BRS: Static Axisymmetric Gravitationless Massless Scalar Field Solutions

[Chris Hillman]

BRS: Static Axisymmetric "Gravitationless" Massless Scalar Field Solutions

This thread is an (easy and amusing) companion to a previous BRS, "The Weyl Vacuums"

www.physicsforums.com/showthread.php?t=378662

I. The Family of "Gravitationless" Solutions

I will describe a family of static axisymmetric minimally coupled massless scalar field (mcmsf) solutions which are "gravitationless" in the sense that

• the Riemann curvature is "purely spatial"
• there is a family of inertial static observers; the congruence of their world lines has vanishing acceleration vector, expansion tensor, and vorticity tensor.

Mcmsf solutions are exact solutions in which the only contribution to the stress tensor comes from the field energy of a massless scalar field, which is minimally coupled to curvature. Sad to say, at present, there are no known physical fields which answer to the theory of classical relativistic massless scalar fields, but they are often used for pedagogical purposes.

Consider the frame field
$$\begin{array}{rcl} \vec{e}_1 & = & \partial_t \\ \vec{e}_2 & = & \exp(-u) \\ \vec{e}_3 & = & \exp(-u) \\ \vec{e}_4 & = & \frac{1}{r} \; \partial_\phi \end{array}$$
where the massless scalar field is $\zeta = \sqrt{2} \, w$. Here, w and the metric function u are functions of z,r only, such that:
$$\framebox{\begin{array}{rcl} 0 &= & w_{zz} + w_{rr} + \frac{w}{r} \\ u_z & = & 2 r \, w_z \, w_r \\ u_r & = & r \, \left( u_r^2 - u_z^2 \right) \end{array}}$$
To solve this system of nonlinear coupled PDEs, you should first choose any axisymmetric static harmonic function for w. This then determines the massless scalar field and also determines by quadrature the metric function u.

In case you are wondering why I wrote $\zeta = \sqrt{2} \, w$ in defining these solutions: the reason is that I wanted the PDEs to bear an obvious close relationship to the PDEs which define the Weyl vacuums (and some other close relatives of that family of exact solutions).

The line element in our Weyl canonical chart is very simple:
$$ds^2 = -dt^2 + \exp(2u) \, (dz^2 + dr^2) + r^2 \, d\phi^2$$
This is valid on a region which is some open subset of
$$-\infty < t, \, z < \infty, \; 0 < r < \infty, \; -\pi < \phi < \pi$$
We have a two dimensional abelian Lie algebra of Killing vector fields
$$\partial_t, \; \partial_\phi$$
Exponentiating these "infinitesimal rigid motions" gives respectively time translation and rotation about the symmetry axis r=0. This shows these solutions are indeed static axisymmetric. (Some examples of solutions in this family have a larger Lie algebra of infinitesimal symmetries; see the next two posts.)

The Laplace-Beltrami operator (generalized wave operator) on our spacetime turns out to be
$$\Box = -\partial_t^2 + \exp(-2u^2) \; \left( \partial_z^2 + \partial_r^2 + \frac{\partial_r}{r} \right) + \frac{\partial_\phi^2}{r&2}$$
Thus, as happens for the Weyl vacuums, a time-independent function is axisymmetric harmonic in our spacetime, written using the canonical chart, if and only if it is axisymmetric harmonic on E^3, written in a cylindrical chart. In particular, our massless scalar field $\zeta$ satisifies $\Box \zeta = 0$, which is the field equation for a massless scalar field. Because the contribution this field makes to the stress-energy tensor (see Hawking and Ellis) matches the Einstein tensor of our Lorentzian manifold, we do indeed have an exact mcmsf solution.

The given frame obviously corresponds to static observers, and it has
$$\nabla_{\vec{e}_1} \vec{e}_1 = 0$$
so it is inertial; the Fermi derivatives of the spatial unit vector fields $\vec{e}_2, \ldots \vec{e}_4$ along $\vec{e}_1$ also vanish, so this is in fact an inertial non-spinning frame (closest analog in curved spacetime to a "Lorentz frame" in flat spacetime!). The expansion and vorticity of the congruence of world lines of static observers vanishes identically. That should remind you of a congruence of comoving inertial observers in Minkowski vacuum!--- our static observers comprise a kind of curved spacetime analog of the elementary notion of a "Lorentz frame" from special relativity.

Recall that, writing $\vec{X} = \vec{e}_1$ for notational convenience, the Bel decomposition of the Riemann tensor wrt X is in general
$$\begin{array}{rcl} {E\left[\vec{X}\right]}_{ab} & = & R_{ambn} \, X^m \, X^n \\ &&\\ {B\left[\vec{X}\right]}_{ab} & = & {{}^\ast \! R}_{ambn} \, X^m \, X^n \\ &&\\ {L\left[\vec{X}\right]}_{ab} & = & {{}^\ast \! R^\ast}_{ambn} \, X^m \, X^n \end{array}$$
(left and right Hodge duals denoted by asterisks). These three dimensional tensors are respectively symmetric, traceless, and symmetric, so they have respectively 6,8,6 algebraically independent components, which accounts for all 20 algebraically independent components of the Riemann tensor. The three tensors describe respectively tidal accelerations of nearby nonspinning test particles, spin-spin forces on gyroscopes, and purely spatial curvature effects. The Bel decomposition is the gravitational analog of the familiar decomposition of the EM field tensor (two-form) into electric and magnetic vector fields; both decompositions are defined wrt some given family of observers (a given timelike congruence).

