What is the Derivative of the Scale Factor in Cosmology?

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SUMMARY

The discussion centers on calculating the derivative of the scale factor in cosmology, specifically the expression $$\frac{d}{da}(\dot{a}(t)^{-2})$$. Participants confirm that $$\dot{a}$$ represents the time derivative of the scale factor, leading to the application of the chain rule and power rule in differentiation. The conversation reveals that setting $$\frac{d}{da}(\dot{a}(t)^{-2}) = 0$$ may not be accurate, as it contradicts the results derived from the first order Friedmann equation, which relates $$\dot{a}$$ to the scale factor $$a$$.

PREREQUISITES
  • Understanding of cosmological concepts, particularly the scale factor $$a(t)$$.
  • Familiarity with calculus, specifically differentiation techniques such as the chain rule and power rule.
  • Knowledge of the Friedmann equations in cosmology.
  • Basic understanding of the energy density parameters, such as $$\Omega$$.
NEXT STEPS
  • Study the derivation and implications of the first order Friedmann equation in cosmology.
  • Explore the relationship between the scale factor $$a(t)$$ and the time derivative $$\dot{a}$$ in detail.
  • Learn about energy density parameters and their role in cosmological models, focusing on $$\Omega(t)$$.
  • Practice differentiation of complex functions involving time-dependent variables in cosmological contexts.
USEFUL FOR

Students and researchers in cosmology, physicists working on theoretical models of the universe, and anyone interested in the mathematical foundations of cosmological dynamics.

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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