I am reading chapter 7 in the 3rd edition of Goldstein's Classical mechanics textbook and the expression for the Lorentz force is confusing me. I cannot scan it so I am just going to write it out verbatim and formulate my question afterwards. Here is the extract of page 298 from the text:

In terms of $\phi$ and $\mathbf{A}$, the Lorentz force is $$\mathbf{F} = q\{-\nabla\phi+\frac{1}{c}\frac{\partial \mathbf{A}}{\partial t} + 1[v \times(\nabla\times \mathbf{A})]\}\tag{7.67c}.$$ This suggests that we should generalize the force law to $$\frac{dp_{\mu}}{d\tau} = q\left(\frac{\partial (u^\nu A_\nu)}{\partial x^\mu}-\frac{dA_\mu}{d\tau}\right).\tag{7.68}$$

The first equation is the three three dimensional Lorentz force express using the vector and scalar potentials (As a note I think the second term should be $-\frac{\partial A}{\partial t}$ but the above is as written.)

I am unsure howhow you reach the second equation from the first expression, I would appreciate any help in understanding this problem.

  • $\begingroup$ I forgot the details, but there's the mechanical momentum and then a more generalized momentum, $\vec{p}=\vec{p}-q\vec{A}$. This helps account for momentum in the EM field. Now that you have a generalized momentum, you can derive a compatible Laplacian or Hamiltonian. From there, you can probably derive what you seek. $\endgroup$
    – R. Romero
    Nov 10, 2023 at 18:59
  • $\begingroup$ More details: physics.stackexchange.com/questions/356063/… $\endgroup$
    – R. Romero
    Nov 10, 2023 at 19:02

1 Answer 1


TL;DR: The total derivative term $$\frac{dA_\mu}{d\tau}~=~\gamma\frac{dA_\mu}{dt} ~=~\gamma\left(\vec{v}\cdot\vec{\nabla} A_\mu+ \frac{\partial A_\mu}{\partial t}\right)$$ in eq. (7.68) is correct. It should not be a partial derivative.

Before trying to read the relativistic formulation in section 7.6, I would strongly recommend you to fully understand the non-relativistic derivation in section 1.5, which essentially features the same issue.

  • $\begingroup$ Hi, thanks for your response, sorry I took so long to respond, I had a lot on last week. I appreciate your your edits as it now looks a lot clearer but you changed some of the coefficient terms in the first equation. I have re-read my copy and I was correct the first time, so I don't know why you changed it. Is your version different to mine? $\endgroup$ Nov 17, 2020 at 17:59
  • $\begingroup$ Hi Kristian Stokkereit. I re-check my 3rd edition, and it agrees with what I wrote, so you must indeed have a different version. Anyway, feel free to edit it back. $\endgroup$
    – Qmechanic
    Nov 17, 2020 at 18:11
  • $\begingroup$ Thanks is this given in cgs units as in SI, the units are not the same throughout the equation unless the 1/c coefficient is removed. $\endgroup$ Nov 17, 2020 at 18:19

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