Blog Post 7 - The Lyapunov Exponent

Welcome back to another blog post! In this post, I’m going to finish reviewing the derivation presented in the Staelens thesis we have been following for the Lyapunov exponent, and then present not only a final analytical expression, but also some graphs of how the Lyapunov exponent might look for Kerr-Newman black holes with different values for its parameters.

Constants of Motion

A key note we haven’t discussed yet is how to actually find the constants of motion $\lambda$ and $\chi$ for bound orbits. We do this by placing two conditions on the radial integral: firstly, that the bound radius is a root of the radial potential (so that the radius is actually fixed), and secondly that the bound radius is also a root of the derivative of the radial potential (this is less obvious immediately, but follows from the fact that the radius remains fixed). Thus we have that $\mathcal{R}(r)={\mathcal{R}}^{\prime }(r)=0$. Finding the roots of our normalized radial potential and its derivative and solving for $\lambda$ and $\chi$, we find that

$$\lambda =\frac{M(r^2-a^2)-r\Delta }{a(r-M)}$$ $$\chi =\frac{4r^2\Delta }{(r-M{)}^2}$$

Interestingly, if you try this yourself, you may notice a $\chi =0$ solution; it turns out you can show that such a solution is unphysical by showing that it leads to the angular integral being devoid of roots.

Now that we have solved for the constants of motion in terms of a bound radius, we can now express our potentials in terms of these values and begin solving the integral equation in the last post.

Deriving the Exponent

We now begin deriving the Lyapunov exponent by simplifying both sides of this integral equation. Let us begin first with the radial side,

$${\int }_{r_B+\delta r_0}^{r_B+\delta r_n}\frac{dr}{\sqrt{\mathcal{R}(r)}}$$

At first, it may appear like this integral is not well-defined; after all, we have stated that the condition $\mathcal{R}=0$ must hold for bound orbits. However, it actually makes sense that this integral is not well-defined for a bound orbit; after all, a bound orbit’s radius will never grow (at least, not a fixed-orbit radius, which we can show that all bound orbits are for Kerr-Newman black holes). Rather, we are interested in evaluating this integral for a nearly bound orbit. To find the value of $\mathcal{R}$ very near a bound orbit but not quite on it, we can perform a Taylor expansion about some bound radius $r_B$ to lowest nonzero order (which is second order in this case) to obtain $\mathcal{R}(r)\approx \frac{1}{2}{\mathcal{R}}^{\prime \prime }(r_B)(r-r_B{)}^2$. Plugging this into our integral expression above, we find that

$${\int }_{r_B+\delta r_0}^{r_B+\delta r_n}\frac{dr}{\sqrt{\mathcal{R}(r)}}=\sqrt{\frac{2}{{\mathcal{R}}^{\prime \prime }(r_B)}}{\int }_{r_B+\delta r_0}^{r_B+\delta r_n}\frac{dr}{r-r_B}=\sqrt{\frac{2}{{\mathcal{R}}^{\prime \prime }(r_B)}}\ln \left(\frac{\delta r_n}{\delta r_0}\right)$$

Computing $R^{\prime \prime }(r)$, we find that for an orbit with bound radius $r$ the radial integral simplifies to

$${\int }_{r_B+\delta r_0}^{r_B+\delta r_n}\frac{dr}{\sqrt{\mathcal{R}(r)}}=\frac{1}{2}\frac{1}{\sqrt{r^2-\frac{Mr\Delta }{(r-M{)}^2}}}\ln \left(\frac{\delta r_n}{\delta r_0}\right)$$

Now we can move on to the angular integral, which you may recall took this form:

$$2n{\int }_{{\theta }_{+}}^{\frac{\pi }{2}}\frac{d\theta }{\sqrt{\Theta (\theta )}}$$

[to be updated soon]