4.1. Force, moment, and power on elementary charge distributions

4.1.1. Force, moment and power on a point electric charge

Point electric charge with charge \(q\) in a point \(\vec{r}_P(t)\) at time \(t\) where electromagnetic field is \(\vec{e}(\vec{r},t)\), \(\vec{b}(\vec{r},t)\):

• Lorentz’s force

  \[\vec{F} = q \left( \vec{e}(\vec{r}_P(t), t) - \vec{b}(\vec{r}_P(t),t) \times \vec{v}_P(t) \right) \ ,\]

• zero moment, since it has no dimension (and assumed uniform or symmetric or… distribution of electric charge)

• power

  \[\begin{split}\begin{aligned} P & = \vec{v}_P(t) \cdot \vec{F} = \\ & = \vec{v}_P(t) \cdot \, q \, \left( \vec{e}(\vec{r}_P(t), t) - \vec{b}(\vec{r}_P(t), t) \times \vec{v}_P(t) \right) = q \, \vec{v}_P(t) \cdot \vec{e}(\vec{r}_P(t),t) \ . \end{aligned}\end{split}\]

4.1.2. Force, moment and power on a electric dipole

Electric dipole with center \(\vec{r}_C(t)\), axis \(\vec{\ell}\), so that the positive charge \(q\) is in \(P_+ = C + \dfrac{\vec{\ell}}{2}\) and the negative charge is in \(P_- = C - \dfrac{\vec{\ell}}{2}\), with \(q \rightarrow +\infty\), \(|\vec{\ell}| \rightarrow 0\), s.t. \(q|\vec{\ell}| = |\vec{d}|\) finite.

Kinematics and expansion of the field

\[\vec{v}_{\pm} = \vec{v}_C \pm \vec{\omega} \times \frac{\vec{\ell}}{2}\]

\[\vec{e}(P_{\pm}) = \vec{e}\left( C \pm \dfrac{\vec{\ell}}{2} \right) = \vec{e}(C) \pm \dfrac{\vec{\ell}}{2} \cdot \nabla \vec{e}(C) + o(|\vec{\ell}|)\]

\[\vec{b}(P_{\pm}) = \vec{b}\left( C \pm \dfrac{\vec{\ell}}{2} \right) = \vec{b}(C) \pm \dfrac{\vec{\ell}}{2} \cdot \nabla \vec{b}(C) + o(|\vec{\ell}|)\]

Net force.

\[\begin{split}\begin{aligned} \vec{F} & = \vec{F}_+ + \vec{F}_- = \\ & = q \left[ \vec{e}(P_+) - \vec{b}(P_+) \times \vec{v}_{+} \right] - q \left[ \vec{e}(P_-) - \vec{b}(P_-) \times \vec{v}_{-} \right] = \\ & = q \left[ \vec{e}_C + \dfrac{\vec{\ell}}{2} \cdot \nabla \vec{e}_C - \left( \vec{b}_C + \dfrac{\vec{\ell}}{2} \cdot \nabla \vec{b}_C \right) \times \left( \vec{v}_C + \vec{\omega} \times \dfrac{\vec{\ell}}{2} \right) \right] + \\ & - q \left[ \vec{e}_C - \dfrac{\vec{\ell}}{2} \cdot \nabla \vec{e}_C - \left( \vec{b}_C - \dfrac{\vec{\ell}}{2} \cdot \nabla \vec{b}_C \right) \times \left( \vec{v}_C - \vec{\omega} \times \dfrac{\vec{\ell}}{2} \right) \right] = \\ & = q \vec{\ell} \cdot \nabla \vec{e}(C) - \left( q \vec{\ell} \cdot \nabla \vec{b}(C) \right) \times \vec{v}_C - \vec{b}(C) \times \left( \vec{\omega} \times q \vec{\ell} \right) + o(|\vec{\ell}|) \end{aligned}\end{split}\]

