I'm puzzled with the following question: find an analog of the Hubble's law for the Kasner solution.

Kasner metric is a solution to the vacuum Einstein equations $$ds^2=dt^2-\sum_{i=1}^3a^2_i(t)(dx^i)^2$$ with coefficients $a_i(t)=t^{p_i}$, where $p_i$ are constant parameters satisfying $\sum_ip_i=1$ and $\sum_i (p_i)^2=1$. This solution describes anisotropic space.

It is known that in the Robertson-Walker metric, when coefficients along all axes are the same $a_i=a$ wavelengths of a photon at times $t$ and $t_0$ are related by the scale parameter $a(t)$ as $$\frac{\lambda(t)}{\lambda(t_0)}=\frac{a(t)}{a(t_0)}$$
this is true if the rate if the space expanding $\dot{a}/a$ is significantly smaller then the frequency of the photon.

The question is how to generalize this statement in the case of the Kasner metric. A natural idea is to suppose that the statement is generalized directly to the propagation along "main" axes $$\frac{\lambda_i(t)}{\lambda_i(t_0)}=\frac{a_i(t)}{a_i(t_0)}$$ and "linearly" for a generic direction, i.e. for $\vec{\lambda}=\lambda_i \vec{e_i}$ $$\frac{|\vec{\lambda}(t)|}{|\vec{\lambda}(t_0)|}=\frac{\dot{a}_i(t)\lambda_i}{a_i(t)\lambda_i}$$

However, I'm not able to confirm this conjecture by any quantitative argument. I've tried to show that the EOM in a curved space-time $\nabla^\mu F_{\mu\nu}=0$ admit a solution of a kind $A_\mu(x)\sim e_{\mu}\exp{\left(i g_{\mu\nu}q^\mu x^\nu\right)}$ with $q^\mu$ being a rescaled "minkowskian" momentum $q^i(t)=\frac{k_i}{a_i(t)}, \eta^{\mu\nu}k_\mu k_\nu=0$ but failed to succeed.

After some time I've realized that one should probably look not for an exact solution of this type (even in the Robertson-Walker case the relation $\lambda(t)\sim a(t)$ is approximate) but for a "high-frequency" approximate solution. But still I'm not currently able to find it.

So my questions are:

1) Is the suggested generalization of the Hubble's law correct?

2) If so, how to provide quantitative evidence for it? I would be satisfied even with calculations for a massless scalar field if they are simpler to perform in order to achieve the goal.

  • $\begingroup$ Why not use Hamilton-Jacobi equation directly, rather than trying to take high-frequency limit of Maxwell equations? HJ equation is easily solvable, since $\partial_i$ are Killing vectors. $\endgroup$ – user23660 Dec 18 '13 at 18:19
  • $\begingroup$ Intuitively, if we consider a length in a direction given by a unit vector $\vec \alpha$, so such as $dx^i= \alpha^i~ dl$, with $dx^2=dl^2$ and $\sum \alpha_i^2=1$, this gives a proper lengh : $dl_{proper}^2 = \sum \alpha_i^2 ~ a_i^2(t) ~ dl^2$ So, we may hope to have a law like : $\dfrac{\lambda(t)}{\lambda(t_0)} =\dfrac{\sqrt{\sum \alpha_i^2 ~ a_i^2(t)}}{\sqrt{\sum \alpha_i^2 ~ a_i^2(t_0)}}$ $\endgroup$ – Trimok Dec 18 '13 at 19:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.