Here's the problem and corresponding question.

Let's consider a uniform magnetic field $\vec{B}=B_0\hat{z}$. Looking for the solution of vector potential satisfying $\vec{B}=\vec{\nabla}\times \vec{A}$, we can find that its condition is $\partial _x A_y - \partial _y A_x = B_0$. We can have either $\vec{A}=(-y,0,0)$ or $\vec{A}=(0,x,0)$ or a linear combination $\vec{A}=(-y,x,0)$. All these three give the same magnetic field $\vec{B}=B_0\hat{z}$.

On the other hand, we can easily check that the Lagrangian $L=\frac{1}{2}m(\dot{x}^2+\dot{y}^2)+qB_0(x\dot{y}-\dot{x}y)$ is consistent with $\vec{F}=q\vec{v}\times\vec{B}$ by substituting it into Euler-Lagrange equation.

Taking $\vec{A}=(-y,x,0)$, we conclude that the Lagrangian in terms of vector potential should take the form of $L=\frac{1}{2}m(\dot{x}^2+\dot{y}^2)+q\vec{v} \cdot \vec{A}$. However, this form seems not consistent when we choose our vector potential to be $\vec{A}=(-y,0,0)$ or $\vec{A}=(0,x,0)$. So my questions are:

  1. Is $\vec{A}=(-y,x,0)$ the only vector potential that is physically reasonable to choose, and if so, why?

  2. If other choices of vector potentials are reliable choices as well, how should we construct our Lagrangian such that we can be consistent with the Lorentz force description?


1 Answer 1


The Lagrangian $$ L=\frac{m}{2}\mathbf{\dot x}^2 +q\mathbf{\dot x}\cdot \mathbf{A}(\mathbf{x}) $$ works fine for any time-independent $\mathbf{A}$ with $\nabla\times \mathbf{A}=\mathbf{B}$. To confirm this, use $$ \frac{\delta L}{\delta x_k} = q\,\mathbf{\dot x}\cdot (\nabla_k\mathbf{A}) $$ and $$ \frac{\delta L}{\delta \dot x_k} = m\dot x_k + q\, A_k $$ to get the Euler-Lagrange equation of motion $$ m\ddot x_k + q\, \mathbf{\dot x}\cdot\nabla A_k = q\,\mathbf{\dot x}\cdot (\nabla_k\mathbf{A}). $$ (I used the identity $\dot A_k=\mathbf{\dot x}\cdot\nabla A_k$.) Re-arranging and making all of the indices explicit gives $$ m\ddot x_k = q\,\sum_J\big(\dot x_j\nabla_k A_j - \dot x_j\nabla_j A_k\big), $$ which may also be written $$ m\mathbf{\ddot x}\propto q\,\mathbf{\dot x}\times\mathbf{B}. $$

  • $\begingroup$ Is there a reason why you used delta-symbol instead of partial derivatives? $\endgroup$
    – Nugi
    Aug 21, 2019 at 4:56
  • $\begingroup$ @YeonwookJung One reason for this convention comes from field theory, where $L$ is a lagrangian density that can be regarded as a function of both $\phi$ and $\partial_\mu\phi$ for some field $\phi$. Then the Euler-Lagrange equation involves $\delta L/\delta(\partial_\mu\phi)$, which could look confusing if we wrote $\partial L/\partial(\partial_\mu\phi)$. Writing $\partial L/\partial\cdots$ would be fine when the dynamic variablers are $x$ and $\dot x$, but I tend to always write $\delta L/\delta\cdots$ because I often work with fields. $\endgroup$ Aug 22, 2019 at 12:46

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.