Net moment, w.r.t. \(C\).

\[\begin{split}\begin{aligned} \vec{M}_C & = \frac{\vec{\ell}}{2} \times \vec{F}_+ - \frac{\vec{\ell}}{2} \times \vec{F}_- = \\ & = q \frac{\vec{\ell}}{2} \times \left[ \vec{e}(P_+) - \vec{b}(P_+) \times \vec{v}_{+} \right] + q \frac{\vec{\ell}}{2} \times \left[ \vec{e}(P_-) - \vec{b}(P_-) \times \vec{v}_{-} \right] = \\ & = q \frac{\vec{\ell}}{2} \times \left[ \vec{e}_C + \dfrac{\vec{\ell}}{2} \cdot \nabla \vec{e}_C - \left( \vec{b}_C + \dfrac{\vec{\ell}}{2} \cdot \nabla \vec{b}_C \right) \times \left( \vec{v}_C + \vec{\omega} \times \dfrac{\vec{\ell}}{2} \right) \right] + \\ & + q \frac{\vec{\ell}}{2} \times \left[ \vec{e}_C - \dfrac{\vec{\ell}}{2} \cdot \nabla \vec{e}_C - \left( \vec{b}_C - \dfrac{\vec{\ell}}{2} \cdot \nabla \vec{b}_C \right) \times \left( \vec{v}_C - \vec{\omega} \times \dfrac{\vec{\ell}}{2} \right) \right] = \\ & = q \vec{\ell} \times \left[ \vec{e}_C - \vec{b}_C \times \vec{v}_C \right] + o(|\vec{\ell}|) \ . \end{aligned}\end{split}\]

Power. Check, discuss the effects of motion, and uncomment

4.1.3. Force, moment and power on a magnetic dipole

On an elementary magnetic dipole, modeled as a “small” circuit with current \(i\) enclosing area \(S\) and center \(C\), with \(S \rightarrow 0\), \(i \rightarrow + \infty\) so that \(i S \hat{n} := \vec{m}\) finite. Here, a circular loop with center \(C\) and radius \(r\), lying in plane with normal \(\hat{n}\) is considered.

Force. The force acting on such an elementary loop immersed in a magnetic field \(\vec{b}(P)\) reads,

\[\vec{F} = \nabla \vec{b}(C) \cdot \vec{m} \ ,\]

being \(\nabla \vec{b}(C)\) the gradient of the magnetic field, evaluated in the point \(C\).

Force on an Amperian loop

The force is evaluated as the integral of the elementary contributions acting on all the points of the loop. The elementary force contribution acting on an elementary segment \(d \vec{\ell}(P)\) containing point \(P\) and tangent to the loop, with direction \(\hat{t}(P)\), follows Biot-Savart elementary law

\[d \vec{F}(P) = - i \vec{b}(P) \times d \vec{\ell}(P) \ . \]

Here, a Cartesian basis \(\{ \hat{l}, \hat{m}, \hat{n}\}\) is introduced, and the points of the Amperian loop are described with the angle \(\theta\) between the \(\hat{l}\) vector and the radius \(\vec{r} = P - C\) connecting the point \(P\) of the loop with the center \(C\). The magnetic field is approximated with its Taylor expansion as the dimension of the loop goes to zero, and everything is expressed using the angle \(\theta\) and the Cartesian reference frame.

