- #51

Chris Hillman

Science Advisor

- 2,345

- 8

Continuing (my post grew too long!):

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

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

However,

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

Continuing with Carl's program, his transformation yields the line element

[tex]

ds^2 = -\left(1-\frac{2m}{\sqrt{x^2+y^2+z^2}} \right) \, dT^2

+ 2 \, \frac{\sqrt{2m}}{(x^2+y^2+z^2)^{3/4}} \, \left( x \, dx + y \, dy + z \, dz \right)

\; dT + dx^2 + dy^2 + dz^2, [/tex]

[tex] -\infty < T, \, x, \, y, \, z < \infty, \; x^2+y^2+z^2 \neq 0 [/tex]

The Lemaitre frame field becomes:

[tex]

\vec{e}_0 = \partial_T - \frac{\sqrt{2m}}{(x^2+y^2+z^2)^{3/4}} \;

\left( x \, \partial_x + y \, \partial_y + z \, \partial_z \right)

[/tex]

[tex]

\vec{e}_1 = \frac{1}{\sqrt{x^2+y^2+z^2}} \;

\left( x \, \partial_x + y \, \partial_y + z \, \partial_z \right)

[/tex]

[tex]

\vec{e}_2 = \frac{z}{\sqrt{x^2+y^2+z^2} \, \sqrt{x^2+y^2}} \;

\left( x \, \partial_x + y \, \partial_y \right)

- \frac{\sqrt{x^2+y^2}}{\sqrt{x^2+y^2+z^2}} \; \partial_z

[/tex]

[tex]

\vec{e}_3 = \frac{1}{\sqrt{x^2+y^2}} \; \left( -y \partial_x + x \partial_y \right)

[/tex]

To see this, we simply write down the coordinate vectors of the old chart in terms of the new one and plug these into our previous expressions for the frame field. The easiest way to transform a coordinate vector is to simply apply it to the new coordinates, which are simply functions on our manifold! For example,

[tex]

\partial_r z = \partial_r (r \cos(\theta)) = \cos(\theta) = \frac{z}{\sqrt{x^2+y^2+z^2}}

[/tex]

and so forth, so that

[tex]

\partial_r = \frac{1}{\sqrt{x^2+y^2+z^2}} \;

\left( x \, \partial_x + y \, \partial_y + z \, \partial_z \right)

[/tex]

and so on.

The geodesic equations turn out to be a bit messy. It's much easier to state what the four standard Killing vectors look like:

[tex] \partial_T, \;

-y \partial_x + x \partial_y, \;

-z \partial_y + y \partial_z, \;

-x \partial_z + z \partial_x

[/tex]

We should in fact take this simple result as the defining characteristic of "the cartesian-type Painleve chart" for our static spherically symmetric vacuum solution-- maybe that was Carl's point in one of his remarks above.

The first two geodesic equations are

[tex]

0 = \ddot{T}

+ \frac{\sqrt{2m^3}}{(x^2+y^2+z^2)^{5/4}} \, \dot{T}^2

+ \frac{2 m \, \dot{T}}{(x^2+y^2+z^2)^{3/2}} \;

\left( x \dot{x} + y \dot{y} + z \dot{z} \right)

[/tex]

[tex] \hspace{0.5in}

-\frac{\sqrt{m/2}}{(x^2+y^2+z^2)^{7/4}} \;

\left( (x^2-2\,(y^2+z^2)) \, \dot{x}^2

+ (y^2-2\,(z^2+x^2)) \, \dot{y}^2

+ (z^2-2\,(x^2+y^2)) \, \dot{z}^2 \right)

[/tex]

[tex] \hspace{0.5in}

+ \frac{3 \, \sqrt{2m}}{(x^2+y^2+z^2)^{7/4}} \;

\left( x y \, \dot{x} \dot{y} + y z \, \dot{y} \dot{z} + z x \, \dot{z} \dot{x} \right)

[/tex]

and

[tex]

0 = \ddot{x} - m x \,

\frac{2m - \sqrt{x^2+y^2+z^2}}{(x^2+y^2+z^2)^2} \, \dot{T}^2

- \frac{2 \sqrt{2 m^3} \, x \, \dot{T}}{(x^2+y^2+z^2)^{9/4}} \;

\left( x \dot{x} + y \dot{y} + z \dot{z} \right)

[/tex]

[tex] \hspace{0.5in}

-\frac{m x}{(x^2+y^2+z^2)^{5/2}} \;