In our zero-g mcmsf solution,
$$\begin{array}{rcl} E\left[\vec{X}\right]}_{ab} & = & 0 \\ &&\\ {B\left[\vec{X}\right]}_{ab} & = & 0 \\ &&\\ {L\left[\vec{X}\right]}_{ab} & = & \left[ \begin{array}{ccc} -(w_r^2-w_z^2) \, \exp(-2u) & 2 \, w_z \, w_r \exp(-2u) & 0 \\ 2 \, w_z \, w_r \exp(-2u) & (w_r^2-w_z^2) \, \exp(-2u) & 0 \\ 0 & 0 & -(w_r^2+w_z^2) \, \exp(-2u) \end{array} \right] \end{array}$$
The last represents "purely spatial curvature"; the first two show that the electroriemann tensor (tidal tensor) and magnetoriemann tensor vanish identically. So: no tidal accelerations of nonspinning test particles, and no spin-spin forces on gyroscopes.

As with Weyl vacuums, at first glance there appears to be a one-one correspondence between axisymmetric static harmonic functions and solutions from our family. As we should expect from the example of the Weyl vacuums (where you recall that Minkowski vacuum shows up in more than one way), this expectation turns out to be somewhat naive.

The Weyl tensor is Petrov D, if you are keeping track.

Exercise: show that every asymptotically vanishing axisymmetric harmonic function yields an asymptotically flat solution in this family. Show that such a solution has vanishing Komar mass and spin (about r=0).

In the next two posts, I will study two very simple but amusing examples of solutions in this family. The first is cylindrically symmetrical, and the second is spherically symmetrical.

 

[Chris Hillman]

BRS: Static Axisymmetric "Gravitationless" Mcmsf Slns. II. Cylindrically Symmetric

II. A Cylindrically Symmetric Example

The simplest example of a solution in this family is obtained by choosing w to be the planar symmetric harmonic function $w = a \, z$. Then we obtain a very simple solution with line element
$$ds^2 = -dt^2 + \exp(a^2 r^2) (dz^2+dr^2) + r^2 \, d\phi^2$$
with massless scalar field
$$\zeta = \sqrt{2} \, a \, z$$
Notice a remarkable feature: the metric depends only on r, while the scalar field depends only on z!

This spacetime has a three dimensional abelian Lie algebra of Killing vector fields
$$\partial_t, \; \partial_z, \; \partial_\phi$$
corresponding respectively to time translation, axial translation along r=0, and rotation about r=0. Thus, although this solution arose by choosing a plane symmetric harmonic function, the resulting spacetime is static cylindrically symmetric, as promised.

The frame of the static observers is
$$\begin{array}{rcl} \vec{e}_1 & = & \partial_t \\ \vec{e}_2 & = & \exp(a^2 \, r^2/2) \; \partial_z \\ \vec{e}_3 & = & \exp(a^2 \, r^2/2) \; \partial_r \\ \vec{e}_4 & = & \frac{1}{r} \; \partial_\phi \end{array}$$

The characteristic polynomial of the Riemann tensor is very simple and yields the eigenvalues

• triple: zero
• double: $a^2 \, \exp(a^2 r^2)$
• simple: $-a^2 \, \exp(a^2 r^2)$

So the curvature is bounded at r=0 but grows rapidly as r increases.

Exercise: using Noetherian magic, solve the geodesic equations. Show that every null geodesic fits into a null geodesic congruence of form
$$\vec{\ell} = E \; \partial_t + A \; \exp(a^2 r^2) \; \partial_z \pm \sqrt{\exp(a^2 r^2) (E^2-L^2/r^2) - A^2 \, \exp(2a^2r^2)} \; \partial_r + \frac{L}{r^2} \; \partial_\phi$$
Describe the geometry of this congruence of null geodesics. What is the physical interpretation of the constants of motion E, A, L shared by all the geodesic curves in this congruence?

Because our mcmsf solution is well behaved at r=0 but has curvature diverging as r grows, it is natural to try to match across some cylindrical surface r=r_0 to some other static cylindrically symmetric spacetime. Suppose that we want to match a portion of this solution across a cylindrical shell at $r=r_0$ to a portion of Levi-Civita's cylindrically symmetric static vacuum solution
$$ds^2 = -r^{4m} \; dt^2 + r^{-4m \, (1-2m)} (dz^2 + dr^2) + r^{2 \, (1-2m)} \; d\phi^2$$
Is such a matching possible, and if so what conditions on m and r_0 are needed? The data we require to answer these questions are:

• for our mcmsf solution, the induced metric on the -++ signature slice r =r_0 is
  $$d\sigma^2 = -dt^2 + \exp(-a^2 r_0^2) \, dz^2 + r_0^2 \,d\phi^2$$
  with coordinate basis components of extrinsic curvature tensor:
  $$K_{ab} = \left[ \begin{array}{ccc} 0 & 0 & 0 \\ 0 & -a^2 r_0 \, \exp(-a^2 r_0^2/2) & 0 \\ 0 & 0 & r_0 \, \exp(a^2 r_0^2/2) \end{array} \right]$$
• for the Levi-Civita static cylindrically symmetric vacuum, the induced metric on the -++ signature slice r =r_0 is
  $$d\bar{\sigma}^2 = -r_0^{4m} \; dt^2 + r_0^{-4m \, (1-2m)} \; dr^2 + r_0^{2 \, (1-2m)} \; d\phi^2$$
  with coordinate basis components of extrinsic curvature tensor:
  $$\bar{K}_{ab} = \left[ \begin{array}{ccc} -2m \; r_0^{-(1-6m+4m^2)} & 0 & 0 \\ 0 & -2m \, (1-2m) \, r_0^{-(1-2m-4m^2)} & 0 \\ 0 & 0 & (1-2m) \, r_0^{1-2m-4m^2} \end{array} \right]$$

To satisfy the Darmois matching conditions the induced metrics should agree, and either the extrinsic curvatures should also agree, or else we must introduce a material shell at r=r_0 and then the difference between the extrinsic curvatures will then indicate the stress tensor in the material shell. (See Poisson, A Relativist's Toolkit for a very nice introduction to matching conditions.)

We see that to match the induced metrics at r=r_0, we must take m=0 in the vacuum region (i.e. Minkowski vacuum written in a cylindrical chart), and then we must have a thin material shell at r=r_0. But then this thin material shell must have vanishing mass density but nonzero stresses, which we reject as unphysical. Therefore, our mcmsf solution cannot be matched across a static cylindrical surface to a static cylindrically symmetric vacuum.

Excercise: find a charged generalization of Levi-Civita's vacuum solution and show that we cannot match to a portion of this static cylindrically symmetric non-null electrovacuum either.

 

[Chris Hillman]

BRS: Static Axisymmetric "Gravitationless" Mcmsf Slns. III. Spherically Symmetric

III. The Spherically Symmetric Example

If you recall some curious features explored in the "BRS: The Weyl Vacuums", you will not be surprised that choosing the spherically symmetric harmonic function
$$w=\frac{b}{\sqrt{z^2+r^2}}$$
does not yield a spherically symmetrical solution in our family!

Such a solution does exist, but to find it, I think it will be easier to start over with a new spherically symmetrical static Ansatz. The result is the remarkably simple line element
$$ds^2 = -dt^2 + \frac{dr^2}{1+b/r^2} + r^2 \; \left( d\theta^2 + \sin(\theta)^2 \; d\phi^2 \right), \; \; -\infty < t < \infty, \; 0 < r < \infty, \; 0 < \theta < \pi, \; -\pi < \phi < \pi$$
where we are of course using a Schwarzschild style chart, with Schwarzschild radial coordinate r. The massless scalar field is
$$\zeta = \sqrt{2} \, \operatorname{arctan} \left( \frac{1}{\sqrt{1+b/r^2}} \right)$$
where we must require b > 0 in order to obtain a positive energy density for the massless scalar field. Then the contribution of this field to the stress-energy tensor matches the Einstein tensor
$$G^{ab} = \frac{b}{r^4} \; \operatorname{diag}(1,1,-1,-1)$$
where the sign pattern is diagnostic of a massless scalar field solution, and where--- this almost goes without saying!--- the components are evaluated wrt the frame field corresponding to a family of static observers:
$$\begin{array}{rcl} \vec{e}_1 & = & \partial_t \\ \vec{e}_2 & = & \sqrt{1+b/r^2} \; \partial_r \\ \vec{e}_3 & = & \frac{1}{r} \; \partial_\theta \\ \vec{e}_4 & = & \frac{1}{r \, \sin(\theta)} \; \partial_\phi \end{array}$$

This spacetime has a four dimensional Lie algebra of Killing vector fields
$$\partial_t, \; \partial_\phi, \; \; \cos(\phi) \; \partial_\theta - \cot(\theta) \, \sin(\phi) \; \partial_\phi, \; \; \sin(\phi) \; \partial_\theta + \cot(\theta) \, \cos(\phi) \; \partial_\phi$$
corresponding to time translation and three rotations about the origin.

Exercise: using Noetherian magic, solve the geodesic equations.

The characteristic polynomial of the Riemann tensor is remarkably simple:
$$\frac{\xi^3 \; (r^4 \, \xi + b) \; (r^4 \, \xi - b)^2}{r^{12}}$$
which gives the eigenvalues

• triple: zero
• double: b/r^4
• simple: -b/r^4

As happens for all the solutions in the family under discussion, the curvature of this spacetime is "entirely spatial", in fact the Riemann tensor of spacetime is entirely contained in the three-dimensional Riemann tensor of the spatial hyperslices $t=t_0$, which in this example is simply:
$$r_{2323} = r_{2424} = -r_{3434} = b/r^4$$
Notice that we have a strong scalar timelike and naked curvature singularity at r=0.

Exercise: suppress one angular coordinate and embed a hyperslice as a radially symmetric surface in E^3.

Obviously, our spacetime is asymptotically flat, as well as static spherically symmetric so we can compute the Komar mass and spin (about any axis through the origin). The result is that these quantities both vanish. Thus, while our massless scalar field has energy density
$$\varepsilon = \frac{b}{8 \, \pi \, r^4}$$
(strongly concentrated near the origin) our spacetime has no mass or angular momentum as measured at conformal spatial infinity.

