- #1
binbagsss
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<<Mentor note: Edited for readability.>>
<<Follow-up: futher edited to fix LaTeX tags>>
Context: FRW Metric /universe, perfect comological fluid , dark/vacuum energy
So the equation of state ## w \rho=p, w=-1, \rho=-p##
so this clealy implies either ##p## or ##\rho## is negative.
Am I correct in thinking which it is, depends on the cosmological constant?
So ##cosmo constant >0, \rho>0, p<0##
##cosmo constant <0, \rho <0, p>0?##
Now I look at a Friedmann equation given by: ##\Omega -1 = \frac{k}{H^2a^2} ##, where ## \Omega = \frac{\rho}{\rho_{c}} ##, ##H=\frac{\dot{a^{2}}}{a^{2}} ##
and solving for ##a## as a function of ##t## for ##cosmo constant >0##, my book says that in this case ##\Omega <0 ## ( which I understand if my above reasoning is correct) and so from the Friedmann equation this is only possible if ##k=-1 ##. So this is fine, I agree , but the denominator needs to be greater than zero.
Anyway, it than solves for ## cosmo constant >0 ##, and says all ## k=-1,0,1 ## are fine. This is my QUESTION. I don't understand how ##k=-1## can be okay for both ##cosmo constant >0, <0 ## which in turn say different things above which of ## \rho ## and ##p## are negative and positive.
Thanks very much for your help.
<<Follow-up: futher edited to fix LaTeX tags>>
Context: FRW Metric /universe, perfect comological fluid , dark/vacuum energy
So the equation of state ## w \rho=p, w=-1, \rho=-p##
so this clealy implies either ##p## or ##\rho## is negative.
Am I correct in thinking which it is, depends on the cosmological constant?
So ##cosmo constant >0, \rho>0, p<0##
##cosmo constant <0, \rho <0, p>0?##
Now I look at a Friedmann equation given by: ##\Omega -1 = \frac{k}{H^2a^2} ##, where ## \Omega = \frac{\rho}{\rho_{c}} ##, ##H=\frac{\dot{a^{2}}}{a^{2}} ##
and solving for ##a## as a function of ##t## for ##cosmo constant >0##, my book says that in this case ##\Omega <0 ## ( which I understand if my above reasoning is correct) and so from the Friedmann equation this is only possible if ##k=-1 ##. So this is fine, I agree , but the denominator needs to be greater than zero.
Anyway, it than solves for ## cosmo constant >0 ##, and says all ## k=-1,0,1 ## are fine. This is my QUESTION. I don't understand how ##k=-1## can be okay for both ##cosmo constant >0, <0 ## which in turn say different things above which of ## \rho ## and ##p## are negative and positive.
Thanks very much for your help.
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