Expansion of the magnetic field.

\[\vec{b}(P) = \vec{b}(C) + \vec{r} \cdot \nabla \vec{b}(C) + o(|\vec{r}|) \ ,\]

or, using Einstein notation for Cartesian components,

\[b_i(P) = b_i(C) + r_k \partial_k b_i(C) + o(|\vec{r}|) \ .\]

Integral of elementary contribution. The force acting on the Amperian loop is evaluated through integration of elementary contributions on segments $\(d \vec{\ell} = \hat{t} \, ds = \hat{t} \, r \, d\theta\),

\[\begin{split}\begin{aligned} \vec{F} & = \oint_{\alpha} d \vec{F} = \\ & = \oint_{\alpha} - i \, \vec{b}(P) \times \hat{t}(P) \, ds = \\ & = -i r \int_{\theta=0}^{2 \pi} \left[ \vec{b}(C) + \vec{r} \cdot \nabla \vec{b}(C) + o(|d \vec{r}) \right] \times \hat{t}(P) \, d \theta \ . \end{aligned}\end{split}\]

The first contribution goes to zero, as it’s proportional to the loop integral

\[\oint_{\alpha} \hat{t}(P) = \int_{\theta=0}^{2\pi} \left[ - \hat{l} \sin \theta + \hat{m} \cos \theta \right] r \, d \theta = \vec{0} \ .\]

Using the Cartesian reference frame, the second contribution can be written as

\[\begin{split}\begin{aligned} & -i r \int_{\theta=0}^{2 \pi} \left[ \vec{r} \cdot \nabla \vec{b}(C) \right] \times \hat{t}(P) \, d \theta = \\ & \quad = - i r^2 \int_{\theta=0}^{2\pi} \left[ \left( \hat{l} \cos \theta + \hat{m} \sin \theta \right) \cdot \nabla \vec{b}(C) \right] \times \left( - \hat{l} \sin \theta + \hat{m} \cos \theta \right) \, d\theta = \\ & \quad = - i r^2 \int_{\theta=0}^{2\pi} \left[ \cos \theta \, \partial_l \vec{b}(C) + \sin \theta \, \partial_m \vec{b}(C) \right] \times \left( - \hat{l} \sin \theta + \hat{m} \cos \theta \right) \, d\theta = \\ & \quad = - i r^2 \int_{\theta=0}^{2\pi} \left\{ \hat{l} \left( - \partial_l b_n \cos^2 \theta - \partial_m b_n \sin \theta \cos \theta \right) + \hat{m} \left( - \partial_l b_n \sin \theta \cos \theta - \partial_m b_n \sin^2 \theta \right) + \hat{n} \left( \partial_l b_l \cos^2 \theta + \partial_l b_m \cos \theta \sin \theta + \partial_m b_l \sin \theta \cos \theta + \partial_m b_m \sin^2 \theta \right) \right\} \, d\theta = \\ & \quad = - i \underbrace{ \pi r^2}_{S} \left\{ -\partial_l b_n \hat{l} - \partial_m b_n \hat{m} - \partial_n b_n \hat{n} \right\} = \\ & \quad = i S \, \nabla \vec{b}(C) \cdot \hat{n} = \\ & \quad = \nabla \vec{b}(C) \cdot \vec{m} \ . \end{aligned}\end{split}\]

being \(\partial_i \vec{b} = \hat{l} \partial_i b_l + \hat{m} \partial_i b_m + \hat{n} \partial_i b_n\), having used \(\int_{0}^{2\pi} \sin \theta \cos \theta d \theta = 0\), and \(\int_{0}^{2\pi} \sin^2 \theta d \theta = \int_{0}^{2\pi} \cos^2 \theta d \theta = \pi\), and the Gauss’ law for the magnetic field, \(0 = \nabla \cdot \vec{b} = \partial_l b_l + \partial_m b_m + \partial_n b_n\).

Higher order terms in \(\vec{r}\) vanish as the dimension of the loop goes to zero.