\left( (x^2-2 \, (y^2+z^2)) \, \dot{x}

+ (y^2-2 \, (z^2+x^2)) \, \dot{y}

+ (z^2-2 \, (x^2+y^2)) \, \dot{z} \right)

[/tex]

[tex]\hspace{0.5in}

- \frac{6 m x}{(x^2+y^2+z^2)^{5/2}} \;

\left( x y \, \dot{x}\dot{y} + y z \, \dot{y}\dot{z} + z x \, \dot{z}\dot{x} \right)

[/tex]

(similarly for [itex]\ddot{y}, \, \ddot{z}[/itex]), where the dots denote differentiation wrt the affine parameter. These expressions are not particularly simple, but they are at least symmetric between the coordinates x,y,z (as must be the case). As a check, note that in relativistic units, the right hand sides have the dimensions of inverse length.

If I understand correctly, Carl's intention is not simply to write down the geodesic equations and numerically integrate them and plot results in illuminating ways (I enthusiastically second his recommendation of Andrew Hamilton's website, BTW; see the link at RelWWW below), but to express affine parameterized null geodesics (or their "tracks" in a constant time slice) in closed form. Most textbooks show how to do this (using the Schwarzschild exterior chart) in the

As a check, hardy readers can compute the connection one-forms and curvature two-forms, read off the Riemann components with respect to the Lemaitre frame (don't forget that this is an anholonomic frame!), and compute the characteristic polynomial, which is of course an invariant. If you do this correctly, you will obtain

[tex]

\lambda^6

- \frac{6 m^2}{(x^2+y^2+z^2)^3} \, \lambda^4

+ \frac{4 m^3}{(x^2+y^2+z^2)^{9/2}} \, \lambda^3

+ \frac{9 m^4}{(x^2+y^2+z^2)^6} \, \lambda^2

[/tex]

[tex] \hspace{0.5in}

- \frac{12 m^5}{(x^2+y^2+z^2)^{15/2}} \, \lambda

+ \frac{4 m^6}{(x^2+y^2+z^2)^9}

[/tex]

Of course this should agree with the result of simplying plugging Carl's transformation into the characteristic computed in the original Schwarzschild chart, and it does! Similarly, the principle Lorentz invariants of the Riemann tensor are

[tex] R_{abcd} \, R^{abcd} = \frac{48 m^2}{(x^2+y^2+z^2)^3}, \; \;

R_{abcd} \, (\star R)^{abcd} = 0

[/tex]

(no intrinsic gravitomagnetism in the sense of Ciufolini and Wheeler).

By the way, it is easy to rationalize the original Painleve chart. Set

[tex]\tau=T, \; \rho = \sqrt{r}, \; \eta = \sin(\theta)[/tex]

Writing [itex]M=\sqrt{2m}[/itex], the line element becomes

[tex]

ds^2 = -(1-M^2/\rho^2) \, d\tau^2

+ 4 M \, d\tau d\rho

+ 4\rho^2 \, d\rho^2

+ \rho^4 \; \left( \frac{d\eta^2}{1-\eta^2} + \eta^2 \, d\phi^2 \right),

[/tex]

[tex]

-\infty < \tau < \infty, \; 0 < \rho < \infty, \; 0 < \eta < 1, \; -\pi < \phi < \pi

[/tex]

The geodesic equations become

[tex]

\ddot{\tau} + \frac{2 M^2}{\rho^3} \, \dot{\tau} \dot{\rho}

+ \frac{M^3}{2 \rho^5} \, \dot{\tau}^2 + \frac{2M}{\rho} \, \dot{\rho}^2

- M \rho \; \left( \frac{\dot{\eta}^2}{1-\eta^2} + \eta^2 \dot{\phi}^2 \right) = 0

[/tex]

[tex]

\ddot{\rho} - \frac{M^3}{\rho^5} \, \dot{\tau} \dot{\rho}

+ (\rho^2-M^2) \; \left(

\frac{M^2 \, \dot{\tau}^2}{4 \rho^7}

+ \frac{\dot{\rho}^2}{\rho^3}

+ \frac{1}{2 \rho} \; \left( \frac{\dot{\eta}^2}{1-\eta^2}

+ \frac{\dot{\phi}^2}{\eta^2} \right)

\right) = 0

[/tex]

[tex]

\ddot{\eta} + \frac{4}{\rho} \, \dot{\rho} \dot{\eta}

+ \eta \; \left( \frac{\dot{\eta}^2}{1-\eta^2}

- (1-\eta)^2 \, \dot{\phi}^2 \right) = 0

[/tex]

[tex]

\ddot{\phi} + 2 \, \dot{\phi} \;

\left( \frac{2 \dot{\rho}}{\rho} + \frac{\dot{\eta}}{\eta} \right) = 0

[/tex]

In the second equation, note that the horizon is located at [itex]\rho=M[/itex], so this equation simplifies greatly there!

In the equatorial plane [itex]\eta=1[/itex], the geodesic equations become

[tex]

\ddot{\tau} + \frac{2 M^2}{\rho^3} \, \dot{\tau} \dot{\rho}

+ \frac{M^3}{2 \rho^5} \, \dot{\tau}^2 + \frac{2M}{\rho} \, \dot{\rho}^2

- M \rho \, \dot{\phi}^2 \right) = 0

[/tex]

[tex]

\ddot{\rho} - \frac{M^3}{\rho^5} \, \dot{\tau} \dot{\rho}

+ (\rho^2-M^2) \; \left(

\frac{M^2 \, \dot{\tau}^2}{4 \rho^7}

+ \frac{\dot{\rho}^2}{\rho^3}

+ \frac{\dot{\phi}^2}{2 \rho} \right)

\right) = 0

[/tex]

[tex]

\ddot{\phi} + \frac{4}{\rho} \, \dot{\rho} \dot{\phi} = 0

[/tex]

where the last immediately yields the first integral

[tex]\dot{\phi} = \frac{L}{\rho^4}[/tex]

Now plug this into the first two equations. We obtain a coupled system which is first order in [itex]\dot{\tau}, \dot{\rho}[/itex] provided that we pretend to forget the relationship between [itex]\rho[/itex] and [itex]\dot{\rho}[/itex]. Thus, near some fixed locus of constant "rootradius", we can apply a phase plane analysis.

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

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

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

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

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

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

However,

the Hamilton/Lisle eprint (I seem to have overlooked this--- thanks for bringing it to my attention!) seems, at a glance, to refer to the flat metric inherited by the constant Painleve time hyperslices. They write "picture space as flowing like a river into the Schwarzschild black hole", which suggests to me they are thinking of projecting [itex]\vec{e}_0[/itex] into [itex]T=0[/itex] to form a flow (in the sense of differential equations on manifolds).An example of a recent paper in the literature using "Gullstrand-Painleve-Cartesian" coordinates is:

http://arxiv.org/abs/gr-qc/0411060

The above has 12 citations and gives easily understood reasons for looking at things this way that can be understood at a high-school level.

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

Continuing with Carl's program, his transformation yields the line element

[tex]

ds^2 = -\left(1-\frac{2m}{\sqrt{x^2+y^2+z^2}} \right) \, dT^2

+ 2 \, \frac{\sqrt{2m}}{(x^2+y^2+z^2)^{3/4}} \, \left( x \, dx + y \, dy + z \, dz \right)

\; dT + dx^2 + dy^2 + dz^2, [/tex]

[tex] -\infty < T, \, x, \, y, \, z < \infty, \; x^2+y^2+z^2 \neq 0 [/tex]

The Lemaitre frame field becomes:

[tex]

\vec{e}_0 = \partial_T - \frac{\sqrt{2m}}{(x^2+y^2+z^2)^{3/4}} \;

\left( x \, \partial_x + y \, \partial_y + z \, \partial_z \right)

[/tex]

[tex]

\vec{e}_1 = \frac{1}{\sqrt{x^2+y^2+z^2}} \;

\left( x \, \partial_x + y \, \partial_y + z \, \partial_z \right)

[/tex]

[tex]

\vec{e}_2 = \frac{z}{\sqrt{x^2+y^2+z^2} \, \sqrt{x^2+y^2}} \;

\left( x \, \partial_x + y \, \partial_y \right)

- \frac{\sqrt{x^2+y^2}}{\sqrt{x^2+y^2+z^2}} \; \partial_z

[/tex]

[tex]

\vec{e}_3 = \frac{1}{\sqrt{x^2+y^2}} \; \left( -y \partial_x + x \partial_y \right)

[/tex]

To see this, we simply write down the coordinate vectors of the old chart in terms of the new one and plug these into our previous expressions for the frame field. The easiest way to transform a coordinate vector is to simply apply it to the new coordinates, which are simply functions on our manifold! For example,

[tex]

\partial_r z = \partial_r (r \cos(\theta)) = \cos(\theta) = \frac{z}{\sqrt{x^2+y^2+z^2}}

[/tex]

and so forth, so that

[tex]

\partial_r = \frac{1}{\sqrt{x^2+y^2+z^2}} \;

\left( x \, \partial_x + y \, \partial_y + z \, \partial_z \right)

[/tex]

and so on.

