Relating PDEs: Showing Laplace's Eq in Cylindrical Coordinates

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In summary: The problem is solvable in closed form, and the solution is not hard to find in the literature. It's been discussed in sci.physics.relativity, and the best account I know of is in Harvey Reall's "Exact Solutions of Einstein's Field Equations" chapter 6. IMHO his book is (like Stephani's) not so easy to use unless you already know the material. But it's nice to have as a reference. I would say that a good modern account is in the paper by Bini et. al. that I recommended in the "torque on a gyroscope" thread, gr-qc/0104034. Plus there are a few gems in the ancient literature
  • #1
Spinny
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This problem may be very easy or very difficult (probably the first), but I can't seem to make sense of it, and that annoys me. It's not all that important (at least not yet), but I just can't seem to let it go. Anyways, here it is.

Consider the following PDE:

[tex]\frac{\partial^2 f}{\partial r^2}+\frac{\partial^2 f}{\partial z^2}-\frac{1}{r}\frac{\partial f}{\partial r} = 0[/tex]

What I would like to do is to be able to show explicitly how we can go from the above PDE to this one:

[tex]\frac{\partial^2 g}{\partial r^2}+\frac{\partial^2 g}{\partial z^2}+\frac{1}{r}\frac{\partial g}{\partial r} = 0[/tex]

by using the transformation

[tex]g = \int \frac{f}{r}dr \Rightarrow f = r\frac{\partial g}{\partial r}[/tex]

The functions are [tex]f = f(r,z)[/tex] and [tex]g = g(r,z)[/tex]. And yes, the second PDE is Laplace's equation in cylindrical coordinates, [tex]\nabla^2 g(r,z) = 0[/tex].

Any help would be greatly appreciated!
 
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  • #2
Spinny said:
And yes, the second PDE is Laplace's equation in cylindrical coordinates, [tex]\nabla^2 g(r,z) = 0[/tex].
No it is not.
 
  • #3
Kummer said:
No it is not.

Indeed it is, but the function [tex]g[/tex] as I wrote explicitly, depends only on [tex](r,z)[/tex]. Consider equations (98)-(99) here: http://mathworld.wolfram.com/CylindricalCoordinates.html

[tex]\nabla^2 f = \frac{\partial^2 f}{\partial r^2}+\frac{1}{r}\frac{\partial f}{\partial r} + \frac{1}{r^2}\frac{\partial^2 f}{\partial \theta^2}+\frac{\partial^2 f}{\partial z^2}[/tex]

And since in my case, there's no angular dependence, and so it reduces to the second equation in my OP.

Or am I missing something...?
 
  • #4
Kummer: I believe that it actually is... [tex]\nabla^2 = \frac{1}{r}\frac{\partial}{\partial r}\left(r \frac{\partial}{\partial r} \right) + \frac{\partial^2}{\partial z^2} + \frac{1}{r^2}\frac{\partial^2}{\partial \phi^2}[/tex]

For the axisymmetric case, the last term drops out.

Spinny: I've attempted this, but couldn't see that they are equal. In fact, I'm pretty sure you'll need more than just the substitution that you've used.
 
  • #5
Citation, please

What's the source of the problem, please, Spinny? (I wonder if you might not be misunderstanding something I wrote at PF in another context.)

Kummer, Spinny is trying to map the solution space of a certain PDE to the space of axisymmetric harmonic functions, just as he said. The first PDE arises for example in studying weak-field Ernst vacuums (gtr model of an axisymmetric rotating object, Spinny), as the equation constraining the off-diagonal component when one writes the metric tensor in cylindrical coordinates; when one writes it in Cartesian coordinates all components are axisymmetric harmonic.
 
  • #6
Chris Hillman said:
What's the source of the problem, please, Spinny? (I wonder if you might not be misunderstanding something I wrote at PF in another context.)

Kummer, Spinny is trying to map the solution space of a certain PDE to the space of axisymmetric harmonic functions, just as he said. The first PDE arises for example in studying weak-field Ernst vacuums (gtr model of an axisymmetric rotating object, Spinny), as the equation constraining the off-diagonal component when one writes the metric tensor in cylindrical coordinates; when one writes it in Cartesian coordinates all components are axisymmetric harmonic.