It is natural to try to match a portion of this solution across a spherical shell $r=r_0$ to a portion of some other solution. Suppose in particular that we wish to match to a portion of the Schwarzschild vacuum. The data we require are:

• for our mcmsf solution, the induced metric on the hyperslice r=r_0 (signature -++) is
  $$d\sigma^2 = -dt^2 + r_0^2 \; \left( d\theta^2 + \sin(\theta)^2 \; d\phi^2 \right)$$
  while the coordinate basis components of the extrinsic curvature are
  $$K_{ab} = \left[ \begin{array}{ccc} 0 & 0 & 0 \\ 0 & r_0 \; \sqrt{1 + b/r_0^2} & 0 \\ 0 & 0 & r_0 \; \sqrt{1 + b/r_0^2} \; \sin(\theta)^2 \end{array} \right]$$
• for the Schwarzschild vacuum, the induced metric on the slice is
  $$d\bar{\sigma}^2 = -(1-2m/r_0) \, dt^2 + r_0^2 \; \left( d\theta^2 + \sin(\theta)^2 \; d\phi^2 \right)$$
  while the coordinate basis components of the extrinsic curvature tensor are
  $$\bar{K}_{ab} = \left[ \begin{array}{ccc} \frac{-m}{r_0^2} \; \sqrt{1-2m/r_0} & 0 & 0 \\ 0 & r_0 \; \sqrt{1 -2m/r_0} & 0 \\ 0 & 0 & r_0 \; \sqrt{1-2m/r_0} \; \sin(\theta)^2 \end{array} \right]$$

We see immediately see that to match the induced metrics we must take m=0 (i.e. Minkowski vacuum). Then, the extrinsic curvatures show we must have some kind of material shell at $r=r_0$. This shell must have the striking property that it completely confines the minimally coupled massless scalar field to the interior, and this would have to be explained on physical grounds, presumably using the theory of a particular massless scalar field, if such a thing even existed, which apparently is not the case. Furthermore, the stress energy tensor inside this thin spherical shell would have zero mass density but nonzero stresses, which we reject as unphysical. Even worse, we would need to take b < 0 which means that our massless scalar field would have negative energy density in the interior. In a word: nuts. Tant pis!

To be sure, this discussion is somewhat context dependent. I rejected negative energy densities, but they can occur in classical models arising from QFT phenomena (but only in a small region or for a short time), e.g. in the Casimir effect. Indeed, Ellis 1973 introduced this solution with b < 0 as a simple model of a static wormhole. By coincidence, a new arXiv eprint--- only a few hours old as I write!--- discusses the Ellis wormhole, using an arc length chart instead of the Schwarzschild chart; see 1009.6084v1. If b < 0, notice that we require r larger than the "radius" of the throat of the wormhole. If b >0, instead of a static wormhole, we obtain an mcmsf solution with a timelike curvature singularity at r=0.

Those of you who use GRtensor under Maple can easily verify the results mentioned above. For those of you who use Maxima, here is a Ctensor file you can run in batch mode under wxmaxima which verifies the Petrov type of the Weyl tensor, the fact that the electroriemann and magnetoriemann tensors vanish, and the fact that the Einstein tensor has the correct form for a mcmsf solution:

/* 
Aysmpotically flat 
static spherically symmetric 
"gravitationless"
minimally coupled massless scalar field solution
*/
load(ctensor);
cframe_flag: true;
ratchristof: true;
ctrgsimp: true;
/* define the dimension */
dim: 4;
/* list the coordinates */
ct_coords: [t,r,theta,phi];
/* define background metric */
lfg: ident(4);
lfg[1,1]: -1;
/* Declare the dependent and independent variables */
constant(b);
/* Define the coframe covectors */
/* only need enter the nonzero components */
fri: zeromatrix(4,4);
fri[1,1]: -1;
fri[2,2]:  1/sqrt(1+b/r^2);
fri[3,3]:  r;
fri[4,4]:  r*sin(theta);
/* setup the spacetime definition */
cmetric();
/* compute a matrix whose rows give frame vectors */
fr;
/* metric tensor g_(ab) */
lg;
/* compute g^(ab) */
ug: trigsimp(invert(lg));
christof(false);
/* Compute fully covariant Riemann components R_(mijk) = riem[i,k,j,m] */
lriemann(true);
/* Compute R^(mijk) */
uriemann(false);
/* Compute Ricci componets R_(jk) */
ricci(true);
/* Compute trace of Ricci tensor */
tracer;
/* Compute R^(jk) */
uricci(false);
/* Compute and display MIXED Einstein tensor G^a_b */
/* For (-1,1,1,1) sig Flip sign of top row to get G^(ab) */
einstein(false);
/* WARNING! leinstein(false) only works for metric basis! */
/* Compute Kretschmann scalar */
rinvariant();
expand(factor(%));
/* 
Einstein tensor as matrix
NOTE: flip sign of top row because we want G^(ab) 
*/
matrix([-ein[1,1],-ein[1,2],-ein[1,3],-ein[1,4]],
[ein[2,1],ein[2,2],ein[2,3],ein[2,4]],
[ein[3,1],ein[3,2],ein[3,3],ein[3,4]],
[ein[4,1],ein[4,2],ein[4,3],ein[4,4]]);
/* electroriemann tensor */
matrix([lriem[2,2,1,1], lriem[2,3,1,1],lriem[2,4,1,1]],
[lriem[3,2,1,1],lriem[3,3,1,1],lriem[3,4,1,1]],
[lriem[4,2,1,1],lriem[4,3,1,1],lriem[4,4,1,1]]);
/* magnetoriemann tensor */
matrix([lriem[2,4,3,1],lriem[2,2,4,1],lriem[2,3,2,1]],
[lriem[3,4,3,1],lriem[3,2,4,1],lriem[3,3,2,1]],
[lriem[4,4,3,1],lriem[4,2,4,1],lriem[4,3,2,1]]);
/* Construct NP tetrad for our frame, compute Weyl spinors and Petrov type */
nptetrad(true);
weyl(false);
psi(true);
factor(psi[0]);
factor(psi[1]);
factor(psi[2]);
factor(psi[3]);
factor(psi[4]);
petrov();

