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Timeline for Dark energy and neutrinos

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Dec 20, 2022 at 19:16 comment added Andrew @qatch These notes might help: astro.princeton.edu/~gk/A403/state.pdf
Dec 20, 2022 at 18:13 comment added qatch I am more familiar with the classical EOS, so may be that is a misleading analogy: for an ideal gas p = nRT/V holds. So if you double the volume you get half the pressure – in an isolated system. You can double the volume by doubling n as well. So having a source of particles, such as fusion for neutrinos, could -in theory- contribute to a change in volume.
Dec 18, 2022 at 12:31 comment added Andrew @qatch I'm not sure what you mean. You can calculate the equation of state for a relativistic gas of particles and you find $p=\frac{1}{3}\rho$.
Dec 18, 2022 at 10:44 comment added qatch w will give me something to think about. However, the equation of state does not take into account possible source terms, does it? Could it change the game?
Dec 18, 2022 at 10:33 vote accept qatch
Dec 18, 2022 at 2:19 comment added Andrew @DaveTheWave Essentially it follows from the symmetry of FRW spacetimes (homogeneity and isotropy) that the stress energy tensor has the form $T^{i}_{\ \ j} = {\rm diag}(-\rho, p, p, p)$, so there's only really one relationship needed in the equation of state. You can also calculate the stress energy explicitly for special cases, like pressureless dust, the Maxwell stress-energy tensor for the electromagnetic field, or for the cosmological constant, and $\rho=wp$ falls out naturally in those cases.
Dec 17, 2022 at 23:36 comment added user353451 Andrew said: we assume that the relationship between density ρ and pressure p is given by an equation of state of the form p=wρ .Why is this assumed?
Dec 17, 2022 at 20:12 comment added niels nielsen excellent! thanks for this explanation, I had not seen it before. -NN
Dec 17, 2022 at 17:24 history answered Andrew CC BY-SA 4.0