What Does a Negative Value for the Deceleration Parameter Imply in Cosmology?

In summary, the conversation discusses the general Taylor series for ##a(t)##, which can be rewritten in terms of the Hubble constant ##H(t_0)## and the deceleration parameter ##q(t_0)##. The equation for ##a(t)## shows that a negative value for ##q(t_0)##, as observed in the data, implies a positive value for the term ##\left[- \frac{1}{2} q(t_0) H^2(t_0) (t-t_0)^2\right]##.
  • #1
redtree
332
15
TL;DR Summary
Understanding the definition
I note the general Taylor series for ##a(t)## as:

\begin{equation}

\begin{split}

a(t)&\approx a(t_0) + a'(t_0) (t-t_0) + \frac{1}{2!} a''(t_0) (t-t_0)^2 ...

\end{split}

\end{equation}
which I rewrite as:

\begin{equation}

\begin{split}

a(t)&\approx a(t_0)\left(1 + \frac{a'(t_0)}{a(t_0)} (t-t_0) + \frac{1}{2} \frac{a''(t_0)}{a(t_0)} (t-t_0)^2 + ... \right)

\end{split}

\end{equation}
In this context, the Hubble constant ##H(t_0)## is defined as follows:

\begin{equation}

\begin{split}

H(t_0) &\doteq \frac{a'(t_0)}{a(t_0)}

\end{split}

\end{equation}
and the deceleration parameter ##q(t_0)## is defined as follows:

\begin{equation}

\begin{split}

q(t_0)&\doteq -\frac{a''(t_0)}{a(t_0) H^2(t_0)}

\end{split}

\end{equation}
such that

\begin{equation}

\begin{split}

a(t)&\approx a(t_0)\left(1 + H(t_0) (t-t_0) - \frac{1}{2} q(t_0) H^2(t_0) (t-t_0)^2 + ... \right)

\end{split}

\end{equation}

My question: Does a negative value for ##q(t_0)##, which is what is observed in the data, therefore imply that the term ##\left[- \frac{1}{2} q(t_0) H^2(t_0) (t-t_0)^2\right]## is positive?
 
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  • #2
It seems that way yes
 

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