 

[Chris Hillman]

BRS: Static Axisymmetric "Gravitationless" Mcmsf Slns. IV. Another Babka?

IV. Another Cylindrically Symmetric Example

Suppose we take our axisymmetric harmonic function to be $w= b \, \log r$. Then we obtain
$$ds^2 = -dt^2 + r^{8b^2} \; \left( dz^2 + dr^2 \right) + r^2 \; d\phi^2$$
with scalar field $\zeta = \sqrt{2} \, b \, \log r$. The energy density of the massless scalar field turns out to be
$$\varepsilon = \frac{b^2}{2 \pi \, r^{2 \, (1+4b^2)}}$$
so this solution is physically acceptable--- assuming you are not perturbed by the strong scalar timelike curvature singularity at r=0! (Notice that the scalar field also diverges there.)

This spacetime is static cylindrically symmetric; that is, it admits a three-dimensional abelian Lie algebra of Killing vector fields
$$\partial_t, \; \partial_z, \; \partial_\phi$$
but it is obviously distinct from the static cylindrically symmetric example previously discussed! (Because the radial coordinate has the same geometric meaning in both example, we can easily compare the curvature dependence and we see they are not the same. In addition, we can compare coordinate-free algebraic relations between differential curvature invariants, and find that these show that the two spacetimes are not locally isometric.)

Can we match this to the Levi-Civita static cylindrically symmetric vacuum solution? The data we require to answer this question are:

• for our mcmsf solution, the induced metric on the -++ signature slice r =r_0 is
  $$d\sigma^2 = -dt^2 + r_0^{8 b^2} \; dz^2 + r_0^2 \; d\phi^2$$
  with coordinate basis components of extrinsic curvature tensor:
  $$K_{ab} = \left[ \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 4 b^2 r_0^{-(1-4b^2)} & 0 \\ 0 & 0 & r_0^{1-4b^2} \end{array} \right]$$
• for the Levi-Civita static cylindrically symmetric vacuum, the induced metric on the -++ signature slice r =r_0 is
  $$d\bar{\sigma}^2 = -r_0^{4m} \; dt^2 + r_0^{-4m \, (1-2m)} \; dr^2 + r_0^{2 \, (1-2m)} \; d\phi^2$$
  with coordinate basis components of extrinsic curvature tensor:
  $$\bar{K}_{ab} = \left[ \begin{array}{ccc} -2m \; r_0^{-(1-6m+4m^2)} & 0 & 0 \\ 0 & -2m \, (1-2m) \, r_0^{-(1-2m-4m^2)} & 0 \\ 0 & 0 & (1-2m) \, r_0^{1-2m-4m^2} \end{array} \right]$$

Once again, we find that we must take m=0 (Minkowski vacuum) in the Levi-Civita region, and then we would need to take b=0 even to match the metric. Stymied again!

 

[Chris Hillman]

II. A Cylindrically Symmetric Example. Addendum.

In my Post #2 above, I forgot to mention another striking aspect of this example: the metric (depends only on r, static and cylindrically symmetrical) appears to break the symmetry of the scalar field (depends only on z, static and nominally plane symmetrical). So how was the locus r=0 chosen as distinguished by the gravitational field due to the energy of a nominally planar symmetric massless scalar field?

This conundrum is partially resolved by considering more carefully what "plane symmetric" might mean in a curved manifold.

The Lie algebra of Killing vector fields shows that the spacetime is static (since $\partial_t$ is a timelike hypersurface orthogonal Killing vector field, i.e. if you normalize it, the resulting timelike unit vector field has vanishing vorticity) and cylindrically symmetric (since $\partial_z, \; \partial_\phi$ allows the expected rigid motions within the cylinders $t=t_0, \; r=r_0$).

Compare the Lie algebra of Killing vector fields of a plane symmetric spacetime, such as this example of a Kasner vacuum:
$$ds^2 = -dt^2 + t^{4/3} \; (dx^2+dy^2) + t^{-2/3} \; dz^2, \; \; 0 < t < \infty, \; -\infty < x, \, y, \, z < \infty$$
which is spanned by
$$\partial_z, \; \partial_x, \; \partial_y, \; -y \, \partial_x + x \, \partial_y$$
The last three give the subalgebra e(2), corresponding to the Euclidean group E(2) (generated by translations and rotations) acting on the planes $t=t_0, \; z=z_0$. In cylindrical coordinates, the three generators (two translations and one rotation) become
$$\cos(\phi) \; \partial_r - \frac{\sin(\phi)}{r} \; \partial_\phi, \; \sin(\phi) \; \partial_r + \frac{\cos(\phi)}{r} \; \partial_\phi, \; \partial_\phi$$
But in the cylindrically symmetrical example of a "gravitationless" mcmsf under discussion here, only the last is a Killing vector. I want to say that the Lie algebra of Killing vector fields does not contain a subalgebra e(2), but this gets into something a bit tricky, since locally the cylindrical symmetry resembles the plane symmetry one obtains by "lifting" the (geometrically flat) cylinders to flat two-planes.

Subtleties aside, the point is that despite appearances, the scalar field $\zeta = \sqrt{2} \, a z$ is not plane symmetric in terms of the Lorentzian geometry of our spacetime manifold. The values are indeed constant on the coordinate two-planes $t=t_0, \; z=z_0$, but these surfaces (treated as Riemannian two-manifolds using the metric inherited from the spacetime) are not locally isometric to E^2.

Contrast the cylinders $t=t_0, \; r=r_0$, which are locally isometric to E^2, and globally isometric to ordinary cylinders embedded in E^3, so this manifold is (static and) cylindrically symmetric.

Similarly, in Schwarzschild vacuum, the surfaces $t=t_0, \; r=r_0$ are round spheres, and they are nested, so this Lorentzian manifold is (static and) spherically symmetric.

There's another general lesson lurking here: when solving the EFE via the Ansatz method, one more or less guarantees that the solution will be expressed in a coordinate chart respecting various geometrical symmetries, but this "encourages" further mathematically convenient phenomena, which can however cause momentary confusion, as with the "condundrum" discussed in this post.

 

[Chris Hillman]

III. A Spherically Symmetric Example. (cont'd)

I forgot to mention some important points about the spherically symmetric examples in the family of static axisymmetric mcmsf solutions under discussion:

• The "Ellis" named in the Ellis wormhole is H.G. Ellis, not the mathematical cosmologist G.F.R. Ellis, and K. Bronnikov independently introduced it in the same year, 1973. Later, Morris and Thorne reconsidered it in 1988, whereupon the study "traversable wormholes" took off, and there are now many other models of traversable wormholes, not all involving a mcmsf, but the others employ some kind of exotic matter (e.g. negative mass dust), possibly in addition to mundane stuff like a magnetic field.
• In Part I (my Post #1 above), I discussed a family of static axisymmetric mcmsf solutions, in which our scalar field is real, a solution of the (curved-spacetime) wave equation, and has positive energy density; physicists sometimes call this as a Klein-Gordon field. To obtain a traversable wormhole from this setup, you need to take a pure imaginary scalar field, and then your field has negative energy density. Physicists sometimes call this a "ghost Klein-Gordon field".
• The Penrose diagram for both the static spherically symmetric KG (positive energy density) solution and the static spherically symmetric "ghost KG" (negative energy density) solution (Ellis-Bronnikov-Morris-Thorne wormhole) is easily obtained, and is quite instructive.
• It turns out that the Ellis wormhole is not stable against small perturbations, but some other models are. (But AFAIK no known model has any direct experimental support; all employ some kind of wildly speculative hypothetical field or exotic matter.)
• In writing down the massless scalar field, I should have used b^2 instead of b. This ensures that b has the units of length in geometric units, but requires treating seperately the case of positive and negative energy density--- but that is appropriate because the character of the KG fields and the Penrose diagrams differ.
• The Bel decomposition of the Riemann tensor is mathematically analogous to the usual decomposition of the EM field tensor into electric and magnetic vectors, but we obtain three second rank tensor fields instead of two vector fields. In electromagnetism, the principle Lorentz invariants of the field tensor tell us when we have "no intrinsic magnetism", i.e. when some family of observers measure no magnetic field. In gravitation, analogous invariants tell us when we have "no intrinsic electric/magnetic parts", as happens for the "gravitationless" solutions discussed here. But as the electromagnetic analogy suggests, while some observers experience no tidal effects and no gravitomagnetic effects, in these models most other observers do experience such effects. Just as if one observer measures no magnetic field, another typically will measure a nonzero magnetic field.

The Ellis wormhole is sufficiently interesting that it seems worthwhile to discuss it in more detail. The frame of static observers is
$$\begin{array}{rcl} \vec{e}_1 & = & \partial_t \\ \vec{e}_2 & = & \sqrt{1-b^2/r^2} \; \partial_r \\ \vec{e}_3 & = & \frac{1}{r} \; \partial_\theta \\ \vec{e}_4 & = & \frac{1}{r \, \sin(\theta)} \; \partial_\phi \end{array}$$
where are using a Schwarzschild chart with line element
$$ds^2 = -dt^2 + \frac{dr^2}{1-b^2/r^2} + r^2 \; d\Omega^2, \; \; b < r < \infty$$
Here, r=b is the "throat" of the wormhole. Pop quiz: is this locus null or timelike (-++ signature)?

