Help me derive the relativistic equation of motion for the Universe

AI Thread Summary
The discussion revolves around deriving the relativistic equation of motion for the Universe from the equation adot^2 = Λc^2a^2/3. Participants clarify that the time derivative of the scale factor, da/dt, relates to the Hubble constant, leading to the expression da/dt = Ha. To arrive at the scale factor function a(t) = a(t0)e^Ht, they manipulate the equations, integrating to find that log(a) = Ht + constant, which ultimately leads to the desired exponential form. The discussion emphasizes the relationship between the scale factor and the accelerating expansion of the Universe. The final expression showcases the mathematical foundation of cosmic expansion dynamics.
QuantumX
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Guys,

My calculus is really rusty and I need help solving this equation using a time derivative (denoted with a dot) in order to get the relativistic equation of motion for the Universe

The equation is:

adot^2 = Λc^2a^2/3

where adot is the time derivative of the scale factor, lambda is a cosmological constant, and c is the speed of light.

The answer should be:

a(t) = a(t0)e^Ht

where H is Hubble's constant = sqrt(c^2Λ/3)

But I don't know how to get there. Any help is appreciated!
 
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Hi QuantumX! Welcome to PF! :smile:

(try using the X2 button just above the Reply box :wink:)
QuantumX said:
adot2 = Λc2a2/3

So da/dt = √(Λc2/3)a = Ha :smile:
 
Thanks Tiny-Tim,

So then the first equation is just the relationship between the scale factor and the Hubble constant...

So how do I then get a(t) = a(t0)e^Ht ? (that is for t>t0). That's what i need to ultimately arrive at. It's supposed to showcase that the Universe is accelerating in its expansion.
 
tiny-tim said:
So da/dt = √(Λc2/3)a = Ha :smile:
QuantumX said:
So how do I then get a(t) = a(t0)e^Ht ?

da/dt = Ha

so dt/da = 1/Ha, so t = loga/H + constant

so loga = Ht + constant

so a = eHt + constant = econstanteHt

and the econstant is a(to)
 
Thanks!
 
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