Equations of motions for a spin interacting with a magnetic field.

Question

Consider the Lagrangian

$$L(\theta, \phi) = (-1+\cos \theta) \dot{\phi} + \mathbf{B}\cdot \mathbf{n}$$

Where $\mathbf{B}$ is a constant magnetic field and $\mathbf{n}$ denotes a point on the unit sphere with spherical coordinates $(\theta,\phi)$. This arises in a path integral for spin coherent states. I'm attempting to show that the Euler Lagrange equations reduce to (at worst, up to a few constants):

$$\mathbf{\dot{n}} = \mathbf{B}\times \mathbf{n}$$

One way to do this is to show that:

$$\mathbf{n}\times \mathbf{\dot{n}}=\mathbf{B}$$

As then I could use "cross through" on the left with $\mathbf{n}$ and use a famous vector identity to get the result. So the question is, which one is it best to go for?

After a quick computation, I found the Euler Lagrange equations to be of the form:

$$\sin \theta \dot{\phi} + B_i \dfrac{\partial n_i}{\partial \theta} = 0$$

$$\sin \theta \dot{\theta} - B_i\dfrac{\partial n_i}{d\phi}=0$$

Moreover

$$\epsilon_{ijk}n_j\dot{n}_k = \epsilon_{ijk}n_j\dfrac{\partial n_k}{\partial \theta} \dot{\theta} +\epsilon_{ijk} n_j \dfrac{\partial n_k}{\partial \phi}\dot{\phi}$$

In which I attempted to substitute $\dot{\theta}, \dot{\phi}$ to little success. Any thoughts on how to progress to show the RHS of the above is proportional to $\mathbf{B}$?

Answer 1

There is this (very brute-force type of calculations) way to do it. We let that $\hat{n} = \sin(\theta)\cos(\phi)\hat{x} + \sin(\theta)\sin(\phi)\hat{y}+\cos(\theta)\hat{z}$ and $\vec{B} = B_x\hat{x}+B_y\hat{y}+B_z\hat{z}$ so we will have that

$$\dot{\hat{n}} = \left(\begin{matrix}\color{blue}{\dot\theta} \cos(\theta)\cos(\phi) - \color{red}{\dot \phi \sin(\theta )} \sin(\phi) \\ \color{blue}{\dot \theta} \cos(\theta)\sin(\phi) + \color{red}{\dot \phi \sin(\theta)} \cos(\phi)\\-\color{blue}{\dot \theta} \sin(\theta)\end{matrix}\right)$$ And we also have that

$$\vec{B}\times \hat{n} = \left(\begin{matrix}B_y\cos(\theta) - B_z\sin(\theta )\sin(\phi) \\ B_z\sin(\theta)\cos(\phi) - B_x\cos(\theta)\\ B_x \sin(\theta)\sin(\phi) - B_y\sin(\theta)\cos(\phi)\end{matrix}\right)$$

And we have the lagrangian

$$\mathcal{L}(\theta,\phi,\dot \theta, \dot \phi) = \cos(\theta)\dot\phi - \dot \phi +B_x\sin(\theta)\cos(\phi) + B_y\sin(\theta)\sin(\phi) + B_z\cos(\theta)$$

Wich gives two Euler-Lagrange equations

$$\color{red}{\dot \phi \sin(\theta)} = B_x \cos(\theta)\cos(\phi) + B_y\cos(\theta)\sin(\phi) - B_z\sin(\theta)\tag{1}$$

$$\color{blue}{\dot\theta } = B_x\sin(\phi) - B_y\cos(\phi)\tag{2}$$

Then, substituting equations $(1)$ and $(2)$ in the vector $\dot{\hat{n}}$ we have that

$$\dot{\hat{n}} = \left(\begin{matrix}\color{blue}{\dot\theta} \cos(\theta)\cos(\phi) - \color{red}{\dot \phi \sin(\theta )} \sin(\phi) \\ \color{blue}{\dot \theta} \cos(\theta)\sin(\phi) + \color{red}{\dot \phi \sin(\theta)} \cos(\phi)\\-\color{blue}{\dot \theta} \sin(\theta)\end{matrix}\right) =$$

\begin{equation} \left(\begin{matrix}\color{blue}{(B_x\sin\phi - B_y\cos\phi)} \cos\theta\cos\phi - \color{red}{(B_x \cos\theta\cos\phi + B_y\cos\theta\sin\phi - B_z\sin\theta)} \sin\phi \\ \color{blue}{(B_x\sin\phi - B_y\cos\phi)} \cos\theta\sin\phi + \color{red}{(B_x \cos\theta\cos\phi + B_y\cos\theta\sin\phi - B_z\sin\theta)} \cos\phi\\-\color{blue}{(B_x\sin\phi - B_y\cos\phi)} \sin\theta\end{matrix}\right) = \end{equation}

$$\left(\begin{matrix} -B_y\cos(\theta ) + B_z\sin(\theta)\sin(\phi)\\B_x\cos(\theta) - B_z\sin(\theta)\cos(\phi) \\-B_x\sin(\phi)\sin(\theta) + B_y\sin(\theta)\cos(\phi))\end{matrix}\right) = - \vec{B}\times \hat{n}$$

We then proved that the Lagrange equations reduce to

$$\dot{\hat{n}} = - \vec{B}\times \hat{n} = \hat{n}\times \vec{B}$$