- #1

Chris Hillman

Science Advisor

- 2,345

- 8

The simplest frame Ansatz consistent with Hubble's discovery is:

[tex]

\begin{array}{rcl}

\vec{e}_1 & = & \partial_t - f \, \left( x \, \partial_x + y \, \partial_y + z \, \partial_z \right) \\

\vec{e}_2 & = & \partial_x \\

\vec{e}_3 & = & \partial_y \\

\vec{e}_4 & = & \partial_z

\end{array}

[/tex]

where f is a function of t only. This says:

[tex]

\partial_\rho = \frac{1}{\sqrt{x^2+y^2+z^2}} \; \left( x \, \partial_x + y \, \partial_y + z \, \partial_z \right)

[/tex]

so that

[tex]

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

[/tex]

is a vector field pointing away from the origin, with magnitude [itex]\rho[/itex]. Needless to say, the rate factor here is Hubble's "constant", which depends on time but is constant over each spatial hyperslice.

Now, computing the Einstein tensor we find we have a perfect fluid, and the condition that the pressure vanish gives a simple ODE which we can easily solve, finding [itex]f = 2/3/t[/itex].

To verify this, we need to first find the dual coframe, which only requires some trivial linear algebra:

[tex]

\begin{array}{rcl}

\sigma^1 & = & -dt \\

\sigma^2 & = & dx - f \, x \, dt \\

\sigma^3 & = & dy - f \, y \, dt \\

\sigma^4 & = & dz - f \, z \, dt

\end{array}

[/tex]

Now, we verify the claims above using Maxima! Here is a file you can run in wxmaxima's batchmode:

Next, you can modify the above file to replace f with the solution we just found, [itex]f=2/3/t[/itex].

So we have a dust solution with line element

[tex]

\begin{array}{rcl}

ds^2 & = & -\left( 1- \frac{4 \, (x^2+y^2+z^2)}{9 \, t^2} \right) \; dt^2

- \frac{4 \, dt}{3 \, t} \; \left( x \, dx + y \, dy + z \, dz \right)

+ dx^2+dy^2+dz^3 \\

&& 0 < t < \infty, \; -\infty < x, \, y, \, z < \infty

\end{array}

[/tex]

The frame of observers riding on dust particles is

[tex]

\begin{array}{rcl}

\vec{e}_1 & = & \partial_t - \frac{2}{3 \, t} \; \left( x \, \partial_x + y \, \partial_y + z \, \partial_z \right) \\

\vec{e}_2 & = & \partial_x \\

\vec{e}_3 & = & \partial_y \\

\vec{e}_4 & = & \partial_z

\end{array}

[/tex]

We know this because the Einstein tensor vanishes except for a single component, which also shows that the density of the dust, as measured by our observers, is

[tex]

\mu = \frac{1}{6 \pi \, t^2}

[/tex]

The Riemann tensor has two triple eigenvalues, [itex]4/9/t^2, \; -2/9/t^2[/itex], which shows that we have a strong scalar curvature singularity at t=0. This locus is of course familiarly known as "the Big Bang".

The world lines of the dust particles are the integral curves of our timelike unit vector field [itex]\vec{e}_1[/itex], namely

[tex]

x = x_0 \, (t/t_0)^{2/3}, \;

y = y_0 \, (t/t_0)^{2/3}, \;

z = z_0 \, (t/t_0)^{2/3}

[/tex]

It is a good idea to draw some of them and to "decorate" them with some light cones (drawn to scale) at some events on these world lines. Note that the light cones appear to be sheared radially outwards, with (for a given time [itex]t=t_0[/itex], the shear being larger for larger radii [itex]\rho = \sqrt{x^2+y^2+z^2}[/itex]. (See sketch below.) Note too that the volume form is

[tex]

dt \wedge dx \wedge dy \wedge dz

[/tex]

the same as in flat spacetime, so our chart correctly represents volumes and distances within each constant time slice, but at least some timelike geodesics are not represented as coordinate lines, and due to the shearing of the light cones, Lorentzian "angles" are not correctly represented (we say that this is not a conformal chart). Because of the "cross-terms", is also not an orthogonal coordinate chart.

If you use GRTensorII running under Maple, you can easily find the Killing vector fields of this solution. It turns out that we have a six dimensional Lie algebra of Killing vector fields, consisting of three "infinitesimal rotations"

[tex]

-y \, \partial_x + x \, \partial_y \; -z \, \partial_y + y \, \partial_z, \; -x \, \partial_z + z \, \partial_x

[/tex]

and three "infinitesimal translations"

[tex]

t^{2/3} \, \partial_x, \; t^{2/3} \, \partial_y, \; t^{2/3} \, \partial_z

[/tex]

In other words, the isometry group of our manifold is the six dimensional euclidean group, a Lie group which acts transitively on each spatial hyperslice t=t_0. In the next post, we'll use this information to solve the geodesic equations.

