Wikipedia:Reference desk/Archives/Science/2022 December 10

= December 10 =

Maxwell's third and fourth equations
Using Formulation in SI units convention and ignoring the conventional current term of Ampere-Maxwell equation (focusing solely on the displacement current term), we have: $$ \begin{align} \oint_{\partial \Sigma} & \mathbf{B} \cdot \mathrm{d}\boldsymbol{\ell} = = \mu_0 \varepsilon_0 \frac{\mathrm{d}}{\mathrm{d}t} \iint_{\Sigma} \mathbf{E} \cdot \mathrm{d}\mathbf{S} \\ \end{align} $$

So my question is: why is the Maxwell-Faraday equation in the following form: $$\oint_{\partial \Sigma} \mathbf{E} \cdot \mathrm{d}\boldsymbol{\ell} = - \frac{\mathrm{d}}{\mathrm{d}t} \iint_{\Sigma} \mathbf{B} \cdot \mathrm{d}\mathbf{S} $$ and not in the following form: $$\oint_{\partial \Sigma} \mathbf{E} \cdot \mathrm{d}\boldsymbol{\ell} = - \mu_0 \varepsilon_0\frac{\mathrm{d}}{\mathrm{d}t} \iint_{\Sigma} \mathbf{B} \cdot \mathrm{d}\mathbf{S} $$ ?

Many thanks, 173.209.130.10 (talk) 21:29, 10 December 2022 (UTC)
 * The units wouldn't match. The SI unit of the electric field is $$[E] = 1\,\mathrm{\frac{N}{As}}$$, of the magnetic field $$[B] = 1\,\mathrm{\frac{N}{Am}}$$. With that the unit of the left hand side is $$\mathrm{\frac{N}{As}\cdot m} = \mathrm{\frac{Nm}{As}}$$, of the right hand side it is $$\mathrm{\frac{1}{s}\frac{N}{Am}\cdot m^2} = \mathrm{\frac{Nm}{As}}$$, i.e. the same, as it must be. The factor $$\varepsilon_0 \mu_0 = 1/c^2$$ with unit $$\mathrm{\frac{s^2}{m^2}}$$, and if you insert that as you propose, you mess up the units of the right hand side. --Wrongfilter (talk) 21:47, 10 December 2022 (UTC)
 * Using natural units in which $$c=1,$$ the identities take on a more symmetric appearance. --Lambiam 05:13, 11 December 2022 (UTC)