Sorry, this is not intentionally related to anything you've written. It is, however, one of the equations that arises from studying an axisymmetric rotating object in gtr, using cylindrical coordinates. The first equation in my OP is one of the components of the Einstein tensor. It is then stated that that equation can be expressed as Laplace's equation (second equation in my OP) by defining the second function as described ibid.

The way it is stated it's almost as if it should be clear as day, although I don't see it. I could of course just take it for given (which seems to be done by many others), but I'd like to understand it and have a useful derivation of it.

As I'm sitting here writing this, I realize there's one thing I haven't tried yet, and that is to try to actually solve the first equation, perhaps by using sep. of variables as with the actual Laplace equation. Maybe in doing so, something might pop up. That, however, will have to wait until tomorrow for my part.

Finally let me just say, I appreciate all your help and input. Thanks a lot everyone :)
 
  • #7
But can you cite your source, Spinny?

Spinny said:
Sorry, this is not intentionally related to anything you've written. It is, however, one of the equations that arises from studying an axisymmetric rotating object in gtr, using cylindrical coordinates. The first equation in my OP is one of the components of the Einstein tensor. It is then stated that that equation can be expressed as Laplace's equation (second equation in my OP) by defining the second function as described ibid.

Whaddya mean, ibid--- you still haven't cited your source! Remarks like "it is then stated that..." suggest that you are reading something you don't understand. Not only that, but comparing my guess (got it in one!) with your description, it is quite clear that we are talking about the same thing. Except that you said "one of the components of the Einstein tensor" (which makes no sense as stated) instead of "the constraint governing the off-diagonal component of the metric tensor".

Spinny said:
The way it is stated it's almost as if it should be clear as day, although I don't see it. I could of course just take it for given (which seems to be done by many others), but I'd like to understand it and have a useful derivation of it.

Well, evidently you've come to the right place, since you seem to asking about something I wrote, but can you link to or otherwise cite what you are reading so that we can all have a look at it?

E.g. it might be this (look for paragraph beginning "Another interesting point is that one can write out ..."):

Code:
From: *** Chris Hillman 
Subject:  Re: GR: Spun-up mass shell spins up particle at shell's center?
Organization: Pythagorean Illuminati
Newsgroups: sci.physics.research
Date: 22 Aug 2003 

Long, long ago, Gene Partlow asked:

> I am hoping that someone having an easy familiarity with general
> relativity might help me with these questions:
>
> I was reading in Abraham Pais' book "Subtle is the Lord", a reference
> (in Chap. 15, pg. 284) to an early 1912 paper by Einstein (presaging the
> Lense-Thirring effect), where if a massive hollow spherical shell is
> "accelerated", apparently along the straight axis passing through its
> center, a mass point-particle at that center will experience an increase
> in its inertial mass.
>
> I want to know something slightly different:
>
> 1) If the central particle has spatial extent (say a tiny dust grain?),
> and the hollow massive shell is SPUN UP around its central axis (though
> not otherwise moved), will the central particle also show a change in
> its angular momentum (ie: will it spin up also), in addition to
> (possibly?) increasing its inertial mass?
>
> 2) Will its spin vector point in the same direction as that of the
> spinning mass shell?
>
> 3) If so, is there a not too technical source (using no more than
> calculus) which would give quantitative details on this (so that I might
> better understand the relative importance of the variables involved,
> i.e.: shell mass and thickness, its rate of change of spin, mass and
> size of particle, etc)?

These are good questions.  Unfortunately, I think it will prove hard to
obtain good answers!  None of the sources I have seen explain this stuff
the way I think it should be presented.  Unfortunately, I will be unable
to elaborate on this, but I can give a few quick pointers for the benefit
of serious and well-prepared students.

First, at a glance it looks like Greg Egan rederived the original
Lens-Thirring precession.  This is indeed a good place to begin, but I'd
back up even further and start with an electromagnetic analog.

Consider a spherical shell (nonconducting) of radius R carrying an
electric charge Q, with the charge uniformly distributed. Let the shell be
rotating about the z-axis of the lab frame with constant angular velocity
Omega.  A classical exercise in Maxwell's theory of EM is to show that
this induces a magnetic field which inside the shell is uniform and
parallel to the z-axis, and outside looks like a dipole field.  Here is a
outline of how this computation proceeds:

1. Start with Maxwell's equations in the form

  Div E = 4 pi rho

  Curl E = -@B/@t

  Div B = 0

  Curl B = 4 pi J + @E/@t

where rho is the charge density and J is the current density vector.  Put

  E = -Grad phi - @A/@t

  B = Curl A

where phi is the electric potential and A is the magnetic vector
potential.  For a stationary EM field as in our problem, the time
derivatives all vanish.  Use this observation to derive

  Lap phi = -4 pi rho

  Lap A = -4 pi J

which we can consider separately.

