Stress Energy Tensor - Confusion about signs and metrics

AI Thread Summary
The discussion focuses on deriving Friedmann's Equations from the Einstein Field Equations, specifically addressing the stress-energy tensor components. The user is working with the FLRW metric and is confused about the correct form of the stress-energy tensor for a perfect fluid, particularly regarding the spatial indices 'ii'. They note that while the diagonal form of the tensor is often presented as T^{\alpha \beta} = diag(ρ, P, P, P), the conversion to lowered indices introduces scale factors from the metric. Ultimately, they conclude that the correct expression for the spatial components is T_{ii} = P a^2, aligning with their metric signature. This clarification aids in their calculations for Einstein's Field Equations.
Allday
Messages
164
Reaction score
1
Hi everyone. I'm working on deriving Friedmanns Equations from the Einstein Field Equations. I've got the '00' components worked out but I'm having some trouble with the spatial indices 'ii' of the stress energy tensor ## T_{ii} ##. I'm the FLRW metric with c=1 and signature (-,+,+,+) so that ##g_{\mu \nu} = (-1,a^2,a^2,a^2) ##. My question is what is the stress energy tensor for a perfect fluid and how does it change with raised and lowered indices. I know how the metric operates to raise and lower indices but I don't know what to start with for ## T_{ii} ##. For example, I've seen

$$ T^{\alpha \beta} = diag(\rho,P,P,P) $$

but also the same thing for ## T^{\alpha}_{\beta} ##. The trouble comes in with factors of the scale factor ## a ## when I try to calculate the diagonal spatial components for Einsteins Field Equations. For example if ## T^{\alpha \beta} = diag(\rho,P,P,P) ## then ## T_{ii} ## gets two factors from the metric meaning ## T_{ii} = P a^4 ##. Is that correct? Any help would be appreciated.

thanks,
Allday
 
Last edited:
Space news on Phys.org
Thanks yenchin, those notes do help. It seems I want to use the form
$$ T_{\alpha \beta} = (\rho + P) u_{\alpha} u_{\beta} + P g_{\alpha \beta} $$
along with the fact that the spatial components of ##u_{\alpha} ## are zero so that with my choice of metric sign convention
$$ T_{i i} = P a^2 $$
 
https://en.wikipedia.org/wiki/Recombination_(cosmology) Was a matter density right after the decoupling low enough to consider the vacuum as the actual vacuum, and not the medium through which the light propagates with the speed lower than ##({\epsilon_0\mu_0})^{-1/2}##? I'm asking this in context of the calculation of the observable universe radius, where the time integral of the inverse of the scale factor is multiplied by the constant speed of light ##c##.
Why was the Hubble constant assumed to be decreasing and slowing down (decelerating) the expansion rate of the Universe, while at the same time Dark Energy is presumably accelerating the expansion? And to thicken the plot. recent news from NASA indicates that the Hubble constant is now increasing. Can you clarify this enigma? Also., if the Hubble constant eventually decreases, why is there a lower limit to its value?
Back
Top