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

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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: ${\begin{aligned}\oint _{\partial \Sigma }&\mathbf {B} \cdot \mathrm {d} {\boldsymbol {\ell }}==\mu _{0}\varepsilon _{0}{\frac {\mathrm {d} }{\mathrm {d} t}}\iint _{\Sigma }\mathbf {E} \cdot \mathrm {d} \mathbf {S} \\\end{aligned}}$

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)