Again using GRTensorII (or in a pinch, computing by hand as per Flanders, Applications of Differential Forms to the Physical Sciences), you can show that the expansion tensor of the timelike congruence generated by our timelike unit vector field [itex]\vec{e}_1[/itex] is

[tex]

{H\left[ \vec{e}_1 \right]}_{ab} = \frac{2}{3 \, t} \; \operatorname{diag} \left( 1,1,1 \right)

[/tex]

shows uniform expansion, as expected--- in particular, while in our chart the geodesic [itex]x=y=z=0[/itex] appears to play the role of a dust particle from which all the others are expanding, in fact the expansion tensor and the action by the euclidean group shows that all dust particles are equivalent in this solution. Also, the acceleration and vorticity vectors vanish. (The acceleration vector of a dust particle always vanishes in a dust solution, so this result was expected.) The electroriemann tensor (aka tidal tensor) is

[tex]

{E\left[ \vec{e}_1 \right]}_{ab} = \frac{2}{9 \, t^2} \; \operatorname{diag} \left( 1,1,1 \right)

[/tex]

which shows uniform tidal compression--- the gravitational attraction of the dust inside a small sphere of dust particles slows the expansion, so this is also, qualitatively, the expected result. The magnetoriemann tensor vanishes, as we should expect because the dust particles are not swirling about each other, so there is no "rotating matter" in that sense.

You might recognize that we have been using a Painleve type chart for the

Figure: Sketch of some world lines of dust particles and some light cones drawn to scale.

[tex]

\begin{array}{rcl}

\vec{e}_1 & = & \partial_t - f \, \left( x \, \partial_x + y \, \partial_y + z \, \partial_z \right) \\

\vec{e}_2 & = & \partial_x \\

\vec{e}_3 & = & \partial_y \\

\vec{e}_4 & = & \partial_z

\end{array}

[/tex]

where f is a function of t only. This says:

- we have a family of observers whose world lines are the integral curves of the timelike unit vector field [itex]\vec{e}_1[/itex],
- the coordinate line x=y=z=0 is one of these world lines,
- the coordinate planes [itex]t=t_0[/itex] are spatial hyperslices orthogonal to the world lines, and they have euclidean geometry,
- the time coordinate t corresponds to elapsed proper time according to clocks carried by our observers,
- the other world lines are curves which expand from the origin at a rate proportional to distance [itex]\rho = \sqrt{x^2+y^2+z^2}[/itex], and also depending on time t via the undetermined function f.

[tex]

\partial_\rho = \frac{1}{\sqrt{x^2+y^2+z^2}} \; \left( x \, \partial_x + y \, \partial_y + z \, \partial_z \right)

[/tex]

so that

[tex]

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

[/tex]

is a vector field pointing away from the origin, with magnitude [itex]\rho[/itex]. Needless to say, the rate factor here is Hubble's "constant", which depends on time but is constant over each spatial hyperslice.

Now, computing the Einstein tensor we find we have a perfect fluid, and the condition that the pressure vanish gives a simple ODE which we can easily solve, finding [itex]f = 2/3/t[/itex].

To verify this, we need to first find the dual coframe, which only requires some trivial linear algebra:

[tex]

\begin{array}{rcl}

\sigma^1 & = & -dt \\

\sigma^2 & = & dx - f \, x \, dt \\

\sigma^3 & = & dy - f \, y \, dt \\

\sigma^4 & = & dz - f \, z \, dt

\end{array}

[/tex]

Now, we verify the claims above using Maxima! Here is a file you can run in wxmaxima's batchmode:

Code:

```
/*
Derive the FRW dust with E^3 hyperslices!
Simplest Anstaz consistent with Hubble law is
e_1 = @_t + f(t) ( x @_x + y @_y + z @_x)
e_2 = @_x
e_3 = @_y
e_4 = @_z
Computing the Einstein tensor, we find this gives a perfect fluid.
We solve a simple ODE to obtain a dust solution.
*/
load(ctensor);
cframe_flag: true;
ratchristof: true;
ctrgsimp: true;
/* define the dimension */
dim: 4;
/* list the coordinates */
ct_coords: [t,x,y,z];
/* variables */
depends(f,t);
/* define background metric */
lfg: ident(4);
lfg[1,1]: -1;
/* define the coframe */
fri: zeromatrix(4,4);
fri[1,1]: -1;
fri[2,1]: -x*f;
fri[2,2]: 1;
fri[3,1]: -y*f;
fri[3,3]: 1;
fri[4,1]: -z*f;
fri[4,4]: 1;
/* setup the spacetime definition */
cmetric();
/* display matrix whose rows give coframe covectors */
fri;
/* compute a matrix whose rows give frame vectors */
fr;
/* metric tensor g_(ab) */
lg;
/* compute g^(ab) */
ug: 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);
cdisplay(ein);
/* solve */
ode2(ein[2,2],f,t);
solve([%],[f]);
subst(0,%c,%);
```

So we have a dust solution with line element

[tex]

\begin{array}{rcl}

ds^2 & = & -\left( 1- \frac{4 \, (x^2+y^2+z^2)}{9 \, t^2} \right) \; dt^2

- \frac{4 \, dt}{3 \, t} \; \left( x \, dx + y \, dy + z \, dz \right)

+ dx^2+dy^2+dz^3 \\

&& 0 < t < \infty, \; -\infty < x, \, y, \, z < \infty

\end{array}

[/tex]

