- #1

- 2,355

- 10

**BRS: Static Axisymmetric "Gravitationless" Massless Scalar Field Solutions**

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

Code:

`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.

Consider the frame field

[tex]

\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}

[/tex]

where the massless scalar field is [itex]\zeta = \sqrt{2} \, w[/itex]. Here, w and the metric function u are functions of z,r only, such that:

[tex]

\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}}

[/tex]

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 [itex]\zeta = \sqrt{2} \, w[/itex] 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:

[tex]

ds^2 = -dt^2 + \exp(2u) \, (dz^2 + dr^2) + r^2 \, d\phi^2

[/tex]

This is valid on a region which is some open subset of

[tex]

-\infty < t, \, z < \infty, \; 0 < r < \infty, \; -\pi < \phi < \pi

[/tex]

We have a two dimensional abelian Lie algebra of Killing vector fields

[tex]

\partial_t, \; \partial_\phi

[/tex]

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

[tex]

\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}

[/tex]

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 [itex]\zeta[/itex] satisifies [itex]\Box \zeta = 0[/itex], 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

[tex]

\nabla_{\vec{e}_1} \vec{e}_1 = 0

[/tex]

so it is inertial; the Fermi derivatives of the spatial unit vector fields [itex]\vec{e}_2, \ldots \vec{e}_4[/itex] along [itex]\vec{e}_1[/itex] 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 [itex]\vec{X} = \vec{e}_1[/itex] for notational convenience, the Bel decomposition of the Riemann tensor wrt X is in general

[tex]

\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}

[/tex]

(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,

[tex]

\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}

[/tex]

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.

Last edited: