# What does it mean to have infinite negative conformal time?

In the context of Inflationary Cosmology, it is postulated that there was a period of shrinking Hubble Sphere radius $(aH)^{-1}$.

$$\frac{d}{dt} (aH)^{-1} < 0$$

Then the regions of the universe which appear to us to not be in causal contact can indeed have been in causal contact at some (conformal? physical? I'm not sure. My intuition would be physical) time in the past.

This solves the Horizon Problem, namely that we observe a homogenous isotropic universe to one part in 10,000, yet it doesn't seem that all regions of the universe were in casual contact to achieve such a uniformity.

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Anyway, when this shrinking Hubble sphere radius $(aH)^{-1}$ is invoked, it has the result of pushing the Hot Big Bang singularity back to negative infinity conformal time.

It is this statement that I am having trouble grasping.

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Explicitly

$$\chi_{PH}(\tau) = \tau - \tau_i = \int_{t_i}^{t} \frac{1}{a(t)} dt = \int_{a_i}^a (a H)^{-1} d\log(a)$$

where by Friedmann

$$(aH)^{-1} \propto a^{\frac{1+3w}{2}}$$

giving

$$\chi_{PH}(a(\tau)) = \tau - \tau_i \propto \frac{2}{1+3w} \left[ a(\tau)^{\frac{2}{1+3w}} - a_i^{\frac{2}{1+3w} } \right]$$

So for a strong energy violating fluid such that

$$1 + 3w < 0$$

this becomes

$$\tau_i \propto \frac{2}{1+3w} a_i^{\frac{2}{1+3w}} \rightarrow -\infty$$

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I understand the maths behind it, and can't argue against that the initial conformal time goes to minus infitity, but I don't know how to think about that.

In a standard Hot Big Bang the singularity occurs at $\tau_i = 0$ conformal time. This seems fine to me.

I am aware that conformal time is unphysical, but unless the scale factor goes negative, then negative conformal time suggests to me negative physical time.

And a negative scale factor would bring a whole other conceptual problem!

Namely, they are directly related by

$$d\tau = \frac{1}{a(t)} dt$$

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So my questions are;

What does it mean to have negative conformal time?

And what does it mean to have negative infinity conformal time?

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Thanks

• Why negative time should be different from, let's say, negative x coordinate? (btw, I don't know the answer) – Antonio Ragagnin May 9 '14 at 13:31
• @AntonioRagagnin : Well, for a start we know that the scale factor satisfies $$a(t) \propto t^{\frac{2(1+w)}{3}}$$ or $$a(\tau) \propto \tau^{\frac{2}{(1+3w}}$$ and so having negative physical time or conformal time would give a negative scale factor. I do not think this should be the case. Certainly I have never seen graphs or data for example that consider such a thing. Moreover, the idea of the scale of the universe being negative does not make sense, as I understand it... – Flint72 May 9 '14 at 14:19
• @AntonioRagagnin : ...The scale of the universe is a measure of the expansion of the universe, that is, a measure of the conformal distance and physical distance between bodies. Distance is a scalar quantity, which cannot be negative. Take for example the simple case of a radial null geodesic $$ds^2 = a^2(\tau) \left[ d\tau^2 - d\chi^2 \right]$$. Then a negative scale would be like a sphere with negative radius. It is not defined. – Flint72 May 9 '14 at 14:27

Conformal space is nice because in it, photons have straight world-lines, so we can easily see what we must do to achieve causal contact between two points in the CMB, after the physical time $t_i=0$ of the initial singularity, but before the physical time $t_{\text{CMB}}$ of decoupling.

Since we have

$$d\tau=\frac{dt}{a(t)},$$

then if we make causal contact between world-lines in physical time such that $\Delta t=0$ we also have $\Delta \tau=0$. So the intersection points between straight lines in conformal time tell us all we need to know about where causal contact is achieved.

However, this relation does not mean that if we have negative $\tau$ then we also have negative $t$. It tells us only the relationship between changes in physical and conformal time.

It tells us that we can have a small change in $t$ (between $t_i=0$ and $t_{\text{CMB}}$) but an infinite change in $\tau$ (between $\tau \rightarrow -\infty$ and $\tau=\tau_{\text{CMB}}$), when $a(t)$ is very small near the singularity. This turns out to be exactly what we need inflation to do, to get the world-lines of sections at opposite ends of the CMB to meet (causal connection).

Therefore in cosmology, by requiring that the Hubble sphere shrinks before the CMB is produced such that

$$\frac{d(aH)^{-1}}{dt}<0,$$

we are placing a requirement on the evolution of $a(t)$ such that between the physical times $t_i=0$ and $t_{\text{CMB}}$, $\tau$ goes from $\tau=-\infty$ to $\tau=\tau_{\text{CMB}}$, and causal contact is achieved.

A nice diagram with straight world-lines in conformal time (like the one below from the cosmology lectures of Daniel Baumann at DAMTP Cambridge) shows why this negative conformal time is required to achieve causal contact, and how it is an equivalent requirement to that of the shrinking Hubble sphere.

The lecture notes themselves are not available on arXiV but another set of Baumann's lecture notes on Inflation is: http://arxiv.org/abs/0907.5424.