is it posible to express an integral version of Biot-Savart- Laplace law $$\vec B
=\frac{\mu_0}{4\pi}\int\frac{I\mathrm{d}\vec{l}\times (\vec {r }-\vec {l})}{|\vec {r }-\vec {l} |^3}
=\frac{\mu_0}{4\pi}\int_V\frac{\vec{J}\times (\vec {r }-\vec {r^\prime})}{|\vec {r }-\vec {r^\prime} |^3}\mathrm{d}^3\vec {r^\prime },
$$
by (exterior) differential forms?
I know that J is 2-form and that cross product using differential forms must be 3-form, so that integral is over volume. So the integrand should be $J\wedge r$, where r is 1-form obtained from vector field $(\vec {r }-\vec {r^\prime})$.
BUT: Integral of differential form is real number and not vector/form.
Thanks for any ideas.
Best Answer
Hint: You need to use the 'fibre-wise' integration of forms: if $p:\mathbb R^m\to\mathbb R^n$ is the projection ($m\geq n$), or more generally $p:M\to N$ is a submersion of oriented manifolds, and if $\alpha\in\Omega^k(M)$, then its fibre-wise integral $p_*\alpha\in\Omega^{k-m+n}(N)$ can be defined e.g. via ($\forall\beta\in\Omega(N)$) $\int_N(p_*\alpha)\wedge\beta=\int_M\alpha\wedge p^*\beta$. In your case you have $p,q:\mathbb R^3\times\mathbb R^3\to\mathbb R^3$ the projection to $\vec r$ and to $\vec{r'}$ respectively (or $p:\mathbb R^3\times\Gamma\to\mathbb R^3$ and $q:\mathbb R^3\times\Gamma\to\Gamma$ where the curve $\Gamma$ is the wire) and you're looking for a form $\kappa$ on $\mathbb R^3\times\mathbb R^3$ such that $B = p_*(\kappa\wedge q^* J)$.
I will, however, let you figure out the formula for $\kappa$ and the degrees of all the forms appearing here ;)