Schwarzschild Orbits in Cartesian coordinates

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


  • Total voters
    14
CarlB
Science Advisor
Homework Helper
Messages
1,246
Reaction score
45
My Java applet gravity simulator http://www.gaugegravity.com/testapplet/SweetGravity.html
draws beautiful orbits, however the GR simulation is very badly broken as one can tell when comparing it with Newton at long distances. The source code is at:
http://www.gaugegravity.com/testapplet/SwGrav_Top.java

Test bodies are required to all lie in a plane. The Newtonian physics is produced by writing the x and y accelerations in terms of the phase space parameters of the test body, that is, in terms of its (x,y) position and its (Vx,Vy) velocity. With Newton, the acceleration has no dependence on velocity of course.

The GR "physics" is produced by a complicated calculation that follows the artificial potential that is discussed with respect to the Schwarzschild orbits in MTW and in various places all over the web. These methods use energy and angular momentum as constants of the motion to describe the orbits but they are less than optimal when used in an integration form.

I've seen a computer method for computing GR orbits that should work better, but I am hesitating to implement it because it is based on polar coordinates. This means that the calculation has to use trig functions to convert back and forth and these are slow.

So I'm wondering how hard it is to put the Schwarzschild orbits into phase space form in Cartesian coordinates. 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 d^2x/dt^2 and d^2y/dt^2 from the two 2nd order DEs.

Now for the poll: Do you think that this can be done reasonably easily? Part of the reason I am working on it is that I want to get a better idea on how these gravity theories differ one from another. I will post calculations as I make them, if convenient.

Carl
 
Last edited by a moderator:
Physics news on Phys.org
Step (1), write the Schwarzschild metric in Cartesian coordinates.

x = r \cos(\phi)
y = r \sin(\phi)

dx = \cos(\phi) dr - r\sin(\phi) d\phi
dy = \sin(\phi) dr + r\cos(\phi) d\phi

dr = \cos(\phi) dx + \sin(\phi) dy
rd\phi = -\sin(\phi) dx + \cos(\phi) dy

dr^2 = \cos^2(\phi) dx^2 + \sin^2(\phi) dy^2 + \sin(2\phi) dx\;dy
r^2d\phi^2 = \sin^2(\phi) dx^2 + \cos^2(\phi) dy^2 - \sin(2\phi) dx\;dy

dr^2 = (x^2dx^2 + y^2dy^2 + 2xy dx\;dy)/r^2
r^2d\phi^2 = (y^2dx^2 + x^2dy^2 - 2xy dx\;dy)/r^2

ds^2 = -\frac{(r-2)}{r}dt^2 + \frac{r}{r-2}dr^2 + r^2d\phi^2

ds^2 = -\frac{(r-2)}{r}dt^2 + \frac{1}{r(r-2)}(x^2dx^2 + y^2dy^2 + 2xy dx\;dy)<br /> + \frac{1}{r^2}(y^2dx^2 + x^2dy^2 - 2xy dx\;dy)

ds^2 = -\frac{(r-2)}{r}dt^2 + \frac{1}{r^2(r-2)}((rx^2+ ry^2 -2y^2)dx^2 + (ry^2+rx^2-2x^2)dy^2 + (2rxy -2rxy +4rxy) dx\;dy)

ds^2 = -\frac{(r-2)}{r}dt^2 + \frac{1}{r^2(r-2)}((r^3 -2y^2)dx^2 + (r^3-2x^2)dy^2 + (4rxy) dx\;dy)

ds^2 = -\frac{(r-2)}{r}dt^2 + <br /> \frac{(r^3-2y^2)}{(r^3-2r^2)}dx^2 + \frac{(r^3-2x^2)}{(r^3-2r^2)}dy^2 +\frac{4rxy}{(r^3-2r^2)}dx\;dy
 
Last edited:
CarlB said:
...
Now for the poll: Do you think that this can be done reasonably easily? Part of the reason I am working on it is that I want to get a better idea on how these gravity theories differ one from another. I will post calculations as I make them, if convenient.
Carl
Interesting excercise Carl, but how do you think having the orbital equation in Cartesian form (as against polar form) will help you get a "better idea on how these gravity theories differ one from another."?
 
There isn't any such thing as "Cartesian coordinates" for a black hole. Cartesian coordinates work only for flat space-times. The space-time around a black hole is not flat. In particular, r is not a radial coordinate.

The closest you can come to "cartesian" coordiantes are isotropic coordinates.

Howver, from experience, I can tell you it will be far far easier to get your program working for Schwarzschild coordinates than isotropic cooridinates.

As I mentioned before, you can just use the geodesic equations to express your set of equations as a second order system that look pretty much like the second order set you have already for Newton's equations.

This is the "geodesic" equation

http://en.wikipedia.org/wiki/Solving_the_geodesic_equations

(Wiki somewhat confusingly uses "t" instead of "tau" for the variables.

The necessary Christoffel symbols are simplest for the schwarzschild metric, they are

Code:
                        r                 m
                      CC   [r r] = - -----------
                                     (r - 2 m) r                         theta
                       CC       [theta r] = 1/r                           phi
                         CC     [phi r] = 1/r                         t               m
                       CC   [t r] = -----------
                                    (r - 2 m) r                         theta
                       CC       [r theta] = 1/r                      r
                    CC   [theta theta] = -r + 2 m                     phi                cos(theta)
                   CC     [phi theta] = ----------
                                        sin(theta)                           phi
                         CC     [r phi] = 1/r                     phi                cos(theta)
                   CC     [theta phi] = ----------
                                        sin(theta)                 r                                   2
               CC   [phi phi] = -(r - 2 m) sin(theta)               theta
             CC       [phi phi] = -sin(theta) cos(theta)                         t               m
                       CC   [r t] = -----------
                                    (r - 2 m) r                         r          (r - 2 m) m
                       CC   [t t] = -----------
                                         3
                                        r

As I also mentioned, you should be able to obtain the same set of geodesic equations from the conserved quantites, as well, by replacing "conserve quantity = constant" with "d/dtau (conserved quantity) = 0"
 
Last edited:
Let's try to make things really, really, really simple.

theta will always be 90 degrees. Therfore d theta / dtau will be zero.

We then have for our variables

r(tau), t(tau), phi(tau).

The second order geodesic equations of motion (the geodesic equations) will therefore be

<br /> \frac{d^2 r}{d tau^2} + \Gamma^r{}_{rr} \left( \frac{dr}{dtau} \right) ^2 + \Gamma^r{}_{tt} \left( \frac{dt}{d tau} \right)^2 + \Gamma^r{}_{\phi\phi} \left( \frac{d \phi}{d tau} \right)^2 = 0<br />

<br /> \frac{d^2 t}{d tau^2} + \left( \Gamma^t{}_{tr} + \Gamma^t{}_{rt} \right) \left( \frac{dr}{d tau} \right) \left( \frac{dt}{dtau} \right) = 0<br />

<br /> \frac{d^2 \phi}{d tau^2} + \left( \Gamma^{\phi}{}_{r\phi} + \Gamma^{\phi}{}_{\phi r} \right) \left( \frac{dr}{d tau} \right) \left( \frac{d \phi} {d tau} \right) = 0<br />

See the previous post for the values for the Christoffel symbols.

Note for instance that since r^2 {d \phi}/ {d tau} = constant, when differentiated, gives by the chain rule

<br /> 2 r \frac{dr}{d\tau} \frac{d \phi}{d tau} + r^2 \frac{d^2 \phi} {d tau^2} = 0<br />

and that this is equivalent to the last equation above when you substitute in the correct value for the Christoffel symbol.

The conseved energy when differentiated will give one of the other geodesic equations (the one for time).

The last geodesic equation is a consequence of the others plus the equation for the metric.
 
Last edited:
Cartesian coordinates

pervect said:
There isn't any such thing as "Cartesian coordinates" for a black hole. Cartesian coordinates work only for flat space-times. The space-time around a black hole is not flat. In particular, r is not a radial coordinate.

The closest you can come to "Cartesian" coordinates are isotropic coordinates.
Are you not a bit harsh on CarlB? I understood that he simply wants to express Schwarzschild coordinates in terms of x,y,z rather than in terms of r,phi,theta. Nothing fundamentally wrong with that, I think! Maybe his (Cartesian) choice of words was wrong, but let’s not get too semantic!

Nevertheless, I agree that we should rather stick to the tried and tested procedure of treating black holes.
 
pervect, one of the things I'm trying to get away from is differential equations written in proper time. It makes for a more complicated simulation and also makes it a lot easier for the simulation to fail in various ways. So I'm continuing with the process of writing the equations in Cartesian coordinates. I will allow you to continue calling them isotropic coordinates if that's what you want to do.

Carl
 
Carl (and Jorrie, too) I think you've totally missed the point.

The 'r' in the Scwarzschild coordinate system is not the same 'r' in the isotropic coordinate system.

In fact, one can see that r = r'(1+2m/r')^2 from the webpage, where r' is the isotropic radial coordinate and r is the Schwarzschild r coordinate.

Neither of these 'r' coordinates can be interpreted as 'radial distance'. The isotropic coordinates, though, define a set of coordinates where light cones will apear to be circles.

Thus assuming that r^2 = x^2 + y^2 does not make any physical sense. This is assumed implicitly when one writes x = r cos(phi) y = r sin(phi), one assumes that x^2 + y^2 = r^2. One can think of it as a purely formal mathematical expression, I suppose.

As such a formal exercise, unless Carl has made an algebraic error he's got a new and totally non-standard metric that doesn't even have a name in terms of new and toatally non-standard coordiantes that don't have names either (except what Carl has christened them). They definitely shouldn't be confused with Cartesian coordinates, which is what Carl has chosen to call his variables.

One can easily see that Carl's metric is not the same as the isotropic metric either by comparing the line elements.

Aside from the fact that it is non-standard, and thus will be hard to check solutions for, even deriving the correct solutions is going to be a giant pain. Re-writing a simple metric in a more complicated form is not going to get anywyere as far as generating a solution for the metric.

If the problem is with the factor of 'tau' in the standard equations, just eliminate that tau. This shouldn't be too hard since

(1-2m/r) dt/dtau = E

The most annoying thing is going to be this

If we write r as a function of t : r = rr(t) then

r(tau) = rr(t(tau))

and dr/dtau = drr/dt dt/dtau

but by the chain rule

<br /> \frac{d^2 r} {d tau^2} = \frac{d^2 rr}{d t^2} \left( \frac{dt}{d tau} \right) ^2 + \frac{dr}{dt} \frac{d^2 t}{d tau^2}<br />

Still, a little bit of algebra should eliminate tau as a variable if this is really what is desired.

Solving for phi shouldn't be terriby difficult, since

r^2 dphi/dtau =L

therfore

r^2 dphi/dt * dt/dtau = L

And we've already noted that dt/dtau is easy to eliminate
 
Unless I've made a mistake (quite possible, even with computer assistance) the equations with proper time eliminated for the Schwarzschld metric are just

r(t), phi(t)

d phi/dt = K*(1-2*m/r) / r^2

K is a constant of motion, equal to "L/E" in the old notation. Knowing dphi/dt and r at one point will determine the value of K

The second order equation for r(t) is

<br /> \frac{d^2 r}{dt^2} = <br /> \frac{2\,m}{r^2 \left( 1 - \frac{2m}{r} \right) } \left( \frac{dr}{dt} \right) ^2<br /> - \left( 1-{\frac {2 m}{r}} \right) \frac{m}{r^2}+ \left( 1-{\frac {2m}{r}} \right) ^{3}\frac{{K}^{2}}{r^3}<br />

It looks sensible at first glance, dr/dt is forced to zero, however I have not tested it.

Compare this to the Newtonian equations in polar coordinates (r, phi)

r^2 * d phi/dt = K

where K = L/m

d^2 r / dt^2 = -m/r^2 + r (dphi/dt)^2 = -m/r^2 + K^2 / r^3

we can also see that the Schwarzschild equations reduce to the Newtonian equations when dr/dt << 1 and r >> 2m as they should, another sign that they are probably right.
 
Last edited:
  • #10
The Christoffel symbols for the Painleve metric are:

Code:
                                   1/2      1/2
                       T          2    (M/r)    M
                     CC   [T T] = ---------------
                                         2
                                        r                         r          (r - 2 M) M
                       CC   [T T] = -----------
                                         3
                                        r                            T           M
                          CC   [r T] = ----
                                         2
                                        r                                    1/2      1/2
                      r            2    (M/r)    M
                    CC   [r T] = - ---------------
                                          2
                                         r                            T           M
                          CC   [T r] = ----
                                         2
                                        r                                    1/2      1/2
                      r            2    (M/r)    M
                    CC   [T r] = - ---------------
                                          2
                                         r                                       1/2
                        T             2    M
                      CC   [r r] = -------------
                                          1/2  2
                                   2 (M/r)    r                           r             M
                         CC   [r r] = - ----
                                          2
                                         r                         theta
                       CC       [theta r] = 1/r                           phi
                         CC     [phi r] = 1/r                         theta
                       CC       [r theta] = 1/r                  T                    1/2      1/2
                CC   [theta theta] = -2    (M/r)    r                      r
                    CC   [theta theta] = -r + 2 M                     phi                cos(theta)
                   CC     [phi theta] = ----------
                                        sin(theta)                           phi
                         CC     [r phi] = 1/r                     phi                cos(theta)
                   CC     [theta phi] = ----------
                                        sin(theta)              T                1/2      1/2             2
            CC   [phi phi] = -2    (M/r)    r sin(theta)                 r                                   2
               CC   [phi phi] = -(r - 2 M) sin(theta)               theta
             CC       [phi phi] = -sin(theta) cos(theta)

a similar analysis using the geodesic equations and eliminating tau should be possible.
 
  • #11
pervect said:
Carl (and Jorrie, too) I think you've totally missed the point.

The 'r' in the Scwarzschild coordinate system is not the same 'r' in the isotropic coordinate system.

In fact, one can see that r = r'(1+2m/r')^2 from the webpage, where r' is the isotropic radial coordinate and r is the Schwarzschild r coordinate.

Neither of these 'r' coordinates can be interpreted as 'radial distance'. The isotropic coordinates, though, define a set of coordinates where light cones will apear to be circles.

Thus assuming that r^2 = x^2 + y^2 does not make any physical sense. This is assumed implicitly when one writes x = r cos(phi) y = r sin(phi), one assumes that x^2 + y^2 = r^2. One can think of it as a purely formal mathematical expression, I suppose.

As such a formal exercise, unless Carl has made an algebraic error he's got a new and totally non-standard metric that doesn't even have a name in terms of new and toatally non-standard coordiantes that don't have names either (except what Carl has christened them). They definitely shouldn't be confused with Cartesian coordinates, which is what Carl has chosen to call his variables.

Aside from the fact that it is non-standard, and thus will be hard to check solutions for, even deriving the correct solutions is going to be a giant pain. Re-writing a simple metric in a more complicated form is not going to get anywyere as far as generating a solution for the metric.


It is my intention to work only in the Schwarzschild metric, but with the polar coordinates replaced by Cartesian coordinates. I realize that you think this is unphysical, but here in the United States computers are built with screens that have rectangular arrays of pixels, and so any computation I make must be eventually so converted. Now that you've explained what they are, I don't mean to use "isotropic" coordinates at all. It's just extra confusion you've brought in, not me. I thought it was you calling my coordinates "isotropic". Recall, it was I who "allowed" you to call them that if you like.

Nor am I the only person in the world who calls these things I'm using "Cartesian". The United States is a very primitive place, filled with horses, saloons and cowboys. Other obscure, amateurish authors here use the same confusing and incorrect terminology. For example see:

Numerical Relativity for Inspiraling Binaries in Co-Rotating Coordinates: Test Bed for Lapse and Shift Equations
Kip S. Thorne, 1998
http://arxiv.org/abs/gr-qc/9808024

By the way, I should mention that while I only took one graduate level General Relativity class at the University of California, Irvine, I was the top student, the only student in the class receiving an "A+". I don't think that taking one class makes you an expert in a subject, but you can be sure that I understand the principles of general relativity at least a little better than some.

[edit] But I see that it is not only on this side of the Atlantic that one finds the use of Cartesian coordinates in describing black holes:

Shortcut Method of Solution of Geodesic Equations for Schwarzschild Black Hole
Jean-Alain Marck,
D´epartement d’Astrophysique Relativiste et de Cosmologie
Observatoire de Paris - Section de Meudon
F-92195 Meudon Cedex, France

Classical and Quantum Gravity 13 (1996) 393-402

r^2 = x^2 + y^2 + z^2\;\;\;\;\;\;\;(6)

http://arxiv.org/abs/gr-qc/9505010
[/edit]

Carl
 
Last edited:
  • #12
pervect said:
Unless I've made a mistake (quite possible, even with computer assistance) the equations with proper time eliminated for the Schwarzschld metric are just:

This is exactly what I need and much thanks. To convert it into Cartesian form, I need only split the radial acceleration into x and y accelerations. I should have a new simulation running later today, but Seattle is sooooooo hot (about 90 F) that everyone has left for the beach and very little work is getting done. Maybe I'll wait until it begins to cool down.

I was in mathematics grad school at the University of Washington 10 years ago and got a student priced copy of Mathematica. I played around with it, but never worked at it long enough to get around to messing with the symbolic calculus. Since then, most of my work has been in Clifford algebra and for that I wrote Java programs. But after this fiasco I'm convinced that I need to get some sort of a symbolic math package.

Carl
 
Last edited:
  • #13
Symbolic math really rocks. Couldn't live without it.

I've also added horizon crossing solutions to the Painleve thread (but only in terms of conseved quantites) I haven't eliminated dt/dtau, it appears to be a royal pain to do so. dt/dtau will range in values from 1 to E for the horizon crossing solutions.

I note that non-horizon crossing alternate solutions exist, but I haven't explored them very far.

I've also computed d^2 r / dtau^2 only in the special case where dr/dtau = 0.
 
Last edited:
  • #14
pervect said:
<br /> \frac{d^2 r}{dt^2} = <br /> \frac{2\,m}{r^2 \left( 1 - \frac{2m}{r} \right) } \left( \frac{dr}{dt} \right) ^2<br /> - \left( 1-{\frac {2 m}{r}} \right) \frac{m}{r^2}+ \left( 1-{\frac {2m}{r}} \right) ^{3}\frac{{K}^{2}}{r^3}<br />


Unfortunately, I couldn't get it to work. This could always be a programming error of course. It acts like there is too much energy in the angular kinetic energy term, that is the K^2 term. In fact, the simulation doesn't match Newton at large r because of this term. To get it to work, I have to divide by an extra factor of r.

Take a look at the units for large r, in the first and third terms of the above and see if there is a transcription error, most likely in the last term of the above, but at least conceivably in the formula for K = \dot{\phi} r^2 / (1-2m/r) which can be used to simplify the third term.

Having a differential equation in r and a constant of the motion in phi is unsatisfying to me. I realize that I can put the Newtonian equation into the same form, but I am one of those crazy guys who thinks that the equations of motion are closer to the physics than the constants of the motion. Yes, I believe that the constants of the motion are nothing more than convenient methods we use for simplifying the original physical problem, which to me are always defined in terms of positions and velocities. So it looks like I will continue my process of deriving the Newton style equations of motion for the Schwarzschild metric.

Carl
 
  • #15
Step (2), write the proper length of a path as an integral over coordinate time.

Here's where I begin:

CarlB said:
ds^2 = \frac{(r-2)}{r}dt^2 - <br /> \frac{(r^3-2y^2)}{(r^3-2r^2)}dx^2 - \frac{(r^3-2x^2)}{(r^3-2r^2)}dy^2 -\frac{4rxy}{(r^3-2r^2)}dx\;dy

Note that in the above I've swapped over to the other coast metric in order to make s time like instead of spacelike. Sorry for starting the problem with the spacelike signature, it's a habit from Euclidean relativity where s is interpreted as a little curled up spatial dimension. Dividing by (dt)^2 gives:

\left(\frac{ds}{dt}\right)^2 = \frac{(r-2)}{r} -<br /> \frac{(r^3-2y^2)}{(r^3-2r^2)}\dot{x}^2 - \frac{(r^3-2x^2)}{(r^3-2r^2)}\dot{y}^2 -\frac{4rxy}{(r^3-2r^2)}\dot{x}\dot{y}

where the dots indicate derivatives with respect to coordinate time t rather than the more usual proper time. Thus the proper time along the path is:

S = \int_{t_1}^{t_2} \frac{ds}{dt}\; dt

= \int_{t_1}^{t_2} \sqrt{\frac{(r-2)}{r} -<br /> \frac{(r^3-2y^2)}{(r^3-2r^2)}\dot{x}^2 - \frac{(r^3-2x^2)}{(r^3-2r^2)}\dot{y}^2 -\frac{4rxy}{(r^3-2r^2)}\dot{x}\dot{y} }\;\; dt
 
Last edited:
  • #16
The next step is to vary the path by the Euler-Lagrange equations. Discretion being the better part of valor, I will wait until I have a symbolic mathematics engine before I attempt that slope. Basically, I need to calculate the partial derivatives with respect to the coordinates and velocities. The equations of motion are then the time derivatives of the partials with respect to velocities set equal to the partials with respect to positions. That is, for example,

\frac{d}{dt} \left(\frac{\partial ds/dt}{\partial x}\right) = \frac{\partial ds/dt}{\partial \dot{x}}

Carl

Okay, now I've got Maxima running on my laptop. Pretty cool. And cheap.
 
Last edited:
  • #17
If you use Maxima to calculate the Christoffel symbols, watch out for the reversed ordering of the indices. I fell afoul of that issue in GrTensor. Maxima does it to, I believe (I've only used that briefly).

If you have mathematica I think grtensor version M will work with it. Grtensor is better than Maxima in my experience, and Maxima is much better than nothing at all.

I've recalculated the formula for the Schwarzschild metric, and gotten a different answer. The main difference is that I removed some manual copying steps, collecting the values of the Christoffel symbols automatically instead of transcribing them by hand. The new formula seems to conserve energy and momentum in the numerical test case I ran, though some variance can be noted near the horizon. This variance is improved by tightning up the tolearances of the integration routine, though, so I think it's a numerical issue.

The new formula are:

<br /> \frac{d^2 r}{d t^2} = \frac {3 m{{\it vr}}^{2}}{ \left( r-2\,m \right) r} + \left( r-2\,m \right) \left( {{\it vphi}}^{2}-{\frac {m}{{r}^{3}}} \right) <br /> <br />

<br /> \frac{d^2 phi}{d t^2} = -\frac {2 {\it vr}\,{\it vphi}\, \left( r -3\,m \right) }{ \left( r-2\,m \right) r}<br />

vr = dr/dt and vphi = dphi/dt
 
Last edited:
  • #18
Here's what the solution looks like at the code level, minus some manual simplifications of the formula.

manual simplifications are checked by computing simplify(manual - auto) and making sure it is zero.

CC(up,dn,dn) calculates the Christoffel symbols - it's a macro that uses grtensor's built-in routine and changes the order to the standard order.


>qload(schw);
>grcalc(CC(up,dn,dn));
> grdisplay(_);

> tmp := grarray(CC(up,dn,dn));
> vlist := [rdot,0,phidot,tdot];

> eq1 := 0;
> eq2 := 0;
> eq3 := 0;
> eq4 := 0;

> # sum geodesic equations
> for j from 1 by 1 to 4 do
> for k from 1 by 1 to 4 do
> eq1 := eq1 + tmp[1,j,k]*vlist[j]*vlist[k];
> eq2 := eq2 + tmp[2,j,k]*vlist[j]*vlist[k];
> eq3 := eq3 + tmp[3,j,k]*vlist[j]*vlist[k];
> eq4 := eq4 + tmp[4,j,k]*vlist[j]*vlist[k];
> end do;
> end do;
>
> eq1 := subs(sin(theta)=1,eq1);
> eq2 := subs(cos(theta)=0,eq2);
> eq3 := subs(sin(theta)=1,eq3);
> eq4 := subs(sin(theta)=1,eq4);

> # we have t(tau), r(tau), phi(tau)
> # tdot = dt/dtau, rdot = dr/dtau, phidot = dphi/dtau
> # we want vr(t) = dr/dt, vphi(t) = dphi/dt
> # vr * tdot = rdot; vphi * tdot = phidot
> # rdotdot = ar*tdot^2+vr*tdotdot
> # but tdotdot = -eq4, thus

> # equation for r, ar is d^2 r / dt^2
> eq1a := ar*tdot^2 -vr*eq4 + eq1;
> eq1b := subs(rdot = vr*tdot, phidot = vphi*tdot, eq1a);
> eq1c:= simplify(solve(eq1b,ar));
> collect(eq1c,vr);

> # phi equation
> eq3a := aphi*tdot^2 -vphi*eq4 + eq3;
> eq3b := subs(rdot = vr*tdot, phidot = vphi*tdot, eq3a);
> eq3c := simplify(solve(eq3b,aphi));
> collect(eq3c,vr);
 
  • #19
CarlB said:
The next step is to vary the path by the Euler-Lagrange equations. Discretion being the better part of valor, I will wait until I have a symbolic mathematics engine before I attempt that slope.

The standard textbook approach is to write the Lagrangian as

<br /> L = g_{\mu \nu} \frac{d x^\mu}{d \lambda} \frac{d x^\nu}{d \lambda} d \lambda<br />

(Summation is implied).

Applyin the Euler-Lagrange equations to this Lagrangian will yield the geodesic equations.

Of course this approach yields equations parameterized in terms of proper time - i.e. the standard geodesic equations.

AFAIK the easiest way to eliminate proper time is to use the approach I outlined earlier. It is not immediately obvious why one doesn't need to solve for "tdot", but it seems to factor itself out automatically, so one need not solve for it.
 
  • #20
Finally, the painleve metric equations with proper time eliminated. They also seemed to conserve energy and angular momentum in the one test case I ran

<br /> \frac{d^2 r}{d t^2} = \sqrt {{\frac {2 M}{r}}}{\it vr}\, \left( {\frac {{{\it vr<br /> }}^{2}}{2 r}}+3\,{\frac {M}{{r}^{2}}}-r{{\it vphi}}^{2} \right) +3\,{<br /> \frac {M{{\it vr}}^{2}}{{r}^{2}}}+ \left( r-2\,M \right) \left( {{<br /> \it vphi}}^{2}-{\frac {M}{{r}^{3}}} \right) <br />

<br /> \frac{d^2 phi}{d t^2} = {\it vphi}\, \left( \sqrt {{\frac {2 M}{r}}} \left( {\frac {{{\it vr}}^{2}}{2 r}}-r{{\it vphi}}^{2}+{\frac {M}{{r}^{2}}}<br /> \right) -2\,{\frac {{\it vr}\, \left( r-M \right) }{{r}^{2}}}<br /> \right) <br />

One can see that while the Painleve equations are simple in the Newtonian limit when vr=vphi is almost zero they are not as simple as the Schwarzschild equations in the fully relativistic case.

This could be anticipated by the "count the number of Christoffel symbols" method.
 
Last edited:
  • #21
Finally, since I have the automaton working, here's the sort of result the above procedure gets when applied to the psueudo-cartesian metric you suggested earlier.

d^2x/dt^2 :=
-(-6*x^3*vy^2*y^4+6*x^5*vy^2*y^2+4*vr*y^7*vy-8*x^5-6*y*vx*vy*r*x^6-4*x
^3*vx^2*y^6-8*x^3*y^2+2*vr*r*y^3*vy*x^4+4*x*vy^2*z^6*y^2+2*vr*r*x^5*vx
*y^2-6*vr*r*x^3*vx*z^4-8*x^3*z^2-8*x^5*y^2+2*x^3*y^4-16*x^5*z^2+2*x^9*
vy^2-6*x^7-12*x^3*y^2*z^2+8*x^7*vy^2+4*x^7*vy^2*y^2+8*x^7*vy^2*z^2+4*v
r*x^7*vx-4*x*vx^2*y^8-4*x^3*vx^2*y^4+4*x^5*vx^2*z^2+2*x^3*vx^2*z^4+4*x
^7*vx^2*y^2+4*x^5*vx^2*y^4-4*x^3*vy^2*y^6-2*x*vy^2*y^8+6*x^5*vy^2*z^2-
6*x^3*vy^2*z^4+12*x^5*vy^2*z^4+8*x^3*vy^2*z^6+12*y^3*vx*vy*x^4-8*y^5*v
x*vy*x^2+12*vr*x^5*vx*y^2+8*vr*x^5*vx*z^2+12*vr*x^3*vx*y^4+4*vr*x^3*vx
*z^4+12*vr*y^3*vy*x^4+12*vr*y^5*vy*x^2+8*vr*y^5*vy*z^2+4*vr*y^3*vy*z^4
-4*x*vx^2*y^6-12*x*vx^2*y^6*z^2+6*x^3*vx^2*z^2*y^2+2*x^5*vx^2*y^2+4*x^
5*vx^2*y^2*z^2-4*x*vy^2*y^6-4*x*vy^2*y^6*z^2-12*x^3*vy^2*y^2*z^2-12*x*
vy^2*y^4*z^2+12*x^5*vy^2*y^2*z^2+12*x^3*vy^2*z^4*y^2-4*x*vy^2*z^6-12*x
*vy^2*z^4*y^2+12*y^3*vx*vy*x^2*z^2+4*y^6*x-14*x^3*z^4-4*x*z^6+2*x^7*vx
^2+16*vr*x^3*vx*y^2*z^2+16*vr*y^3*vy*x^2*z^2+2*x*vy^2*z^8+4*y^4*x*z^2-
4*x*z^4*y^2+4*vr*x*vx*y^6+8*vr*x*vx*y^4*z^2+4*vr*x*vx*y^2*z^4+4*vr*y*v
y*x^6+4*x*vx^2*y^2*z^4-8*x^3*vx^2*y^4*z^2-12*x*vx^2*y^4*z^4-4*x^3*vx^2
*y^2*z^4-4*x*vx^2*y^2*z^6+20*y*vx*vy*x^6+8*y^5*vx*vy*z^2+16*y^3*vx*vy*
z^4+8*y*vx*vy*z^6+8*vr*y*vy*x^4*z^2+4*vr*y*vy*x^2*z^4+32*y*vx*vy*x^4*z
^2+20*y*vx*vy*x^2*z^4-8*vr*r*y^3*vy*x^2+13*x*vx^2*r*y^4*z^2-6*y*vx*vy*
r*z^6-16*y*vx*vy*r*x^4+9*x*vy^2*r*y^4*z^2+3*x*vy^2*r*y^2*z^4-6*vr*r*x^
5*vx*z^2+2*vr*r*x^3*vx*y^4-2*vr*r*x^7*vx-2*vr*r*y^7*vy-3*x^5*vx^2*r*z^
2+5*x^3*vx^2*r*y^4-3*x^3*vx^2*r*z^4-6*x^5*vx^2*r*y^2+10*x*vx^2*r*y^6-x
*vx^2*r*z^6+8*x^3*vx^2*r*y^2+8*x^3*vy^2*r*z^2-17*x^5*vy^2*r*z^2+4*x^3*
vy^2*r*y^4-10*x^3*vy^2*r*z^4-9*x^5*vy^2*r*y^2+5*x*vy^2*r*y^6-x*vy^2*r*
z^6+8*x^3*vy^2*r*y^2+2*r*x^3*z^2*y^2-y^4*r*x*z^2+x*r*z^4*y^2+3*r*x^5*z
^2+8*y^2*r*x^3-4*y^4*r*x-8*x^7*vy^2*r+16*r*x^3*z^2+4*x*r*z^4-x^7*vx^2*
r+3*r*x^3*z^4+x^5*r*y^2-y^4*r*x^3-y^6*r*x+x*r*z^6+12*x^5*r+x^7*r-4*x^3
*vx^2*r*y^2*z^2-8*vr*r*x^3*vx*y^2-2*vr*r*x*vx*y^6-6*vr*r*x*vx*y^4*z^2-
4*vr*r*x^3*vx*y^2*z^2-6*vr*r*x*vx*y^2*z^4-6*vr*r*y*vy*x^4*z^2-4*vr*r*y
^3*vy*x^2*z^2-6*vr*r*y*vy*x^2*z^4-2*vr*r*x*vx*z^6-2*vr*r*y*vy*x^6+2*vr
*r*y^5*vy*x^2-6*vr*r*y^5*vy*z^2-6*vr*r*y^3*vy*z^4-2*vr*r*y*vy*z^6+2*x*
vx^2*r*y^2*z^4-16*y*vx*vy*r*x^2*z^2-6*y*vx*vy*r*x^4*z^2+6*y^5*vx*vy*r*
x^2-6*y*vx*vy*r*x^2*z^4-6*y^5*vx*vy*r*z^2-12*y^3*vx*vy*r*z^4-6*x^3*vy^
2*r*y^2*z^2)/(5*y^8*z^2+y^8*x^2-2*y^4*x^6+44*x^4*y^2*z^2+44*x^2*y^4*z^
2+28*x^2*y^2*z^4+4*x^8+4*y^8+20*x^6*y^2+12*x^6*z^2+32*x^4*y^4+12*x^4*z
^4+20*x^2*y^6+4*x^2*z^6+12*y^6*z^2+12*y^4*z^4+4*y^2*z^6+y^10+x^10+z^10
+6*y^4*x^4*z^2+8*y^6*x^2*z^2+18*y^4*x^2*z^4+8*x^6*z^2*y^2+18*x^4*z^4*y
^2+16*x^2*z^6*y^2+x^8*y^2+10*x^6*z^4-2*y^6*x^4+10*x^4*z^6+5*x^8*z^2+5*
x^2*z^8+10*y^4*z^6+5*y^2*z^8+10*y^6*z^4-8*y^2*r*x^2*z^2-28*r*x^2*y^2*z
^4-26*r*x^2*y^4*z^2-26*r*x^4*y^2*z^2-8*y^2*r*x^4-8*y^4*r*x^2-14*r*x^6*
z^2-8*r*x^4*y^4-18*r*x^4*z^4-10*r*x^2*z^6-14*r*y^6*z^2-18*r*y^4*z^4-10\

I haven't checked this for accuracy, and I don't plan to - actually, it looks to me like the whole thing didn't paste correctly, it was probably too long for the paste buffer...
 
  • #22
pervect said:
Finally, since I have the automaton working, here's the sort of result the above procedure gets when applied to the psueudo-cartesian metric you suggested earlier.

It appears I've got MAXIMA running. To crunch out the equations of motion by varying a path, the square root is somewhat ugly. Consequently, it helps to substitute L^2 for L in the Euler-Lagrange equations. One begins by defining:

I = \left(\frac{ds}{dt}\right)^2

then the thing which is to be minimized is \int \sqrt{I} \;dt. The usual Euler-Lagrange equations minimize \int L\;dt. To convert them to the square root form, simply substitute and obtain:

\frac{d}{dt}\left(\frac{\partial I}{\partial \dot{x}}\right)<br /> = \frac{1}{2I}\frac{\partial I}{\partial \dot{x}} + \frac{\partial I}{\partial x},

and similarly for y. The equations work out a lot simpler if one abbreviates sqrt(x^2+y^2) as r. While I've computed the equations of motion once, I'm not going to post the results until tomorrow as the tools are new to me and I'd like to go through the exercise again. But I have no doubt that the Maxima software will save me much grief:

http://maxima.sourceforge.net/

Here it is the next day and I'm too busy to deal with this. One thing I'd like to mention, working out the equations of motion in coordinate time this way has the advantage that the resulting geodesics can be massless as well as massive.

Two of Einstein's principles were simplicity and the use of geometry. It seems that there are three different things that one could require to be simple. One could require the equations of motion, the energies / lagrangian, or one could require an overall principle be simple. What Einstein did was to require the overall principle be simple.

I'm not at all convinced that this is the correct choice. Maybe the Gravity Probe B will give us more information on this. Newton's gravitation is simple in all three parts, the equations of motion, the Lagrangian, and the overall principle (linearity). I know that the equations of motion have to be a part of reality, but the others might just be parts of our mathematics. So my inclination is to require that the equations of motion be simple. Accordingly, I'll be looking for equations of motion that are simple rather than Lagrangians or principles that are simple.

Carl
 
Last edited:
  • #23
CarlB said:
It appears I've got MAXIMA running. To crunch out the equations of motion by varying a path, the square root is somewhat ugly. Consequently, it helps to substitute L^2 for L in the Euler-Lagrange equations. One begins by defining:

I = \left(\frac{ds}{dt}\right)^2

then the thing which is to be minimized is \int \sqrt{I} \;dt. The usual Euler-Lagrange equations minimize \int L\;dt. To convert them to the square root form, simply substitute and obtain:

\frac{d}{dt}\left(\frac{\partial \sqrt{I}}{\partial \dot{x}}\right)<br /> = \frac{d}{d t} \left[ \frac{1}{2 \sqrt{I}} \left( \frac{\partial I}{\partial \dot{x}} \right) \right] = \frac{1}{2 \sqrt{I}} \frac{\partial I}{\partial x}

Good luck with this approach. (I think there was a small typo which I changed, see above).

Minimizing the action sqrt(I) [in some sign conventions, sqrt(-I)] gives you a parameter independent path. However, I think you'll find when you actually try this direct route is that it gets very messy.

Minimizing the action I (rather than the square root) is the usual route to the geodesic equation. While the path generated by the previous approach was independent of the parametrization of the path, the path generated by the approach using the action I (rather than its square root) requires that the path be parameterized by proper time or an affine transformation of proper time.

You might want to consult a textbook on GR for more verification if you are interested in the details.

I expect that you'll find that minimzing I, and then eliminating proper time as a separate step, will be computationally much more convenient than your direct approach using sqrt(I).
 
Last edited:
  • #24
BTW, Maxima's ctensor() package will calculate the geodesic equations for you:

For example (/* are comments */):

load(ctensor);
ct_coordsys(exteriorschwarzschild);
//* calculate and setup stuff such as inverse metric */
cmetric();
//* display some stuff */
//* coordinates */
ct_coords;
//* metric */
lg;
//* inverse metric */
ug;
//* the geodesic equations */
cgeodesic(true);
 
  • #25
pervect said:
Good luck with this approach. (I think there was a small typo which I changed, see above).

Thanks for pointing this out. I've run out of all possible other things worth doing at the moment, so time to run this calculation out. Let me begin where you left off:

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

Multiply by 2 to get rid of the halves. I want to eliminate the sqrt(I). Accordingly, use the product rule on the time derivative:

-\frac{1}{2 I^{3/2}}\left(\frac{\partial I}{\partial \dot{x}}\right)\frac{dI}{dt} + \frac{1}{I^{1/2}}\frac{d}{dt}\left(\frac{\partial I}{\partial \dot{x}}\right) = \frac{1}{I^{1/2}}\frac{\partial I}{\partial x}

Bring all the terms to the LHS and multiply by I sqrt(I) to get:

I\frac{d}{dt}\left(\frac{\partial I}{\partial \dot{x}}\right) -\frac{1}{2}\left(\frac{\partial I}{\partial \dot{x}}\right)\frac{dI}{dt} - I\frac{\partial I}{\partial x} = 0

Write out the derivatives with respect to time in terms of partial derivatives:

I\left(\frac{\partial^2 I}{\partial \dot{x}\partial x}\right)\dot{x} +<br /> I\left(\frac{\partial^2 I}{\partial \dot{x}^2}\right) \ddot{x} +<br /> I\left(\frac{\partial^2 I}{\partial \dot{x}\partial y}\right) \dot{y} +<br /> I\left(\frac{\partial^2 I}{\partial \dot{x}\partial \dot{y}}\right) \ddot{y}
-\frac{1}{2}\left(\frac{\partial I}{\partial \dot{x}}\right)\left(\frac{\partial I}{\partial x}\right)\dot{x}<br /> -\frac{1}{2}\left(\frac{\partial I}{\partial \dot{x}}\right)\left(\frac{\partial I}{\partial \dot{x}}\right)\ddot{x} <br /> -\frac{1}{2}\left(\frac{\partial I}{\partial \dot{x}}\right)\left(\frac{\partial I}{\partial y}\right)\dot{y}<br /> -\frac{1}{2}\left(\frac{\partial I}{\partial \dot{x}}\right)\left(\frac{\partial I}{\partial \dot{y}}\right)\ddot{y} <br /> -I\left(\frac{\partial I}{\partial x}\right) = 0

Collect terms according to type in the phase coordinates:

\begin{array}{rcl}<br /> &amp;&amp;\left[I\left(\frac{\partial^2 I}{\partial \dot{x}^2}\right)<br /> -\frac{1}{2}\left(\frac{\partial I}{\partial<br /> \dot{x}}\right)\left(\frac{\partial I}{\partial<br /> \dot{x}}\right)\right] \ddot{x} + \left[I\left(\frac{\partial^2<br /> I}{\partial \dot{x}\partial \dot{y}}\right)<br /> -\frac{1}{2}\left(\frac{\partial I}{\partial<br /> \dot{x}}\right)\left(\frac{\partial I}{\partial<br /> \dot{y}}\right)\right] \ddot{y}\\<br /> &amp;=&amp;\left[\frac{1}{2}\left(\frac{\partial I}{\partial<br /> \dot{x}}\right)\left(\frac{\partial I}{\partial x}\right)<br /> -I\left(\frac{\partial^2 I}{\partial \dot{x}\partial x}\right)<br /> \right]\dot{x} + \left[\frac{1}{2}\left(\frac{\partial I}{\partial<br /> \dot{x}}\right)\left(\frac{\partial I}{\partial y}\right)<br /> -I\left(\frac{\partial^2 I}{\partial \dot{x}\partial y}\right)<br /> \right] \dot{y} + I\left(\frac{\partial I}{\partial x}\right)<br /> \end{array}

A similar equation obtains for y:

\begin{array}{rcl}<br /> &amp;&amp;\left[I\left(\frac{\partial^2<br /> I}{\partial \dot{y}\partial \dot{x}}\right)<br /> -\frac{1}{2}\left(\frac{\partial I}{\partial<br /> \dot{y}}\right)\left(\frac{\partial I}{\partial<br /> \dot{x}}\right)\right] \ddot{x} +<br /> \left[I\left(\frac{\partial^2 I}{\partial \dot{y}^2}\right)<br /> -\frac{1}{2}\left(\frac{\partial I}{\partial<br /> \dot{y}}\right)\left(\frac{\partial I}{\partial<br /> \dot{y}}\right)\right] \ddot{y}\\<br /> &amp;=&amp;\left[\frac{1}{2}\left(\frac{\partial I}{\partial<br /> \dot{y}}\right)\left(\frac{\partial I}{\partial y}\right)<br /> -I\left(\frac{\partial^2 I}{\partial \dot{y}\partial y}\right)<br /> \right]\dot{y} + \left[\frac{1}{2}\left(\frac{\partial I}{\partial<br /> \dot{y}}\right)\left(\frac{\partial I}{\partial x}\right)<br /> -I\left(\frac{\partial^2 I}{\partial \dot{y}\partial x}\right)<br /> \right] \dot{x} + I\left(\frac{\partial I}{\partial y}\right)



Carl
 
Last edited:
  • #26
The above equations can be written as:

\begin{array}{rcl}<br /> A &amp;=&amp; I\left(\frac{\partial^2 I}{\partial \dot{x}^2}\right)<br /> -\frac{1}{2}\left(\frac{\partial I}{\partial<br /> \dot{x}}\right)\left(\frac{\partial I}{\partial \dot{x}}\right) \\<br /> B &amp;=&amp; I\left(\frac{\partial^2 I}{\partial \dot{x}\partial<br /> \dot{y}}\right) -\frac{1}{2}\left(\frac{\partial I}{\partial<br /> \dot{x}}\right)\left(\frac{\partial I}{\partial \dot{y}}\right) \\<br /> C &amp;=&amp; \left[\frac{1}{2}\left(\frac{\partial I}{\partial<br /> \dot{x}}\right)\left(\frac{\partial I}{\partial x}\right)<br /> -I\left(\frac{\partial^2 I}{\partial \dot{x}\partial x}\right)<br /> \right]\dot{x} + \left[\frac{1}{2}\left(\frac{\partial I}{\partial<br /> \dot{x}}\right)\left(\frac{\partial I}{\partial y}\right)<br /> -I\left(\frac{\partial^2 I}{\partial \dot{x}\partial y}\right)<br /> \right] \dot{y} + I\left(\frac{\partial I}{\partial x}\right) \\<br /> D &amp;=&amp; I\left(\frac{\partial^2 I}{\partial \dot{y}\partial<br /> \dot{x}}\right) -\frac{1}{2}\left(\frac{\partial I}{\partial<br /> \dot{y}}\right)\left(\frac{\partial I}{\partial \dot{x}}\right) \\<br /> E &amp;=&amp; I\left(\frac{\partial^2 I}{\partial \dot{y}^2}\right)<br /> -\frac{1}{2}\left(\frac{\partial I}{\partial<br /> \dot{y}}\right)\left(\frac{\partial I}{\partial \dot{y}}\right) \\<br /> F &amp;=&amp; \left[\frac{1}{2}\left(\frac{\partial I}{\partial<br /> \dot{y}}\right)\left(\frac{\partial I}{\partial y}\right)<br /> -I\left(\frac{\partial^2 I}{\partial \dot{y}\partial y}\right)<br /> \right]\dot{y} + \left[\frac{1}{2}\left(\frac{\partial I}{\partial<br /> \dot{y}}\right)\left(\frac{\partial I}{\partial x}\right)<br /> -I\left(\frac{\partial^2 I}{\partial \dot{y}\partial x}\right)<br /> \right] \dot{x} + I\left(\frac{\partial I}{\partial y}\right)<br />

so that

A \ddot{x} + B \ddot{y} = C
D \ddot{x} + E \ddot{y} = F

The solution is:

\begin{array}{rcl}<br /> \ddot{x} &amp;=&amp; (B*F - C*E)/(B*D-A*E),\\<br /> \ddot{y} &amp;=&amp; (C*D - A*F)/(B*D-A*E).<br /> \end{array}

