advanced18 min studyLesson 852 in Physics

Cosmic time, conformal time, and horizons

Integrate light propagation to distinguish particle horizons, event horizons, and the Hubble length.

Before you begin

On this page

The question for today

What you’ll understand

Transform radial light propagation into conformal coordinates.
Compute particle and event horizons for power-law expansion.
Explain why the Hubble length is not generally a causal boundary.

A local expansion rate does not by itself tell us how far signals have traveled. Causal contact depends on an integral over the expansion history. Cosmic time measures comoving clocks, while conformal time rearranges the coordinates so radial light rays have constant coordinate slope. We restrict calculations to a spatially flat FLRW background and keep its scale factor dimensionless. Natural units set light speed to one in equations; the examples restore a factor of $c$ when reporting distances in millions of light-years. No horizon in this lesson is a material surface that a photon collides with.

A clock adapted to null rays

For radial null propagation, setting the spacetime interval to zero gives a comoving speed inversely proportional to the scale factor. The photon always moves locally at light speed; the changing coordinate speed reflects the expanding spatial ruler. Introduce conformal time by integrating inverse scale factor against cosmic time. In the resulting coordinates the radial metric is an overall positive scale factor times a Minkowski metric, which preserves its null directions.

The additive origin of conformal time is arbitrary, but differences determine comoving light-travel distances. A finite cosmic age need not correspond to a finite conformal interval. To decide whether a horizon exists, inspect convergence at the actual initial or future boundary of the model. Replacing the early expansion history by a convenient late-time power law can change that convergence and therefore invent or erase a horizon.

ds^2=-dt^2+a^2(t)d\chi^2=a^2(\eta)(-d\eta^2+d\chi^2),\qquad d\eta=\frac{dt}{a(t)},\qquad \frac{d\chi}{d\eta}=\pm1

The flat radial coordinate

\chi

measures comoving length, and

\eta

is conformal time. Restoring light speed replaces the last equation by

d\chi/d\eta=\pm c

.↔ Scroll wide equations horizontally.

Past and future integrals answer different questions

A particle horizon marks the greatest comoving distance from which a signal could have reached an observer since the chosen initial boundary. Multiplying that distance by the current scale factor gives its proper radial distance on the current spatial slice. This is not the physical distance traversed through a static universe. Expansion stretches the separation between the emitting matter and the observer while the signal travels.

An event horizon instead asks which comoving emitters can ever send a signal that reaches the observer after the present time, given the assumed complete future. It exists when the future conformal-time interval is finite. Persistent accelerated power-law expansion supplies a simple example. A temporary period of acceleration does not establish the required infinite-future convergence; the expansion could later change. Cosmological event horizons therefore depend on an extrapolation beyond currently observed epochs.

D_{\rm p}(t)=a(t)\int_{t_i}^{t}\frac{dt'}{a(t')},\qquad D_{\rm e}(t)=a(t)\int_t^{t_f}\frac{dt'}{a(t')},\qquad D_H=H^{-1}

These are proper radial distances in units

c=1

. The past and future integrals define horizons only when finite, with boundaries

t_i

and

t_f

specified by the cosmological model.↔ Scroll wide equations horizontally.

The Hubble scale is a local comparison length

The Hubble length compares expansion with light propagation at one time. In a flat background the proper separation of an inward radial photon from a comoving observer obeys a competition between expansion and inward light motion. A photon initially beyond the Hubble length can increase its proper separation while still moving inward in comoving coordinates. If the Hubble length later grows sufficiently, that same photon can ultimately arrive; no local violation of special relativity occurs.

For a power law $a\propto t^p$ starting at zero cosmic time, a finite past horizon exists for $0<p<1$ and equals $t/(1-p)$ . The Hubble length is instead $t/p$ . For an expanding power law with $p>1$ maintained forever, the future event horizon is $t/(p-1)$ . These formulas expose which quantity uses the expansion derivative and which uses an integral. Their numerical coincidences in special models should not be promoted to definitions.

Work it through

A matter universe has two different distance scales

Assume an ideal flat matter-only universe has $a(t)=(t/t_*)^{2/3}$ from its initial boundary, with $t_*=1\,\mathrm{Myr}$ . At cosmic time $t=8\,\mathrm{Myr}$ , calculate its scale factor, conformal age, proper particle-horizon distance, and Hubble length. Treat the extrapolation to zero time as part of this mathematical model.

Evaluate the dimensionless scale factor: $a=8^{2/3}=4$ . Integrate $dt/a$ from zero to the chosen time, obtaining $\eta=3t_*^{2/3}t^{1/3}=6\,\mathrm{Myr}$ . The lower endpoint converges because the exponent is less than one.
Multiply conformal age by the present scale factor and restore light speed. The proper particle horizon is $D_{\rm p}=4c(6\,\mathrm{Myr})=24$ million light-years. Its size exceeds cosmic age times light speed because proper separations evolve during propagation.
Differentiate the expansion law to obtain $H=2/(3t)$ . Thus $c/H=12$ million light-years, one half of the particle-horizon distance. A present local expansion scale cannot replace the accumulated past light-travel integral.

The result

The scale factor is $4$ , conformal age is $6\,\mathrm{Myr}$ , the particle horizon is $24$ million light-years, and the Hubble length is $12$ million light-years. These numbers describe the matter-only extrapolation rather than our universe's measured horizon.

The living notebook

Explore the model

Interactive lab

Interactive lab · function

A derivative and an integral contain different information

Use $y=x^2$ as a dimensionless toy scale factor, with $x>0$ interpreted as time. Reset and set both controls to coefficient $a=1$ , position $x=0.5$ : $y=0.25$ and slope $1$ , so slope divided by height is $4$ . Reset and set both controls to $a=1$ , $x=1$ : height is $1$ , slope $2$ , and that ratio is $2$ . The lab displays height and slope, not horizon integrals or null geodesics.

y = ax²TangentSigned integral

y at chosen x: 0.608
Tangent slope · 2ax: 1.35
Signed area · ax³/3: 0.182

Coefficient a0.75Position x0.9

t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forAn expansion-rate analogue comes from a derivative divided by the scale factor; a horizon still requires a separate historical or future integral.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

An event horizon depends on the future integral of inverse scale factor. A finite past age concerns a different endpoint, and the instantaneous Hubble length is generally a different scale.

Your turn

Try it yourself

Now assume $a(t)=(t/t_*)^2$ persists for all future time. At $t=8\,\mathrm{Myr}$ , find the proper event-horizon distance and Hubble length. Explain which assumption makes the first calculation possible.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The future integral of $t'^{-2}$ converges; keep the scale-factor normalization until it cancels.

Compare with the solution

The comoving future interval is $t_*^2/t$ . Multiplication by $a(t)=t^2/t_*^2$ gives $D_{\rm e}=ct=8$ million light-years. Since $H=2/t$ , the Hubble length is $ct/2=4$ million light-years. Their factor-of-two difference is consistent. The event-horizon result requires this accelerated law to continue indefinitely; observing its validity over a finite interval would not determine the future integral.

Keep the idea

Take it with you

Conformal-time intervals determine comoving null travel distances.
Particle and event horizons depend on opposite boundaries of cosmic history.
The Hubble length is generally distinct from both causal horizons.

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