Conservation Laws & Poynting Theorem

Step 1: The rate of work done by EM fields on charges (per unit volume)

$$\frac{dW}{dt\,d\tau} = \mathbf{F} \cdot \mathbf{v} / d\tau = \rho(\mathbf{E} + \mathbf{v}\times\mathbf{B})\cdot\mathbf{v} = \rho\mathbf{v}\cdot\mathbf{E} = \mathbf{J}\cdot\mathbf{E}$$

Note: $\mathbf{v}\times\mathbf{B}$ is perpendicular to $\mathbf{v}$, so the magnetic force does no work.

Step 2: Eliminate J using the Ampere-Maxwell law

$$\mathbf{J} = \frac{1}{\mu_0}\nabla\times\mathbf{B} - \epsilon_0\frac{\partial\mathbf{E}}{\partial t}$$

$$\mathbf{J}\cdot\mathbf{E} = \frac{1}{\mu_0}(\nabla\times\mathbf{B})\cdot\mathbf{E} - \epsilon_0\mathbf{E}\cdot\frac{\partial\mathbf{E}}{\partial t}$$

Step 3: Use the product rule identity: ∇·(E×B) = B·(∇×E) - E·(∇×B)

$$(\nabla\times\mathbf{B})\cdot\mathbf{E} = \mathbf{B}\cdot(\nabla\times\mathbf{E}) - \nabla\cdot(\mathbf{E}\times\mathbf{B})$$

Step 4: Substitute Faraday's law: ∇×E = -∂B/∂t

$$(\nabla\times\mathbf{B})\cdot\mathbf{E} = -\mathbf{B}\cdot\frac{\partial\mathbf{B}}{\partial t} - \nabla\cdot(\mathbf{E}\times\mathbf{B})$$

Step 5: Recognize the time derivatives as derivatives of squared fields

$$\mathbf{E}\cdot\frac{\partial\mathbf{E}}{\partial t} = \frac{1}{2}\frac{\partial E^2}{\partial t}, \qquad \mathbf{B}\cdot\frac{\partial\mathbf{B}}{\partial t} = \frac{1}{2}\frac{\partial B^2}{\partial t}$$

Step 6: Combine all terms

$$\mathbf{J}\cdot\mathbf{E} = -\frac{1}{2}\frac{\partial}{\partial t}\left(\epsilon_0 E^2 + \frac{B^2}{\mu_0}\right) - \frac{1}{\mu_0}\nabla\cdot(\mathbf{E}\times\mathbf{B})$$

Step 7: Define energy density u and Poynting vector S

$$u = \frac{1}{2}\left(\epsilon_0 E^2 + \frac{B^2}{\mu_0}\right), \qquad \mathbf{S} = \frac{1}{\mu_0}(\mathbf{E}\times\mathbf{B})$$

Step 8: Poynting's theorem in differential form

$$\boxed{\frac{\partial u}{\partial t} + \nabla\cdot\mathbf{S} = -\mathbf{J}\cdot\mathbf{E}}$$

The rate of decrease of EM energy density equals the energy flux out (∇·S) plus the rate of work done on charges (J·E).

Step 1: Integrate Poynting's theorem over a volume V bounded by surface S

$$\frac{d}{dt}\int_V u\,d\tau = -\oint_{\partial V} \mathbf{S}\cdot d\mathbf{a} - \int_V \mathbf{J}\cdot\mathbf{E}\,d\tau$$

Step 2: Interpret each term physically

$\frac{d}{dt}\int_V u\,d\tau$ = rate of change of total EM energy in V

$\oint_{\partial V} \mathbf{S}\cdot d\mathbf{a}$ = rate of energy flowing out through the boundary

$\int_V \mathbf{J}\cdot\mathbf{E}\,d\tau$ = rate of work done on charges (Ohmic dissipation)

Step 3: For a plane wave propagating in the z-direction with E = E₀cos(kz-ωt) x̂

$$\mathbf{B} = \frac{E_0}{c}\cos(kz - \omega t)\,\hat{\mathbf{y}}, \quad \mathbf{S} = \frac{1}{\mu_0}\mathbf{E}\times\mathbf{B} = \frac{E_0^2}{\mu_0 c}\cos^2(kz-\omega t)\,\hat{\mathbf{z}}$$

Step 4: The time-averaged Poynting vector (intensity)

$$\langle\mathbf{S}\rangle = \frac{E_0^2}{2\mu_0 c}\,\hat{\mathbf{z}} = \frac{1}{2}\epsilon_0 c E_0^2\,\hat{\mathbf{z}}$$

Step 5: Verify: S = u·c for a plane wave (energy density times propagation speed)

$$u = \epsilon_0 E_0^2\cos^2(kz-\omega t), \quad S = uc \implies \boxed{\mathbf{S} = \frac{1}{\mu_0}(\mathbf{E}\times\mathbf{B})\;\text{[W/m}^2\text{]}}$$

S points in the direction of energy propagation and its magnitude gives the power per unit area.

Step 1: The electrostatic energy stored in a charge distribution

$$W_E = \frac{1}{2}\int \rho V\,d\tau = \frac{\epsilon_0}{2}\int E^2\,d\tau$$

Derived using $\nabla\cdot\mathbf{E} = \rho/\epsilon_0$ and integration by parts.

Step 2: The magnetostatic energy stored in a current distribution

$$W_B = \frac{1}{2}\int \mathbf{J}\cdot\mathbf{A}\,d\tau = \frac{1}{2\mu_0}\int B^2\,d\tau$$

Derived using $\nabla\times\mathbf{B} = \mu_0\mathbf{J}$ and integration by parts.

Step 3: Identify the electric energy density

$$u_E = \frac{\epsilon_0}{2}E^2$$

Step 4: Identify the magnetic energy density

$$u_B = \frac{1}{2\mu_0}B^2$$

Step 5: For a plane wave, verify u_E = u_B (equipartition)

$$B = E/c \implies u_B = \frac{E^2}{2\mu_0 c^2} = \frac{\epsilon_0 E^2}{2} = u_E \;\checkmark$$

Step 6: The total electromagnetic energy density

$$\boxed{u = u_E + u_B = \frac{1}{2}\left(\epsilon_0 E^2 + \frac{B^2}{\mu_0}\right)}$$

This expression is valid for any electromagnetic field configuration, not just plane waves.

Step 1: The Lorentz force per unit volume on a charge/current distribution

$$\mathbf{f} = \rho\mathbf{E} + \mathbf{J}\times\mathbf{B}$$

Step 2: Eliminate ρ and J using Maxwell's equations

$$\rho = \epsilon_0(\nabla\cdot\mathbf{E}), \qquad \mathbf{J} = \frac{1}{\mu_0}(\nabla\times\mathbf{B}) - \epsilon_0\frac{\partial\mathbf{E}}{\partial t}$$

Step 3: Substitute and add a term involving ∇·B = 0 for symmetry

$$\mathbf{f} = \epsilon_0(\nabla\cdot\mathbf{E})\mathbf{E} + \frac{1}{\mu_0}(\nabla\times\mathbf{B})\times\mathbf{B} - \epsilon_0\frac{\partial\mathbf{E}}{\partial t}\times\mathbf{B}$$

Add $-\frac{1}{\mu_0}(\nabla\cdot\mathbf{B})\mathbf{B} = 0$ and use Faraday's law to rewrite the time derivative.

Step 4: Use the identity E×(∂B/∂t) + (∂E/∂t)×B = ∂(E×B)/∂t, and Faraday's law

$$\mathbf{f} = \epsilon_0\left[(\nabla\cdot\mathbf{E})\mathbf{E} - \mathbf{E}\times(\nabla\times\mathbf{E})\right] + \frac{1}{\mu_0}\left[-(\nabla\cdot\mathbf{B})\mathbf{B} + (\nabla\times\mathbf{B})\times\mathbf{B}\right] - \epsilon_0\frac{\partial}{\partial t}(\mathbf{E}\times\mathbf{B})$$

Step 5: Apply the vector identity (∇×A)×A = (A·∇)A - ½∇(A²), and similarly for the divergence terms

$$(\nabla\cdot\mathbf{E})\mathbf{E} + (\mathbf{E}\cdot\nabla)\mathbf{E} - \frac{1}{2}\nabla E^2 = \nabla\cdot\left(E_iE_j - \frac{1}{2}\delta_{ij}E^2\right)$$

The right side is the divergence of a tensor — this is the key step.