The characteristic equation of the Riemann tensor is
$$\xi^3 \; (b^2 - \xi \, r^4) \; (b^2 + \xi \, r^4)^2$$
so the eigenvalues are

• triple: zero
• double: $-b^2/r^4$
• simple: $b^2/r^4$

(note the sign change from the positive energy density KG version!). The scalar field is
$$\zeta = i \; \sqrt{2} \; \operatorname{arctan} \left( \frac{1}{\sqrt{r^2/b^2-1}} \right)$$
which satisfies the wave equation and gives a stress-energy tensor matching the Einstein tensor, whose components wrt the given frame field are
$$G^{ab} = \frac{b^2}{r^4} \; \operatorname{diag} (-1,-1,1,1)$$
Note the negative energy density $-b^2/8/\pi/r^4$ measured by our static observers! These observers measure vanishing electroriemann and magnetoriemann tensors, but the toporiemann tensor is nonzero:
$${L\left[\vec{e}_1\right]}_{ab} = \frac{b^2}{r^4} \; \operatorname{diag} (-1,1,1)$$
Our spacetime admits a four-dimensional Lie algebra of Killing vector fields (time translation and three rotations about the origin), so it is indeed static spherically symmetric. By Noetherian magic and some clever organization, we find that every null geodesic belongs to some null geodesic congruence of form
$$\vec{\ell} = E \; \partial_t \, \pm \sqrt{1-b^2/r^2} \, \sqrt{E^2 - \frac{L^2}{r^2}} \; \partial_r \, \pm \frac{\sqrt{L^2-J^2/\sin(\theta)^2}}{r^2} \; \partial_\theta \, + \, \frac{J}{r^2 \, \sin(\theta)} \; \partial_\phi$$
where the "energy of a photon" measured by our static observers is E, and where L is the total angular momentum of the photon while J is the component of angular momentum in the equatorial plane. The given null geodesic congruence consists of the world lines of all photons sharing the constants of motion E, L, J; geometrically they all share the same impact parameter (radius of closest approach).

For comparision, in the Schwarzschild chart on one static exterior sheet of the Schwarzschild vacuum, we would obtain instead
$$\vec{\ell} = \frac{E}{1-\frac{2m}{r}} \; \partial_t \, \pm \sqrt{E^2 - \left( 1-\frac{2m}{r} \right) \; \left( 1-\frac{L^2}{r^2} \right)} \; \partial_r \, \pm \frac{\sqrt{L^2-J^2/\sin(\theta)^2}}{r^2} \; \partial_\theta \, + \, \frac{J}{r^2 \, \sin(\theta)} \; \partial_\phi$$
which has an analogous geometric interpretation.

The principle null geodesic congruences (radially ingoing or outgoing) are obtained when we take $E=1, \; L=J=0$. Then from
$$dt = \frac{\pm dr}{\sqrt{1-b^2/r^2}}$$
we immediately obtain
$$t-t_0 = \pm \sqrt{r^2-b^2}$$
To obtain an analog of the Kruskal-Szekeres chart we need only put
$$p = \frac{t + \sqrt{r^2-b^2}}{\sqrt{2}}, \; q = \frac{t - \sqrt{r^2-b^2}}{\sqrt{2}}$$
which gives the line element
$$ds^2 = -2 \, dp \, dq + \left( b^2 + \frac{(p-q)^2}{2} \right) \; d\Omega^2, \; \; -\infty < p, \, q < \infty$$
This clearly shows that the throat $r=b$ is the timelike sheet $p=q$. Finally, we can compactify the null coordinates by taking
$$u = \operatorname{arctan}(p) = \operatorname{arctan} \left( \frac{t+\sqrt{r^2-b^2}}{\sqrt{2}} \right), \; \; v = \operatorname{arctan}(q) = \operatorname{arctan} \left( \frac{t-\sqrt{r^2-b^2}}{\sqrt{2}} \right)$$
which gives the line element
$$ds^2 = \frac{ 2 \, du\, dv}{\cos(u)^2 \, \cos(v)^2} + r^2 \; d\Omega^2$$
where
$$r = \sqrt{b^2 + \frac{(\tan(u)-\tan(v))^2}{2}}$$
is now considered a metric function (a function of u,v only).

The Penrose diagram (see figure below) shows clearly that light signals can pass from one exterior sheet to the other. In this diagram, as usual, points represents round Riemannian two-spheres. The throat is the timelike locus u=v and the two exterior static regions meet at this locus (left and right "triangles" in the diagram). The world sheets of some spheres of static observers are also depicted.