Moment. The moment w.r.t. the center \(C\) acting on such an elementary loop immersed in a magnetic field \(\vec{b}(P)\) reads,

\[\vec{M}_C = \vec{m} \times \vec{b}(C) \ .\]

Moment on an Amperian loop

Following the very same procedure done for evaluating the force acting on an Amperian loop immersed in a magnetic field, the moment is evaluated integrating elemenatry moment contributions,

\[\begin{split}\begin{aligned} d \vec{M}_C & = \vec{r}(P) \times d \vec{F}(P) = \\ & = - i \vec{r}(P) \times \left( \vec{b}(P) \times \hat{t}(P) \right) \, ds = \\ & = - i \vec{r}(P) \times \left[ \left( \vec{b}(C) + \vec{r}(P) \times \nabla \vec{b}(C) + o(\vec{r}) \right) \times \hat{t}(P) \right] \, r \, d\theta = \\ & = - i r^2 \hat{r}(P) \times \left( \vec{b}(C) \times \hat{t}(P) + O(r) \right) \, d\theta \ . \end{aligned}\end{split}\]

The double cross product can be evaluated as

\[\begin{split}\begin{aligned} \hat{r} \times \left(\vec{b} \times \hat{t} \right) & = \left( \hat{l} \, \cos \theta + \hat{m} \, \sin \theta \right) \times \left[ \left( b_l \hat{l} + b_m \hat{m} + b_n \hat{n} \right) \times \left( - \hat{l} \, \sin \theta + \hat{m} \, \cos \theta \right) \right] = \\ & = \left( \hat{l} \, \cos \theta + \hat{m} \, \sin \theta \right) \times \left[ \left( - b_n \cos \theta \right) \hat{l} + \left( - b_n \sin \theta \right) \hat{m} + \left( b_l \cos \theta + b_m \sin \theta \right) \hat{n} \right] = \\ & = \left( b_l \sin \theta \cos \theta + b_m \sin^2 \theta \right) \hat{l} + \left( -b_l \cos^2 \theta - b_m \sin \theta \cos \theta \right) \hat{m} + \left( - b_n \cos\theta \sin\theta + b_n \sin \theta \cos \theta \right) \hat{n} = \\ \end{aligned}\end{split}\]

Integration over the Amperian loop then gives

\[\begin{split}\begin{aligned} \vec{M}_C & = \oint_{\alpha} d\vec{M}_C = \\ & = - i \, \pi r^2 \, \left( b_m \hat{l} - b_l \hat{m} \right) = \\ & = - i \, S \, \vec{b}(C) \times \hat{n} = \\ & = \vec{m} \times \vec{b}(C) \ . \end{aligned}\end{split}\]

Power. Check, discuss the effects of motion, and uncomment

4.1.4. Energy balance

todo Check and put charges, currents, and dipoles together with the electromagnetic field

Ispirati dalle dimensioni fisiche dei campi elettromagnetici,

\[\begin{split}\begin{aligned} \left[\mathbf{e}\right] = \frac{\text{force}}{\text{charge}} \qquad & , \qquad [\mathbf{d}] = \frac{\text{charge}}{\text{length}^2} \\ [\mathbf{b}] = \frac{\text{force}\cdot\text{time}}{\text{charge}\cdot\text{length}} \qquad & , \qquad [\mathbf{h}] = \frac{\text{charge}}{\text{time} \cdot \text{length}} \end{aligned}\end{split}\]

\[\begin{split}\begin{aligned} \left[\mathbf{e} \cdot \mathbf{d}\right] & = \frac{\text{force}}{\text{length}^2} = \frac{\text{energy}}{\text{length}^3} = [u] \\ [\mathbf{b} \cdot \mathbf{h}] & = \frac{\text{force}}{\text{length}^2} = \frac{\text{energy}}{\text{length}^3} = [u] \end{aligned}\end{split}\]

si può costruire la densità di volume di energia (todo trovare motivazioni più convincenti, non basandosi solo sull’analisi dimensionale ma sul lavoro)

\[u = \frac{1}{2} \left( \mathbf{e} \cdot \mathbf{d} + \mathbf{b} \cdot \mathbf{h} \right) \ .\]

