I am asked to compute Hamilton's equations and check that the solutions to said equations are circular paths, centered at the origin, with angular velocity $\xi \in R^3$. The Hamiltonian is given by $H(q,p) = (q \times p)\cdot \xi$ in this example. So, after letting $ q=(q_1,q_2,q_3)$ and $p=(p_1,p_2,p_3)$, we end up with $H(q,p) = \xi_1(q_2p_3-p_2q_3)+\xi_2(p_1q_3-q_1p_3)+\xi_3(q_1p_2-p_1q_2)$. So Hamilton's equations end up being

$q_1'(t)= (\xi_2q_3-\xi_3q_2)(t)$

$q_2'(t)= (\xi_3q_1-\xi_1q_3)(t)$

$q_3'(t)= (\xi_1q_2-\xi_2q_1)(t)$

$p_1'(t)= (\xi_2p_3-\xi_3p_2)(t)$

$p_2'(t)= (\xi_3p_1-\xi_1p_3)(t)$

$p_3'(t)= (\xi_1p_2-\xi_2p_1)(t)$

Now, I am tasked in showing that the solutions to the above equations are circular paths about the origin. Not surprisingly, you get two linear 1st order ODEs:

$p'(t) = Ap(t)$

$q'(t) =Aq(t)$

where $A = \begin{pmatrix} 0 &-\xi_3 &\xi_2 \\\xi_3 &0 &-\xi_1 \\ -\xi_2&\xi_1&0 \end{pmatrix}$

My next step would be to calculate the solution using the matrix exponential, but that seems long and cumbersome. Is there an easier way to show that the solutions are circular paths?


2 Answers 2



  1. Show that if $\vec p$ satisfies the differential equation you wrote, then \begin{align} \frac{d}{dt}|\vec p| = 0. \end{align}
  2. Show that if $\vec p$ satisfies the differential equation you wrote, then \begin{align} \frac{d}{dt}\left|\frac{d\vec p}{dt}\right| = 0. \end{align}
  3. Think about what these two equations mean.
  • $\begingroup$ Not sure what you mean when you say "if p, and if p satisfies..." $\endgroup$ Jan 29, 2019 at 5:27
  • $\begingroup$ @LordVader007 Typo. Fixed. $\endgroup$ Jan 29, 2019 at 5:35
  • $\begingroup$ Hi thanks for the hints, I just realized that this is easier than I was making it. $\endgroup$ Jan 30, 2019 at 1:55

$$\vec{\dot{p}}=A\,\vec{p}\tag 1$$

$$\vec{\dot{q}}=A\,\vec{q}\tag 2$$

The solutions of equation (1) and (2) is:

$\vec{p}=e^{A\,t}\,\vec{p}_0\quad,\vec{q}=e^{A\,t}\,\vec{q}_0$ ,where $\vec{p}_0\,,\vec{q}_0$ the initial condition

the requirement for a circular motion :

$$\vec{p}^2+\vec{q}^2=\text{const}\tag 3$$

so is equation(3) fulfill ?


$\vec{p}^2=\vec{p}^T\,\vec{p}=\vec{p}_0^T\,e^{A^T\,t}\,e^{A\,t}\,\vec{p}_0= \vec{p}_0^T\,e^{\left(A^T+A\right)t}\,\vec{p}_0=\vec{p}_0^T\,\vec{p}_0=\text{const}$


$A^T+A=\mathbb{0}\quad $ ,$A\quad$ is skew matrix!

  • $\begingroup$ I get that they should be constant, but how do you know that $(e^{At})^2 = const.$? After plugging in $p, q$ into the equation, you'd get $(p_0^2 +q_0^2)(e^{At})^2$ , and may not be constant because of the exponential term. $\endgroup$ Jan 30, 2019 at 8:48
  • 1
    $\begingroup$ see the new documentation $\endgroup$
    – Eli
    Jan 30, 2019 at 9:24

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