# What are these physical quantites in electrodynamics?

1. Aug 16, 2009

### mma

What physical quantities are these differential forms in classical electrodynamics?

I read the paper ofhttp://arxiv.org/abs/physics/0005084" [Broken] (A gentle introduction to the foundations of classical electrodynamics: The meaning of the excitations (D,H) and the field strengths (E, B)).

The equation of charge conservation (eq. 2.5 in the paper) and the Faraday's induction law (eq. 5.2) are formally the same, except in the charge conservation appear 3-forms, while in Faraday's induction law 2-forms. I write these equations here.

(2.5)​
$dj + \partial_t \rho = 0$​
(5.2)​
$dE + \partial_t B = 0$​

Because of $d\rho=0$, from the Pioncaré lemma (cited in the paper as de Rham theorem) follows that (on a a contractible 3d manifold) the 3-form $\rho$ is exact, i.e. there is a 2-form D that
(3.1)​
$\rho = dD$.​
The analogy would be that because of $dB=0$ (eq. 5.3 in the paper), from the Pioncaré lemma (cited in the paper as de Rham theorem) follows that (on a a contractible 2d manifold) the 2-form $B$ is exact, i.e. there is a 1-form $$X[/itex] that (?.1) ​ $B = dX$.​ Later on, substititing (3.1) into (2.5) and using again the Poincaré lemma (alias de Rham theorem) we get that the $j + \partial_tD$ 2-form (on a contractible 2d manifold) has an 1-form potential $H$, that is (3.2)​ $j + \partial_tD = dH$.​ The analogy would be that substituting (?.1) into (5.2) and using again the Poincaré lemma (alias de Rham theorem) we get that the $E + \partial_tX$ 1-form (on a contractible 1d manifold) has an 0-form potential $Y$, that is (?.2)​ $E + \partial_tX = dY$.​ What physical quantities correspond to $X$ and $Y$? Are they in use in electrodynamics? If not, then why not? Last edited by a moderator: May 4, 2017 2. Aug 16, 2009 ### mma I found them. $X$ is the vector potential $A$, and $Y$ is the scalar potential $\phi$. Thanks. 3. Aug 16, 2009 ### Ben Niehoff X would be the vector potential A. Y is the scalar potential phi. These equations are usually written [tex]\vec B = \vec \nabla \times \vec A$$

$$\vec E = - \nabla \Phi - \frac{\partial \vec A}{\partial t}$$

Edit: Oh, I see I am a few hours too late. :P