<p>Under the assumptions of large-scale homogeneity and isotropy, cosmic evolution can be described by the <em>scale factor</em> $a(t)$ that describes by what ratio the large-scale structure of the universe has expanded or contracted. The scale factor is unitless, and by convention taken to be $1$ in the present epoch, and the rate of change of $a$ can be denoted $\dot{a}$, so the the universe is expanding whenever $\dot{a}&gt;0$ and contracting whenever $\dot{a}&lt;0$. The existence of a past singularity is then the claim that $a\to 0$ for some finite amount of time in the past.</p>

<p>The scale factor is related to the <em>Hubble parameter</em> by $H \equiv \dot{a}/a$. The Hubble parameter is very frequently and erroneously also called the <em>Hubble constant</em>, which is probably the source of your confusion. It is true that if $H$ is taken to be constant, then trivially $a = a_0e^{Ht}$ is the only possible solution, which only asymptotically tends to $0$ in the infinite past. If spatially flat, as our universe is, this is the <em><a href="http://en.wikipedia.org/wiki/De_Sitter_universe" rel="nofollow">de Sitter solution</a></em> describing a universe dominated by a cosmological constant and negligible matter or radiation.</p>

<p>But that's simply not our universe. In the past, matter was dominant (and prior to that, radiation was dominant), and the de Sitter model simply isn't applicable. Simply put, the Hubble constant is not constant, and so should properly be called Hubble parameter.</p>

<p>In general, the evolution of the scale factor or the Hubble parameter is described by the <a href="http://en.wikipedia.org/wiki/Friedmann_equations" rel="nofollow">Friedmann equations</a>:
$$\begin{eqnarray*}
H^2\equiv\left(\frac{\dot{a}}{a}\right)^2 &amp;=&amp; \frac{8\pi G}{3}\rho - \frac{kc^2}{a^2}\text{,}\\
\dot{H}+H^2\equiv\frac{\ddot{a}}{a} &amp;=&amp; -\frac{4\pi G}{3}\left(\rho+\frac{3p}{c^2}\right)\text{,}
\end{eqnarray*}$$
where $\ddot{a}$ denotes the second time-derivative of $a$, the cosmological constant (dark energy) is included in the density $\rho$ and pressure $p$ of the universe, and for our spatially flat universe, $k=0$.</p>

<p>You may also wish to look up the <a href="http://en.wikipedia.org/wiki/Deceleration_parameter" rel="nofollow">deceleration parameter</a> $q\equiv -\ddot{a}a/\dot{a}^2$ (so defined because the universe was once thought to be decelerating). But the main point is that $H$ is in fact not a constant, so it is consistent for the scale factor to become $0$ a finite amount of time in the past. How much time depends on the changing density $\rho$ and pressure $p$, which is studied in detail in the standard <a href="http://en.wikipedia.org/wiki/Lambda-CDM_model" rel="nofollow">ΛCDM</a> model.</p>

CC BY-SA 3.0
2014-09-11T09:59:24.330Z
2014-09-11T09:59:24.330Z
Stan Liou
<p>If the speed at which the universe expands is constantly increasing (just like a derivative), then the opposite must also be true: the speed at which the universe contracts in reverse time is constantly decreasing, but, never reaches zero (like an asymptote). </p>

<p>If there is no exact point at which the speed of the contraction of the universe (in reverse time) reaches zero (because an asymptote goes on infinitely), then how did scientists arrive at a "date" for the "big bang"? </p>

<p>Since the speed of the expansion of the universe will go on and on ad infinitum, how can one arrive at a specific beginning point of an infinite function? </p>

CC BY-SA 3.0
2014-09-10T23:18:03.723Z
Why could it never be zero? The change in the slope of, say, a parabola is always increasing (for $x > 0$), but at one point it was 0.
HDE 226868
<big-bang-theory><expansion><astrophysics>
If the speed at which the universe contracts in reverse time decreases over time, then how did scientists arrive at a date for "The Big Bang"?
reuven
# If the speed at which the universe contracts in reverse time decreases over time, then how did scientists arrive at a date for "The Big Bang"?

(Asked by: username_0 on 2014-09-10)

If the speed at which the universe expands is constantly increasing (just like a derivative), then the opposite must also be true: the speed at which the universe contracts in reverse time is constantly decreasing, but, never reaches zero (like an asymptote).

If there is no exact point at which the speed of the contraction of the universe (in reverse time) reaches zero (because an asymptote goes on infinitely), then how did scientists arrive at a "date" for the "big bang"?

Since the speed of the expansion of the universe will go on and on ad infinitum, how can one arrive at a specific beginning point of an infinite function?

**Tags:** big-bang-theory, expansion, astrophysics

> **Comments:**
> **username_1**: Why could it never be zero? The change in the slope of, say, a parabola is always increasing (for $x > 0$), but at one point it was 0.  

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## 1 answers

### Answer by username_2 on 2014-09-11. Score: 3

Under the assumptions of large-scale homogeneity and isotropy, cosmic evolution can be described by the *scale factor* $a(t)$ that describes by what ratio the large-scale structure of the universe has expanded or contracted. The scale factor is unitless, and by convention taken to be $1$ in the present epoch, and the rate of change of $a$ can be denoted $\dot{a}$, so the the universe is expanding whenever $\dot{a}>0$ and contracting whenever $\dot{a}<0$. The existence of a past singularity is then the claim that $a\to 0$ for some finite amount of time in the past.

The scale factor is related to the *Hubble parameter* by $H \equiv \dot{a}/a$. The Hubble parameter is very frequently and erroneously also called the *Hubble constant*, which is probably the source of your confusion. It is true that if $H$ is taken to be constant, then trivially $a = a\_0e^{Ht}$ is the only possible solution, which only asymptotically tends to $0$ in the infinite past. If spatially flat, as our universe is, this is the *[de Sitter solution](http://en.wikipedia.org/wiki/De_Sitter_universe)* describing a universe dominated by a cosmological constant and negligible matter or radiation.

But that's simply not our universe. In the past, matter was dominant (and prior to that, radiation was dominant), and the de Sitter model simply isn't applicable. Simply put, the Hubble constant is not constant, and so should properly be called Hubble parameter.

In general, the evolution of the scale factor or the Hubble parameter is described by the [Friedmann equations](http://en.wikipedia.org/wiki/Friedmann_equations):
$$\begin{eqnarray\*}
H^2\equiv\left(\frac{\dot{a}}{a}\right)^2 &=& \frac{8\pi G}{3}\rho - \frac{kc^2}{a^2}\text{,}\\
\dot{H}+H^2\equiv\frac{\ddot{a}}{a} &=& -\frac{4\pi G}{3}\left(\rho+\frac{3p}{c^2}\right)\text{,}
\end{eqnarray\*}$$
where $\ddot{a}$ denotes the second time-derivative of $a$, the cosmological constant (dark energy) is included in the density $\rho$ and pressure $p$ of the universe, and for our spatially flat universe, $k=0$.

You may also wish to look up the [deceleration parameter](http://en.wikipedia.org/wiki/Deceleration_parameter) $q\equiv -\ddot{a}a/\dot{a}^2$ (so defined because the universe was once thought to be decelerating). But the main point is that $H$ is in fact not a constant, so it is consistent for the scale factor to become $0$ a finite amount of time in the past. How much time depends on the changing density $\rho$ and pressure $p$, which is studied in detail in the standard [ΛCDM](http://en.wikipedia.org/wiki/Lambda-CDM_model) model.


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