3
$\begingroup$

If we have one ideal dipole $\mathbf{p}$ in one electric field $\mathbf{E} = E_0\mathbf{\hat{z}}$ we know that the potential energy is:

$$U = -\mathbf{p}\cdot \mathbf{E}.$$

Once we know that the dipole has a initial orientation, one can derive that with time it will evolve until it aligns with $\mathbf{E}$.

I wanted to describe this in the Hamiltonian formalism. The reason for that is to latter use this in the context of statistical mechanics, to compute the partition function.

For that I thought on using the dipole orientation $(\theta,\phi)$ as generalized coordinates, since one ideal dipole is just a vector and since its magnitude is fixed.

In that setting since $\mathbf{E}$ is uniform, in the direction $\mathbf{\hat{z}}$ we have directly that

$$U = -E_0 p\cos \theta.$$

Because of that we could infer that

$$H(\theta,\phi,p_{\theta},p_{\phi})=-E_0p\cos \theta.$$

But this doesn't seem right, because when I try to derive the equations for the evolution of the system we have:

$$\dfrac{dp_{\theta}}{dt}=-\dfrac{\partial H}{\partial \theta}=-E_0p\sin\theta,$$

$$\dfrac{dp_{\phi}}{dt}=-\dfrac{\partial H}{\partial \phi}=0,$$

$$\dfrac{d\theta}{dt}=\dfrac{\partial H}{\partial p_{\theta}}=0,$$

$$\dfrac{d\phi}{dt}=\dfrac{\partial H}{\partial p_{\phi}}=0.$$

Now, this tells that $\theta = \theta_0$ which is certainly wrong, since with time the dipole tends to align with the field.

I've also tried to start from the Lagrangian $L = T - V$, but that's no good. The dipole is not actualy moving, indeed the dipole here is just a vector fixed somewhere with the orientation changing, so it seems that $T = 0$. With that if we were to derive the momentum $p_\theta$ and $p_\phi$ from the Lagrangian we would get just $p_\theta = p_\phi = 0$.

What am I doing wrong here? How does one treat one ideal electric dipole in Hamiltonian Mechanics?

$\endgroup$
2
$\begingroup$

Your 'ideal' dipole is a uniform rigid rod with no mass (and hence its moment of inertia about any axis, through any point, is zero). As such, $p_{\theta}$ and its derivative $\dot{p_{\theta}}$ are trivially (always) zero since, assuming the dipole is fixed at its centre but is free to rotate in $\theta$ and $\phi$, $p_{\theta} = I_{CoM}\dot{\theta}$. Since $I_{CoM}$ is zero, so is $p_{\theta}$. The linear analogue of this would be trying to derive the equations of motion for a massless particle.

If you give your dipole a moment of inertia about an axis through its centre of mass and perpendicular to its length $I \equiv I_{Rod,CoM}$ (alternatively, you could model the dipole as a massless rod with two point masses on either end, giving a different $I$), and call the dipole moment $w$ (I won't use $p$ to avoid confusion with the momenta) one obtains the following Lagrangian (defining the $z$-axis (polar axis) to lie in the direction of $\vec{E_0}$):

$$\mathcal{L} = \frac{I}{2}[\dot{\theta}^2+sin(\theta)^2 \dot{\phi}^2] + wEcos(\theta)cos(\phi)$$

Note the presence of $cos{(\phi)}$ in the dipole interaction term; as per your original post, we're allowing freedom in both $\phi$ and $\theta$. This gives rise to the following conjugate momenta:

$$p_{\theta} = I \dot{\theta}\ ,\ p_{\phi} = Isin(\theta)^2 \dot{\phi}$$

This gives the Hamiltonian, expressed in $p_{\theta}$, $\theta$, $p_{\phi}$ and $\phi$ only:

$$H = \frac{p_{\theta}^2}{2I} + \frac{p_{\phi}^2}{2Isin(\theta)^2} - wEcos(\theta)cos(\phi)$$

From this, one can turn the handle and obtain the expressions for $\dot{\theta}$, etc, as desired, using Hamilton's equations as stated in your original post.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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