The Friedmann equation in a lambda-dominated universe

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Homework Help Overview

The discussion revolves around demonstrating that a lambda-dominated Euclidean universe leads to exponential expansion, specifically through the Friedmann equation. The original poster presents equations and expresses confusion regarding the relationship between terms in the Friedmann equation.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to manipulate the Friedmann equation but encounters difficulties in relating the terms correctly. Some participants question the necessity of using both the constant term lambda and the vacuum energy source term, suggesting redundancy.

Discussion Status

The discussion is ongoing, with participants providing clarifications about the relationship between lambda and the energy density. There is an acknowledgment of the redundancy in using both terms, but no consensus has been reached on the best approach to take.

Contextual Notes

Participants are navigating the implications of using constants in the Friedmann equation and the assumptions underlying the definitions of energy density and pressure in a lambda-dominated universe.

_Andreas
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Homework Statement



Show that a lambda-dominated euclidean universe entails an exponential expansion.

Homework Equations



Equation of state is P=-E (E=epsilon)
There is no curvature (i.e. a flat universe)

The Friedmann equation is then

((a(dot)/a)^2)=((8piG)/3c^2)E + lambda/3 (eq. 1)

which -- and this is where I get lost -- is equivalent to

((a(dot)/a)^2)=((8piG)/3c^2)Elambda (eq. 2)

The Attempt at a Solution



lambda/3 = ((8piG)/3c^2)Elambda, but if I make this substitution into the Friedmann equation, I get

((a(dot)/a)^2)=((8piG)/3c^2)E + ((8piG)/3c^2)Elambda

What am I doing wrong? I'm after (eq. 2).
 
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You have two different representations of the same thing in the Friedmann equation. A constant term lambda is the same thing as a vacuum energy source term satisfying rho=-p. Which one do you want to work with? You don't need both.
 
I'm not sure I understand. Could you elaborate further?
 
_Andreas said:
I'm not sure I understand. Could you elaborate further?

What part don't you understand? A fluid with E=-p satisfies dE/dt=0 by the conservation equation, so is a constant. lambda is also a constant. You have two constants. It's a little redundant.
 
I just wanted to say thank you for your efforts, Dick. They're appreciated. I forgot to say that. Sorry!
 

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