$d(\iota_v\rho)=0 \implies d(\phi\iota_v\rho)=d\phi(v)\rho$?

Question

My motivation is physical, but my question is purely mathematical.

Everybody knows, that the power of the electric current in a piece of wire is $$P=UI$$ where

• the wire is regular domain $V$ in a 3-dimensional orientable differentiable manifold M: it is a flow tube between two level set pieces $A$ and $B$ of a function $\phi: M\to \mathbb R$,
• $U = \phi(B)-\phi(A)$.
• $I = \int_B\iota_v\rho$, where
  • $v$ is a vector field: ths is the velocity field of the moving charges.
  • $\rho$ is a 3-form: this is the charge density 3-form, so
  • $\iota_v\rho$ is the current density 2-form.

We suppose that

1. the current is stationary, i.e. $d(\iota_v\rho) = 0$
2. the velocity field $v$ of the moving charges in the wire is everywhere tangent to the side wall of the tube.

Because of these assumptions $$0=\int_V d(\iota_v\rho) =\int_{\partial V}\iota_v\rho=\int_B\iota_v\rho+\int_A\iota_v\rho = 0$$

hence $\int_A\iota_v\rho = -I$.

Because of assumption 2. $$\int_{\partial V}\phi\iota_v\rho=\int_B\phi\iota_v\rho+\int_A\phi\iota_v\rho = \phi(B)\int_B\iota_v\rho+\phi(A)\int_A\iota_v\rho = (\phi(B)-\phi(A))I=UI=P$$ By applying Stokes theorem, this means that $$ P = \int_Vd(\phi\iota_v\rho)$$ On the other hand, we know also, that the power of the electric field $E=d\phi$ is $$ P = \int_V E(v)\rho = \int_V d\phi(v)\rho$$ That's why I suspect that for any vector field $v$ and 3-form $\rho$ having $d(\iota_v\rho)=0$ and for any function $\phi$
$$d(\phi\iota_v\rho)=d\phi(v)\rho$$ Is this really true?

Answer 1

For a function $f$ and a form $\omega$, we have the product rule $d(f \omega) = df \wedge \omega + f d\omega$. Thus $$d (\phi \iota_v \rho ) = d\phi\wedge\iota_v \rho + \phi d(\iota_v \rho)=d \phi\wedge \iota_v \rho.$$

Now in general this is different to $d\phi(v) \rho$; but in 3 dimensions they are equal. To see this, let's choose coordinates so that at the point of interest, $v = \partial_1$ and $\rho = dx^1 \wedge dx^2 \wedge dx^3$. (Note that this is only possible because $\rho$ is a three-form in three dimensions.) Then we have

$$d\phi \wedge \iota_v \rho = d\phi \wedge dx^2 \wedge dx^3 = \partial_i \phi ~dx^i \wedge dx^2 \wedge dx^3.$$

Note that $dx^i\wedge dx^2 \wedge dx^3$ is nonzero only when $i=1$ (in three dimensions!), so this is equal to

$$\partial_1 \phi ~ dx^1 \wedge dx^2 \wedge dx^3.$$

But $\partial_1 \phi = d\phi(\partial_1) = d\phi(v)$, so recognizing $\rho$ we see that this is $d \phi(v) \rho$ as desired.