-1
$\begingroup$

In Friedman equation:

the case k < 0, with a Universe containing only matter (pressure = 0) so that $\rho = \rho_0/ a^3$ . What is the solution a(t) in a situation where the final term of the Friedmann equation dominates over the density term? How does the density of matter vary with time?

The question from an introduction to modern cosmology from Andrew Liddle.

Friedmann Equation:

$$(\dot {a}/a)^2 = 8\pi G \rho / 3 - k/a^2$$ My problem is I don't even understand the question? what does the first question mean? the answer of the book is $a(t) \alpha t$. I am not even sure that it is the only answer. Can anyone say what the question wants and is the answer of the book, the only answer?

$\endgroup$
4
$\begingroup$

It wants to know what the general solution of the equation is in the circumstances given.

$$\left(\frac{\dot{a}}{a}\right)^2 = \frac{8\pi G\rho}{3} - \frac{k}{a^2}$$ $$ \left(\frac{\dot{a}}{a}\right)^2 = \frac{8\pi G \rho_0}{a^3} - \frac{k}{a^2}$$

If the "second term dominates over the density term" (which it will when $a$ is large), then $$ \dot{a} = -k$$ (but $k$ has a negative value)

Therefore $a(t) \propto t$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.