Step 6: Define the Maxwell stress tensor T_ij

$$\boxed{T_{ij} = \epsilon_0\left(E_iE_j - \frac{1}{2}\delta_{ij}E^2\right) + \frac{1}{\mu_0}\left(B_iB_j - \frac{1}{2}\delta_{ij}B^2\right)}$$

Step 7: The equation of motion for the EM field + charges

$$\mathbf{f} = \nabla\cdot\overleftrightarrow{T} - \mu_0\epsilon_0\frac{\partial\mathbf{S}}{\partial t}$$

Integrating: $\mathbf{F} = \oint_S \overleftrightarrow{T}\cdot d\mathbf{a} - \epsilon_0\mu_0\frac{d}{dt}\int_V \mathbf{S}\,d\tau$. The stress tensor gives the force per unit area exerted by the field across a surface.

Step 1: The total force on charges in volume V equals the rate of change of mechanical momentum

$$\mathbf{F}_{\text{mech}} = \frac{d\mathbf{p}_{\text{mech}}}{dt} = \int_V \mathbf{f}\,d\tau$$

Step 2: From the stress tensor derivation, substitute f = ∇·T - μ₀ε₀ ∂S/∂t

$$\frac{d\mathbf{p}_{\text{mech}}}{dt} = \oint_{\partial V}\overleftrightarrow{T}\cdot d\mathbf{a} - \mu_0\epsilon_0\frac{d}{dt}\int_V \mathbf{S}\,d\tau$$

Step 3: Rearrange — identify the EM momentum stored in volume V

$$\frac{d}{dt}\left(\mathbf{p}_{\text{mech}} + \mu_0\epsilon_0\int_V \mathbf{S}\,d\tau\right) = \oint_{\partial V}\overleftrightarrow{T}\cdot d\mathbf{a}$$

Step 4: Define the electromagnetic momentum density

$$\mathbf{g} = \mu_0\epsilon_0\mathbf{S} = \epsilon_0(\mathbf{E}\times\mathbf{B})$$

Step 5: Express in terms of the speed of light

$$\boxed{\mathbf{g} = \frac{\mathbf{S}}{c^2} = \epsilon_0(\mathbf{E}\times\mathbf{B})}$$

Step 6: For a plane wave, relate momentum density to energy density

$$|\mathbf{g}| = \frac{S}{c^2} = \frac{uc}{c^2} = \frac{u}{c}$$

This is consistent with relativistic energy-momentum: for a massless particle (photon), $E = pc$, hence $p = E/c$ per unit volume gives $g = u/c$.

Step 1: Recall the EM momentum density

$$\mathbf{g} = \epsilon_0(\mathbf{E}\times\mathbf{B}) = \mu_0\epsilon_0\mathbf{S}$$

Step 2: By analogy with mechanics (L = r×p), the angular momentum density of the EM field is

$$\boldsymbol{\ell} = \mathbf{r}\times\mathbf{g} = \epsilon_0\left[\mathbf{r}\times(\mathbf{E}\times\mathbf{B})\right]$$

Step 3: The total angular momentum stored in the electromagnetic field

$$\mathbf{L}_{\text{em}} = \int_V \boldsymbol{\ell}\,d\tau = \epsilon_0\int_V \mathbf{r}\times(\mathbf{E}\times\mathbf{B})\,d\tau$$

Step 4: The conservation law — total angular momentum (mechanical + field) is conserved

$$\frac{d}{dt}(\mathbf{L}_{\text{mech}} + \mathbf{L}_{\text{em}}) = -\oint_{\partial V} \mathbf{r}\times(\overleftrightarrow{T}\cdot d\mathbf{a})$$

The right side is the torque exerted on the volume by the external field through the surface.

Step 5: Example — static fields carrying angular momentum (Feynman disk paradox)

$$\text{For crossed static E and B: } \mathbf{g} = \epsilon_0(\mathbf{E}\times\mathbf{B}) \neq 0$$

Even static fields can store angular momentum if E and B are not parallel. When the fields are turned off, mechanical angular momentum appears to conserve the total.

Step 6: The complete expression

$$\boxed{\mathbf{L}_{\text{em}} = \epsilon_0\int \mathbf{r}\times(\mathbf{E}\times\mathbf{B})\,d\tau = \mu_0\epsilon_0\int \mathbf{r}\times\mathbf{S}\,d\tau}$$

Electromagnetic angular momentum is essential for the quantum theory of photon spin — each photon carries angular momentum $\pm\hbar$ along its direction of propagation.