What is the Derivative of the Scale Factor in Cosmology?

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Homework Help Overview

The discussion revolves around the derivative of the scale factor in cosmology, specifically examining the expression $$\frac{d}{da}(\dot{a}(t)^{-2})$$ where $$\dot{a}$$ represents the time derivative of the scale factor $$a(t)$$. Participants are exploring the implications of this derivative in the context of cosmological equations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are questioning whether the derivative $$\frac{d}{da}(\dot{a}(t)^{-2})$$ equals zero or something else, with some suggesting the use of the chain rule and power rule for differentiation. There is also a discussion about the meaning of $$\dot{a}$$ and its implications for the derivative.

Discussion Status

Some participants are exploring different interpretations of the derivative and its implications for related equations, such as the Friedmann equation. There is an ongoing examination of whether the derivative can be assumed to be zero, with some suggesting that it may not be, which could affect the outcomes of other calculations.

Contextual Notes

Participants are working within the constraints of cosmological principles and equations, and there is a mention of needing further clarification on the meaning of certain terms and their roles in the equations being discussed.

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

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

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