The frame of observers riding on dust particles is

[tex]

\begin{array}{rcl}

\vec{e}_1 & = & \partial_t - \frac{2}{3 \, t} \; \left( x \, \partial_x + y \, \partial_y + z \, \partial_z \right) \\

\vec{e}_2 & = & \partial_x \\

\vec{e}_3 & = & \partial_y \\

\vec{e}_4 & = & \partial_z

\end{array}

[/tex]

We know this because the Einstein tensor vanishes except for a single component, which also shows that the density of the dust, as measured by our observers, is

[tex]

\mu = \frac{1}{6 \pi \, t^2}

[/tex]

The Riemann tensor has two triple eigenvalues, [itex]4/9/t^2, \; -2/9/t^2[/itex], which shows that we have a strong scalar curvature singularity at t=0. This locus is of course familiarly known as "the Big Bang".

The world lines of the dust particles are the integral curves of our timelike unit vector field [itex]\vec{e}_1[/itex], namely

[tex]

x = x_0 \, (t/t_0)^{2/3}, \;

y = y_0 \, (t/t_0)^{2/3}, \;

z = z_0 \, (t/t_0)^{2/3}

[/tex]

It is a good idea to draw some of them and to "decorate" them with some light cones (drawn to scale) at some events on these world lines. Note that the light cones appear to be sheared radially outwards, with (for a given time [itex]t=t_0[/itex], the shear being larger for larger radii [itex]\rho = \sqrt{x^2+y^2+z^2}[/itex]. (See sketch below.) Note too that the volume form is

[tex]

dt \wedge dx \wedge dy \wedge dz

[/tex]

the same as in flat spacetime, so our chart correctly represents volumes and distances within each constant time slice, but at least some timelike geodesics are not represented as coordinate lines, and due to the shearing of the light cones, Lorentzian "angles" are not correctly represented (we say that this is not a conformal chart). Because of the "cross-terms", is also not an orthogonal coordinate chart.

If you use GRTensorII running under Maple, you can easily find the Killing vector fields of this solution. It turns out that we have a six dimensional Lie algebra of Killing vector fields, consisting of three "infinitesimal rotations"

[tex]

-y \, \partial_x + x \, \partial_y \; -z \, \partial_y + y \, \partial_z, \; -x \, \partial_z + z \, \partial_x

[/tex]

and three "infinitesimal translations"

[tex]

t^{2/3} \, \partial_x, \; t^{2/3} \, \partial_y, \; t^{2/3} \, \partial_z

[/tex]

In other words, the isometry group of our manifold is the six dimensional euclidean group, a Lie group which acts transitively on each spatial hyperslice t=t_0. In the next post, we'll use this information to solve the geodesic equations.

Again using GRTensorII (or in a pinch, computing by hand as per Flanders, Applications of Differential Forms to the Physical Sciences), you can show that the expansion tensor of the timelike congruence generated by our timelike unit vector field [itex]\vec{e}_1[/itex] is

[tex]

{H\left[ \vec{e}_1 \right]}_{ab} = \frac{2}{3 \, t} \; \operatorname{diag} \left( 1,1,1 \right)

[/tex]

shows uniform expansion, as expected--- in particular, while in our chart the geodesic [itex]x=y=z=0[/itex] appears to play the role of a dust particle from which all the others are expanding, in fact the expansion tensor and the action by the euclidean group shows that all dust particles are equivalent in this solution. Also, the acceleration and vorticity vectors vanish. (The acceleration vector of a dust particle always vanishes in a dust solution, so this result was expected.) The electroriemann tensor (aka tidal tensor) is

[tex]

{E\left[ \vec{e}_1 \right]}_{ab} = \frac{2}{9 \, t^2} \; \operatorname{diag} \left( 1,1,1 \right)

[/tex]

which shows uniform tidal compression--- the gravitational attraction of the dust inside a small sphere of dust particles slows the expansion, so this is also, qualitatively, the expected result. The magnetoriemann tensor vanishes, as we should expect because the dust particles are not swirling about each other, so there is no "rotating matter" in that sense.

You might recognize that we have been using a Painleve type chart for the

*expanding*FRW dust with E^3 hyperslices orthogonal to the world lines of the dust particles. It is not hard to immediately write down the analogous chart for the*contracting*FRW dust with E^3 hyperslices, which can be matched across a contracting sphere of dust particles to a region of the Schwarzschild vacuum to obtain the famous Oppenheimer-Snyder collapsing dust ball solution, which models the formation of a black hole by a dust (pressure-free fluid) which undergoes complete gravitational collapse. This can be generalized to LTB dusts and to various fluid solutions, in addition to solutions exhibiting further possibilities for the stress-energy tensor, such as contributions from a hypothetical minimally coupled massless scalar field or a Lambda term. We can also pass from E^3 hyperslices to S^3 or H^3 hyperslices, and we can take discrete quotients (for example, we can obtain a model with T^3 hyperslices by quotienting the solution we have been discussing).Figure: Sketch of some world lines of dust particles and some light cones drawn to scale.

#### Attachments

Last edited: