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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?

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  • $\begingroup$ $A_\mu J^\mu$ is not called potential energy, only $q\phi$ is. $\endgroup$ Jul 18, 2022 at 23:34
  • $\begingroup$ @JánLalinský why not? isn't $U$ called the potential energy? can you give a source? $\endgroup$
    – Rd Basha
    Sep 16, 2023 at 7:53
  • $\begingroup$ 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. $\endgroup$ Sep 16, 2023 at 12:19

1 Answer 1

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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.


So, your main error was the mis-definition of the electric field and in the following statement:

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.

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  • $\begingroup$ Yep. Thanks! :) $\endgroup$
    – Rd Basha
    Jul 17, 2022 at 19:06

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