Wednesday, November 21, 2012

redundancy in the Friedmann equations

under which circumstances is the $\dot{a}$-Friedmann equation the integral of the $\ddot{a}$-equation? what acts as the integration constant? why would that particular variable not be able to introduce de/ac-celeration?

bonus question: why is (in contrast to real analysis) the derivative of a complex differentiable function always again complex differentiable?

3 comments:

  1. bonus question: that's really a very interesting property of holomorphic/analytical/complex differentiable/regular functions: using the residue theorem, you can compute the value of a derivative from an integration process,
    \begin{equation}
    f(z)=\frac{1}{2\pi\mathrm{i}}\oint_C\frac{f(\zeta)}{\zeta-z}\mathrm{d}\zeta
    \end{equation}
    and this relation be integrated by parts for the first derivative and then be generalised by induction to $n$th derivatives,
    \begin{equation}
    f^{(n)}(z)=\frac{n!}{2\pi\mathrm{i}}\oint_C\frac{f(\zeta)}{(\zeta-z)^{n+1}}\mathrm{d}\zeta
    \end{equation}
    and you'd just need complex differentiability for computing derivatives of any order.

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  2. one needs both Friedmann-equations for treating fluids with an arbitrary $p(\rho)$-relation, or equivalently, one of the Friedmann-equations and the adiabatic equation, assuming that there is no energy exchange between fluids. once the equation of state is fixed, the $\dot{a}$-equation is the integral of the $\ddot{a}$-equation, with the curvature appearing as the integration constant. this makes sense, too, because the equation of state of curvature (although it is not a physical substance, but hey, dark energy might not be either) is $w=-1/3$ and can therefore not introduce any acceleration, but rather leads to a constant $\dot{a}$.

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  3. side remark: this can even be understood in Newtonian cosmology, where the curvature directly appears as an integration constant with the interpretation as the total energy content of the Universe.

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