User:Eric Kvaalen/Notes/A coordinate system for an expanding universe

The usual coordinate system for the universe uses comoving coordinates, meaning that galaxies stay (approximately) at a constant spatial coordinate point, and the age of any galaxy is given by the time coordinate. In this system, the "proper distances" to distant galaxies are increasing with time faster than the speed of light. People think of this as "space expanding". But there's nothing in General Relativity that says whether space "is expanding". So I was wonderin' what would happen if one devised a coordinate system in which space does not "expand". My idea is that any point in space-time can be connected by a geodesic to "here" at some point in time, such that the geodesic is totally spatial where it hits "here". So then the coordinates of that point would be given by the time when the geodesic goes off at infinite speed from here and the distance of the point from here along the geodesic. The distance is "proper distance" in the other sense, namely the invariant integral of distance along the geodesic. One question I had was whether distant galaxies would still seem to be moving faster than light, or indeed whether light may move "faster than the speed of light".

I will develop equations relating my coordinates to hyperspherical coordinates t and x (flat variety, that is, k=0). I use x to mean r along a constant angular direction. First, let's derive the equation for a geodesic in the comoving system. We'll treat the geodesic as a curve of t versus x. The invariant length of some curve $t(x)$ is

$\int {\sqrt {a(t(x))^{2}-(dt/dx)^{2}}}dx$

where $a(t)$ is the scale factor, which for the moment we won't specify. To be a geodesic, this must be invariant to small changes $\delta (x)$ that are added to $t(x)$. The change in the invariant length is given by

$\int {\frac {a(t(x))a'(t(x))\delta (x)-(dt/dx)\delta '(x)}{\sqrt {a(t(x))^{2}-(dt/dx)^{2}}}}dx$

Integration by parts transforms this to

$\int \left[{\frac {a(t(x))a'(t(x))}{\sqrt {a(t(x))^{2}-(dt/dx)^{2}}}}+{\frac {d^{2}t/dx^{2}}{\sqrt {a(t(x))^{2}-(dt/dx)^{2}}}}-{\frac {a(t(x))a'(t(x))(dt/dx)^{2}-(dt/dx)^{2}d^{2}t/dx^{2}}{{\sqrt {a(t(x))^{2}-(dt/dx)^{2}}}^{3}}}\right]\delta (x)dx$

or

$\int {\frac {[a(t(x))^{2}-(dt/dx)^{2}][a(t(x))a'(t(x))+d^{2}t/dx^{2}]-a(t(x))a'(t(x))(dt/dx)^{2}+(dt/dx)^{2}d^{2}t/dx^{2}}{{\sqrt {a(t(x))^{2}-(dt/dx)^{2}}}^{3}}}\delta (x)dx$

The numerator of the fraction must therefore be zero for a geodesic. We can simplify it and divide by $a(t(x))$ to give this equation for a geodesic:

$a(t(x))^{2}a'(t(x))-2a'(t(x))(dt/dx)^{2}+a(t(x))d^{2}t/dx^{2}=0$

or

$d^{2}t/dx^{2}=2{\frac {a'(t(x))}{a(t(x))}}(dt/dx)^{2}-a(t(x))a'(t(x))$

(Note that we can write this as

${\tfrac {1}{2}}{\frac {d(dt/dx)^{2}}{dt}}=2{\frac {a'(t)}{a(t)}}(dt/dx)^{2}-a(t)a'(t)$

which gives a first-order differential equation for $(dt/dx)^{2}$ as a function of t. This gives an alternative route to solution instead of what follows. By the way, if instead of $t$ we had a spatial dimension $y$, our equation would be

$d^{2}y/dx^{2}=2{\frac {a'(y)}{a(y)}}(dy/dx)^{2}+a(y)a'(y).$

or

${\tfrac {1}{2}}{\frac {d(dy/dx)^{2}}{dy}}=2{\frac {a'(y)}{a(y)}}(dy/dx)^{2}+a(y)a'(y)$ Eric Kvaalen (talk) 14:50, 20 January 2022 (UTC))

We can check this for an empty universe having $a(t)=t.$ Then the equation for a geodesic becomes

$d^{2}t/dx^{2}={\frac {2}{t(x)}}(dt/dx)^{2}-t(x)$

This universe corresponds to flat space time with coordinates:

$\xi =t\sinh x$

$\tau =t\cosh x$

The geodesic with τ equal to some constant C is

$t=C/\cosh x.$

One can verify that this satisfies our differential equation for a geodesic.

