Does Lyapunov exponent equate to exponential inflation?

Physics can be modeled by dynamical systems $f^t(x)$ as well as by PDEs. The most common dynamical system has hyperbolic fixed point and can be an attractor or a repellor. The dynamics at repellors and attractors are simple exponential expansion or contraction and are represented by a Lyapunov exponent. My question is how could one prove or disprove that exponential inflation is just an example of a repellor with a Lyapunov exponent?

• Do you have something particular in mind for exponential inflation? – Wrzlprmft Apr 2 '14 at 7:17
• I can't say I fully understand this question, but there are several papers which use a Lyapunov exponent to describe inflation: ms.mcmaster.ca/craig/BrandenbergerCraig08c.pdf and sbfisica.org.br/bjp/files/v31_131.pdf – DavePhD Apr 8 '14 at 16:30
• you might want to change your title, as 'equating' is not the same as 'is just an example of'. – chris Apr 9 '14 at 8:35
• @DavePhD your reply qualifies as an answer as the citations are a useful starting place for me. – Daniel Geisler Apr 9 '14 at 9:38
• @DavePhD it is worth noting that neither of the papers you cited describe inflation with a Lyapunov exponent. They use the exponent to describe the exponential growth in the equations of motion of low-density matter and they both explicitly state that this description is done in a non-expanding universe – Jim Apr 9 '14 at 19:05

I suspect that you cannot prove it purely from experimentally/observational, and I think that approach is useless.

If you think that the whole universe as a single state in state space, then there is no way you can compare with any other state. You may think that there might be a phase space with (macroscopic) parameters $(V,\{\alpha_i\})$, so we can have a maximum Lyapunov exponents $\xi$ and a vector $\delta Z$ deviated from a state $Z_0$. The resulting state might be described by $Z_0+e^{\xi t}\delta Z$.

However, why do you need a phase space description? The other phase space states do not matter to us at all since we can't observe their dynamic. An exponential inflating universe can be simply described by $\dot{V}=kV$. In comparison, most dynamic system can actually be reproduced by repeating the experiments and this will give us the dynamic of the whole phase space.

• Cosmological inflation would be something that happens to a single dynamical system (the universe).
• The exponential divergence of phase-space trajectories (as described with the Lyapunov exponent) is something seen in the comparison of two (nearby) trajectories in a dynamical system. Translated to the universe example, we would look at two almost identical universes and see whether their states diverge from each other.

The two have roughly as much in common as the phenomenon of a single balloon exploding and two balloons drifting away from each other in the wind¹. Equating the two without describing how does not make any sense whatsoever and there isn’t an obvious how. It feels a bit like equating exponential population growth and cosmic inflation, just because they are both exponential (and those have at least in common that they both phenomena of a single trajectory of a system).

Also note that:

• A temporarily expanding system can still have no positive Lyapunov exponent. And temporarily can mean time scales of the total age of the universe. If our universe is destined to have a big crunch, it would still only be temporarily expanding.
• A temporarily shrinking system can still have a positive Lyapunov exponent.
• At most, some mechanism for exponential inflation could induce a big time-localised Lyapunov exponent in a system, but for purposes of determining the time-averaged Lyapunov exponent (which one usually does for the questions that are usually addressed with Lyapunov exponents), this would just be an annoying artefact.

¹ Mind that this is not a perfect analogy, but should suffice to illustrate the absurdity of equating the two.