The geodesic equations turn out to be a bit messy. It's much easier to state what the four standard Killing vectors look like:

[tex] \partial_T, \;

-y \partial_x + x \partial_y, \;

-z \partial_y + y \partial_z, \;

-x \partial_z + z \partial_x

[/tex]

We should in fact take this simple result as the defining characteristic of "the cartesian-type Painleve chart" for our static spherically symmetric vacuum solution-- maybe that was Carl's point in one of his remarks above.

The first two geodesic equations are

[tex]

0 = \ddot{T}

+ \frac{\sqrt{2m^3}}{(x^2+y^2+z^2)^{5/4}} \, \dot{T}^2

+ \frac{2 m \, \dot{T}}{(x^2+y^2+z^2)^{3/2}} \;

\left( x \dot{x} + y \dot{y} + z \dot{z} \right)

[/tex]

[tex] \hspace{0.5in}

-\frac{\sqrt{m/2}}{(x^2+y^2+z^2)^{7/4}} \;

\left( (x^2-2\,(y^2+z^2)) \, \dot{x}^2

+ (y^2-2\,(z^2+x^2)) \, \dot{y}^2

+ (z^2-2\,(x^2+y^2)) \, \dot{z}^2 \right)

[/tex]

[tex] \hspace{0.5in}

+ \frac{3 \, \sqrt{2m}}{(x^2+y^2+z^2)^{7/4}} \;

\left( x y \, \dot{x} \dot{y} + y z \, \dot{y} \dot{z} + z x \, \dot{z} \dot{x} \right)

[/tex]

and

[tex]

0 = \ddot{x} - m x \,

\frac{2m - \sqrt{x^2+y^2+z^2}}{(x^2+y^2+z^2)^2} \, \dot{T}^2

- \frac{2 \sqrt{2 m^3} \, x \, \dot{T}}{(x^2+y^2+z^2)^{9/4}} \;

\left( x \dot{x} + y \dot{y} + z \dot{z} \right)

[/tex]

[tex] \hspace{0.5in}

-\frac{m x}{(x^2+y^2+z^2)^{5/2}} \;

\left( (x^2-2 \, (y^2+z^2)) \, \dot{x}

+ (y^2-2 \, (z^2+x^2)) \, \dot{y}

+ (z^2-2 \, (x^2+y^2)) \, \dot{z} \right)

[/tex]

[tex]\hspace{0.5in}

- \frac{6 m x}{(x^2+y^2+z^2)^{5/2}} \;

\left( x y \, \dot{x}\dot{y} + y z \, \dot{y}\dot{z} + z x \, \dot{z}\dot{x} \right)

[/tex]

(similarly for [itex]\ddot{y}, \, \ddot{z}[/itex]), where the dots denote differentiation wrt the affine parameter. These expressions are not particularly simple, but they are at least symmetric between the coordinates x,y,z (as must be the case). As a check, note that in relativistic units, the right hand sides have the dimensions of inverse length.

If I understand correctly, Carl's intention is not simply to write down the geodesic equations and numerically integrate them and plot results in illuminating ways (I enthusiastically second his recommendation of Andrew Hamilton's website, BTW; see the link at RelWWW below), but to express affine parameterized null geodesics (or their "tracks" in a constant time slice) in closed form. Most textbooks show how to do this (using the Schwarzschild exterior chart) in the

*exterior region*, but to my knowledge none discuss how to do it in the*future interior*region (using for example ingoing Eddington or Painleve charts).As a check, hardy readers can compute the connection one-forms and curvature two-forms, read off the Riemann components with respect to the Lemaitre frame (don't forget that this is an anholonomic frame!), and compute the characteristic polynomial, which is of course an invariant. If you do this correctly, you will obtain

[tex]

\lambda^6

- \frac{6 m^2}{(x^2+y^2+z^2)^3} \, \lambda^4

+ \frac{4 m^3}{(x^2+y^2+z^2)^{9/2}} \, \lambda^3

+ \frac{9 m^4}{(x^2+y^2+z^2)^6} \, \lambda^2

[/tex]

[tex] \hspace{0.5in}

- \frac{12 m^5}{(x^2+y^2+z^2)^{15/2}} \, \lambda

+ \frac{4 m^6}{(x^2+y^2+z^2)^9}

[/tex]

Of course this should agree with the result of simplying plugging Carl's transformation into the characteristic computed in the original Schwarzschild chart, and it does! Similarly, the principle Lorentz invariants of the Riemann tensor are

[tex] R_{abcd} \, R^{abcd} = \frac{48 m^2}{(x^2+y^2+z^2)^3}, \; \;

R_{abcd} \, (\star R)^{abcd} = 0

[/tex]

(no intrinsic gravitomagnetism in the sense of Ciufolini and Wheeler).

By the way, it is easy to rationalize the original Painleve chart. Set

[tex]\tau=T, \; \rho = \sqrt{r}, \; \eta = \sin(\theta)[/tex]

Writing [itex]M=\sqrt{2m}[/itex], the line element becomes

[tex]

ds^2 = -(1-M^2/\rho^2) \, d\tau^2

+ 4 M \, d\tau d\rho

+ 4\rho^2 \, d\rho^2

+ \rho^4 \; \left( \frac{d\eta^2}{1-\eta^2} + \eta^2 \, d\phi^2 \right),

[/tex]

[tex]

-\infty < \tau < \infty, \; 0 < \rho < \infty, \; 0 < \eta < 1, \; -\pi < \phi < \pi

[/tex]

The geodesic equations become

[tex]

\ddot{\tau} + \frac{2 M^2}{\rho^3} \, \dot{\tau} \dot{\rho}

+ \frac{M^3}{2 \rho^5} \, \dot{\tau}^2 + \frac{2M}{\rho} \, \dot{\rho}^2

- M \rho \; \left( \frac{\dot{\eta}^2}{1-\eta^2} + \eta^2 \dot{\phi}^2 \right) = 0

[/tex]

[tex]

\ddot{\rho} - \frac{M^3}{\rho^5} \, \dot{\tau} \dot{\rho}

+ (\rho^2-M^2) \; \left(

\frac{M^2 \, \dot{\tau}^2}{4 \rho^7}

+ \frac{\dot{\rho}^2}{\rho^3}

+ \frac{1}{2 \rho} \; \left( \frac{\dot{\eta}^2}{1-\eta^2}

+ \frac{\dot{\phi}^2}{\eta^2} \right)

\right) = 0

[/tex]

[tex]

\ddot{\eta} + \frac{4}{\rho} \, \dot{\rho} \dot{\eta}

+ \eta \; \left( \frac{\dot{\eta}^2}{1-\eta^2}

- (1-\eta)^2 \, \dot{\phi}^2 \right) = 0

[/tex]

[tex]

\ddot{\phi} + 2 \, \dot{\phi} \;

\left( \frac{2 \dot{\rho}}{\rho} + \frac{\dot{\eta}}{\eta} \right) = 0

[/tex]

In the second equation, note that the horizon is located at [itex]\rho=M[/itex], so this equation simplifies greatly there!

In the equatorial plane [itex]\eta=1[/itex], the geodesic equations become

[tex]

\ddot{\tau} + \frac{2 M^2}{\rho^3} \, \dot{\tau} \dot{\rho}

+ \frac{M^3}{2 \rho^5} \, \dot{\tau}^2 + \frac{2M}{\rho} \, \dot{\rho}^2

- M \rho \, \dot{\phi}^2 \right) = 0

[/tex]

[tex]

\ddot{\rho} - \frac{M^3}{\rho^5} \, \dot{\tau} \dot{\rho}

+ (\rho^2-M^2) \; \left(

\frac{M^2 \, \dot{\tau}^2}{4 \rho^7}

+ \frac{\dot{\rho}^2}{\rho^3}

+ \frac{\dot{\phi}^2}{2 \rho} \right)

\right) = 0

[/tex]

[tex]

\ddot{\phi} + \frac{4}{\rho} \, \dot{\rho} \dot{\phi} = 0

[/tex]

where the last immediately yields the first integral

[tex]\dot{\phi} = \frac{L}{\rho^4}[/tex]

Now plug this into the first two equations. We obtain a coupled system which is first order in [itex]\dot{\tau}, \dot{\rho}[/itex] provided that we pretend to forget the relationship between [itex]\rho[/itex] and [itex]\dot{\rho}[/itex]. Thus, near some fixed locus of constant "rootradius", we can apply a phase plane analysis.

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

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

Last edited: