# A Cosmological horizon at the Hubble radius?

I have calculated that if one extends a rigid ruler into space by a fixed proper distance $D$ then a clock at the end of the ruler, running on proper time $\tau$, will run more slowly than time $t$ at the origin by a time dilation factor:

$$\frac{dt}{d\tau} = \frac{1}{\sqrt{1 - H^2 D^2 / c^2}}$$

where $H$ is the Hubble parameter.

If one substitutes in Hubble's law, $v = H D$ (the theoretical law which is exact), one finds the following satisfying result that

$$\frac{dt}{d\tau} = \frac{1}{\sqrt{1 - v^2 / c^2}}.$$

Although this looks like a result from special relativity I derived it by combining the FRW metric line element and an equation for the path of the end of the ruler, $\chi = D / R(t)$, where $\chi$ is the radial comoving coordinate of the end of the ruler, $D$ is a fixed proper distance and $R(t)$ is the scalefactor.

Does this prove that there is a cosmological horizon at the Hubble radius $D=c/H$ where the proper time $\tau$ slows down to a stop compared to our time $t$?

This seems to be a general result which is true regardless of cosmological model.

Details of Calculation

The general FRW metric is given by:

$$ds^2 = -c^2 dt^2 + R(t)^2\left[d\chi^2+S^2_k(\chi)d\psi^2 \right]$$

where $d\psi^2 = d\theta^2 + \sin^2 \theta d\phi^2$ and $S_k(\chi)=\sin \chi$,$\chi$ or $\sinh \chi$ for closed ($k=+1$), flat ($k=0$) or open ($k=-1$) universes respectively. The scale factor $R(t)$ has units of length.

Consider a ruler of fixed proper length $D$ extending out radially from our position at the origin. The path of the far end of the ruler in comoving coordinates is

$$\chi = \frac{D}{R(t)}$$

Differentiating this equation by proper time $\tau$ gives us:

$$\frac{d\chi}{d\tau} = - \frac{D}{R^2} \frac{dR}{dt} \frac{dt}{d\tau}.$$

Using the FRW metric we can find a differential equation for the path of the far end of the ruler. We substitute in $ds^2=-c^2d\tau^2$ (end of ruler has a time-like path), $d\psi=0$ (the ruler is radial) and divide through by $d\tau^2$ to obtain:

$$c^2\left(\frac{dt}{d\tau}\right)^2 - R(t)^2 \left(\frac{d\chi}{d\tau}\right)^2 = c^2.$$

Substituting the expression for $d\chi/d\tau$ into the above equation we find:

$$c^2\left(\frac{dt}{d\tau}\right)^2 - D^2 \left(\frac{\dot R}{R}\right)^2 \left(\frac{dt}{d\tau}\right)^2 = c^2.$$

Using the definition of the Hubble parameter $H=\dot{R}/R$ we finally obtain:

$$\frac{dt}{d\tau} = \frac{1}{\sqrt{1 - H^2 D^2 / c^2}}.$$

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No, the Hubble radius is not a horizon. It would be if the Hubble parameter $H$ were constant; however, $H$ decreases over time. I'll write a detailed answer when I have more time, but for the moment I can refer you to this post. –  Pulsar Jul 8 '13 at 17:38
The Hubble sphere is not a horizon - see physics.stackexchange.com/q/41058/10851. –  Chris White Jul 8 '13 at 17:39
@Pulsar You beat me by a few seconds! I'll leave this cosmic expansion question to your expertise :) (For the record, I'm inclined to not say this is a duplicate, since there is a specific calculation here that can be addressed.) –  Chris White Jul 8 '13 at 17:41
If the proper distance $D$ is fixed, that means that the light can travel from one extremity to the other in a finite fixed time interval. So, it seems incompatible with an horizon. –  Trimok Jul 8 '13 at 17:44
@Chris Great minds think alike :-) Yes, I wouldn't label it a duplicate either, as the calculation in the OP deserves a proper response. By all means, write an answer if you want :-) –  Pulsar Jul 8 '13 at 17:55
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The problem is that your calculation has no real physical meaning. It is only meaningful to compare two quantities within the same inertial frame. But there is no global inertial frame that connects us to a co-moving galaxy at the Hubble radius: such a galaxy is at rest in its own local cosmological inertial frame (ignoring local peculiar motions). So, due to the Cosmological Principle, its own cosmic proper time is similar to ours.

To obtain meaningful information, we need to analyse the light of that galaxy as we observe it. And as you indicate in your own answer, the Hubble radius is not a cosmological horizon in terms of light paths.

So see why, let us look at the Standard ΛCDM-model in more detail. The expansion of the universe can be expressed in terms of the scale factor $a(t)$ and its derivatives, with $a=1$ today. From the Friedmann equations, it can be shown that the Hubble parameter $H$ has the form $$H(a) = \frac{\dot{a}}{a} = H_0\sqrt{\Omega_{R,0}\,a^{-4} + \Omega_{M,0}\,a^{-3} + \Omega_{K,0}\,a^{-2} + \Omega_{\Lambda,0}},$$ with $H_0$ the present-day Hubble constant, $\Omega_{R,0}, \Omega_{M,0}, \Omega_{\Lambda,0}$ the relative present-day radiation, matter and dark energy density, and $\Omega_{K,0}=1-\Omega_{R,0}- \Omega_{M,0}- \Omega_{\Lambda,0}$. I will assume the values $$\begin{gather} H_0 = 67.3\;\text{km}\,\text{s}^{-1}\text{Mpc}^{-1},\\ \Omega_{R,0} = 0, \quad\Omega_{M,0} = 0.315, \quad\Omega_{\Lambda,0} = 0.685,\quad\Omega_{K,0} = 0. \end{gather}$$ It's important to note that, even though the expansion of the universe is currently accelerating ($\ddot{a}>0$), the Hubble parameter $H$ is in fact always decreasing. The cosmic time can then be calculated from $$\text{d}t = \frac{\text{d}a}{\dot{a}} = \frac{\text{d}a}{aH(a)},$$ so that $$t(a) = \int_0^a\frac{\text{d}a}{aH(a)}.$$ Likewise, a photon travels on a null-geodesic $$0 = c^2\text{d}t^2 - a^2(t)\text{d}\ell^2,$$ with $\text{d}\ell$ a co-moving displacement, so that the co-moving distance travelled by a photon is $$D_\text{c} = c\int_{a_\text{em}}^{a_\text{ob}}\frac{\text{d}a}{a\dot{a}} = c\int_{a_\text{em}}^{a_\text{ob}}\frac{\text{d}a}{a^2H(a)},$$ with $a_\text{em} = a(t_\text{em})$ and $a_\text{ob} = a(t_\text{ob})$ the scale-factors at the moments of emission and observation. At any moment $t$, a co-moving distance $D_\text{c}(t)$ can be converted into a proper distance $D(t) = a(t) D_\text{c}(t)$.

The distance to the current Hubble radius is $$D_\text{H}=\frac{c}{H_0}\approx 14.5\;\text{Gly}.$$ Now, we can define two important cosmological horizons: the first is the particle horizon, which marks the edge of the observable universe. It is the maximum distance that light has been able to travel to us between $t=0$ and the present day, i.e. it is as far as we can see: $$D_\text{ph} = c\int_0^1\frac{\text{d}a}{a^2H(a)}.$$ Since $H(a) < a^{-2}H_0\$ for $a<1$, we get $$D_\text{ph} > \frac{c}{H_0}\int_0^1\text{d}a = D_\text{H},$$ it follows that the Hubble radius is smaller than the particle horizon, and thus part of the observable universe. In fact, $D_\text{ph}\approx 46.2\;\text{Gly}$.

The second horizon is the event horizon. It is the region of space from which a photon that is emitted 'today' will still be able to reach us at some point in the future. Thus $$D_\text{eh} = c\int_1^\infty\frac{\text{d}a}{a^2H(a)}.$$ Since $H(a) < H_0\$ for $a>1$, we get $$D_\text{eh} > \frac{c}{H_0}\int_1^\infty\frac{\text{d}a}{a^2} = D_\text{H},$$ so the Hubble radius is also smaller than the event horizon: a galaxy on the Hubble radius can still send signals to us (or we to them). $D_\text{eh}\approx 16.7\;\text{Gly}$.

The situation is displayed in these graphs (click on 'view image' for a larger version): the first is with co-moving coordinates, the second with proper coordinates.

The black solid lines indicate our present position. The blue line is the particle horizon through time, the red line is the event horizon, the green area is the Hubble sphere. The dotted black line is a co-moving galaxy that is currently located on the Hubble radius. Its photons that we observe today have travelled on the cyan path (they were emitted at $t=4.3\;\text{Gy}$). Its photons that it emits today ($t=13.8\;\text{Gy}$) will travel on the purple path (they will reach us at $t=49\;\text{Gy}$).

So can we say anything about time dilation? The answer is yes. The light that we observe is also redshifted $$1 + z = \frac{\lambda_\text{ob}}{\lambda_\text{em}} = \frac{\nu_\text{em}}{\nu_\text{ob}}.$$ Within a small time $\delta t_\text{em}$ the source emits a light wave with $\nu_\text{em}\delta t_\text{em}$ oscillations. Those same oscillations are observed within time $\delta t_\text{ob}$ with frequency $\nu_\text{ob}$, in other words $\nu_\text{em}\delta t_\text{em} = \nu_\text{ob}\delta t_\text{ob}$, so that $$1 + z = \frac{\delta t_\text{ob}}{\delta t_\text{em}}.$$ In other words, cosmic redshift is directly related to time dilation. Also, within a small time $\delta t_\text{em}$, the distance that light has to travel doesn't change: $$\int_{t_\text{em}}^{t_\text{ob}}\frac{c\,\text{d} t}{a(t)} = \int_{t_\text{em} + \delta t_\text{em}}^{t_\text{ob} + \delta t_\text{ob}}\frac{c\,\text{d} t}{a(t)},$$ or $$\int_{t_\text{ob}}^{t_\text{ob} + \delta t_\text{ob}}\frac{c\,\text{d} t}{a(t)} = \int_{t_\text{em}}^{t_\text{em} + \delta t_\text{em}}\frac{c\,\text{d} t}{a(t)}.$$ in these small intervals, the integrands remain constant, so that $$\frac{\delta t_\text{ob}}{a(t_\text{ob})} = \frac{\delta t_\text{em}}{a(t_\text{em})}$$ and $$1 + z = \frac{a(t_\text{ob})}{a(t_\text{em})}.$$ The light from the co-moving galaxy at the current Hubble radius that we observe today was emitted when $a(t_\text{em})=0.403$, so that $z=1.48$, and the observed events from this galaxy are time-dilated by a factor $2.48$. And the light it emits today will be observed when $a(t_\text{ob})=8.07$, with redshift $z=7.07$.

Such time dilations have indeed been observed: we literally see distant supernovae explode in slow motion!

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Thanks very much for the comprehensive answer! –  John Eastmond Jul 18 '13 at 23:00
@John My pleasure (you can upvote it if you want :-)) –  Pulsar Jul 18 '13 at 23:11
I wonder if I could put forward a rather heretical view. If the Hubble redshift can be viewed as a time dilation effect then that implies that clocks now tick at a faster rate than similar clocks in the past. This would also imply that atomic frequencies now are higher than the frequencies of the same atoms in the past. Thus it seems that the energy of say a hydrogen atom now is higher than it was in the past. –  John Eastmond Jul 19 '13 at 14:24
@John No, time dilation is purely an observer effect when signals are sent from one inertial frame to another. We see those signals as time-dilated because the light had to 'struggle' against the expansion of space to reach us. Nothing changed about the proper time in the source restframe, or the proper time in our restframe. It is similar to the Doppler effect in special relativity: events from a moving source appear time-dilated to us, but that doesn't mean time itself behaved differently in the past. –  Pulsar Jul 20 '13 at 11:47
If an observer a long way away from a large mass observes signals sent from a source near the mass then he observes those signals gravitationally redshifted. Many people argue that one can think of the light "struggling" to get out of the potential well and thus losing energy on the journey. But the associated gravitation time dilation effect is real. A clock lowered into the potential well and then pulled back will have slowed compared to an identical clock that stays with the observer. Thus it seems the frequency of an emitting atom must have been lower in the well. –  John Eastmond Jul 22 '13 at 10:16

A first check to see that your demonstration is not correct is that, if the proper distance $D$ is fixed, that means that the light can travel from one extremity to the other in a finite fixed time interval. So, it seems clearly incompatible with an horizon.

The problem, in fact, is that it is not correct to state that $v=HD$, where $D$ is a constant.

In fact, if we consider two comoving galaxies, with corresponding comoving coordinates $\chi_1$ and $\chi_2$, we may define a "distance" $D$ between them, by supposing that time is "stopped", that is $D(t) = R(t) |\chi_2 - \chi_1 |$

So, it is not a true distance between the galaxies, to obtain a true distance, we would have to emit a light ray from $1$ towards $2$, then back to $1$, compute the total time, divide by $2$ and multiply by $c$, so the coordinate $t$ does change in this case.

Now, it is clear that $v(t) =\frac{dD(t)}{dt}$ has the dimension of a speed, and it is called the recessional velocity. But because $D(t)$ is not a true distance, $v_c(t)$ is not a true speed.

Of corse, we may write : $v(t) = \frac{dD(t)}{dt} = R'(t) |\chi_2 - \chi_1| =\frac{R'(t)}{R(t)}D(t) = H(t) D(t)$, and this is the Hubble law.

[EDIT 1]

If we consider a fixed distance $D$, we can calculate the limit value, for a given t, such as $DH(t)=c$. At present time, with $H(t)=67.8$ km/(s. Mpc), this gives:

$D = 14.4$ billion light-year. (Hubble distance)

This is to be compared with the current radius of observable universe : $46$ billion light-year

[EDIT 2]

Looking once more time at the graph (presented by Pulsar), You see that, at present time, this Hubble distance (corresponding to green line $v_{rec} = c$) is smaller than the present event horizon ($\sim 16$ billion light-year), so any light ray emitted from the extremity of the ruler, at present time, will always reach the origin of the ruler in the future.

In fact, at all times, the Hubble distance is smaller than the event horizon.

You see, in the graph, that the redshift can be read directly, at the intersection of the green line $v_{rec} = c$, and the horizontal black line, the redshift (dashed purple line) is between $z=1$ and $z=3$

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Light can travel the distance $D$ to us but its frequency will be red-shifted due to the time dilation. If $D=c/H$ then the light will be red-shifted to zero. –  John Eastmond Jul 9 '13 at 9:25
The pseudo-distance $D(t)$ defined in the answer, does not correspond to a real distance (It is not actually a distance traveled by light), so the recessional velocity, which is not a true velocity, could be greater than the speed of light, see for instance, the graph given by Pulsar (constant recessional velocity are green lines) in a previous post, or this graph in the Hubble's law Wikipedia article –  Trimok Jul 9 '13 at 10:35
But D could be the fixed proper length of a ruler. A light source at the far end could send photons along a long optic fibre attached to the ruler to us at the origin. Those photons would be red-shifted. –  John Eastmond Jul 9 '13 at 17:28
@JohnEastmond : I have made an edit in the answer –  Trimok Jul 10 '13 at 8:37