Lorentz force from potential- extra term?

I'm trying to verify the E.M potential energy $$U= \int{A_\mu J^\mu} = q(\phi - A_j v^j )$$ by using the connection: $$F= - \frac{\partial U}{\partial r} + \frac{d}{dt} \frac{\partial U}{\partial v}$$ with $$F=q(E+v \times B)$$.

I seem to have some extra term.

We work in units where $$q=1$$.

The L.H.S: $$F_i=E_i + (v \times B)_i = E_i + \epsilon_{ijk} v_j B_k = \\= - \frac{\partial \phi}{\partial r^i}-\dot{A}_i + \epsilon_{kij} \cdot v_j \cdot \epsilon_{klm}\partial_lA_m = \\ = - \frac{\partial \phi}{\partial r^i}-\dot{A}_i + v_j \partial_lA_m \cdot \left( \delta^l_i \delta^m_j - \delta^m_i \delta^l_j \right) = \\ = - \frac{\partial \phi}{\partial r^i}-\dot{A}_i + v_j \partial_i A_j - v_j \partial_jA_i .$$

Now, the last term is:

$$v_j \partial_jA_i= \frac{dr^j}{dt} \frac{ \partial A^i}{\partial r^j }= \frac{dA_i}{dt} = \dot{A_i}$$

So we get the L.H.S: $$- \frac{\partial \phi}{\partial r^i}-\dot{A}_i + v_j \partial_i A_j - \dot{A_i}$$

The R.H.S (first term):

$$- \frac{\partial U}{\partial r^i} = - \frac{\partial (\phi-A_j v_j )}{\partial r^i} \\ = - \frac{\partial \phi}{\partial r^i} + v_j \partial_i A_j$$

The R.H.S (second term):

$$\frac{d}{dt} \frac{\partial U}{\partial v^i} = \frac{d}{dt} \frac{\partial }{\partial v^i} \left( -A_j v_j \right) = -\frac{d}{dt} \left( A_i \right) = -\dot{A}_i$$

So the R.H.S gives: $$- \frac{\partial \phi}{\partial r^i} + v_j \partial_i A_j -\dot{A}_i$$

and there is a $$-\dot{A}_i$$ term difference. What am I missing?

• $A_\mu J^\mu$ is not called potential energy, only $q\phi$ is. Jul 18, 2022 at 23:34
• @JánLalinský why not? isn't $U$ called the potential energy? can you give a source? Sep 16, 2023 at 7:53
• It isn't. "Potential energy" refers to a function of coordinates of the system, from which forces can be derived as partial derivatives with respect to coordinates, so kinetic plus potential energy is conserved. Your $U$ is a function of velocity too, and kinetic energy plus your $U$ is not conserved. It's a Lagrangian term, not a potential energy. Sep 16, 2023 at 12:19

You're messing up partial vs total derivatives with the $$\dot{A}_i$$ term. The electric field is \begin{align} \mathbf{E}&=-\nabla\phi-\frac{\partial \mathbf{A}}{\partial t}. \end{align} Recall that these fields depend on $$(t,x,y,z)$$, on time and position; also when calculating $$\frac{d}{dt}$$, you need to keep in mind that you're evaluating along a particle's trajectory so you have $$x,y,z$$ as functions of time as well. So, in components, \begin{align} E_i=-\frac{\partial \phi}{\partial r^i}-\frac{\partial A_i}{\partial t}. \end{align} So, with this, the Lorentz force we expect is given in components by \begin{align} F_i=q\left[\left(-\frac{\partial \phi}{\partial r^i}-\frac{\partial A_i}{\partial t}\right) + \left(\frac{\partial A_j}{\partial r^i}v^j-\frac{\partial A_i}{\partial r^j}v^j\right)\right]. \end{align} You pretty much have this expression written down as your LHS, I just wanted to point out that instead of $$\dot{A}_i$$, you should have written $$\frac{\partial A_i}{\partial t}$$.
Now, we go to the RHS . You actually have all the right expressions, but you're not using the chain rule correctly. We have \begin{align} -\frac{\partial U}{\partial r^i}+\frac{d}{dt}\left(\frac{\partial U}{\partial v^i}\right)&=-q\left(\frac{\partial \phi}{\partial r^i}-\frac{\partial A_j}{\partial r^i}v^j\right)+ (-q\dot{A_i})\\ &=-q\left(\frac{\partial \phi}{\partial r^i}-\frac{\partial A_j}{\partial r^i}v^j\right) -q\left(\frac{\partial A_i}{\partial t}+\frac{\partial A_i}{\partial r^j}v^j\right)\\ &=q\left[\left(-\frac{\partial \phi}{\partial r^i}-\frac{\partial A_i}{\partial t}\right) + \left(\frac{\partial A_j}{\partial r^i}v^j-\frac{\partial A_i}{\partial r^j}v^j\right)\right]\\ &=F_i, \end{align} where it is the second equal sign that the chain rule is used.
Now, the last term is: \begin{align} v_j \partial_jA_i= \frac{dr^j}{dt} \frac{ \partial A^i}{\partial r^j }= \frac{dA_i}{dt} = \dot{A_i} \end{align}
You're missing the $$\frac{\partial A_i}{\partial t}$$ term.