What is the Derivative of the Scale Factor in Cosmology?

[Arman777]

Homework Statement

Derivative of Scale factor

Relevant Equations

None

In cosmology we have a scale factor that depends only on time ##a(t)##. Now how can I solve this thing

$$\frac{d}{da}(\dot{a}(t)^{-2}) = ?$$

Is it 0 ? Or something else ?

 

[Mark44]

Homework Statement: Derivative of Scale factor
Homework Equations: None

In cosmology we have a scale factor that depends only on time ##a(t)##. Now how can I solve this thing

$$\frac{d}{da}(\dot{a}(t)^{-2}) = ?$$

Is it 0 ? Or something else ?

What is ##\dot a##? The usual meaning of dot notation is the time derivative of something.
If so, then ##\frac{d}{da}(\dot{a}(t)^{-2}) = \frac{d}{da}\left( (\frac {da}{dt})^{-2}\right)##, and the chain rule and power rule would be applicable.

If ##\dot a## means something else, then you'll need to give more of an explanation.

 

[Arman777]

The usual meaning of dot notation is the time derivative of something

Yes time derivative.

chain rule and power rule would be applicable.

Hmm It seems like it gives something awkward ? When I tried to calculate it on the symbolab I get 0. (If I did not make mistake when I type the equation).

 

[Orodruin]

The first order Friedmann equation should give you ##\dot a## as a function of ##a##.

 

[Arman777]

I am actually trying to prove that

$$\frac{d\Omega}{d\ln{a} } = (1+3\omega)\Omega(\Omega - 1) $$

$$\frac{d\Omega}{dln(a) } = \frac{d\Omega}{da} / \frac{dln(a)} {da}$$

$$\Omega(t)=\frac{ε(t)}{ε_c(t)}= \frac{8\pi Gε_0a^{-3-3w}}{3\dot{a}^2a^{-2}}$$
$$\Omega(t)=\frac{ε(t)}{ε_c(t)}= \frac{8\pi Gε_0a^{-1-3w}}{3\dot{a}^2}$$

$$\frac{d\Omega}{da} =\frac{d}{da}(\frac{8\pi Gε_0a^{-1-3w}}{3\dot{a}^2})$$

If you set

$$\frac{d}{da}(\dot{a}(t)^{-2}) = 0$$

I get $$\frac{d\Omega}{dln(a) } = -(1+3w)\Omega$$

So I was wandering about maybe $$\frac{d}{da}(\dot{a}(t)^{-2}) \ne 0$$

can solve the issue

The first order Friedmann equation should give you ##\dot a## as a function of ##a##.

So you are saying

$$\dot{a} = \sqrt{ \frac{ 8\pi Ga^{-1-3w}} {3}}$$
then

$$\frac{d}{da}(\dot{a}(t)^{-2}) = -2\dot{a}^{-3} \times \frac{a^{-2-3w}(-1-3w)}{2\sqrt{a^{-1-3w}}}$$ ?