Si può calcolare la derivata parziale nel tempo della densità di energia, \(u\), e usare le equazioni di Maxwell per ottenere un’equazione di bilancio dell’energia del campo elettromagnetico. Per un mezzo isotropo lineare, per il quale valgono le equazioni costitutive \(\mathbf{d} = \varepsilon \mathbf{e}\), \(\mathbf{b} = \mu \mathbf{h}\), la derivata parziale nel tempo dell’energia elettromagnetica può essere riscritta sfuttando la regola di derivazione del prodotto e le equazioni di Faraday-Lenz-Neumann e Ampére-Maxwell,

\[\begin{split}\begin{aligned} \dfrac{\partial u}{\partial t} & = \dfrac{\partial}{\partial t}\left( \frac{1}{2} \mathbf{e} \cdot \mathbf{d} + \mathbf{b} \cdot \mathbf{h} \right) = \qquad (...) \\ & = \mathbf{e} \cdot \partial_t \mathbf{d} + \mathbf{h} \cdot \partial_t \mathbf{b} = \\ & = \mathbf{e} \cdot (\nabla \times \mathbf{h} - \mathbf{j}) - \mathbf{h} \cdot \nabla \times \mathbf{e} \ . \end{aligned}\end{split}\]

L’ultimo termine può essere ulteriormente manipolato, usando l’identità vettoriale

\[\begin{split}\begin{aligned} \mathbf{e} \cdot \nabla \times \mathbf{h} - \mathbf{h} \cdot \nabla \times \mathbf{e} & = e_i \varepsilon_{ijk} \partial_j h_k - h_i \varepsilon_{ijk} \partial_j e_k = \qquad (i \rightarrow k, k \rightarrow i)\\ & = e_i \varepsilon_{ijk} \partial_j h_k - h_k \varepsilon_{kji} \partial_j e_i = \\ & = e_i \varepsilon_{ijk} \partial_j h_k + h_k \varepsilon_{ijk} \partial_j e_i = \\ & = \partial_j (\varepsilon_{ijk} e_i h_k ) = \\ & = \partial_j (\varepsilon_{jki} e_i h_k ) = \\ & = \nabla \cdot (\mathbf{h} \times \mathbf{e}) = - \nabla \cdot (\mathbf{e} \times \mathbf{h}) \end{aligned}\end{split}\]

che permette di scrivere l’equazione del bilancio di energia elettromagnetica come,

\[\frac{\partial u }{\partial t} + \nabla \cdot \mathbf{s} = - \mathbf{e} \cdot \mathbf{j} \ ,\]

dove è stato definito il vettore di Poynting, o meglio il campo vettoriale di Poynting,

\[\mathbf{s}(\mathbf{r},t) := \mathbf{e}(\mathbf{r},t) \times \mathbf{h}(\mathbf{r},t) \ ,\]

che può essere identificato come un flusso di potenza per unità di superficie, comparendo sotto l’operatore di divergenza nel bilnacio di energia.

todo. Rimandare a una sezione in cui si mostra questa ultima affermazione passando dal bilancio differenziale al bilancio integrale e si usa il teorema della divergenza, \(\int_V \nabla \cdot \mathbf{s} = \oint_{\partial V} \mathbf{s} \cdot \hat{\mathbf{n}}\).

Bilancio di energia di cariche nel vuoto, o i materiali senza polarizzazione o magnetizzazione

Moto di cariche puntiformi. L’equazione del moto di carica puntiforme \(q_k\) nella posizione \(\mathbf{r}_k(t)\) al tempo \(t\) è

\[m_k \ddot{\mathbf{r}}_k = \mathbf{f}_k + \mathbf{f}_k^{em} \ ,\]

avendo riconosciuto i contributi di forza dovuti al campo elettromagnetico come \(\mathbf{f}_k^{em}\) dagli altri. L’espressione della forza dovuta al campo elettromagnetico sulla carica \(k\) è data dalla forza di Lorentz,

\[\mathbf{f}_k^{em}(t) = q_k \left[ \mathbf{e}(\mathbf{r_k}(t), t) - \mathbf{b}(\mathbf{r}_k(t), t) \times \dot{\mathbf{r}}_k(t) \right]\]

Continuità della carica elettrica. La densità di carica e di corrente elettrica di un insieme di cariche libere puntiformi macroscopiche può essere scritta come

\[\begin{split}\begin{aligned} \rho(\mathbf{r},t) & = \sum_k q_k \delta(\mathbf{r} - \mathbf{r}_k(t)) \\ \mathbf{j}(\mathbf{r},t) & = \sum_k q_k \dot{\mathbf{r}}_k(t) \delta(\mathbf{r} - \mathbf{r}_k(t)) \ . \end{aligned}\end{split}\]

L’equazione di continuità della carica, \(\partial_t \rho + \nabla \cdot \mathbf{j} = 0\), risulta quindi soddisfatta,

\[\begin{split}\begin{aligned} \partial_t \rho & = - \sum_k q_k \, \partial_i \delta(\mathbf{r} - \mathbf{r}_k(t)) \, \dot{r}_{k,i} \\ \partial_i j_i & = \sum_k q_k \, \dot{r}_{k,i} \, \partial_i \delta(\mathbf{r} - \mathbf{r}_k(t)) \\ \end{aligned}\end{split}\]

Procedimento alternativo (e più generale?)

todo In caso questo procedimento sia più generale, o più corretto, sostituire il procedimento precedente.

La carica elementare in un volumetto \(\Delta V\) è data da dal prodotto tra il volume e la densità volumetrica di carica, \(\rho \Delta V\); la velocità media locale della carica elettrica è \(\mathbf{v}\); la forza agente sulla carica elementare immersa in un campo elettromagnetico è determinata dalla formula di Lorentz, \(\mathbf{f} \Delta V = \Delta V \rho \left( \mathbf{e} - \mathbf{b} \times \mathbf{v} \right)\). La potenza di questa forza è il prodotto scalare con la velocità media delle cariche, \(\Delta V \mathbf{f} \cdot \mathbf{v}\)

La potenza del campo elettromagnetico sul moto della carica elettrica per unità di volume è quindi

\[\mathbf{v} \cdot \mathbf{f} = \rho \mathbf{v} \cdot \left( \mathbf{e} - \mathbf{b} \times \mathbf{v} \right) = \rho \mathbf{v} \cdot \mathbf{e} = \mathbf{j} \cdot \mathbf{e} \ .\]

todo

• discutere questo termine del bilancio di energia cinetica nel moto della carica elettrica

• questo termine compare con segno opposto nel bilancio dell’energia elettromagnetica del sistema

• dove compare la non-conservatività del problema in presenza di materiali dissipativi (come resistenza elettrica con \(\mathbf{e} = \rho_R \mathbf{j}\)?

Il termine \(\mathbf{e} \cdot \mathbf{j}\) può essere manipolato usando le equazioni di Maxwell, e le relazioni

\[\begin{split}\begin{cases} \mathbf{d} = \varepsilon_0 \mathbf{e} + \mathbf{p} \\ \mathbf{h} = \frac{\mathbf{b}}{\mu_0} - \mathbf{m} \\ \end{cases}\end{split}\]

\[\begin{split}\begin{aligned} \mathbf{e} \cdot \mathbf{j} & = \mathbf{e} \cdot \left( \nabla \times \mathbf{h} - \partial_t \mathbf{d} \right) = \\ & = - \nabla \cdot \left( \mathbf{e} \times \mathbf{h} \right) + \mathbf{h} \cdot \nabla \times \mathbf{e} - \mathbf{e} \cdot \partial_t \mathbf{d} = \\ & = - \nabla \cdot \left( \mathbf{e} \times \mathbf{h} \right) - \mathbf{h} \cdot \partial_t \mathbf{b} - \mathbf{e} \cdot \partial_t \mathbf{d} \end{aligned}\end{split}\]

