What is the Derivative of the Scale Factor in Cosmology?

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Arman777
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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 ?
 
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Arman777 said:
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.
 
Mark44 said:
The usual meaning of dot notation is the time derivative of something
Yes time derivative.

Mark44 said:
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).
 
The first order Friedmann equation should give you ##\dot a## as a function of ##a##.
 
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

Orodruin said:
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

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