0
$\begingroup$

In part II, lecture 18, of Feynman's Lectures on Physics, on Table 18-1 Feynman writes Ampère's law as $$ c^2 \nabla \times \vec{B} = \frac{\vec{j}}{\epsilon_0} + \frac{\partial \vec{E}}{\partial t}. $$

What unit system is this in? It's not obviously Gaussian, which states, $$ c \nabla \times \vec{B} = 4 \pi \vec{j} + \frac{\partial \vec{E}}{\partial t}. $$

Nor does it appear to be formulated in SI, which states,

$$ \frac{1}{\mu_0} \nabla \times \vec{B} = \vec{j} + \epsilon_0 \frac{\partial \vec{E}}{\partial t}. $$

$\endgroup$
  • 3
    $\begingroup$ This appears to be using SI convention, taking advantage of the relation $\epsilon_0\mu_0 c^2 = 1$ $\endgroup$ – By Symmetry Oct 21 at 17:49
  • 1
    $\begingroup$ @BySymmetry Shouldn't that be an answer instead of a comment? $\endgroup$ – Emilio Pisanty Oct 21 at 18:03
2
$\begingroup$

Multiply $$ c^2 \nabla \times \vec{B} = \frac{\vec{j}}{\epsilon_0} + \frac{\partial \vec{E}}{\partial t} $$ through by $\epsilon_0$, and substitute in the identity $$ c^2 = \frac1{\epsilon_0\mu_0} , $$ and it will be quickly revealed to be in the SI form you quote.

$\endgroup$

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.