Gli ultimi due termini possono essere manipolati in diverse maniere,

\[\begin{split}\begin{aligned} \mathbf{e} \cdot \partial_t \mathbf{d} = \mathbf{e} \cdot \partial_t \left( \varepsilon_0 \mathbf{e} + \mathbf{p} \right) & = \partial_t \left( \frac{1}{2} \varepsilon_0 \mathbf{e} \cdot \mathbf{e} \right) + \mathbf{e} \cdot \partial_t \mathbf{p} \\ & = \partial_t \left( \frac{1}{2} \mathbf{e} \cdot \mathbf{d} \right) + \frac{1}{2} \left( \mathbf{e} \cdot \partial_t \mathbf{p} - \mathbf{p} \cdot \partial_t \mathbf{e} \right) \\ & = \partial_t \left( \frac{1}{2 \varepsilon_0} \mathbf{d} \cdot \mathbf{d} \right) - \frac{\mathbf{p}}{\varepsilon_0} \cdot \partial_t \mathbf{d} \\ \end{aligned}\end{split}\]

\[\begin{split}\begin{aligned} \mathbf{h} \cdot \partial_t \mathbf{b} = \mathbf{h} \cdot \partial_t \left( \mu_0 \mathbf{h} + \mu_0 \mathbf{m} \right) & = \partial_t \left( \frac{1}{2} \mu_0 \mathbf{h} \cdot \mathbf{h} \right) + \mu_0 \mathbf{h} \cdot \partial_t \mathbf{m} \\ & = \partial_t \left( \frac{1}{2} \mathbf{b} \cdot \mathbf{h} \right) + \frac{1}{2} \mu_0 \left( \mathbf{h} \cdot \partial_t \mathbf{m} - \mathbf{m} \cdot \partial_t \mathbf{h} \right) \\ & = \partial_t \left( \frac{1}{2 \mu_0} \mathbf{b} \cdot \mathbf{b} \right) - \mathbf{m} \cdot \partial_t \mathbf{b} \\ \end{aligned}\end{split}\]

Nel vuoto o in mezzi lineari \(\mathbf{e} \cdot \partial_t \mathbf{p} - \mathbf{p} \cdot \partial_t \mathbf{e} = \mathbf{0}\), \(\mathbf{h} \cdot \partial_t \mathbf{m} - \mathbf{m} \cdot \partial_t \mathbf{h} = \mathbf{0}\). Usando le seconde espressioni, si può riscrivere l’equazione dell’energia del campo elettromagnetico come

\[\begin{split}\begin{aligned} \partial_t \left( \frac{1}{2} \mathbf{e} \cdot \mathbf{d} + \frac{1}{2} \mathbf{b} \cdot \mathbf{h} \right) + \nabla \cdot \left( \mathbf{e} \times \mathbf{h} \right) & = - \ \mathbf{e} \cdot \mathbf{j} \ + \\ & \quad - \frac{1}{2} \left[ \mathbf{e} \cdot \partial_t \mathbf{p} - \mathbf{p} \cdot \partial_t \mathbf{e} + \mu_0 \left( \mathbf{h} \cdot \partial_t \mathbf{m} - \mathbf{m} \cdot \partial_t \mathbf{h} \right) \right] \end{aligned}\end{split}\]

o, usando le definizioni di densità di energia elettromagnetica \(u\) e vettore di Poynting \(\mathbf{s}\),

\[ \partial_t u + \nabla \cdot \mathbf{s} = - \ \mathbf{e} \cdot \mathbf{j} \ - \frac{1}{2} \left[ \mathbf{e} \cdot \partial_t \mathbf{p} - \mathbf{p} \cdot \partial_t \mathbf{e} + \mu_0 \left( \mathbf{h} \cdot \partial_t \mathbf{m} - \mathbf{m} \cdot \partial_t \mathbf{h} \right) \right] \]