1
$\begingroup$

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?

$\endgroup$

1 Answer 1

1
$\begingroup$

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 $p$

$$H'(R)==0$$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.