geodesic equations" but something I'll call "shape equations", which can often be obtained fairly easily from the geodesic equations. It's probably best to explain by example:

Consider the line element of the hyperbolic plane in the upper half space chart familiar from elementary complex analysis:
$$ds^2 = \frac{dx^2 + dy^2}{y^2}, \; -\infty < x < \infty, \; 0 < y < \infty$$
Now the unqualified phrase "the geodesic equations" can only mean one thing:
$$\ddot{x} - \frac{ 2 \, \dot{x} \, \dot{y} } {y} = 0, \; \ddot{y} + \frac{ \dot{x}^2 - \dot{y}^2 }{y} = 0$$
These are easily obtained by reading off what MTW call the dynamic Lagrangian from the line element
$$L = \frac{\dot{x}^2 + \dot{y}^2}{y^2}$$
and applying the operator
$$\left( \frac{d}{d\lambda} \frac{\partial}{\partial \dot{x}} - \frac{\partial}{\partial x}, \, \frac{d}{d\lambda} \frac{\partial}{\partial \dot{y}} - \frac{\partial}{\partial y} \right)$$
to L, which you can check gives the equations I wrote above, after making the second order derivatives "monic". The solution to these equations gives the geodesics of H^2 as affine parameterized curves in H^2.

Sometimes one isn't interested in any parameterization, but just the "shape" of the curves. In our example, we can obtain a differential equation for the "shape" y(x) from the geodesic equations as follows:

First, note that we immediately obtain a first integral from the "logarithmic" form of the first equation: $\dot{x} = A \, y^2$. Next, we can "force" an arc length parameter by writing
$$1 = \frac{\dot{x}^2 + \dot{y}^2}{y^2}$$
whence
$$\dot{y}^2 = y^2 \, \left( 1-A^2 \, y^2 \right)$$
Now we have

(EDIT: removed pseudolatex tags due to apparent cache exhaustion)

\frac{dx}{dy} = \frac{A \, y}{\sqrt{1-A^2 y^2}
Solutions of this equation give the "shape" of the geodesics, and in this case the shape has a nice description: the geodesics look like semicircles orthogonal to the locus y=0.

(EDIT: subsequent posters in this thread look out!--- we might have reached the maximal number of pseudo-latex formatted expressions in this thread.)

The point is that it seems that Carl seeks something like "shape equations", not what are called "the geodesic equations" in all the textbooks, including the course notes he himself cited. Indeed, he apparently seeks to differentiate with respect to the Painleve coordinate time, which I am writing T to avoid confusing with the Schwarschild time t (the other coordinates are the very same nonconstant monotonic functions on our Lorentzian manifold, so we can use the same letters without any possibility of confusion). Just to add to the opportunity for confusion, the Painleve time just so happens to be an arc length parameter for certain timelike geodesics, namely the integral curves of the Lemaitre congruence, i.e. the world lines of observers who fall in freely and radially "from rest at infinity". The point of Box 13.3 in MTW is that solutions of the geodesic equations (obtained as above from the dynamical Lagrangian) are always affine parameterized curves. A slightly tricky point in Lorentzian manifolds: affine parameters make perfect sense for null geodesics, but we can't turn these into arc length parameters, as we can for timelike or spacelike geodesics parameterized by an affine parameter! At times, Carl seems to think that affine parameters are useless for null geodesics, but this is quite wrong: the geodesic equations work perfectly well for describing all geodesics, including null geodesics.

Another possible complication is that Carl is suppressing one spatial coordinate (reasonable because of the spherical symmetry, and a good idea if you want to make a spacetime plot in the end), but at times he also seems to be projecting into a constant Painleve time slice. The eprint by Hamilton and Lisle which he cited visualizes Kerr geometry by studying how the local light cones (see my post #50 above) are sheared from Minkowski background to the Painleve cones, and also how a gyrostabilized spatial frame carried by Doran observers (the analog of the Lemaitre observers) appear to spin in the Doran chart (the analog of the Painleve chart) as the observers fall in. They express this as a combination of Galilei transformation (for the shearing) plus a twist (for the spinning). This is a just a novel way of expressing, for the Kerr vacuum (and so far only for the Kerr vacuum!), the common phenomenon I often mention: gyrostabilized frames often appear to spin in a given coordinate chart. In particular, I pointed out long ago that the obvious frame in the original paper presenting the Doran chart is in fact spinning, but is easily "despun" because the frame is actually spinning about one of the spatial vectors, the one aligned with \partial_\phi.

Let me finish up by saying that new ways to visualize Schwarschild and Kerr are constantly appearing, and the more the merrier, say I, subject to one important injunction: make no errors. I have repeatedly tried to suggest to Carl that he tackle some simpler examples first and compare results with those obtained using standard methods. Being systematic is more important, not less so, if one is prone to making errors. Published papers on visualizations so far tend to be OK, in my experience (but then I don't read trashcan journals), but there are quite a few websites out there which give offer alleged visualizations which are in fact erroneous, and we don't want to add to that plague.

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Um, Chris, regarding "secrets" ... note the use of inverted commas here. You are quite free to read more of Carl's writing, widely available on the crackpot web. And of course we know about Baez's website.

Just trying to help.

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Carl says he is trying to compute "the geodesic equations". Used without qualification, this phrase always means what all the textbooks say it means, indeed what Kip Thorne's notes which he just cited say it means.

"Geodesic equations" does carry a meaning in physics as Chris states. However, if he will bother to read the thread, he will find that I never use the phrase "geodesic equations" to describe what I am looking for.

Most of the rest of the post is based on this failure to carefully read what's been written.

Carl also says he is trying to compute them "the Euler-Lagrange way". To those who have studied variational calculus, this can only mean "write down a Lagrangian and apply the Euler-Lagrange operator".

It is crucial to understand that in this standard approach, "dot" refers to differentiation with respect to an affine parameter for the unknown curve.

Anyone who wants to search for the "Euler-Lagrange" method will find that it does not imply the use of a Lagrangian. Anyone who reads through my posts will find that I have not claimed that I am using a Lagrangian, and have specifically noted otherwise in several places.

Here's a link to Mathworld's definition of Euler-Lagrange:
http://mathworld.wolfram.com/Euler-LagrangeDifferentialEquation.html

If you read the above link, you will find that (a) they use "L" to refer to the function being minimized, and (b) they never mention the word "Lagrangian". I don't know how Chris could have got the impression that Euler-Lagrange is a technique that only works with Lagrangians, and I certainly don't see how he can justify this comment. Furthermore, you will find that they use the "dot" notation to refer to the derivatives with respect to the integration variable, which they label as "t". I've followed these conventions precisely.

After fifty posts, one really ought to be to expect some sense of what he is actually trying to do ...

Really??? You mean that someone really couldn't understand the first post I made on this??? Let me quote myself:

CarlB said:
Here's the basic plan:

(1) Write the Schwarzschild metric in Cartesian coordinates.

(2) Write the proper length of a path as an integral over coordinate time.

(3) Vary the path and use the Euler-Lagarange equation to determine a pair of 2nd order differential equations that the orbits solve.

(4) Find and from the two 2nd order DEs.

What part of "vary the path" is confusing here? Most of the rest of Chris' post just continues to boringly beat the same drum over and over. Yes, we all know how to calculate geodesic equations the standard way. So what.

Before I forget, this concept of writing GR in Cartesian coordinates is not a heresy that I invented. Here's a handout for students at U. Colorado:

ASTR 3740. Spring 2004.
Using the River Model to Draw Geodesics around Black Holes

Fancy creating a computer program, maybe a Java applet, to draw the orbits of particles or photons around black holes? The river model provides a nice way to implement this.

...

Rewrite the river metric (1) in Cartesian coordinates $$x^\mu \equiv (x^0,x^1,x^2,x^3) \equiv (t_{ff},x,y,z)$$ with origin at the center of the black hole:

$$ds^2 = \eta_{\mu\nu}(dx^\mu-v^\mu dt_{ff}(dx^\nu - v^\nu dt_{ff})$$

where $$\eta_{\mu\nu} \equiv$$ diag(-1,1,1,1) is the Minkowski metric and $$v^\mu$$ are the components of the river velocity

$$v^\mu = v(0,-x/r,-y/r,-z/r)$$

Another possible complication is that Carl is suppressing one spatial coordinate (reasonable because of the spherical symmetry, and a good idea if you want to make a spacetime plot in the end), but at times he also seems to be projecting into a constant Painleve time slice.

Cool, a possible error. Could you point it out? By the way, when I'm done, symmetry will indicate how to modify the equations for three spatial dimensions, so there's no reason to carry them around now.

Carl

Update: Looking for the problem. I started plotting the acceleration as a function of position and velocity for the Newton, Schwarzschild, and Painleve coordinates. I was just a little surprised that the calculated acceleration (i.e. Cartesian acceleration with respect to coordinate time) is identical between the Schwarzschild and Painleve metrics for particles that have no velocity (again, in the Cartesian coordinates). The plot does cool things inside the event horizon.

Where the accelerations differ is when the test particle is moving. When I check phase space points with slight velocities, the accelerations diverge quite a bit. Looking at particle speeds from -0.009 to +0.007. Turns out Einstein acceleration changes by 0.000436, but my version of Painleve acceleration changes by 0.2366, probably waaaaay too big. Very likely I'm missing some divisions by r in terms that depend on velocity.

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Chris Hillman
Well, I tried....

Before I forget, this concept of writing GR in Cartesian coordinates is not a heresy that I invented. Here's a handout for students at U. Colorado:

Carl, maybe you should ask yourself why I understand and enjoy reading AH's papers, but despite my good faith effort, I have been unable to avoid an exponential growth of confusion and misery in this thread. Apparently I am not the only one who has similar experiences trying to help you "check your work" or whatever it is you are here for.

Kea, I didn't get the joke, but I'll leave you and Carl to it. Good luck.

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VICTORY!!!

Painleve coordinates are working. The last problem was that my Java was missing part of a term. The orbits are gorgeous, and match the Schwarzschild orbits I earlier computed.

 The new simulation is here:
http://www.gaugegravity.com/testapplet/SweetGravity.html [Broken] [/edit]

The only thing I don't like about them is that the Schwarzschild particles don't follow down the same paths as the Painleve particles even though they have the same initial conditions. This is just due to the difference in time for the two coordinate systems. To fix it, I can use the differences in proper time / coordinate time between the two metrics to make their velocities measure up the same. I might add a switch box to do that.

The other thing that is ugly is that I haven't tried to reduce the equations to lower terms. Instead, the equations of motion are:

$$\left(\frac{ds}{dt}\right)^2 = I =\frac{2}{r}-1 + \dot{x}^2 + \dot{y}^2+\sqrt{8}r^{-1.5}(x\dot{x}+y\dot{y})$$

$$\frac{\partial I}{\partial x} = -2r^{-3}x + \sqrt{8}r^{-1.5}\dot{x} -\sqrt{18}r^{-3.5}x(x\dot{x}+y\dot{y}),$$
$$\frac{\partial I}{\partial y} = -2r^{-3}y + \sqrt{8}r^{-1.5}\dot{y} -\sqrt{18}r^{-3.5}y(x\dot{x}+y\dot{y}),$$
$$\frac{\partial I}{\partial \dot{x}} = 2\dot{x}+\sqrt{8}r^{-1.5}x,$$
$$\frac{\partial I}{\partial \dot{y}} = 2\dot{y}+\sqrt{8}r^{-1.5}y$$

The required second order partial derivatives are:

$$\frac{\partial^2 I}{\partial x\partial\dot{x}} = +\sqrt{8}r^{-1.5} - \sqrt{18}r^{-3.5}x^2$$
$$\frac{\partial^2 I}{\partial x\partial\dot{y}} = -\sqrt{18}r^{-3.5}xy$$
$$\frac{\partial^2 I}{\partial y\partial\dot{x}} = -\sqrt{18}r^{-3.5}xy$$
$$\frac{\partial^2 I}{\partial y\partial\dot{y}} = +\sqrt{8}r^{-1.5} - \sqrt{18}r^{-3.5}y^2$$

$$\frac{\partial^2 I}{\partial \dot{x}^2} = 2$$
$$\frac{\partial^2 I}{\partial \dot{y}^2} = 2$$
$$\frac{\partial^2 I}{\partial \dot{x}\partial\dot{y}} = 0$$

The equations of motion are then defined by computing the following six functions of phase space:

$$A = \left(2I\frac{\partial^2 I}{\partial\dot{x}^2}- \frac{\partial I}{\partial \dot{x}}\frac{\partial I}{\partial\dot{x}} \right)$$

$$B = \left(2I\frac{\partial^2 I}{\partial\dot{x}\partial\dot{y}}- \frac{\partial I}{\partial \dot{y}}\frac{\partial I}{\partial\dot{x}} \right)$$

$$C = 2I\frac{\partial I}{\partial x} + \left(\frac{\partial I}{\partial x}\frac{\partial I}{\partial\dot{x}}- 2I\frac{\partial^2 I}{\partial\dot{x}\partial x}\right)\dot{x} + \left(\frac{\partial I}{\partial y}\frac{\partial I}{\partial\dot{x}}- 2I\frac{\partial^2 I}{\partial\dot{x}\partial y}\right)\dot{y}$$

$$D = \left(2I\frac{\partial^2 I}{\partial\dot{y}\partial\dot{x}}- \frac{\partial I}{\partial \dot{x}}\frac{\partial I}{\partial\dot{y}} \right)$$

$$E = \left(2I\frac{\partial^2 I}{\partial\dot{y}^2}- \frac{\partial I}{\partial \dot{y}}\frac{\partial I}{\partial\dot{y}} \right)$$

$$F = 2I\frac{\partial I}{\partial y} + \left(\frac{\partial I}{\partial y}\frac{\partial I}{\partial\dot{y}}- 2I\frac{\partial^2 I}{\partial\dot{y}\partial y}\right)\dot{y} + \left(\frac{\partial I}{\partial x}\frac{\partial I}{\partial\dot{y}}- 2I\frac{\partial^2 I}{\partial\dot{y}\partial x}\right)\dot{x}$$

and combining them as follows:

$$\frac{d^2x}{dt^2} = (EC-BF)/(AE-BD),$$

$$\frac{d^2y}{dt^2} = (AF-DC)/(AE-BD).$$

This eliminates the need to keep track of proper time of particles when computing simultaneously multiple orbits of relativistic particles in the Painleve metric. I expect that I will be able to simplify it with MAXIMA, but the important thing is that it works.

As an aside, I almost wasted my time programming up the usual geodesic equations. In fact, I actually sat down to do it. What stopped me was the realization of how much more difficult the usual geodesic equations are to use than the simple equations of motion I've found here.

The geodesic equations require that you keep track of the particle's "proper time". But this won't work for photons, so you have to have a different algorithm for massless particles (and your algorithm for massive particles probably blows up when faced with very relativistic massive particles). But the equations I've got cleanly do the job for all particles, massive, massless and tachyonic.

The phase space I'm using (after you add back in the z dimension), has 6 dimensions, x,y,z and their velocities. If you do it with the geodesic equations, you will have 8 dimensions, x, y, z, and t and their velocities with respect to s. This means that in the case of the usual geodesic equations, you have 33% more variables to keep track of, and in addition you need to properly initialize the dt/ds variable. In my equations of motion (note I don't call them "geodesic equations"), all you do is plug and go.

So when I remembered that I'd have to deal with initialization hassle, (which I once did have in my Schwarzschild simulation), plus the hassle of having to a given number of position iterations not correspond to always the same amount of time, I decided to put more effort into debugging the Painleve case.

What I've done, essentially, is integerated out the extra dimension normally used in the geodesic equations.

Carl

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Exiting! When can we see the working applet?

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Exiting! When can we see the working applet?

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

I'd have put it up yesterday, but it still had the debug prints turned on and I didn't have protection against division by zero at the singularity.

The Painleve orbits are labeled as "gauge" because the purpose of the applet is to show off the gauge gravity equations. The Schwarzschild coordinates are labeled "Einstein".

The Schwarzschild and Painleve orbits are similar but not quite identical. I can make them identical by correcting the initial conditions. That is the initial conditions for (coordinate) velocity:

$$\left. \frac{dx}{dt}\right|_0$$

means different things for Schwarzschild and Painleve coordinates because t is different between them. I'm contemplating adding a button to the applet that will change the initial conditions for Schwarzschild to match the Painleve or vice versa.

By the way, if you look around in the literature, you will find formulas for approximate relativistic corrections to Newton's equations of motion. What I've computed here are the exact relativistic corrections to Newton's equations of motion. After I do the rotating black hole in Doran coordinates I will publish this.

Here are some papers talking about 1st order corrections to Newton:

http://www.numdam.org/numdam-bin/fitem?id=AIHPA_1985__43_1_107_0 [Broken]
http://www.obs-azur.fr/gemini/pagesperso/pireaux/proc/SCRMI_integrator.pdf
http://syrte.obspm.fr/journees2004/PDF/Pireaux.pdf [Broken]

[Latest Update]: The equations take the form of a ratio of two quantities. In post #53 I factored the denominator factors into:

$$4 \;I \;\left(4I - \left(\frac{\partial I}{\partial \dot{x}}\frac{\partial I}{\partial \dot{y}}\right)^2\right)$$

As usual, this is wrong. The correct factorization is:

$$4 \;I \;\left(4I - \left(\frac{\partial I}{\partial \dot{x}}\right)^2 - \left(\frac{\partial I}{\partial \dot{y}}\right)^2\right)$$

It turns out that the complicated part of the factorization simplifies to -4. Thus the denominator is:

$$-16 I$$

I've now factored the $$I$$ out of the numerator. Consequently I have a set of equations that have no singularities other than at the origin. The numerator is still messy, but probably can be simplified (which I'm doing).
[/Latest Update]

Carl

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From reducing the equations of motion, it's pretty clear that your life is easier if you write your Painleve metric in the following fashion:

$$ds^2 = \left(dx + \sqrt{2}xr^{-1.5}\;dt\right)^2 + \left(dy + \sqrt{2}yr^{-1.5}\;dt\right)^2 + \left(dz + \sqrt{2}zr^{-1.5}\;dt\right)^2 - dt^2$$

Then the calculations end up using terms like $$\dot{x}+\sqrt{2}xr^{-1.5}$$, which in the river vernacular is the velocity relative to the river. For example, if $$x$$ and $$\dot{x}$$ are both positive, then you are climbing out of the black hole and swimming against the stream. Therefore the relative velocity is increased.

I've got the terms fairly well reduced and well behaved, but I suspect that if I play with them for a little longer I'll get them much better.

Carl

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