I am given the equations of motion: $$\dot{r} = f(r)p_r,$$ $$\dot{p_r} = -\frac{V'(r)}{2f(r)}-\frac{f'(r){p_r}^2}{2}+\frac{V(r)-E^2}{2f^2(r)}f'(r)$$ Along with the conditions that $$E^2-V(r_0)=V'(r_0)=0$$

I need to linearize the equations of motion around a circular orbit at $r=r_0$ so I tried to calculate the Jacobian matrix using Mathematica:

jacobianMatrix = 
 Grad[{f[r] p[r], -V'[r]/(2 f[r]) - (f'[r] p[r]^2)/
        2 + (V[r] - En^2)/(2 f[r]^2) f'[r]}, {r, p[r]}] /. r -> r0 /. 
    En^2 -> V[r0] /. V'[r0] -> 0 // MatrixForm

which gives me the output:

$$\left( \begin{array}{cc} p_{r_0} f'({r_0}) & f({r_0}) \\ -\frac{1}{2} p_{r_0}^2 f''({r_0})-\frac{V''({r_0})}{2 f({r_0})} & -p_{r_0} f'({r_0}) \\ \end{array} \right)$$

But the given solution in the paper on page 13 is:

$$\delta\dot{r}=\delta p_r,$$ $$\delta \dot{p_r}=-\frac{V''(r_0)\delta r}{2f(r_0)}$$

where I am getting the $-\frac{V''(r_0)}{2f(r_0)}$ term along with extra terms, I am not sure why the other terms vanish. How to get the solution by using the concept of linearization?


1 Answer 1


Because the canonical apparatus is made for not doing geometry on tangent spaces:

Its always easier to take the series expansion of the Hamiltonian in the radius $r$ around some fixed $R$ up ou second order in

$$H(R+s) = H(R)+ H'(R) s + 1/2 H''(R) s^2.$$

For a stable circle solve for $p$


The rest is an harmonic oscillator.

By the principles, any function

$$F = f(q) P$$

generates canonically

$$Q = f(q) , p = f'(q) P = f'(f^1(Q)) P$$

In the same way. probably, your $f(r)$ as was induced into the Hamiltonian.


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