To compute A, B, C, D, E, and F, I need the following terms:
\begin{array}{cccc}<br /> \left(\frac{\partial I}{\partial x}\right) &amp;<br /> \left(\frac{\partial I}{\partial \dot{x}}\right) &amp;<br /> \left(\frac{\partial I}{\partial y}\right) &amp;<br /> \left(\frac{\partial I}{\partial \dot{y}}\right) \\<br /> \left(\frac{\partial^2 I}{\partial \dot{x}^2}\right) &amp;<br /> \left(\frac{\partial^2 I}{\partial \dot{y}^2}\right) &amp;<br /> \left(\frac{\partial^2 I}{\partial \dot{x} \partial \dot{y}}\right) &amp; \\<br /> \left(\frac{\partial^2 I}{\partial \dot{x} \partial x}\right) &amp;<br /> \left(\frac{\partial^2 I}{\partial \dot{y} \partial y}\right) &amp;<br /> \left(\frac{\partial^2 I}{\partial \dot{x} \partial y}\right) &amp;<br /> \left(\frac{\partial^2 I}{\partial \dot{y} \partial x}\right)<br /> \end{array}

It remains to compute these from the value of I for the Schwarzschild and Painleve coordinates.

Carl
 
Last edited:
  • #27
As a test of this solution to the equations of motion, let me see what happens when I use it to derive Newtonian gravitation. I need to minimize the following function over the path:

L(x,y,\dot{x},\dot{y}) = T - V = \frac{\dot{x}^2 + \dot{y}^2}{2} + \frac{1}{r}

Writing this in \sqrt{I} form, I have that

I = I_N = \left(\frac{\dot{x}^2 + \dot{y}^2}{2} + \frac{1}{r}\right)^2

The first partial derivatives are:

\frac{\partial I}{\partial x} = -\left(\dot{x}^2 + \dot{y}^2 + \frac{2}{r}\right)\frac{x}{r^3}
\frac{\partial I}{\partial y} = -\left(\dot{x}^2 + \dot{y}^2 + \frac{2}{r}\right)\frac{y}{r^3}
\frac{\partial I}{\partial \dot{x}} = 2\left(\dot{x}^2 + \dot{y}^2 + \frac{2}{r}\right)\dot{x}
\frac{\partial I}{\partial \dot{y}} = 2\left(\dot{x}^2 + \dot{y}^2 + \frac{2}{r}\right)\dot{y}

The second order partial derivative with respect to momenta are:
\frac{\partial^2 I}{\partial \dot{x}^2} = \left(6\dot{x}^2 + 2\dot{y}^2 + \frac{4}{r}\right)
\frac{\partial^2 I}{\partial \dot{y}^2} = \left(6\dot{y}^2 + 2\dot{x}^2 + \frac{4}{r}\right)
\frac{\partial^2 I}{\partial \dot{x} \partial \dot{y}} = 4 \dot{x}\dot{y}

And finally, the four mixed second order partial derivatives are:
\frac{\partial^2 I}{\partial \dot{x} \partial x} = -\frac{4 \dot{x} x}{r^3}
\frac{\partial^2 I}{\partial \dot{y} \partial y} = -\frac{4 \dot{y} y}{r^3}
\frac{\partial^2 I}{\partial \dot{x} \partial y} = -\frac{4 \dot{x} y}{r^3}
\frac{\partial^2 I}{\partial \dot{y} \partial x} = -\frac{4 \dot{y} x}{r^3}

It's clear that these are going to simplify wonderfully when I substitute them in for A, B, C, D, E, and F.

Carl
 
Last edited:
  • #28
Are you still working on this? Given that I presented an answer, quite a long time ago, by using the standard geodesic equations and then eliminating proper time, I would have thought that you would have been able to confirm or deny the correctness of my answer by now.

Specifically, I'm talking about the answers in:
https://www.physicsforums.com/showpost.php?p=1046874&postcount=17
for the Schwarzschild metric and
https://www.physicsforums.com/showpost.php?p=1047041&postcount=20
for the Painleve metric.

The prorcess of eliminating proper time turns out to be quite simple, BTW, if approached in the right manner. It turns out to be a matter of just subtracting a multiple of one quadtratic form from another quadratic form, as you'll see if you actually attempt this method.

I'm reasonably sure the answers I gave in these two posts are correct, BTW. but I haven't done a lot of testing, just one simulation that seemed to conserve the quantities that should be conserved. (And there's always the possibility of typos).

Interestingly enough, there is an alternate method available that was mentioned in another thread, called the Einbein. See Self-adjoint's thread of that title for more info. This accomplishes the same result as using "proper time" by using a different auxillary variable which accomplishes much the same purpose.

I'll also standy by remarks that changing to "cartesian-like" x and y coordinates before solving the equations of motion is Not A Good Idea. See

https://www.physicsforums.com/showpost.php?p=1047056&postcount=21

for why.

So in concluion, I think I've done about all I can possibly do to help you with this problem, to the point of actually giving you a correct solution for the Schwarzschild and Painleve metrics.

The one thing more that I could do is to recommend some specific textbooks, which I'd be quite willing to do. However, as you are not a GR person and probably don't have GR textbooks in your library, I'm not sure that that would be all that helpful. If you have some questions about how the standard geodesic equations are derived, this issue should be covered in most GR textbooks.

I imagine that particle physicists might have their own approach to the problem (like the Einbein I mentioned) as well. In any event, I don't think there's much more I can do for you (except perhaps to give some textbook references, which won't help that much if you don't have the textbooks in question) having already solved the problem and having presented the solution.
 
  • #29
Not to take away from this very interesting discussion, but are the trig functions really that slow? They used to be in older versions of Java but are fairly optimized now in modern JVM's...
 
  • #30
pervect said:
Are you still working on this? Given that I presented an answer, quite a long time ago, by using the standard geodesic equations and then eliminating proper time, I would have thought that you would have been able to confirm or deny the correctness of my answer by now.

Business has been very busy these past few weeks and I haven't spent much time on it, but yes, I'm still working on this. Eventually I will get around to checking your equation.

pervect said:
Interestingly enough, there is an alternate method available that was mentioned in another thread, called the Einbein.

I've been following this with interest.

pervect said:
I'll also standy by remarks that changing to "cartesian-like" x and y coordinates before solving the equations of motion is Not A Good Idea.

You're wrong.

pervect said:
The one thing more that I could do is to recommend some specific textbooks, which I'd be quite willing to do. However, as you are not a GR person and probably don't have GR textbooks in your library, I'm not sure that that would be all that helpful.

Of course I have MTW which covers the problem in the mode you've given at some length. I took a one quarter (graduate) GR class at the Univ. of California Irvine. I had the highest grade in the class and earned the "A+" that were assigned to the top student in a class at that time. Probably the reason I did very well in physics grad school was because I had a masters degree in mathematics before I switched to physics.

Carl
 
Last edited:
  • #31
jbusc said:
Not to take away from this very interesting discussion, but are the trig functions really that slow? They used to be in older versions of Java but are fairly optimized now in modern JVM's...

It's partly speed and partly elementary particles stuff that I am not ready to explain. Java allows you to write subroutines in other languages (i.e. assembler or C), so you can write blindingly fast integration subroutines.

Example, AMD-64 CPU with 64-bit arithmetic. The latency of a 64-bit floating multiply or add (FMUL or FADD) is 4 clocks, while the latency of the 64-bit floating cosine/sine (FSINCOS) is 104. For these numbers, see
http://www.amd.com/us-en/assets/content_type/white_papers_and_tech_docs/25112.PDF

Carl
 
  • #32
CarlB said:
You're wrong.

Carl

The geodesic equation for your metric isn't any ordinary rabbit. It's got teeth like - it can leap like - look at the bones man! Look at the pages of output from Maxima's built in cgeodesic() function!

I warned ye! I try and try and try and tell them, but does anybody ever listen to Tim? Nooooo. "It's just a little bunny rabit", they say. "Go and chop it's head off", they say.

Oh well, I'll go put my "no" vote in on the poll now...
 
  • #33
Looks like an extra factor of 2 appeared in a previous post, let me give the corrected values for the partial derivatives with the Newtonian gravitation (as computed in sqrt(I) form):

I = I_N = \left(\frac{\dot{x}^2 + \dot{y}^2}{2} + \frac{1}{r}\right)^2 = (\sqrt{I})^2

\frac{\partial I}{\partial x} = -\sqrt{I}\left(\frac{2x}{r^3}\right)
\frac{\partial I}{\partial y} = -\sqrt{I}\left(\frac{2y}{r^3}\right)
\frac{\partial I}{\partial \dot{x}} = \sqrt{I}(2\dot{x})
\frac{\partial I}{\partial \dot{y}} = \sqrt{I}(2\dot{y})

The second order partial derivative with respect to momenta are:
\frac{\partial^2 I}{\partial \dot{x}^2} = \left(3\dot{x}^2 + \dot{y}^2 + \frac{2}{r}\right)
\frac{\partial^2 I}{\partial \dot{y}^2} = \left(3\dot{y}^2 + \dot{x}^2 + \frac{2}{r}\right)
\frac{\partial^2 I}{\partial \dot{x} \partial \dot{y}} = 2 \dot{x}\dot{y}

And finally, the four mixed second order partial derivatives are:
\frac{\partial^2 I}{\partial \dot{x} \partial x} = -\frac{2 \dot{x} x}{r^3}
\frac{\partial^2 I}{\partial \dot{y} \partial y} = -\frac{2 \dot{y} y}{r^3}
\frac{\partial^2 I}{\partial \dot{x} \partial y} = -\frac{2 \dot{x} y}{r^3}
\frac{\partial^2 I}{\partial \dot{y} \partial x} = -\frac{2 \dot{y} x}{r^3}
 
Last edited:
  • #34
Now, substituting the above into post #26 gives:

\begin{array}{rcl}A &amp;=&amp; I\left(\frac{\partial^2 I}{\partial \dot{x}^2}\right)-\frac{1}{2}\left(\frac{\partial I}{\partial\dot{x}}\right)\left(\frac{\partial I}{\partial \dot{x}}\right) \\ &amp;=&amp;<br /> I(3\dot{x}^2+\dot{y}^2+2/r)-0.5(2\dot{x}\sqrt{I})(2\dot{x}\sqrt{I}) \\ &amp;=&amp;<br /> I(\dot{x}^2+\dot{y}^2 +2/r)\end{array}

\begin{array}{rcl}B &amp;=&amp; I\left(\frac{\partial^2 I}{\partial \dot{x}\partial \dot{y}}\right) -\frac{1}{2}\left(\frac{\partial I}{\partial \dot{x}}\right)\left(\frac{\partial I}{\partial \dot{y}}\right) \\ &amp;=&amp;<br /> I (2\dot{x}\dot{y})-0.5(2\sqrt{I}\dot{x})(2\sqrt{I}\dot{y}) \\ &amp;=&amp; 0\end{array}

\begin{array}{rcl}C &amp;=&amp; \left[\frac{1}{2}\left(\frac{\partial I}{\partial\dot{x}}\right)\left(\frac{\partial I}{\partial x}\right)-I\left(\frac{\partial^2 I}{\partial \dot{x}\partial x}\right)\right]\dot{x} + \left[\frac{1}{2}\left(\frac{\partial I}{\partial\dot{x}}\right)\left(\frac{\partial I}{\partial y}\right)-I\left(\frac{\partial^2 I}{\partial \dot{x}\partial y}\right)\right] \dot{y} + I\left(\frac{\partial I}{\partial x}\right) \\ &amp;=&amp;<br /> (0.5(2\sqrt{I}\dot{x})(-2\sqrt{I}x/r^3) + I2\dot{x}x/r^3)\dot{x} +<br /> (0.5(2\sqrt{I}\dot{x})(-2\sqrt{I}y/r^3) +I2\dot{x}y/r^3)\dot{y} + I\sqrt{I}(-2x/r^3) \\<br /> &amp;=&amp; -I(\dot{x}^2+\dot{y}^2 +2/r )x/r^3 \end{array}

\begin{array}{rcl} D &amp;=&amp; I\left(\frac{\partial^2 I}{\partial \dot{y}\partial\dot{x}}\right) -\frac{1}{2}\left(\frac{\partial I}{\partial\dot{y}}\right)\left(\frac{\partial I}{\partial \dot{x}}\right) \\ &amp;=&amp; <br /> I(2\dot{x}\dot{y})-0.5(2\sqrt{I}\dot{y})(2\sqrt{I}\dot{x}) \\ &amp;=&amp; 0\end{array}

\begin{array}{rcl} E &amp;=&amp; I\left(\frac{\partial^2 I}{\partial \dot{y}^2}\right)-\frac{1}{2}\left(\frac{\partial I}{\partial\dot{y}}\right)\left(\frac{\partial I}{\partial \dot{y}}\right) \\ &amp;=&amp;<br /> I(6\dot{y}^2+2\dot{x}^2+4/r) -0.5(4\sqrt{I}\dot{y})^2 \\&amp;=&amp;<br /> I(-2\dot{y}^2+2\dot{x}^2+4/r)\end{array}

\begin{array}{rcl} F &amp;=&amp; \left[\frac{1}{2}\left(\frac{\partial I}{\partial\dot{y}}\right)\left(\frac{\partial I}{\partial y}\right)-I\left(\frac{\partial^2 I}{\partial \dot{y}\partial y}\right)\right]\dot{y} + \left[\frac{1}{2}\left(\frac{\partial I}{\partial\dot{y}}\right)\left(\frac{\partial I}{\partial x}\right)-I\left(\frac{\partial^2 I}{\partial \dot{y}\partial x}\right)\right] \dot{x} + I\left(\frac{\partial I}{\partial y}\right) \\ &amp;=&amp;<br /> (0.5(2\sqrt{I}\dot{y})(-2\sqrt{I}y/r^3) + I2\dot{y}y/r^3)\dot{y} +<br /> (0.5(2\sqrt{I}\dot{y})(-2\sqrt{I}x/r^3) + I2\dot{y}x/r^3)\dot{x} +I\sqrt{I}(-2y/r^3) \\ &amp;=&amp;<br /> -I(\dot{x}^2+\dot{y}^2+2/r)y/r^3\end{array}

Since A, B, C, D, E and F are each written in the above as multiples of I, we can factor this out to get A', B', C', D', E' and F':

\begin{array}{rcl}<br /> A&#039; = E&#039; &amp;=&amp; (\dot{x}^2+\dot{y}^2 +2/r) \\<br /> B&#039; = D &amp;=&amp; 0 \\<br /> C&#039; &amp;=&amp; -(\dot{x}^2+\dot{y}^2 +2/r )x/r^3 \\<br /> F&#039; &amp;=&amp; -(\dot{x}^2+\dot{y}^2 +2/r )y/r^3\end{array}


And the equations of motion are now simply:

\begin{array}{rcccl}\ddot{x} &amp;=&amp; C&#039;/A&#039; &amp;=&amp; -x/r^3,\\<br /> \ddot{y} &amp;=&amp; F&#039;/E&#039; &amp;=&amp; -y/r^3,\end{array}

which are the correct equations of motion for Newton's gravitation. Thus the method works. Next, the Schwarzschild metric.

Carl
 
Last edited:
  • #35
Okay, it's time for me to push the Schwarzschild metric of post #15 through the machinery developed in post #26. I was going to do this the other night but PF crashed and I deleted the post.

Rewriting the "I" of post #15 (and removing a factor of r that unnaturally appeared in post #2) gives a particularly simple form:

I = 1-\frac{2}{r}-(\dot{x}^2+\dot{y}^2) -2\frac{(x\dot{x}+y\dot{y})^2}{r^3-r^2}

The first order partial derivatives are:

\left(\frac{\partial I}{\partial x}\right) = <br /> \frac{2x}{r^3}-\frac{4\dot{x}(x\dot{x}+y\dot{y})}{r^3-r^2}<br /> +2\frac{(x\dot{x}+y\dot{y})^2 (3rx-2x)}{(r^3-r^2)^2}

\left(\frac{\partial I}{\partial y}\right) = <br /> \frac{2y}{r^3}-\frac{4\dot{y}(x\dot{x}+y\dot{y})}{r^3-r^2}<br /> +2\frac{(x\dot{x}+y\dot{y})^2 (3ry-2y)}{(r^3-r^2)^2}

\left(\frac{\partial I}{\partial \dot{x}}\right) = <br /> -2\dot{x}-\frac{4x(x\dot{x}+y\dot{y})}{r^3-r^2}

\left(\frac{\partial I}{\partial \dot{y}}\right) = <br /> -2\dot{y}-\frac{4y(x\dot{x}+y\dot{y})}{r^3-r^2}

It doesn't look like the 2nd order partials will be particularly difficult. However, this busy old man is going to call it a night early.

Carl
 
Last edited:
  • #36
Looks like my previous effort dropped a 2 on the r squared term in the denominator, so as usual, let me begin by repairing the previous damage.

I = 1-\frac{2}{r}-(\dot{x}^2+\dot{y}^2) -2\frac{(x\dot{x}+y\dot{y})^2}{r^3-2r^2}

The first order partial derivatives are:

\left(\frac{\partial I}{\partial x}\right) = <br /> \frac{2x}{r^3}-\frac{4\dot{x}(x\dot{x}+y\dot{y})}{r^3-2r^2}<br /> +2\frac{(x\dot{x}+y\dot{y})^2 (3rx-4x)}{(r^3-2r^2)^2}

\left(\frac{\partial I}{\partial y}\right) = <br /> \frac{2y}{r^3}-\frac{4\dot{y}(x\dot{x}+y\dot{y})}{r^3-2r^2}<br /> +2\frac{(x\dot{x}+y\dot{y})^2 (3ry-4y)}{(r^3-2r^2)^2}

\left(\frac{\partial I}{\partial \dot{x}}\right) = <br /> -2\dot{x}-\frac{4x(x\dot{x}+y\dot{y})}{r^3-2r^2}

\left(\frac{\partial I}{\partial \dot{y}}\right) = <br /> -2\dot{y}-\frac{4y(x\dot{x}+y\dot{y})}{r^3-2r^2}


2nd order partials with respect to velocity:

\left(\frac{\partial^2 I}{\partial \dot{x}^2}\right) = -2-\frac{4x^2}{r^3-2r^2}

\left(\frac{\partial^2 I}{\partial \dot{y}^2}\right) = -2-\frac{4y^2}{r^3-2r^2}

\left(\frac{\partial^2 I}{\partial \dot{x}\partial \dot{y}}\right) = -\frac{4xy}{r^3-2r^2}

Mixed 2nd order partials:

\left(\frac{\partial^2 I}{\partial \dot{x}\partial x}\right) = -\frac{8x\dot{x}+4y\dot{y}}{r^3-2r^2}<br /> +\frac{(x\dot{x}+y\dot{y}) (12r-16)x^2}{(r^3-2r^2)^2}

\left(\frac{\partial^2 I}{\partial \dot{y}\partial y}\right) = -\frac{8y\dot{y}+4x\dot{x}}{r^3-2r^2}<br /> +\frac{(x\dot{x}+y\dot{y}) (12r-16)y^2}{(r^3-2r^2)^2}

\left(\frac{\partial^2 I}{\partial \dot{x}\partial y}\right) = -\frac{4x\dot{y}}{r^3-2r^2}<br /> +\frac{(x\dot{x}+y\dot{y}) (12r-16)xy}{(r^3-2r^2)^2}

\left(\frac{\partial^2 I}{\partial \dot{y}\partial x}\right) = -\frac{4y\dot{x}}{r^3-2r^2}<br /> +\frac{(x\dot{x}+y\dot{y}) (12r-16)xy}{(r^3-2r^2)^2}

It remains to substitute these into post #26, and then simplify. In doing this, it is clear that I'll want to cancel out the factors of r^3-2r^2, which was the cause of the problem at r=2 that I wanted to avoid by using coordinate time instead of proper time.

Of the terms A, B, C, D, E, and F of post #26, the number of times r^3-2r^2 appears in the denominator is 2 for A, B, D and E, and 3 for C and F. Thus there will be Consequently, in the formulas for \ddot{x}, \ddot{y}, there will be 4 cancellations of 1/(r^3-2r^2) between the numerator and denominator, and the numerator will have a left over copy of 1/(r^3-2r^2). This means that, at first glance, the solution blows up at r=2. Alternatively, we should be able to factor 2 copies of (r-2) out of the numerator.

It's getting late and I'm going to start making mistakes (and possibly typing with a Texas accent). Let me rewrite out the results of this page in form XXX/(r^n(r-2)^m, probably drop a 2, and then go to bed. [Nope, a man's got to know his limitations.]

Carl
 
Last edited:
  • #37
Factoring out r and r-2 gives:
\begin{array}{rcl}<br /> I &amp;=&amp; (r(r-2)^2-(\dot{x}^2+\dot{y}^2)r^2(r-2)-2(x\dot{x}+y\dot{y})^2)<br /> /r^2(r-2)\\<br /> \left(\frac{\partial I}{\partial x}\right) &amp;=&amp;<br /> (2xr(r-2)^2 - 4\dot{x}(x\dot{x}+y\dot{y})r^2(r-2)<br /> +2x(x\dot{x}+y\dot{y})^2(3r-4) )<br /> /r^4(r-2)^2\\<br /> \left(\frac{\partial I}{\partial y}\right) &amp;=&amp;<br /> (2yr(r-2)^2 - 4\dot{y}(x\dot{x}+y\dot{y})r^2(r-2)<br /> +2y(x\dot{x}+y\dot{y})^2(3r-4) )<br /> /r^4(r-2)^2\\<br /> \left(\frac{\partial I}{\partial \dot{x}}\right) &amp;=&amp; (-2\dot{x}r^2(r-2)<br /> -4x(x\dot{x}+y\dot{y}))/r^2(r-2)\\<br /> \left(\frac{\partial I}{\partial \dot{y}}\right) &amp;=&amp; (-2\dot{y}r^2(r-2)<br /> -4y(x\dot{x}+y\dot{y}))/r^2(r-2)\\<br /> \left(\frac{\partial^2 I}{\partial \dot{x}^2}\right) &amp;=&amp;<br /> (-2r^2(r-2)-4x^2)/r^2(r-2)\\<br /> \left(\frac{\partial^2 I}{\partial \dot{y}^2}\right) &amp;=&amp;<br /> (-2r^2(r-2)-4y^2)/r^2(r-2)\\<br /> \left(\frac{\partial^2 I}{\partial \dot{x}\partial \dot{y}}\right) &amp;=&amp;<br /> (-4xy)/r^2(r-2)\\<br /> \left(\frac{\partial^2 I}{\partial \dot{x}\partial x}\right) &amp;=&amp;<br /> (-(8x\dot{x}+4y\dot{y})r^2(r-2)<br /> + (x\dot{x}+y\dot{y}) (12r-16)x^2)/r^4(r-2)^2\\<br /> \left(\frac{\partial^2 I}{\partial \dot{y}\partial y}\right) &amp;=&amp;<br /> (-(8y\dot{y}+4x\dot{x})r^2(r-2)<br /> + (x\dot{x}+y\dot{y}) (12r-16)y^2)/r^4(r-2)^2\\<br /> \left(\frac{\partial^2 I}{\partial \dot{x}\partial y}\right) &amp;=&amp;<br /> (-4x\dot{y}r^2(r-2) + ( x\dot{x}+y\dot{y}) (12r-16)xy)/r^4(r-2)^2\\<br /> \left(\frac{\partial^2 I}{\partial \dot{y}\partial x}\right) &amp;=&amp;<br /> (-4y\dot{x}r^2(r-2) + ( x\dot{x}+y\dot{y}) (12r-16)xy)/r^4(r-2)^2<br /> \end{array}

Let me first just see what happens to all those 1/r and 1/(r-2) factors. Treating these alone, I have

\begin{array}{rcl}<br /> A &amp;=&amp; (XXX)/r^4(r-2)^2\\<br /> B &amp;=&amp; (XXX)/r^4(r-2)^2\\<br /> C &amp;=&amp; (XXX)/r^6(r-2)^3\\<br /> D &amp;=&amp; (XXX)/r^4(r-2)^2\\<br /> E &amp;=&amp; (XXX)/r^4(r-2)^2\\<br /> F &amp;=&amp; (XXX)/r^6(r-2)^3<br /> \end{array}

where "XXX" are various polynomials in the phase space variables. The equations of motion are then of the form:

\begin{array}{rcl}<br /> \ddot{x} &amp;=&amp; (BF-CE)/(BD-AE)\\<br /> &amp;=&amp;\frac{(XXX)/r^{10}(r-2)^5}{(YYY)/r^8(r-2)^4}\\<br /> &amp;=&amp;\frac{XXX}{YYY}\frac{1}{r^2(r-2)}.\end{array}

So I don't need to keep around all the r and r-2 factors dividing I and the various partial derivatives. Instead, I just keep around the numerators. Let me rewrite these various terms as primed:

\begin{array}{rcl}<br /> I&#039; &amp;=&amp; r(r-2)^2-(\dot{x}^2+\dot{y}^2)r^2(r-2)-2(x\dot{x}+y\dot{y})^2\\<br /> \left(\frac{\partial I}{\partial x}\right)&#039; &amp;=&amp;<br /> 2xr(r-2)^2 - 4\dot{x}(x\dot{x}+y\dot{y})r^2(r-2)<br /> +2x(x\dot{x}+y\dot{y})^2(3r-4)\\<br /> \left(\frac{\partial I}{\partial y}\right)&#039; &amp;=&amp;<br /> 2yr(r-2)^2 - 4\dot{y}(x\dot{x}+y\dot{y})r^2(r-2)<br /> +2y(x\dot{x}+y\dot{y})^2(3r-4)\\<br /> \left(\frac{\partial I}{\partial \dot{x}}\right)&#039; &amp;=&amp; -2\dot{x}r^2(r-2)<br /> -4x(x\dot{x}+y\dot{y})\\<br /> \left(\frac{\partial I}{\partial \dot{y}}\right)&#039; &amp;=&amp; -2\dot{y}r^2(r-2)<br /> -4y(x\dot{x}+y\dot{y})\\<br /> \left(\frac{\partial^2 I}{\partial \dot{x}^2}\right)&#039; &amp;=&amp;<br /> -2r^2(r-2)-4x^2\\<br /> \left(\frac{\partial^2 I}{\partial \dot{y}^2}\right)&#039; &amp;=&amp;<br /> -2r^2(r-2)-4y^2\\<br /> \left(\frac{\partial^2 I}{\partial \dot{x}\partial \dot{y}}\right)&#039; &amp;=&amp;<br /> -4xy\\<br /> \left(\frac{\partial^2 I}{\partial \dot{x}\partial x}\right)&#039; &amp;=&amp;<br /> -(8x\dot{x}+4y\dot{y})r^2(r-2)<br /> + (x\dot{x}+y\dot{y}) (12r-16)x^2\\<br /> \left(\frac{\partial^2 I}{\partial \dot{y}\partial y}\right)&#039; &amp;=&amp;<br /> -(8y\dot{y}+4x\dot{x})r^2(r-2)<br /> + (x\dot{x}+y\dot{y}) (12r-16)y^2\\<br /> \left(\frac{\partial^2 I}{\partial \dot{x}\partial y}\right)&#039; &amp;=&amp;<br /> -4x\dot{y}r^2(r-2) + ( x\dot{x}+y\dot{y}) (12r-16)xy\\<br /> \left(\frac{\partial^2 I}{\partial \dot{y}\partial x}\right)&#039; &amp;=&amp;<br /> -4y\dot{x}r^2(r-2) + ( x\dot{x}+y\dot{y}) (12r-16)xy<br /> \end{array}

The above puts the answer into rational form, that is, the answer is the ratio of two polynomials in x, y, \dot{x}, \dot{y}, r. The only problem is that the answer is going to have a LOT of terms. Fortunately, we have MAXIMA, a free symbolic computing engine.
 