Figure:

• Penrose diagram for the Ellis wormhole ("ghost" type Klein-Gordon minimally coupled massless scalar field with negative energy density)

 

[Chris Hillman]

III. A Spherically Symmetric Example. (cont'd)

It is instructive to consider the frame of a family of nonspinning inertial observers who are moving radially inward with constant velocity (as measured far from the origin) wrt our static observers:
$$\begin{array}{rcl} \vec{f}_1 & = & \frac{1}{\sqrt{1-v^2}} \; \partial_t - \frac{v}{\sqrt{1-v^2}} \, \sqrt{1-b^2/r^2} \; \partial_r \\ \vec{f}_2 & = & \frac{-v}{\sqrt{1-v^2}} \; \partial_t + \frac{1}{\sqrt{1-v^2}} \, \sqrt{1-b^2/r^2} \; \partial_r \\ \vec{f}_3 & = & \vec{e}_3 = \frac{1}{r} \; \partial_\theta \\ \vec{f}_4 & = & \vec{e}_4 = \frac{1}{r \sin(\theta)} \; \partial_\phi \end{array}$$
With respect to the new frame, the components of the Einstein tensor are
$$G^{ab} = \left[ \begin{array}{cccc} \frac{-b^2}{r^4} \, \frac{1+v^2}{1-v^2} & \frac{-b^2}{r^4} \, \frac{2 v}{1-v^2} & 0 & 0 \\ \frac{-b^2}{r^4} \, \frac{2v}{1-v^2} & \frac{-b^2}{r^4} \, \frac{1+v^2}{1-v^2} & 0 & 0 \\ 0 & 0 & \frac{b^2}{r^4} & 0 \\ 0 & 0 & 0 & \frac{b^2}{r^4} \end{array} \right]$$
The acceleration vector and vorticity tensor of the timelike geodesic congruence obtained from the (proper time parameterized) integral curves of the unit timelike vector field $\vec{f}_1$ both vanish, so this congruence is geodesic and hypersurface orthogonal. The expansion tensor shows convergence orthogonal to the direction of infall:
$${\theta\left[\vec{f}_1\right]}_{ab} = \frac{v}{\sqrt{1-v^2}} \, \frac{\sqrt{1-b^2/r^2}}{r} \; \operatorname{diag} (0,-1,-1)$$
(as we would expect for radial infall!) but no divergence in the radial direction. The electroriemann tensor is
$${E\left[\vec{f}_1\right]}_{ab} = \frac{v}{1-v^2} \, \frac{-b^2}{r^4} \; \operatorname{diag} (0,1,1)$$
which shows no tidal tension in the radial direction. The only algebraically independent component of the magnetoriemann tensor is
$${B\left[\vec{f}_1\right]}_{34} = \frac{v}{1-v^2} \, \frac{b^2}{r^4}$$
and the toporiemann tensor is
$${L\left[\vec{f}_1\right]}_{ab} = \frac{b^2}{r^4} \; \operatorname{diag} (-1,1/(1-v^2),1/(1-v^2))$$
Referring to the ingoing null geodesic congruence obtained in the previous post, the "energy of an ingoing photon" measured by these observers is
$$\frac{E - v \, \sqrt{E^2-L^2/r^2}}{\sqrt{1-v^2}}$$
(our static observers would obtain the value E, irrespective of their r coordinate). For an outgoing photon, change the sign of the radical in the numerator.

More generally, consider the frame field
$$\begin{array}{rcl} \vec{g}_1 & = & \frac{\sqrt{1+f^2}}{\sqrt{1-v^2}} \; \partial_t \, - \, \frac{v \, \sqrt{1+f^2}}{\sqrt{1-v^2}} \, \sqrt{1-b^2/r^2} \; \partial_r + \frac{f}{r \, \sin(\theta)} \; \partial_\phi \\ \vec{g}_2 & = & \frac{-v}{\sqrt{1-v^2}} \; \partial_t \, + \, \frac{1}{\sqrt{1-v^2}} \, \sqrt{1-b^2/r^2} \; \partial_r \\ \vec{g}_3 & = & \vec{e}_3 = \frac{1}{r} \; \partial_\theta \\ \vec{g}_4 & = & \frac{f}{\sqrt{1-v^2}} \; \partial_t - \frac{v \, f \, \sqrt{1-b^2/r^2}}{\sqrt{1-v^2}} \; \partial_r \, + \, \frac{\sqrt{1+f^2}}{r \sin(\theta)} \; \partial_\phi \end{array}$$
where v, f are functions of r only. We can kill all but one component of the acceleration vector of $\vec{g}_1$ by putting
$$v = v_o \; \sqrt{1-J_o^2/r^2}, \; \; f = \frac{v_o}{\sqrt{1-v_o^2}} \, \frac{J_o}{r}$$
and then the remaining component vanishes in the equatorial plane. By the spherical symmetry, this suffices to analyze the physical experience of an arbitrary inertial observer in the Ellis wormhole. Well, sort of: the vorticity tensor is nonzero, and this frame does give our observers nonarbitrary spin. But we can "despin" this frame to obtain the physical experience of an arbitrary nonspinning inertial observer in the Ellis wormhole. The components of the Einstein tensor and other curvature tensors turn out to be fairly complicated, despite the simplicity of this example, so I'll omit them.

The issue of whether or not our observers are spinning is irrelevant to certain measurements, for example the energy of a photon from our null geodesic congruence as measured by arbitrary inertial observers in the Ellis wormhole. The expression we obtain depends on b (the scalar field strength parameter), the parameters $E_p, L_p, J_p$ describing the photon, and the parameters $v_o, J_o$ describing the observer (where, recall that by spherical symmetry we can assume the observer is moving in the equatorial plane).

We can also analyze such familiar effects as light bending (far from the throat), Shapiro time delay, precession of pericenters of orbiting particles, etc., by the same methods used in analyzing the Schwarzschild vacuum.