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.


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 '20 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 '20 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 '20 at 18:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.