Last edited:
  • #38
In putting the above into MAXIMA, I discovered that defining r = sqrt(x^2+y^2) is a bad idea, because it prevents keeping the answer in r, which is what you want. Furthermore, I found that reasonably small results could be obtained by using factorout: factorout(expression, x, y, r); This keeps the terms that are going to be reduced by x^2+y^2 = r^2 together. The input program for the partial derivatives is:

ii:expand(r*(r-2)^2-(xx^2+yy^2)*r^2*(r-2)-2*(x*xx+y*yy)^2);
didx:expand(2*x*r*(r-2)^2-4*xx*(x*xx+y*yy)*r^2*(r-2)+2*x*(x*xx+y*yy)^2*(3*r-4));
didy:expand(2*y*r*(r-2)^2-4*yy*(x*xx+y*yy)*r^2*(r-2)+2*y*(x*xx+y*yy)^2*(3*r-4));
didxx:expand(-2*xx*r^2*(r-2)-4*x*(x*xx+y*yy));
didyy:expand(-2*yy*r^2*(r-2)-4*y*(x*xx+y*yy));
d2idxx2:expand(-2*r^2*(r-2)-4*x^2);
d2idyy2:expand(-2*r^2*(r-2)-4*y^2);
d2idxxdyy:expand(-4*x*y);
d2idxxdx:expand(-(8*x*xx+4*y*yy)*r^2*(r-2)+(x*xx+y*yy)*(12*r-16)*x^2);
d2idyydy:expand(-(8*y*yy+4*x*xx)*r^2*(r-2)+(x*xx+y*yy)*(12*r-16)*y^2);
d2idxxdy:expand(-4*x*yy*r^2*(r-2)+(x*xx+y*yy)*(12*r-16)*x*y);
d2idyydx:expand(-4*y*xx*r^2*(r-2)+(x*xx+y*yy)*(12*r-16)*x*y);

Then the A, B, C, D, E, F variables are:

aa:expand(ii*d2idxx2-0.5*didxx^2);
bb:expand(ii*d2idxxdyy-0.5*didxx*didyy);
cc:expand((0.5*didxx*didx-ii*d2idxxdx)*xx+(0.5*didxx*didy-ii*d2idxxdy)*yy+ii*didx);
dd:expand(ii*d2idxxdyy-0.5*didyy*didxx);
ee:expand(ii*d2idyy2-0.5*didyy^2);
ff:expand((0.5*didyy*didy-ii*d2idyydy)*yy+(0.5*didyy*didx-ii*d2idyydx)*xx+ii*didy);

Finally, the x and y numerators and the shared denominator are defined as:

xxxn:expand(bb*ff-cc*ee);
yyyn:expand(cc*dd-aa*ff);
denom:expand(bb*dd-aa*ee);

To get the MAXIMA program to reduce these to sizes that an old forklift driver feels like crunching by hand on, I used:

grind(factorout(denom,x,y,r));
grind(factorout(xxxn,x,y,r));

and did the denominator first because it's smaller. Running the MAXIMA program, we find that the denominator is particularly easy and reduces to:

B&#039;*D&#039;-A&#039;*E&#039; = 8r^6(r-2)^3(x\dot{x}+y\dot{y})^2<br /> +4r^7(r-2)^5+4r^8(r-2)^4(\dot{x}^2+\dot{y})^2

The Schwarzschild coordinates halt at the event horizon (r=2), and we already have an (r-2) factor in the denominator besides the three above, so we expect to factor at least one more (r-2) from the numerator, for a total of at least 5.

Ouch! The numerator is hard to reduce. Time to get some sleep.
 
Last edited:
  • #39
CarlB said:
The Schwarzschild coordinates halt at the event horizon (r=2), and we already have an (r-2) factor in the denominator besides the three above, so we expect to factor at least one more (r-2) from the numerator, for a total of at least 5. Ouch! The numerator is hard to reduce. Time to get some sleep.

Well no matter what i do I can't factor out 3 copies of (r-2) from the numerator. I went back and checked the equations going in. No problems. I abused MAXIMA by rewriting all the equations with r replaced with s = r+2, and then derived the equation for small s. Finally I solved the differential equation for small s and discovered that one doesn't need to suppose that there are 5 copies of (r-2) in order to make particles stall on the event horizon. 3 copies is enough, and 3 is what I found. This means that the final equation for the acceleration is going to have one copy of (r-2) in the denominator.

Most of the terms in the numerator will have 2 copies of (r-2) and so will be finite across the event horizon. The remaining terms will tend to infinity but will be kept finite (and in fact will tend to zero) because they are proportional to 3 or 4 copies of the velocity.

Technically, none of this is much of an issue for the Schwarzschild metric as orbits never reach the event horizon. But it is an issue for the Painleve metric. So my guess is that the Painleve metric will have two more factors of (r-2) in the numerator and so will be a bit more elegant. And it could be a problem that would cause a division by zero in a simulation of the Schwarzschild metric.

To make the division by zero explicit, I will factor the terms according to powers of (r-2). That might have some use in the computer program, if it turns out that orbits that crash onto the event horizon cause divisions by zero.

By the way, the division by zero is not a consequence of doing the problem in x-y coordinates. I know this because I obtained the differential equation in (q,t) by substituting y=0, q = x-2 > 0, so that r-2 = q. These are equivalent to spherical coordinates for a particle on a ray through the origin. I guess these things are obvious to people who mess around with gravity all the time, but I've certainly learned a few things. But if it was obvious to them I would have appreciated their informing me not to expect 5 copies of (r-2) in the numerator. On the other hand, since some of them had voted that I wasn't going to get this completed, not mentioning anything sort of makes sense.

And now, let me finish off the Schwarzschild numerator, cancel what terms I can, and then test the equations in my gravity simulator.

Carl
 
Last edited:
  • #40
As usual, my earlier formula for the numerator is off a bit. The corrected formula is:

B&#039;D&#039;-A&#039;E&#039; = 4 r^6 (r-2)^3 (2(x\dot{x}+y\dot{y})^2 + r^2(r-2)(\dot{x}^2+\dot{y}^2)-r(r-2)^2)

For the numerator, MAXIMA gives me:

+4*r^5*s^7*x

+4*r^6*s^6*(x*yy^2-2*xx*y*yy-x*xx^2)

-4*r^4*s^5*(2*r^3*x*yy^4-2*r^3*xx*y*yy^3+3*r*x*y^2*yy^2-2*x*y^2*yy^2+2*r^3*x*xx^2*yy^2+4*xx*y^3*yy-2*r^3*xx^3*y*yy+6*r*x^2*xx*y*yy+4*x*xx^2*y^2+3*r*x^3*xx^2+2*x^3*xx^2)

+4*r^5*s^4*(y*yy+x*xx)*(3*r*x*y*yy^3-8*x*y*yy^3+8*xx*y^2*yy^2+3*r*x^2*xx*yy^2-4*x^2*xx*yy^2+3*r*x*xx^2*y*yy+4*xx^3*y^2+3*r*x^2*xx^3)

+8*r^3*s^3*(y*yy+x*xx)^3*(3*r*x*y*yy-4*x*y*yy+4*xx*y^2+3*r*x^2*xx)

In the above s = r-2, xx = \dot{x}, yy =\dot{y}. To get the final result, I need to divide by r^2(r-2) times the denominator calculated two posts ago:

r^2(r-2)(B&#039;*D&#039;-A&#039;*E&#039;) = 4 r^8 (r-2)^4 (2(x\dot{x}+y\dot{y})^2 + r^2(r-2)(\dot{x}^2+\dot{y}^2)-r(r-2)^2)

Some of the factors of r and r-2 will cancel between the numerators and denominators. The resulting equation for the accelerations of the Schwarzschild metric in "flat" coordinates is:

\begin{array}{rcl}<br /> \ddot{x} &amp;=&amp; (+r(r-2)^3x<br /> +r^2(r-2)^2(x\dot{y}^2-2\dot{x}y\dot{y}-x\dot{x}^2)\\<br /> &amp;&amp;\;\;-(r-2)(2r^3x\dot{y}^4-2r^3\dot{x}y\dot{y}^3+3rxy^2\dot{y}^2-2xy^2\dot{y}^2+2r^3x\dot{x}^2\dot{y}^2+4\dot{x}y^3\dot{y}-2r^3\dot{x}^3y\dot{y}+6rx^2\dot{x}y\dot{y}+4x\dot{x}^2y^2+3rx^3\dot{x}^2+2x^3\dot{x}^2)\\&amp;&amp;+r(y\dot{y}+x\dot{x})(3rxy\dot{y}^3-8xy\dot{y}^3+8\dot{x}y^2\dot{y}^2+3rx^2\dot{x}\dot{y}^2-4x^2\dot{x}\dot{y}^2+3rx\dot{x}^2y\dot{y}+4\dot{x}^3y^2+3rx^2\dot{x}^3))\\<br /> &amp;&amp;/(4 r^4 (2(x\dot{x}+y\dot{y})^2 + r^2(r-2)(\dot{x}^2+\dot{y}^2)-r(r-2)^2))\\<br /> &amp;+&amp;(2(y\dot{y}+x\dot{x})^3(3rxy\dot{y}-4xy\dot{y}+4\dot{x}y^2+3rx^2\dot{x}))\\<br /> &amp;&amp;/(4 r^5 (r-2) (2(x\dot{x}+y\dot{y})^2 + r^2(r-2)(\dot{x}^2+\dot{y}^2)-r(r-2)^2))<br /> \end{array}

\begin{array}{rcl}<br /> \ddot{y} &amp;=&amp; (+r(r-2)^3y<br /> +r^2(r-2)^2(y\dot{x}^2-2\dot{y}x\dot{x}-y\dot{y}^2)\\<br /> &amp;&amp;\;\;-(r-2)(2r^3y\dot{x}^4-2r^3\dot{y}x\dot{x}^3+3ryx^2\dot{x}^2-2yx^2\dot{x}^2+2r^3y\dot{y}^2\dot{x}^2+4\dot{y}x^3\dot{x}-2r^3\dot{y}^3x\dot{x}+6ry^2\dot{y}x\dot{x}+4y\dot{y}^2x^2+3ry^3\dot{y}^2+2y^3\dot{y}^2)\\&amp;&amp;+r(x\dot{x}+y\dot{y})(3ryx\dot{x}^3-8yx\dot{x}^3+8\dot{y}x^2\dot{x}^2+3ry^2\dot{y}\dot{x}^2-4y^2\dot{y}\dot{x}^2+3ry\dot{y}^2x\dot{x}+4\dot{y}^3x^2+3ry^2\dot{y}^3))\\<br /> &amp;&amp;/(4 r^4 (2(y\dot{y}+x\dot{x})^2 + r^2(r-2)(\dot{y}^2+\dot{x}^2)-r(r-2)^2))\\<br /> &amp;+&amp;(2(x\dot{x}+y\dot{y})^3(3ryx\dot{x}-4yx\dot{x}+4\dot{y}x^2+3ry^2\dot{y}))\\<br /> &amp;&amp;/(4 r^5 (r-2) (2(y\dot{y}+x\dot{x})^2 + r^2(r-2)(\dot{y}^2+\dot{x}^2)-r(r-2)^2))<br /> \end{array}

Let me check this to see if it gives the right answer in the Newtonian limit (i.e. \dot{x}=\dot{y}=0 ). It checks out, that is, I get x r^4/(-r^7) = -x/r^3. The next step is to program this into my all gravity simulator and see if it looks okay. I think that's enough for one night. I can hardly wait to see the Painleve coordinates.

Carl
 
Last edited:
  • #41
CarlB said:
Some of the factors of r and r-2 will cancel between the numerators and denominators. The resulting equation for the accelerations of the Schwarzschild metric in "flat" coordinates is:

It turns out that the three things that are most useful in MAXIMA are "grind", "factorout", and "factor". I applied "factor" to the above equations of motion and eventually got them into a fairly simple form. I managed to cancel out everything in the denominator except r^5(r-2), and put the numerators into a form where I can do the x and y coordinates by a combined calculation. The simplified equations of motion are now:

\begin{array}{rcl}<br /> \ddot{x} &amp;=&amp; (x(3r(x\dot{x}+y\dot{y})^2-r(r-2)^2)\\<br /> &amp;&amp;+4y(y\dot{x}-x\dot{y})(x\dot{x}+y\dot{y})\\<br /> &amp;&amp;+2\dot{y}(y\dot{x}-x\dot{y})r^2(r-2)) /(r^5(r-2)) \\<br /> \ddot{y} &amp;=&amp; \left(y(3r(x\dot{x}+y\dot{y})^2-r(r-2)^2)\\<br /> &amp;&amp;-4x(y\dot{x}-x\dot{y})(x\dot{x}+y\dot{y})\\<br /> &amp;&amp;-2\dot{x}(y\dot{x}-x\dot{y})r^2(r-2)\right) /(r^5(r-2))<br /> \end{array}

For comparison, here are the equations in polar coordinates, from pervect's post #17, with m=1:

\begin{array}{rcl}<br /> \ddot{r} &amp;=&amp; \frac {3 \dot{r}^2}{r(r-2)} + (r-2)(\dot{\phi}^{2}-{\frac {1}{r^3}}) \\<br /> \ddot{\phi} &amp;=&amp; -\frac {2 \dot{r}\dot{\phi}(r-3)}{r(r-2)}<br /> \end{array}

Are these the same? One quick test is to see what accelerations they give for a test mass that is stationary. So set all velocities to zero and get:

\begin{array}{rcl}<br /> \ddot{x} &amp;=&amp; -xr(r-2)^2 /(r^5(r-2)) = -x(r-2)/r^4 \\<br /> \ddot{y} &amp;=&amp; -y(r-2)/r^4\\<br /> \ddot{r} &amp;=&amp; -(r-2)/r^3 \\<br /> \ddot{\phi} &amp;=&amp; 0<br /> \end{array}

Putting x=r, we see that they are identical. In any case, I've programmed the (x,y) version into my computer and sure enough it does draw what seem like reasonable orbits. I've not yet updated the Java applet because I want to include the Painleve coordinates too. These will be a great check of the equations because they should send particles down the same orbits but with different times (and should spiral into the singularity at the center of the black hole rather than getting stuck on the unphysical event horizon). I've also slightly changed the simulation so that it draws the gray "sun" out to the event horizon, r=2, rather than out to r=1. This makes the Schwarzschild tracks get stuck on the edge of the gray region, which is what you expect sort of.

I'm note sure if these are different. Let me run a simulation...

I still can't get pervect's equation to run correctly. This could be because I am converting from polar to rectangular coordinates wrong, but the first parts of the orbits match, so I think it's probably okay.

What happens is that when the orbit goes down near the event horizon (i.e. so d phi /d t is maximum), the test mass ends up picking up huge amounts of energy and then zooms off to infinity. Even when the orbit doesn't get near the event horizon, they tend to pick up energy. It's like the calculation for acceleration in phi is too large.

I've just tested my version against the impact parameter, 3\sqrt{3}/2 which is near 5.2 and it passes with flying colors. That is, when you throw relativistic particles at it from (+1000,+5.2) with velocity (-1,0), they do interesting things. Also, for my programming of both my version and pervect's, the scattering is approximately twice the Newtonian scattering.

Another test is to compare with:
http://www.fourmilab.ch/gravitation/orbits/
but it's late and I haven't figured out how to convert from his proper time coordinates to my system time. His simulation has a particle whose maximum radius is 12 has interesting orbits if its angular momentum L=3.57, I get similar orbits for x=12 when I set dy/dt = 0.26084. Converting this into angular momentum, one first gets dphi/dt = 0.26084/12. This gives:

3.57= L = r^2 \frac{d\phi}{d\tau} = 12^2 \frac{0.26084}{12}\frac{dt}{d\tau}

which gives me that dt/d\tau = 3.57/(12 X .26084) = 1.14055, which seems reasonable, but I don't know if it's close to exact. To check this, I have to substitute into the Schwarzschild metric:

\begin{array}{rcl}<br /> \left(\frac{d\tau}{dt}\right)^2 &amp;=&amp; \frac{r-2}{r} - \frac{r}{r-2}\left(\frac{dr}{dt}\right)^2 - r^2\left(\frac{d\phi}{dt}\right)^2\\<br /> &amp;=&amp; \frac{10}{12} - 0 - \left(\frac{dy}{dt}\right)^2\\<br /> &amp;=&amp; (1/1.1431)^2\\<br /> \end{array}

and the first couple digits of dt/ds match. Therefore, I conclude that I've probably got the correct Cartesian differential equation for the Schwarzschild metric, as desired. Next, on to the Painleve metric, which may or may not be more difficult.

Carl
 
Last edited:
  • #42
CarlB said:
What happens is that when the orbit goes down near the event horizon (i.e. so d phi /d t is maximum), the test mass ends up picking up huge amounts of energy and then zooms off to infinity. Even when the orbit doesn't get near the event horizon, they tend to pick up energy. It's like the calculation for acceleration in phi is too large.
Carl

I have also encountered this phenoma. Total energy of the particle suddenly starts for infinity, while total angular momentum falls a bit less dramatically.
I am working with spherical coordinates in 3D, with no simlifications of the specific metric. I am solving the geodesic equation for all for coordinates (t,r,theta,phi).

and by the way, if you write metric in cartesian coordinates you get something like
g_tt = -1 + 2m/r
g_xx = g_yy=g_zz = 1 +2m/r
for covariant version of metric tensor.
If you put them in geodesic equation, work out the christoffel symbols and there you go...nice, clean and easy.
 
Last edited:
  • #43
I also had numerical stability problems back when I did this. The reason I believed my results is that the problems were lessened when I tightened the tolerance on the integration algorithm. So I put the problem down to numerical integration issues, rather than an incorrect equation.
 
  • #44
Some possibly useful citations

This old thread seems to continue another thread, so at this late date it is rather hard to follow. For one thing, in this thread, Carl Brannen didn't define what he meant by "cartesian-like coordinates", although presumably he did so in the original posts. I conjecture he might have been trying to reinvent one of two classic types of chart: Kerr-Schild charts, or harmonic charts. If you search the arXiv using the search bar at the site in my sig on ""Kerr-Schild numerical gr-qc" and "harmonic coordinates" numerical gr-qc" quite a few results turn up, which might help in tracking down the source of the problem. BTW, see Chandrasekhar, The Mathematical Theory of Black Holes for the exact solution of the geodesic equations in the Kerr vacuum, including diagrams, which you can use as a "benchmark" for testing numerical integration schemes.

Incidently, gr-qc/0704.2508 is a new eprint extending a classical result of Mueller zum Hagen to the effect that a static vacuum solution (i.e. with timelike Killing vector X) can be written in a harmonic chart such that the metric components are analytic in this chart. In the eprint, Tod (coauthor of the LMS student series textbook on gtr) extends this to electrovacuum solutions, irrespective of whether or not the EMF shares the timelike Killing vector (i.e. L_X F might be nonzero). This shows one way in which harmonic charts can be useful for theoretical work, but as the discussion in Chandrasekhar shows, they may not really be the best for searching for closed form solutions of the geodesic equations. (I guess everyone here knows that the fact that such can be found for the Kerr vacuum is in fact one of the more remarkable properties of this solution, which depends upon the existence of a Killing tensor; see D'Inverno's textbook.)
 
  • #45
Carl's version of "cartesian" coordinates are defined back in post #2

x = r cos phi
y = r sin phi

and I believe it is implied that theta =0, as one is interesting in orbital simulations.

Here r and phi are the Schwarzschild r and phi coordinates.

I never understood the motivation for doing this, you'd have to ask Carl if he's still around. I had assumed that Carl simply confused r with radial distance, so I complained a bit (which didn't go over very well with him if you read the thread). But possibly he has some other motivation, who knows.

The thing that is interesting is that Maple gave pages and pages of output for the differential equations for x and y using these coordinates. If Carl is correct, Maple didn't simplify the equations completely (which is easy to believe).

I don't think this has much to do with the references cited by Chris from my quick scan.
 
  • #46
Chris, the whole thing is here. The "cartesian coordinates" are defined in the first few posts, and should be obvious to anyone in the field. All they amount to are the usual (r,theta, phi,t) coordinates converted to (x,y,z,t), which is commonly done in the industry.

I managed to get that part of the program to work and the simulation runs beautifully. It is here:
http://www.gaugegravity.com/testapplet/SweetGravity.html

What is incomplete is the work on Painleve coordinates. After I get those to run, I'll add a whole bunch of initial conditions that the student can select. These will illustrate all the various cool facts about classical and GR gravity.

The reason for including Painleve coordinates (which frankly, I thought would be simpler than the usual Cartesian GR), is so that test particles can hit the singularity at zero, while in the usual Cartesian they get stuck on the event horizon.

Carl
 
Last edited by a moderator:
  • #47
pervect said:
I never understood the motivation for doing this, you'd have to ask Carl if he's still around. I had assumed that Carl simply confused r with radial distance, so I complained a bit (which didn't go over very well with him if you read the thread). But possibly he has some other motivation, who knows.

For now my secrets will remain my own.

Also, I would like to have equations that allow theta non zero. I don't allow that in the simulation now, but it would be nice to have the ability to model Earth orbit crossing comets, etc.

pervect said:
The thing that is interesting is that Maple gave pages and pages of output for the differential equations for x and y using these coordinates. If Carl is correct, Maple didn't simplify the equations completely (which is easy to believe).

I was unfamiliar with MAXIMA before starting this. From what I know now, I've got doubts about it. For example, I can't seem to solve the following six simple quadratic equations in six unknowns:

\begin{equation}<br /> \begin{array}{rcl}<br /> F_I&amp;=&amp;F_IF_I+F_JF_K+F_KF_J+F_RF_R+F_GF_G+F_BF_B,\\<br /> F_J&amp;=&amp;F_IF_J+F_JF_I+F_KF_K+F_RF_G+F_GF_B+F_BF_R,\\<br /> F_K&amp;=&amp;F_IF_K+F_JF_J+F_KF_I+F_RF_B+F_GF_R+F_BF_G,\\<br /> F_R&amp;=&amp;F_IF_R+F_JF_G+F_KF_B+F_RF_I+F_GF_K+F_BF_J,\\<br /> F_G&amp;=&amp;F_IF_G+F_JF_B+F_KF_R+F_RF_J+F_GF_I+F_BF_K,\\<br /> F_B&amp;=&amp;F_IF_B+F_JF_R+F_KF_G+F_RF_K+F_GF_J+F_BF_I,<br /> \end{array}<br /> \end{equation}

I had to solve it by hand. See setion (5) of
http://brannenworks.com/dmfound.pdf

I'm curious to see what a higher dollar program can do with them. Part of the problem is the messy topology. There are about a dozen finite solutions (i.e. 0-manifolds or point solutions), but there are also a couple of 2-manifolds that do not get near the point solutions. If I ever teach college algebra again, I'll assign problems like this as "take home extra credit ".

An example point solution is (F_I,F_J,F_K,F_R,F_G,F_B) = (1,0,0,0,0,0). For the 2-manifold solutions, see the above link.

Carl
 
Last edited:
  • #48
Points to ponder

Hi, Carl,

CarlB said:
Chris, the whole thing is here. The "cartesian coordinates" are defined in the first few posts, and should be obvious to anyone in the field.

I guess you might be saying that you started with the standard "polar spherical" Schwarzschild chart and employed
<br /> x = \sin (\theta) \, \cos(\phi), \;<br /> y = \sin (\theta) \, \sin(\phi), \; <br /> z = \cos(\theta)<br />
to obtain a new chart.

(Additional: I see now that in a post from some months ago, Carl did say that at least at that point in time this is exactly what he was doing.)

Fine, but trust me, that is not neccessarily the first thing an expert might guess "pseudo-cartesian coordinates" should mean in this context. (My first guess would have been something similar, but starting with the standard "spatially isotropic" chart.) Such a "pseudo-cartesian" version of the usual Schwarzschild chart would not neccessarily be a particularly good idea if you want "simple" geodesic equations; depending upon how you define "simple" there may be better choices.

I still have no idea what you were trying to do, BTW. Since this is a very old thread which someone revived, could you indulge me and remind us what you meant by "Cartesian DE form"? If you can express this precisely I guess you will turn out to be placing some demands upon the form of the geodesic equations in the chart you seek.

CarlB said:
All they amount to are the usual (r,theta, phi,t) coordinates converted to (x,y,z,t), which is commonly done in the industry.

Unless I misunderstand what "industry" you have in mind (generic computer graphics? generic engineering?), these applications probably don't involve curved manifolds. For example, if you aren't familiar with the ways of curved manifolds, you are probably at least partially misinterpreting your radial coordinate. Roughly, "spherical-type charts" can preserve various properties of the radial coordinate familar from the usual polar spherical chart on E^3, but never all of them.

CarlB said:
I managed to get that part of the program to work and the simulation runs beautifully.

Did you compare your plots with analytical solutions transformed into your coordinate chart?

CarlB said:
What is incomplete is the work on Painleve coordinates. After I get those to run, I'll add a whole bunch of initial conditions that the student can select. These will illustrate all the various cool facts about classical and GR gravity.

Sounds great--- just make sure your plots are correct!

CarlB said:
The reason for including Painleve coordinates (which frankly, I thought would be simpler than the usual Cartesian GR), is so that test particles can hit the singularity at zero, while in the usual Cartesian they get stuck on the event horizon.

Right, that's one reason why I suggested them. Another is that we have the Killing vector which suggests a quotient space, and we have the slicing into constant Painleve time. These each suggest different ways of reducing to a kind of "three-dimensional space". Of course, projections of geodesics will not be geodesics in the slice metric! (Otherwise there would be no light bending, since the slice metric is euclidean!) However, the Painleve chart is neither orthogonal nor harmonic.

Did you draw local light cones in the Painleve vs Schwarzschild charts? See http://physics.syr.edu/courses/modules/LIGHTCONE/ for local light cones in Minkowski vacuum. Plotting local light cones in your chart is a very good idea if you want to understand what your plots of geodesics as represented in that chart are trying to tell you! Try plotting in both a slice t=t0, and also plotting equatorial plane geodesics in an E^(1,2) picture of Painleve.

Did you look up harmonic charts and Kerr-Schild charts yet? These are very well worth looking into. See http://casa.colorado.edu/~ajsh/schw.shtml for some more useful charts, including Eddington and Kruskal-Szekeres.
 
Last edited by a moderator:
  • #49
Painleve Coordinates, step 1, convert to Cartesian

Okay, the program of writing Newton and Schwarzschild in Cartesian (see post #11 for references to literature) coordinates without trig has been completed and the results test out nicely. It is time to do the same with Painleve coordinates:

ds^2 = -dt^2 + (dr-(2/r)^{0.5} dt)^2 + r^2 d\phi^2.

First step is to convert them to "Cartesian" form. Following post #2, we find:

\begin{array}{rcl}<br /> ds^2 &amp;=&amp; -dt^2 + ((xdx+ydy)r^{-1}-\sqrt{2}\;r^{-0.5} dt)^2 + (y^2dx^2 + x^2dy^2 -2xy\;dx\;dy)r^{-2},\\<br /> &amp;=&amp;-dt^2 + dx^2 + dy^2 + 2r^{-1}dt^2 -2\sqrt{2}\;r^{-1.5}(x\;dx\;dt+y\;dy\;dt),\\<br /> I = \left(\frac{ds}{dt}\right)^2 &amp;=&amp;\frac{2}{r}-1 + \dot{x}^2 + \dot{y}^2-2\sqrt{2}r^{-1.5}(x\dot{x}+y\dot{y})<br /> \end{array}

The idea is to apply the Euler Lagrange equations to the integral over t for s:

s = \int \sqrt{\left(\frac{ds}{dt}\right)^2}\;dt = \int \sqrt{I}\;dt

to obtain the equations of motion in a Cartesian coordinate system. As supplied by pervect in post #22, the resulting equations of motion are:

\frac{d}{d t} \left[ \frac{1}{2 \sqrt{I}} \left( \frac{\partial I}{\partial \dot{x}} \right) \right] = \frac{1}{2 \sqrt{I}} \frac{\partial I}{\partial x}
 
Last edited:
  • #50
Hi, Carl,

CarlB said:
Okay, the program of writing Newton and Schwarzschild in Cartesian (see post #11 for references to literature) coordinates without trig has been completed and the results test out nicely... Chris, I really appreciate your help on this problem, but I don't have time to bring you up to speed when you can do it for yourself with little effort. If you are willing to swear that you've carefully read each post in the rest of the thread, and at least looked at the articles that were linked to it, then I will answer whatever questions you have left, and we won't clutter up the thread going over the same old same old. Actually, after looking through the thread it appears that I really should rewrite the secret reasons for working on this sort of thing.

Well, I don't think you are giving me enough credit for "reasonable effort" (I can see that for your part, you don't think I am giving you sufficient credit for knowledge and careful work, and you may be right!), but it seems that you have now recognized some of the expository deficiencies of your previous posts, and I applaud your attempt to rectify the lacunae you noticed.

By the way, if you said previously that you have a background in geometric algebra, I must have either missed or forgotten that comment. As it happens, many years ago I knew a fair amount about Cayley-Dickson algebras (see for example my posts on noneuclidean trigonometries, which will probably be familiar to you but not to many others here), and I do know the papers by Doran, so I think I can say with some confidence that it was not nearly so apparent as you thought that this was part of the background for your computations. (By the way, there is an important point about the Doran chart for the Kerr vacuum which we can discuss if you like, which illustrates my point about unanticipated subtleties which can easily trip up newbies.)

Also, if you had seen as many inaccurate plots drawn by enthusiastic computer jocks for fun, or as many accurate but terribly misinterpreted plots, as I have done, you would probably appreciate why it is in your own best interests to quickly convince sceptics that your plots do not fall into these categories!
But even if you don't appreciate why I am skeptical by default, please not that following up on my suggestions should be fun and if you have indeed made no errors of math or interpretation should greatly strengthen the presention of your plots.

OK, I hope this clears the air!

Let me try to continue by annotating Carl's post, saying some things which I thing should help others to follow along.

CarlB said:
Painleve coordinates:
ds^2 = -dt^2 + (dr-(2/r)^{0.5} dt)^2 + r^2 d\phi^2.

The Painleve chart is one of the most important charts on the Schwarzschild vacuum solution, and I've been a big fan of it ever since I independently "discovered" it some time around 1984. (Much later I learned it was introduced by Painleve in 1921.)

Like the Eddington chart, it comes in ingoing and outgoing flavors and the ingoing (outgoing) Painleve chart covers the same region as the ingoing (outgoing) Eddington chart. We are interested in ingoing Painleve chart here because Carl is (I take it) planning to study the motion of freely and radially infalling observers who pass from the "right exterior" into the "future interior" region. The Painleve chart is in fact ideally adapted for this purpose. To obtain it from the exterior Schwarzschild chart
ds^2 = -(1-2m/r) \, dt^2 + \frac{dr^2}{1-2m/r} + r^2 \,<br /> \left( d\theta^2 + \sin(\theta)^2 \, d\phi^2 \right),
-\infty &lt; t &lt; \infty, \; 2 m &lt; r &lt; \infty, \; <br /> 0 &lt; \theta &lt; \pi, \; -\pi &lt; \phi &lt;\pi
put
T = t + \int \frac{\sqrt{2m/r}}{1-2m/r} \, dr<br /> = t + \sqrt{8 m r} - 4m \operatorname{arctanh} \left( \sqrt{\frac{r}{2 m}}\right)
which gives
ds^2 = -(1-2m/r) \, dT^2 + 2 \, \sqrt{2m/r} \, dT \, dr + dr^2 + r^2 \, \left( d\theta^2 + \sin(\theta)^2 \, d\phi^2 \right),
-\infty &lt; T &lt; \infty, \; 2m &lt; r &lt; \infty, \; 0 &lt; \theta &lt; \pi, \; -\pi &lt; \phi &lt;\pi

I'm too lazy to draw pictures, and I've told this story before, but let me try to give some indication of how I discovered this transformation before I knew any calculus (!). Looking at the pictures in MTW, my idea was to "pull down" the constant time slices by defining a new time coordinate such that the new constant time slices would be everywhere orthogonal to the world lines of the Lemaitre observers (whom I learned about from the little textbook by Dirac), in order to combine the best features of the ingoing Lemaitre and ingoing Eddington charts. You can actually do this if you know only the appropriate trignometry and trust a table of integrals :-/ and that's what I did.

Now we immediately notice two things. First, the domain can be extended from the right exterior into the "future interior" by increasing the range of the radial coordinate 0 &lt; r &lt; \infty (we'll need to interpret this carefully inside the horizon, but it still deserves the name of "Schwarzschild radial coordinate"). Second, the constant Painleve coordinate time slices T=T_0 have the metric of euclidean three-space! This was my first noteworthy "discovery" in gtr, so you can see why I am so fond of the Painleve chart!

With a bit of insight (or practice!) we can read off an "obvious" coframe field:
<br /> \sigma^0 = -dT, \;<br /> \sigma^1 = \sqrt{\frac{2m}{r}} \, dT + dr, \;<br /> \sigma^2 = r \, d\theta, \;<br /> \sigma^3 = r \, \sin(\theta) \, d\phi<br />
whose dual frame field consists of the unit vector fields (first timelike, last three spacelike)
<br /> \vec{e}_0 = \partial_T - \sqrt{\frac{2m}{r}} \, \partial_r, \;<br /> \vec{e}_1 = \partial_r, \;<br /> \vec{e}_2 = \frac{1}{r} \, \partial_\theta, \;<br /> \vec{e}_3 = \frac{1}{r \, \sin(\theta)} \, \partial_\phi<br />
This frame field is suitable for studying the physical experience of observers who fall in freely and radially "from rest at r=\infty". That is, the world lines of this class of observers are precisely the integral curves of the timelike unit vector field from our frame field, and the three spacelike vector fields then give the "spatial frame vectors". IOW, our frame field attaches the "local Lorentz frame" of these observers to each event. The point of introducing this frame here is to point out that in the Schwarzschild vacuum geometry, the timelike vectors \vec{e}_0 are everywhere orthogonal to the hyperslices T=T_0. IOW, in a sense, for these particular observers, we can consider their "space at a time" to be flat, but of course the spacetime itself is not flat. (A pedantic aside: the spatial frame vectors have vanishing Fermi derivatives, after projection into the orthogonal hyperplane elements, so this frame is in fact inertial nonspinning, which is the closest we can come in a curved Lorentzian manifold to a global Lorentz frame as in the standard formalism of str.

We can simultaneously employ alternative frame fields to study other families of observers, and this is in fact the best way to clear up the kind of confusion so often evident in discussions of alleged "paradoxes". I have done that in great detail elsewhere and it's off topic here, so moving on:

By the spherical symmetry, we can suppress one angular coordinate by restricting attention to the equatorial plane \theta=\pi/2. Also, the mass parameter can be set to one by rescaling coordinates, so let's do that too. This gives the line element Carl wrote down, after renaming T \mapsto t.

Going one step further, we can temporarily suppress \phi also and draw the local light cones at various events using the frame vectors
\vec{e}_0 = \partial_T - \sqrt{\frac{2m}{r}} \, \partial_r, \; \vec{e}_1 = \partial_r
Switching to a more computer graphics friendly notation,
\vec{A}(T,r) = (1, -\sqrt{\frac{2m}{r}} ), \; \vec{B}(T,r) = (0,1)
Fixing event P = (T_0,r_0), and pretending we are working in Minkowski spacetime (really, we are working in the tangent space at our event),
draw a segment to the event P+\varepsilon \, (\vec{A}+\var{B}),
then from that event to the event P+\varepsilon \, (\vec{A}-\var{B}),
then back to P. That's the top half of our local light cone. Draw the frame vectors to the same scale (i.e. scalar multiplied by \varepsilon) to see how our "local Lorentz frame" relates to the shape of our "local light cones". Now draw some more of these gadgets, to scale. Draw some in the exterior, some in the interior, and some on the horizon itself. Notice how the light cones progressively shear inward as the Schwarzschild radius decreases. This is really a way of visualizing how our local light cones, which are can be considered as a feature of the local "conformal structure" although perceptive readers will note that I am treating them as a "metrical structure", relate to the Killing vector field \partial_t, which is timelike on the right exterior (hence our solution is "static" there), but null on the horizon, and spacelike in the future interior.

Next, if we want to study general geodesics, it makes sense to write down the geodesic equations. The easiest way to do this is to read off the geodesic Lagrangian and then compute the Euler-Lagrange equations; putting the result in "monic" form we obtain the geodesic equations in their standard form, from which we can read off the Christoffel symbols, if we so desire. Alternatively, we can use the methods of Cartan (see MTW). I'll spare you the gory mess: suffice it to say that we get something quite a bit messier than what we obtain for the Schwarzschild exterior chart (see the discussion of effective potentials in MTW for a convenient method of analyzing the geodesics in the interior, in terms of the Schwarzschild exterior chart).

Now, anyone who did the "coordinate tutorial" I posted long ago to sci.physics.relativity knows very well that adopting the "right" coordinate chart can greatly simplify the analysis of the geodesics! As the active reader has now seen, while the Painleve chart has many virtues, it does not succeed in simplifying the geodesic equations if that is our goal; indeed, it makes them appear somewhat more complicated.

Another reason for exploring multiple charts is that Lie's symmetry methods can be much easier to crank through, which can enable us to discover "especially symmetrical" geodesics. In particular, if we didn't already know about the circular photon orbits at r=3m, we would discover them by a systematic study of this kind. Actually, Lie's methods are probably even more relevant for the problem of solving the wave equation, Klein-Gordon equation, Dirac equation, etc., on curved spacetimes, where changing charts can again have drastically beneficial effects, and where again this can be detected (even sometimes predicted) using a Lie theoretic symmetry analysis of the equation to be solved. This game is important for physics for various reasons; one which should sound familiar to many readers here is the study of perturbations, e.g. we might want to know how radiation of various kinds is "scattered" by a Schwarzschild object.

At this point, I'd highly recommend that anyone who hasn't much experience comparing say the wave equation in various charts should pause and collect a bunch of familiar charts for the Schwarzschild vacuum (paying careful attention to noting the coordinate ranges), such as Schwarzschild, static and spatially isotropic, Lemaitre, Eddington, Painleve, Kruskal-Szekeres, and figure out the transformations between them. Next review how to compute the wave operator using the elegant methods of Cartan (see MTW, or my long ago expository post to sci.math.relativity called "The Joy of Forms"), and compute for your charts. Check your results by applying the transformations you found earlier.

OK, so now let's confront the question: how can we find charts in which the geodesic equations, or the wave equation, or whatever, look "as simple as possible"? I don't think there's an algorithmic answer to that, but there are some basic things to try.

One approach is to try to "rationalize" our coordinates. For example, the usual polar spherical line element
<br /> ds^2 = dr^2 + r^2 \; \left( d\theta^2 + \sin(\theta)^2 \, d\phi^2 \right)<br />
contains trig functions. We can rationalize this by setting \eta=\cos(\theta), so that the line element becomes
<br /> ds^2 = dr^2 + r^2 \; \left( \frac{d\eta^2}{\eta^2} + (1-\eta^2) \, d\phi^2 \right)<br />
which contains only rational expressions in the coordinates.

A particularly dramatic example of this arises in a much more general context which includes the solution under discussion: in studying the Ernst vacuums, a major breakthrough was the recognition that the "rational prolate spheroidal chart" results in a dramatic simplication of the Ernst equation (that is, the "master equation" in the set of equations to which we can reduce the vacuum EFE in studying the Weyl canonical chart for a stationary axisymmetric spacetime). If you're curious, the rational prolate spheroidal chart for E^3 has the line element
\frac{ds^2}{A^2} = (\chi^2-\xi^2) \, \left( \frac{d\chi^2}{\chi^2-1} + \frac{d\xi^2}{1-\xi^2} \right) + (\chi^2-1) \, (1-\xi^2) \, d\phi^2,
1 &lt; \chi &lt; \infty, \; -1 &lt; \xi &lt; 1, \; -\pi &lt; \phi &lt; \pi
where A &gt; 0 is a scale factor. The transformation from the standard cylindrical chart is
z = A \, \chi \, \xi, \; r = A \, \sqrt{\chi^2-1} \, \sqrt{1-\xi^2}
Using the elegant exterior calculus methods of Cartan, you can verify with surprisingly little effort that the Laplace operator becomes, for an axisymmetric function,
\Delta = \frac{1}{A^2 \, (\chi^2-\xi^2)} \left( \partial_\chi (\chi^2-1) \partial_\chi + \partial_\xi (1-\xi^2) \partial_\xi \right)
Quite remarkably, this is symmetric under interchanging \chi, \xi, which Chandrasekhar took good advantage of in his work on the Ernst vacuums! Even better, you can readily attack this by the elementary method of separation of variables, obtaining a useful series expansion for asymptotically vanishing harmonic functions in terms of Legrendre polynomials/functions. This looks completely different from the analogous expression we'd obtain by analyzing the same equation in a cylindrical or polar spherical chart! (This is far more impressive for the Ernst equation, where as I said it turns out that the rational prolate spheroidal Weyl canonical chart on a stationary axisymmetric spacetime achieves several substantial simplifications of several interesting cases of the EFE for this situation, including vacuum, electrovacuum, minimally coupled massless scalar fields, dusts, and so on.)

Another approach is to try harmonic coordinates. These are charts in which the coordinates, considered as monotonic functions on the domain of our chart, are harmonic functions (in the "Laplace-Beltrami sense": they satisfy the curved spacetime wave equation).

Still another is try geometric algebra, which is apparently what Carl is working on, ultimately in the context of trying to simplify some equation other than the geodesic equation. Geometric algebra is a kind of elaboration of the formalisms of vector calculus, exterior calculus, and quaternionic calculus. So far it has been studied mainly in the UK, especially Cambridge, and I think Carl will agree that, like Hamilton's pursuit of his program of rewriting nineteenth century physics in terms of quaternions, geometric algebra is either ignored or considered by many observers to be of dubious value, given the intricacy of the notation. However, its enthusiastic proponents (and back in the day, I knew someone studying this stuff at Cambridge pretty well) can point to a few notable successes, including the Doran chart for the Kerr vacuum, which generalizes the Painleve chart (aha!) to the Kerr vacuum, which was obtained using the methods of geometric algebra. As a simpler example, it often happens when using frame fields that one wishes to "despin" a spinning frame by rotating by just the right angle about just the right axis of rotation (said angle and axis defined at each event) to kill the projected Fermi derivatives, and this kind of operation would indeed probalby be more conveniently carried out in geometric algebra, except in special cases. For example, the "obvious" coframe read off the Doran chart is in fact dual to a spinning frame, i.e. its timelike unit vector is indeed what we want, but the spacelike unit vectors need to be "despun" to obtain a nonspinning inertial frame analogous to the Lemaitre frame discussed above. However, this Doran-Lemaitre frame can in this case be obtained easily by elementary methods, once one has the Doran chart in hand, and I have little doubt that the Doran chart itself can be obtained without recourse to geometric algebra.

CarlB said:
First step is to convert them to "Cartesian" form.

Looking back, I see that Carl originally referred to "Cartesian coordinates" for the Schwarzschild vacuum solution, which in the context of this forum, or elementary gtr generally, is terribly confusing because those in the know will fear that someone is trying to treat a curved spacetime using coordinates which can exist only in a flat spacetime! However, eventually it turned out that Carl just meant that he is applying a transformation given by a formula familiar from vector calculus:
<br /> x = r \, \sin(\theta) \, \cos(\phi), \;<br /> y = r \, \sin(\theta) \, \sin(\phi), \;<br /> z = r \, \cos(\theta)<br />
Although of course we are working in part of the Schwarzschild vacuum, not Euclidean three-space, this makes perfect sense, so long as we are careful to restrict the ranges of the coordinates to ensure that we obtain a well defined diffeomorphism on the overlap of the domains (in this case, they are almost the same--- as an exercise, what is the "bad locus" which must be excised from the domain of the original Painleve chart but which we can reasonably hope to include in the variant Carl is switching to?)

CarlB said:
Following post #2, we find:
\begin{array}{rcl}<br /> ds^2 &amp;=&amp; -dt^2 + ((xdx+ydy)r^{-1}-\sqrt{2}\;r^{-0.5} dt)^2 + (y^2dx^2 + x^2dy^2 -2xy\;dx\;dy)r^{-2},\\<br /> &amp;=&amp;-dt^2 + dx^2 + dy^2 + 2r^{-1}dt^2 -2\sqrt{2}\;r^{-1.5}(x\;dx\;dt+y\;dy\;dt),\\<br /> I = \left(\frac{ds}{dt}\right)^2 &amp;=&amp;\frac{2}{r}-1 + \dot{x}^2 + \dot{y}^2-2\sqrt{2}r^{-1.5}(x\dot{x}+y\dot{y})<br /> \end{array}

The idea is to apply the Euler Lagrange equations to the integral over t for s:

s = \int \sqrt{\left(\frac{ds}{dt}\right)^2}\;dt = \int \sqrt{I}\;dt

to obtain the equations of motion in a Cartesian coordinate system.

In the standard "geodesic Lagrangian", the dots refer to differentiation with respect to an affine parameter, which for timelike geodesics we can take to be an arc length paramter. Here, however, Carl's dots refer to differentiation with respect to the Painleve time coordinate, which I am denoting by T. But recall that in the Painleve chart, the Painleve time coordinate happens to give the arc length parameterization of the Lemaitre congruence. The expressions Carl quotes (attributing them to pervect)
\frac{d}{d t} \left[ \frac{1}{2 \sqrt{I}} \left( \frac{\partial I}{\partial \dot{x}} \right) \right] = \frac{1}{2 \sqrt{I}} \frac{\partial I}{\partial x}
do agree with the Euler-Lagrange equations obtained via elementary variational calculus from the standard geodesic Lagrangian, except for what would be an inessential scale factor if I didn't depend upon time. Pervect?
 
Last edited:
Back
Top