I will now assume that

$a(t)=t^{\alpha }.$

Using now primes to designate the first and second derivatives of $t(x)$, our equation for a geodesic is:

$t''=2\alpha t^{-1}(t')^{2}-\alpha t^{2\alpha -1}$

Let's now switch to thinking of x as a function of t. Using the fact that

${\ddot {x}}\equiv {\frac {d^{2}x}{dt^{2}}}=-{\frac {d^{2}t}{dx^{2}}}\left({\frac {dx}{dt}}\right)^{3}$

we have:

${\ddot {x}}=\alpha t^{2\alpha -1}{\dot {x}}^{3}-2\alpha t^{-1}{\dot {x}}$

This no longer depends on the dependent variable, so it can be shifted up or down in the x direction.

A light ray moves with speed $dx/dt=1/a(t),$ so (when $\alpha \neq 1$) one solution for a light ray is:

$x(t)={\frac {t^{1-\alpha }}{1-\alpha }}$

To avoid having to write $1-\alpha$, let us designate this as $\beta$ so this light ray becomes

$x(t)=t^{\beta }/\beta .$

The differential equation has another symmetry (again when $\alpha \neq 1$) — if we take a solution to this and multiply each t by a factor and multiply the corresponding x by the factor to the power β, we get another solution. So we can simplify this equation by introducing the variable

$s\equiv t^{\beta }/\beta$

which gives

$t=(\beta s)^{1/\beta }$

${\dot {s}}=t^{-\alpha }=(\beta s)^{-\alpha /\beta }$

or

${\dot {s}}=t^{-\alpha }=(\beta s)^{-\gamma }$

where $\gamma \equiv \alpha /\beta ,$

${\dot {x}}={\dot {s}}{\frac {dx}{ds}}=(\beta s)^{-\gamma }{\frac {dx}{ds}}$

$d{\dot {x}}/ds=(\beta s)^{-\gamma }s^{-1}\left(s{\frac {d^{2}x}{ds^{2}}}-\gamma {\frac {dx}{ds}}\right)$

${\ddot {x}}={\dot {s}}d{\dot {x}}/ds=(\beta s)^{-2\gamma }s^{-1}\left(s{\frac {d^{2}x}{ds^{2}}}-\gamma {\frac {dx}{ds}}\right).$

We can now use our geodesic equation to equate this to something:

$(\beta s)^{-2\gamma }s^{-1}\left(s{\frac {d^{2}x}{ds^{2}}}-\gamma {\frac {dx}{ds}}\right)=\alpha t^{2\alpha -1}{\dot {x}}^{3}-2\alpha t^{-1}{\dot {x}}$

or

$(\beta s)^{-2\gamma }s^{-1}\left(s{\frac {d^{2}x}{ds^{2}}}-\gamma {\frac {dx}{ds}}\right)=\alpha (\beta s)^{-2\gamma -1}\left(\left({\frac {dx}{ds}}\right)^{3}-2{\frac {dx}{ds}}\right).$

This can be simplified to:

${\frac {d^{2}x}{ds^{2}}}=\gamma s^{-1}\left(\left({\frac {dx}{ds}}\right)^{3}-{\frac {dx}{ds}}\right).$

Letting $u\equiv dx/ds$ we can write this simply as:

${\frac {du}{ds}}={\frac {\gamma }{s}}(u^{3}-u)$

We can now solve this for $dx/ds$ by separating variables. We have

$\int {\frac {du}{u^{3}-u}}=\gamma \int {\frac {ds}{s}}$

and it happens that the left integral can be done. We get four solutions (with C an arbitrary positive constant):

$u=0$ (geodesics that don't move with respect to the galaxies)

$u^{2}=1$ (light rays)

$\ln {\sqrt {1-u^{-2}}}=\ln C(\beta s)^{\gamma }$ or $u=\pm (1-C^{2}(\beta s)^{2\gamma })^{-1/2}$ (space-like geodesics arching from $s=0$ up to $s=C^{-1/\gamma }/\beta$ and then back down to zero)

$\ln {\sqrt {u^{-2}-1}}=\ln C(\beta s)^{\gamma }$ or $u=\pm (1+C^{2}(\beta s)^{2\gamma })^{-1/2}$ (time-like geodesics)

At this point our luck runs out and we cannot integrate the last two to find $x(s)$ and $x(t)$ or $t(x).$ But we can think about the solution and answer the question about galaxies moving faster than the speed of light. In a matter-dominated universe, $\alpha =2/3$ and $\gamma =2.$ When α is positive the geodesics that are purely space-like here bend toward the "past" (that is, they encounter galaxies that are younger than ours at the starting point) and if γ is positive they eventually "hit the big bang" (that is, they arrive to $t=0$).

Let us introduce a scaled version of s, namely:

$y\equiv \beta s=t^{\beta }$

Using the fact that

$a(t)=(\beta s)^{\gamma }=y^{\gamma }$

the path length along the geodesic can be written:

$\int {\sqrt {y^{2\gamma }-y^{2\gamma }/u^{2}}}dx=\int y^{\gamma }{\sqrt {u^{2}-1}}ds$

If we take the geodesic which is "here" at time $t=1$ we have

$\int y^{\gamma }{\sqrt {{\frac {1}{1-y^{2\gamma }}}-1}}ds=\int y^{\gamma }{\sqrt {\frac {y^{2\gamma }}{1-y^{2\gamma }}}}ds$

or

${\frac {1}{\beta }}\int y^{2\gamma }{\sqrt {\frac {1}{1-y^{2\gamma }}}}dy$

In order to evaluate this from 0 to 1, let's introduce yet another function of t,

$z\equiv {\sqrt {1-y^{2\gamma +1}}}$

$dy/dz=-{\frac {2}{2\gamma +1}}y^{-2\gamma }{\sqrt {1-y^{2\gamma +1}}}$

This has the magical effect of making the integrand much better behaved, and the integral is now:

${\frac {2}{1+\alpha }}\int {\sqrt {\frac {1-y^{2\gamma +1}}{1-y^{2\gamma }}}}dz={\frac {2}{1+\alpha }}\int {\frac {z}{\sqrt {1-(1-z^{2})^{2\gamma /(2\gamma +1)}}}}dz$

When y is close to 0 and z is close to 1, the integrand is approximately 1, and when y is close to 1 and z is close to 0 the integrand is approximately ${\sqrt {1+1/(2\gamma )}}.$ With positive γ the integrand is always greater than 1, and doing a simple numerical integration in a spreadsheet for the case of $\alpha =2/3$ I found that the path length is about 1.31. That means that the "big bang" and the furthest galaxies are about 18.3 milliard light-years away in my coordinate system, after only 14 milliard years.

To find x as a function of s or t for this geodesic we need to integrate

$x=\int uds={\frac {1}{\beta }}\int (1-y^{2\gamma })^{-1/2}dy$

This time we will use a different z, namely:

$z\equiv {\sqrt {1-y}}={\sqrt {1-t^{\beta }}}$

$dy/dz=-2z$

This gives

$x={\frac {1}{\beta }}\int {\frac {2zdz}{\sqrt {1-y^{2\gamma }}}}={\frac {1}{\beta }}\int {\frac {2zdz}{\sqrt {1-(1-z^{2})^{2\gamma }}}}.$

Numerical integration finds that we get to the big bang at around $x=3.9.$

By the way, we can also use the z defined this way when numerically integrating for the invariant distance, in which case the integral is:

${\frac {2}{\beta }}\int y^{2\gamma }{\sqrt {\frac {1-y}{1-y^{2\gamma }}}}dz={\frac {2}{\beta }}\int (1-z^{2})^{2\gamma }{\frac {z}{\sqrt {1-(1-z^{2})^{2\gamma }}}}dz$

We may also ask how far away light is that started here when the world was very young. The light goes away as $x=s$, so we just need to integrate the path length from $z=0$ down until it equals s. For $\alpha =2/3$ it turns out that this happens when x and s are around 1.9 or 2, and at that point the invariant distance from here is about 1.2. This means that in my coordinate system, this light is "traveling" at about 1.2 times "the speed of light".

In this system, we are at the centre of a sphere of radius about 18 milliard light-years. As one "approaches" that sphere, the density sky-rockets. It's a kind of רקיע.