2. Write the current density in the form

  J^x = - Q Omega/(4 pi R^2) r sin(u) sin(v) delta(r-R)

  J^y =   Q Omega/(4 pi R^2) r sin(u) cos(v) delta(r-R)

  J^z =   0

Guess that the unknown vector potential can be written in the form

  A^x = f(r) sin(u) sin(v)

  A^y = f(r) sin(u) cos(v)

  A^z = o

Plug this into the Laplacian and find that we have reduced the problem to
the ODE

  f''(r) + 2 f'(r)/r - 2 f(r)/r^2 = Q Omega/R^2 r delta(r-R)        (*)

3. This is solved by properly piecing together solutions valid
respectively on 0 < r < R and R < r < infty.  Considering these two
domains seperatly we see that the form of f is

  f(r) = a r H(R-r) + a R^3/r^2 H(r-R)                             (**)

where H(z) is the usual Heaviside step function, and where a is an
undetermined constant.

Note well: we have used some physically plausible boundary conditions
here!  To wit, outside the shell we assume that |A| -> 0 as r -> infty,
and inside the shell we assume that A remains bounded as r -> 0.  This
assumption is innocuous in the EM context, but becomes has crucially
important implications when we pass to weak-field gtr.

4. Plugging (**) into (*), we find that the "generalized functions" drop
out and we are left with a = Q Omega/3/R.  Thus, inside the shell the
magnetic vector potential is

  A^x = Q Omega/3/R r sin(u) sin(v)

  A^y = Q Oemga/3/R r sin(u) cos(v)

  A^z = 0

and outside (you can fill this in!)...

5. Rewrite this as a one-form A, and similarly solve for the electric
potential phi (you should find that it -vanishes- inside and outside
possesses a reasonable Q/r decay).  Rewrite this information as the four
dimensional one-form potential A.

6. Compute the exterior derivative dA and rewrite this in terms of E and B
fields.

This solves the problem of the fields created by a rotating spherical
charged shell.  (The computation takes less time to perform than it did to
explain it!)  Note that we worked componentwise above, and that it was
convenient to compute the Cartesian components, but then to reduce the
problem to an ODE we needed to write these in terms of polar spherical
coordinates, which is admittedly a bit devious!

Now, in weak-field gtr, standard sources explain why a rotating shell of
mass M can be treated almost identically.  The key equation, which again
can be solved componentwise, is now

  Lap h^(ij) = - 16 pi T^(ij)

where h^(ij) is the traceless perturbation of the Cartesian metric tensor.
Thus we have

 Lap h^(tt) = - 16 pi rho

 Lap h^(tx) = -16 pi J^x

etc., where now J is the mass-current density vector.

In setting this up, we neglect any stresses in the shell, so the
components h^(xx) etc. vanish.  We can also use a slow rotation
approximation so that v/sqrt(1-v^2) ~ v, etc.; then the computation really
is almost identical to what we did for the EM analog.

The grand result is an expression for the weak-field metric inside the
shell and another for the metric outside.  The former is locally isometric
to a weak-field approximation to Minkowski spactime, in terms of a
rotating chart, and the latter is locally isometric to the weak-field
approximation fo the Kerr vacuum.

I urge interested readers to fill in the details!  If you get stuck, this
problem is solved in detail in

author = {Alan P. Lightman and William H. Press and Richard H. Price and
Saul A. Teukolsky},
title = {Problem Book in Relativity and Gravitation},
publisher = {Princeton University Press},
year = 1975}

Indeed, every problem in the relevant chapter is highly relevant, so
readers are urged to do them all.

BTW, those who have heard of "gravitomagnetism" will recognize that we
could have reorganized this computation in terms of computing
"gravitoelectric" and "gravitomagnetic" vectors which obey field equations
almost identical to Maxwell's equations; see

author = {Robert M. Wald},
title = {General Relativity},
publisher = {University of Chicago Press},
year = 1984}

Also, for more about manipulating Dirac and Heaviside "functions" see for
example the recent book by Duffy on Green's functions.

At this point, we should explain what the precession rate

  4 M Omega/3/R

is all about.  Due to lack of time I cannot do this, unfortunately, but
savvy readers can think about the sphere as made of glass and think about
the appearance of distant stars.  Note too that we treated h^(ij) and the
other tensors as tensor fields on Minkowski spacetime, and must now
interpret our results in terms of the geometry of a perturbation of
Minkowski vacuum.

Another interesting point is that one can write out a weak-field Ernst
vacuum (see long ago previous posts by me on these) in which the field
equations decouple completely.  One says that phi is axisymmetric
harmonic, the other says that a second potential obeys a very similar
looking equation.  A good exercise is to reconcile this approach with the
above, which of course gives the same result.

But this is only the tip of the iceberg.  We have found a weak-field
solution and interpreted it in terms of precession of gryoscopes carried
by certain observers (see Lightman et al., but think hard about what they
say and do not say about what their computations mean physically).

We next ask if a small object in freefall can be spun up by a
gravitational field.  In particular, the Newtonian potential vanishes
inside our spherical shell, so we can ask if a nonspining object in the
interior will tend to spin up, and we can ask if a small intially
nonspinnning object falling in radially from r = infty (say) will tend to
spin up.  Very roughly speaking the answer is shape dependent; very
roughly there is a coupling between quadrupole moments (and higher) of
aspherical objects and the -tidal tensor- or "electrogravitic tensor"

  E(X)_(ab) = R_(ambn) X^m X^n

where X is the velocity vector field defining our congruence of freely
falling observers.  See the problem in Lightman et al.

There is also the "magnetogravitic tensor" B(X)_(ab), which comes from the
dual of the Riemann tensor.  This has a direct interpretation in terms of
spin-spin acceleration of a spinning test object.  See

author = {Hans Stephani},
title = {General Relativity: An Introduction of the Theory of the
          Gravitational Field},
publisher = {Cambridge University Press},
edition = {Second},
note = {translated by {J}ohn {S}tewart and {M}artin {P}ollock},
year = 1990}

Finally, a keyword: Papapetrou equations.  These govern the motion of
possibly spinning test objects in a general gravitational field.  They can
be put in various forms; there is a very recent preprint on the ArXiV
deriving a new form which allows for multipole structure of the test
particle as well as spin moments, so this is just what Gene wants.
Unfortunately, AFAIK all known forms of the Papapetrou equations are hard
to use.  Look on the ArXiV for another preprint studying their behavior
numerically in the case of the Kerr vacuum.

Oh yeah, Mach's principle.  I'm out of time, but there's a huge literature
on that, almost all misleading.  Note that the precession rate depends on
R, Omega in a particular way and consider limiting cases.

Again, I regret that I won't be able to elaborate or even read any
responses to this post, so I hope I have provided enough information to
get well prepared students started in pursuing these interesting
questions.

Spinny said:
try to actually solve the first equation, perhaps by using sep. of variables as with the actual Laplace equation.

Well, in at least some of my posts on this that's just what I did, but yeah, by all means try it yourself first.
 
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  • #8
Nu?

Spinny? You're not going to leave me hanging, are you? I'm understandably curious to see your citation.
 
  • #9
Chris Hillman said:
Spinny? You're not going to leave me hanging, are you? I'm understandably curious to see your citation.

My apologies for being so slow to reply. The original article where I read it can be found here: http://arxiv.org/abs/astro-ph/0507619v1

On page 6 (Eq. (11)) is the first equation, and on the following page it is re-expressed as Laplace's equation of some other function [tex]\Phi[/tex], which the authors refer to as a "generating potential" (not to be confused with a Newtonian potential). I've looked through subsequent papers on the same topic by the same authors and others, however they all fail to give any more details as to have this transformation is done. The only additional information given is that the RHS of Laplace's equation could be replaced with an arbitrary function depending only on [tex]z[/tex], since [tex]\Phi[/tex] itself is defined by a partial integral.
 
  • #10
OK, I can answer your question now

Hi again, Spinny,

Spinny said:
My apologies for being so slow to reply. The original article where I read it can be found here: http://arxiv.org/abs/astro-ph/0507619v1

Thanks for the cite! I can see what is going on now (note that I use relativistic units in which [itex]c=G=1[/itex] throughout):

0. Cooperstock is sometimes unconventional but experienced in the field, and the paper appears to be mathematically correct, as I would expect.

1. Despite what the authors imply, in the case [itex]w =0[/itex] their solution is an exact dust solution (as is the well-known Winacour dust which they cite). It is stationary and axisymmetric.

2. Write down a coframe from (1) , set [itex]w =0[/itex] (this ensures that lapse of coordinate time agrees with lapse of time measured by the dust particles), compute the Einstein tensor (wrt the frame!) using GRTensorII, set [itex]G^{11} = 8 \, \pi \, \rho[/itex], apply the powerful "casesplit" command. This gives their (11) (12) immediately, with the remaining metric function v given by quadrature from N, in less time than it took me to type this post.

3. Consider the system
[tex]\Phi_{zz} + \Phi_{rr} + \frac{\Phi_r}{r} = 0, \; \; N = r \, \Phi_r[/tex]
That is, we start with a harmonic function and obtain N from it by differentiation. (That's why [itex]\Phi[/itex] can be called, in a loose but standard sense, a "potential" for the metric function N.) Try to eliminate [itex]\Phi[/itex] in favor of N by another application of casesplit:
Code:
[diff(Phi(z,r),z,z)+diff(Phi(z,r),r,r) + diff(Phi(z,r),r)/r, Diff(Phi(z,r),r) = N(z,r)/r ];
casesplit(%,[[Phi],[N]](z,r));
This gives
[tex]
N_{zz} + N_{rr} - \frac{N_r}{r} = 0, \; \; \Phi_{zz} = \frac{-N_r}{r}, \; \; \Phi_r = \frac{N}{r}
[/tex]
This gives [itex]\Phi[/itex] by quadrature in terms of N, a solution of the "mysterious" (11). This establishes the desired mapping between the solution spaces of (11) and (13).

4. The point is of course that axisymmetric harmonic functions can thus be used to generate explict examples of these Cooperstock-Tieu dusts. Choose an axisymmetric harmonic function [itex]\Phi[/itex] and put [itex]N = r \, \Phi_r[/itex]. Then N satisfies
[tex]
N_{zz} + N_{rr} - \frac{N_r}{r} = 0
[/tex]
Now obtain v by quadrature from
[tex]
v_z = \frac{N_z \, N_r}{2 \, r}, \; \; v_r = \frac{N_z^2-N_r^2}{4 \, r}
[/tex]
Define a Lorentzian spacetime using the metric functions N,v as follows:
[tex]
ds^2 = -(dt - N \, d\phi)^2 + \exp(2 \, v) \, \left( dz^2 + dr^2 \right) + r^2 \, d\phi^2,
[/tex]
[tex]
-\infty < t, \, z < \infty, \; 0 < r < \infty, \; -\pi < \phi < \pi
[/tex]
We might need an additional restriction on the domain to keep the vector [itex]\partial_\phi[/itex] spacelike, or equivalently, to keep the covector [itex]dt[/itex] timelike. (Since v plays a minor role, I felt free to redefine it with a factor of two viz. the eprint you cited.) This gives an exact stationary axisymmetric dust solution, with the density of the dust given by
[tex]
\rho = \frac{N_z^2 + N_r^2}{8 \pi \, r^2 \, \exp(2 \, v)}
[/tex]
(Note that this is non-negative.) The expansion tensor of the congruence given by the integral curves of the coordinate vector [itex]\partial_t[/itex] (the world lines of the dust particles) vanishes, so this is a rigidly rotating dust. Question to ponder: can you find asympotically flat examples?

5. The authors take a non-asympotically vanishing axisymmetric harmonic function in order to generate a rotating dust solution with flat velocity curve. Question to ponder: does their example of a Cooperstock-Tieu dust in fact resemble a rotating disklike distribution of dust (idealized stars)? Actually, the authors propose to match a disklike region filled with their rotating dust to an "exterior" vacuum solution, which would be one of the stationary axisymmetric vacuums found by Ernst. Question to ponder: after finding this solution and carrying out the matching, is it asympotically flat?

6. This is quite independent of, but related to, the stuff I mentioned in the post quoted above.

BTW, casesplit works by Groebner-basis-type methods applied to differential rings. It is a powerful command much-beloved by Maple afficionados! To be truthful, using it here is a bit like shooting a fly with an elephant gun; can you see how to transform the equation defining axisymmetric harmonic functions into our master equation "by hand"? (Hint: plug [itex]N = r \, \Phi_r[/itex] into [itex]\Phi_{zz} + \Phi_{rr} + \frac{\Phi_r}{r} = 0[/itex] and derive [itex]\Phi_{zz} = \frac{-N_r}{r}[/itex]. Then derive [itex]N_{zz} + N_{rr} - \frac{N_r}{r} = 0[/itex]. The real point of the casesplit exercise is to see how one can sometimes find such "Backlund transformations" by trial and error. (Useful Backlund transformations are often quite simple, as here, so this isn't as crazy as it might sound.)
 
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  • #11
Some more Backlund Transformations to Play With

Spinny? You still here?

I was waiting for some response before saying more, in case you were working off-line, but since I have more to say, I'll say some more.

You can obtain the family of stationary axisymmetric "rigidly rotating" dust solutions used by Cooperstock and Tieu to build their galactic models in terms of other charts, such as parabolic rational or spherical rational (put [itex]\eta = \cos(\theta)[/itex] in polar spherical to obtain this one). In each case, the Einstein equation then reduces to one "master equation" (second order linear PDE) with the other metric function given by quadratures. You can find Backlund transformations from the axisymmetric harmonic functions to the solution space of our master equation, not the same as the one mentioned by Cooperstock and Tieu.

Here is the example of rational spherical coordinates in more detail. The frame Ansatz in the rational spherical type Weyl-Lewis-Papapetrou canonical chart is
[tex]
\vec{e}_1 = \partial_t, \;
\vec{e}_2 = \exp(-v) \, \partial_\rho, \;
\vec{e}_3 = \frac{\sqrt{1-\eta^2}}{\rho} \, \exp(-v) \, \partial_\eta, \;
\vec{e}_4 = \frac{1}{\rho \, \sqrt{1-\eta^2}} \, \left( \partial_\phi + w \, \partial_t \right)
[/tex]
or
[tex]
ds^2 = -\left( dt - w \, d\phi \right)^2
+ \exp(2v) \, \left( d\rho^2 + \frac{\rho^2}{1-\eta^2} \, d\eta^2 \right)
+ \rho^2 \, (1-\eta^2) \, d\phi^2
[/tex]
where w,v are again functions of [itex]\rho, \, \eta[/tex] only, and where the coordinate ranges are no larger than
[tex]
-\infty < t < \infty, \; 0 < \rho < \infty, \; -1 < \eta < 1, \; -\pi < \phi < \pi
[/tex]
Note that lapse of coordinate time gives the lapse of clock time measured by observers riding on the dust particles. Note too that null circles (the integral curves of the coordinate vector field [itex]\partial_\phi[/itex]) exist at the locus
[tex]
w(\rho,\eta)^2 = \rho^2 \, (1-\eta^2) = \rho^2 \, \sin(\theta)^2
[/tex]
with timelike closed curves existing for larger radii.

Proceeding as before, we immediately find that the Einstein equations reduce to the master equation
[tex]
\rho^2 \, w_{\rho \rho} + (1-\eta^2) \, w_{\eta \eta} = 0
[/tex]
with v given by quadrature from w. Notice that our master equation is a second order linear PDE, and a remarkably simple one at that. The density of the dust is
[tex]
\mu = \frac{1}{8 \pi \, \rho^2 \, \exp(2v)} \; \left( \frac{w_\rho^2}{1-\eta^2} + \frac{w_\eta^2}{\rho^2} \right)
[/tex]
which is evidently everywhere non-negative.

In contrast, adopting the rational spherical chart for [itex]E^3[/itex] (euclidean three-space)
[tex]
ds^2 = d\rho^2 + \frac{\rho^2}{1-\eta^2} \, d\eta^2 + \rho^2 \, (1-\eta^2) \, d\phi^2
[/tex]
we find that axisymmetric harmonic functions satisfy
[tex]
\rho^2 \, \Phi_{\rho \rho} + (1-\eta^2) \, \Phi_{\eta \eta} + 2 \rho \, \Phi_\rho - 2 \eta \, \Phi_\eta = 0
[/tex]
As is well known, this second order linear PDE (with nonlinear coefficients) has two important families of solutions, the ones which are regular on the symmetry axis,
[tex]\Phi(\rho,\eta) = \rho^n \, P_n(\eta)[/tex]
(where n is a positive integer and the [itex]P_n[/itex] are the Legendre polynomials) and the ones which are asymptotically vanishing,
[tex]\Phi(\rho,\eta) = \frac{P_n(\eta)}{\rho^{n+1}}[/tex]
Most PF readers are probably familiar with these, because they are virtually canonical examples of solutions found by separation of variables.

After a bit of playing around, we find that
[tex]w = \rho^2 \, \left( 1- \eta^2 \right) \, \Phi_{\rho \eta}[/tex]
gives a Backlund transformation from the solution space of the axisymmetric Laplace equation into the solution space of our master equation. Thus, every axisymmetric harmonic function yields one of our rigidly rotating dust solutions. We immediately obtain two sequences of solutions of our master equation. One sequence gives rigidly rotating stationary axisymmetric dust solutions which are regular on the symmetry axis,
[tex]
w(\rho,\eta) = n \, (n+1) \, \rho^{n+1} \, \left ( \eta \, P_n(\eta) - P_{n+1}(\eta) \right)
[/tex]
(it is noteworthy that in this case, both w and v are in fact polynomials in [itex]\rho, \, \eta[/itex]), while the other gives rigidly rotating stationary axisymmetric dust solutions which are asympotically flat,
[tex]
w(\rho,\eta) = \frac{(n+1)^2}{\rho^n} \, \left ( \eta \, P_n(\eta) - P_{n+1}(\eta) \right)
[/tex]

This is the same family of stationary axisymmetric rigidly rotating dust solutions as before, but with the solution space parameterized in a different way, so to speak, since the Backlund transformation we are using here is not the same as the one we used last time. Note that for constructing asympotically vanishing solutions, spherical coordinates are more useful than cylindrical ones.

Exercise: determine w, v for some small values of n, and transform to a polar spherical canonical chart. What can you say about coordinate "spheres" [itex]\rho=\rho_0[/itex]? The variation of the density of the dust over a coordinate sphere? A geometric "sphere" (locus of constant radial arc length)? Are either of these true spheres in the sense of having constant Gaussian curvature?

Exercise: the solutions we gave for the axisymmetric Laplace equation form an orthogonal set wrt a suitable inner product on the solution space. What about the induced solutions of our master equation?

Exercise: solve the master equation by separation of variables. Do the solutions form an orthogonal set wrt a suitable inner product? Can you introduce boundary conditions and apply the Sturm-Liouville theory?

Exercise: what happens if you try to extend our dust solution to a perfect fluid solution with nonzero pressure?

I would like to stress that while the application to general relativity is interesting, we are really talking about techniques which apply to a variety of PDEs! For example, Backlund transformations are important in the theory of solitons.
 
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FAQ: Relating PDEs: Showing Laplace's Eq in Cylindrical Coordinates

1) What is Laplace's equation?

Laplace's equation is a second-order partial differential equation that describes the steady-state behavior of a scalar field. In physics, it is commonly used to describe the distribution of heat, electric potential, and fluid flow.

2) What are cylindrical coordinates?

Cylindrical coordinates are a three-dimensional coordinate system that uses the distance from the origin, the angle from a fixed reference direction, and the height above a reference plane to specify the position of a point.

3) How do you show Laplace's equation in cylindrical coordinates?

Laplace's equation in cylindrical coordinates can be derived by using the chain rule to convert the Laplacian operator from Cartesian coordinates to cylindrical coordinates. This results in a differential equation with terms involving the radial and azimuthal derivatives of the scalar field.

4) What is the significance of Laplace's equation in physics?

Laplace's equation is significant in physics because it is a fundamental equation that describes the behavior of many physical systems. It is used to model various phenomena such as heat transfer, electrostatics, and fluid flow, making it an essential tool in many areas of physics and engineering.

5) Are there other coordinate systems where Laplace's equation can be shown?

Yes, Laplace's equation can also be shown in spherical, polar, and elliptical coordinates. Each coordinate system has its own unique form of Laplace's equation, which is derived in a similar manner using the chain rule to convert the Laplacian operator.

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