Energy of the Electromagnetic Field

Author

Sandro Vitenti

These notes follow the conventions fixed earlier in the course: SI units, signature \((-,+,+,+)\), and \(x^\mu=(ct,\mathbf x)\).

1 Energy stored in the electrostatic field

Assembling a charge distribution means repeatedly bringing in a small charge \(\mathrm{d}q\) from infinity against the potential \(V\) already set up by the charge in place, doing work \[ \mathrm{d}W = V\,\mathrm{d}q, \] and choosing the reference such that the potential vanishes at infinity, \[ V(\infty)=0. \] Summing this over the whole distribution \(\rho\), with the factor \(\frac12\) compensating for counting each pair of charges twice, the total electrostatic energy is \[ W=\frac12\int_V V\rho\,\mathrm{d}^3x. \] Using Poisson’s equation, \[ \nabla^2V=-\frac{\rho}{\varepsilon_0}, \] we obtain \[ W = -\frac{\varepsilon_0}{2} \int_V V\nabla^2V\,\mathrm{d}^3x. \] Trading \(\rho\) for a total-derivative term plus \(|\nabla V|^2\) with the product rule, \[ V\nabla^2V = \nabla\cdot(V\nabla V)-|\nabla V|^2, \] and applying the divergence theorem to the first piece gives \[ W= -\frac{\varepsilon_0}{2} \oint_{\partial V} V\nabla V\cdot \mathrm{d}\mathbf a + \frac{\varepsilon_0}{2} \int_V |\mathbf E|^2\,\mathrm{d}^3x. \] Assuming the surface term vanishes, \[ \boxed{ W = \frac{\varepsilon_0}{2} \int_V |\mathbf E|^2\,\mathrm{d}^3x. } \]

2 Building a stationary current

After assembling the electrostatic configuration we slowly establish a stationary current. During this process

  • the charge distribution is stationary,
  • \(\partial\mathbf E/\partial t=0\) for the electrostatic field,
  • the current varies with time,
  • therefore the magnetic field also varies.

The induced electric field satisfies Faraday’s law, \[ \nabla\times\mathbf E_c = -\frac{\partial\mathbf B}{\partial t}. \] Here \(\mathbf E_c\) denotes the electric field induced by the changing magnetic field.

3 Quasi-static approximation

Ampère–Maxwell’s equation reads \[ \nabla\times\mathbf B - \frac1{c^2} \frac{\partial\mathbf E_c}{\partial t} = \mu_0\mathbf J. \] Estimating \(\nabla\sim1/L\) for a source of size \(L\), Ampère’s law with the displacement current dropped gives \(B\sim\mu_0JL\), \[ \nabla\sim\frac1L, \qquad B\sim\mu_0JL, \] and Faraday’s law, with \(T\) the timescale over which the current changes, gives in the same way \(E_c\sim LB/T\), \[ E_c\sim\frac{LB}{T}, \] and therefore \[ \frac1{c^2} \frac{\partial E_c}{\partial t} \sim \mu_0J \left(\frac{L}{cT}\right)^2. \] Hence, if \[ \boxed{\frac{L}{T}\ll c,} \] the displacement current can be neglected.

4 Magnetic energy

For a single charge, the Lorentz force gives the power delivered by the fields as \(\mathrm{d}W/\mathrm{d}t=\mathbf v\cdot\mathbf F=\mathbf v\cdot(q\mathbf E_c+q\mathbf v\times\mathbf B)=q\mathbf v\cdot\mathbf E_c\), since the magnetic force is always perpendicular to \(\mathbf v\) and does no work. For a continuous distribution, \(q\mathbf v\to\mathbf J\,\mathrm{d}^3x\), so the work done on the charges is \[ \frac{\mathrm{d}W}{\mathrm{d}t} = \int \mathbf J\cdot\mathbf E_c\,\mathrm{d}^3x. \] Using the quasi-static approximation, \[ \mathbf J = \frac1{\mu_0}\nabla\times\mathbf B, \] and the identity \[ \nabla\cdot(\mathbf E_c\times\mathbf B) = \mathbf B\cdot(\nabla\times\mathbf E_c) - \mathbf E_c\cdot(\nabla\times\mathbf B), \] together with Faraday’s law, \[ \nabla\times\mathbf E_c = -\frac{\partial\mathbf B}{\partial t}, \] the integrand becomes a divergence plus \(-\mathbf B\cdot\partial_t\mathbf B/\mu_0=-\frac1{2\mu_0}\partial_t|\mathbf B|^2\), so that, assuming the surface term vanishes, \[ \frac{\mathrm{d}W}{\mathrm{d}t} = -\frac1{2\mu_0} \frac{\partial}{\partial t} \int |\mathbf B|^2\,\mathrm{d}^3x. \] This is negative: the induced field \(\mathbf E_c\) opposes the growing current (Lenz’s law), so it does negative work on the charges, and whatever drives the current against \(\mathbf E_c\) must supply the balance. That energy, \(W_B\equiv-W\), ends up stored in the magnetic field; if the field is initially zero, \[ \boxed{ W_B = \frac1{2\mu_0} \int |\mathbf B|^2\,\mathrm{d}^3x. } \]

5 Total electromagnetic energy

Once the current is steady, \(\mathbf E_c\) has done its job and vanishes; what is left is the electrostatic field of the assembled charges plus the magnetic field of the assembled current, and the configuration “remembers” the energy spent building it: \(U\equiv W+W_B\), the total field energy, is \[ \boxed{ U = \frac1{2\mu_0} \int \left( \frac{|\mathbf E|^2}{c^2} + |\mathbf B|^2 \right) \mathrm{d}^3x. } \] Defining the energy density, \[ u = \frac1{2\mu_0} \left( \frac{|\mathbf E|^2}{c^2} + |\mathbf B|^2 \right), \] its time derivative is \[ \frac{\partial u}{\partial t} = \frac{1}{\mu_0} \left( \frac{\mathbf E\cdot\partial_t\mathbf E}{c^2} + \mathbf B\cdot\partial_t\mathbf B \right). \] Now, unlike in the magnetic-energy calculation, the source \(\mathbf J\) is kept: solving Ampère–Maxwell’s law for the displacement current, \(\partial_t\mathbf E/c^2=\nabla\times\mathbf B-\mu_0\mathbf J\), and using Faraday’s law, \(\partial_t\mathbf B=-\nabla\times\mathbf E\), gives \[ \frac{\partial u}{\partial t} = \frac{1}{\mu_0} \Big( \mathbf E\cdot(\nabla\times\mathbf B)-\mathbf B\cdot(\nabla\times\mathbf E) \Big) -\mathbf J\cdot\mathbf E, \] and the same identity used above, now with the roles of \(\mathbf E\) and \(\mathbf B\) swapped, turns the first piece into a divergence: \[ \frac{\partial u}{\partial t} = -\mathbf J\cdot\mathbf E - \nabla\cdot \left( \frac{\mathbf E\times\mathbf B}{\mu_0} \right). \]

6 Poynting theorem

Define the Poynting vector, \[ \boxed{ \mathbf S = \frac{\mathbf E\times\mathbf B}{\mu_0}. } \] The local conservation law is \[ \boxed{ \frac{\partial u}{\partial t} + \nabla\cdot\mathbf S = -\mathbf J\cdot\mathbf E. } \] Integrating over a volume, \[ \frac{\mathrm{d}}{\mathrm{d}t} \int_Vu\,\mathrm{d}^3x + \oint_{\partial V} \mathbf S\cdot \mathrm{d}\mathbf a = -\frac{\mathrm{d}W}{\mathrm{d}t}. \] Thus the change in electromagnetic energy inside a volume equals the energy flux through its boundary plus the work performed on charged matter.

7 Covariant expressions

Using the field components fixed in the conventions, \[ F^{0i}=\frac{E^i}{c}, \qquad F^{ij}=\epsilon^{ijk}B_k, \] summing over all pairs \(\mu\nu\) counts each antisymmetric pair \((0i)\) and \((i0)\) alike, giving a factor of \(2\) for the electric part, and similarly for the magnetic part, \[ F^{\mu\nu}F_{\mu\nu} = 2\left( -\frac{|\mathbf E|^2}{c^2} + |\mathbf B|^2 \right). \] Moreover, since \(F^{00}=0\), only the spatial index contributes to \[ F^{0\alpha}F^0{}_\alpha = \frac{|\mathbf E|^2}{c^2}. \] The energy-momentum tensor is \[ T^{\alpha\beta} = \frac1{\mu_0} \left( F^{\alpha\gamma}F^\beta{}_\gamma - \frac14 \eta^{\alpha\beta} F^{\mu\nu}F_{\mu\nu} \right). \] For \(\alpha=\beta=0\), using \(\eta^{00}=-1\), \[ T^{00} = \frac1{\mu_0} \left( F^{0\alpha}F^0{}_\alpha + \frac14 F^{\mu\nu}F_{\mu\nu} \right) = \frac1{\mu_0} \left( \frac{|\mathbf E|^2}{c^2} + \frac12\left(-\frac{|\mathbf E|^2}{c^2}+|\mathbf B|^2\right) \right) = \frac1{2\mu_0}\left(\frac{|\mathbf E|^2}{c^2}+|\mathbf B|^2\right) = u. \] For \(\alpha=0,\beta=i\), the second term drops out since \(\eta^{0i}=0\), and only \(F^{00}=0\) fails to contribute to the sum over \(\gamma\), leaving \[ T^{0i} = \frac1{\mu_0}F^{0j}F^i{}_j = \frac1{\mu_0}\frac{E^j}{c}\epsilon^{ijk}B_k = \frac1{\mu_0c}(\mathbf E\times\mathbf B)^i = \frac{S^i}{c}. \] Thus \(T^{00}=u\) and \(T^{0i}=S^i/c\) are, respectively, the time-time and time-space components of a single covariant object.

8 Exercises

8.1 Exercise 1: Self-Energy of a Uniformly Charged Sphere (Essential)

a) For a uniformly charged solid sphere of radius \(R\) and total charge \(Q\), use Gauss’s law to write \(\mathbf E(r)\) for \(r<R\) and \(r>R\).

b) Compute \(W=\frac{\varepsilon_0}{2}\int|\mathbf E|^2\,\mathrm{d}^3x\), splitting the integral at \(r=R\), and show that \[ W=\frac{3}{20\pi\varepsilon_0}\frac{Q^2}{R}=\frac35\left(\frac{Q^2}{4\pi\varepsilon_0R}\right). \]

c) Using \(V(r)=\dfrac{Q}{4\pi\varepsilon_0}\left(\dfrac{3}{2R}-\dfrac{r^2}{2R^3}\right)\) for \(r<R\) (the potential inside a uniformly charged sphere) and \(\rho=Q/(\tfrac43\pi R^3)\), evaluate \(W=\frac12\int_VV\rho\,\mathrm{d}^3x\) directly from the assembling-work definition and confirm it agrees with (b).

8.2 Exercise 2: The Surface Term at Infinity

a) For a localized charge distribution, estimate how \(V\) and \(|\nabla V|\) fall off with distance \(r\) far from the source, and how the area of a bounding sphere grows with \(r\). Use these estimates to show that \(\oint_{\partial V}V\nabla V\cdot\mathrm{d}\mathbf a\to0\) as the bounding surface is taken to infinity.

b) The same assumption is used again (for \(\mathbf E_c\times\mathbf B\)) in the magnetic-energy derivation, and once more (for \(\mathbf S\)) in the Poynting-theorem derivation. In the radiation zone studied in Radiation from a Moving Charge, \(\mathbf S\) falls off only as \(1/r^2\) while the area grows as \(r^2\): explain why this is exactly the marginal case where the surface term does not vanish, and why that is precisely what allows a source to radiate energy to infinity.

8.3 Exercise 3: The Magnetic-Energy Identity (Essential)

a) Starting from the identity \(\nabla\cdot(\mathbf E_c\times\mathbf B)=\mathbf B\cdot(\nabla\times\mathbf E_c)-\mathbf E_c\cdot(\nabla\times\mathbf B)\), substitute Faraday’s law and the quasi-static Ampère law \(\mathbf J=\frac1{\mu_0}\nabla\times\mathbf B\) term by term to reproduce \[ \frac{\mathrm{d}W}{\mathrm{d}t}=-\frac1{2\mu_0}\frac{\partial}{\partial t}\int|\mathbf B|^2\,\mathrm{d}^3x. \]

b) For a long solenoid with \(n\) turns per unit length and radius \(a\), carrying current \(I(t)\), with \(\mathbf B=\mu_0nI\,\hat{\mathbf z}\) inside and \(\approx0\) outside, evaluate \(W_B\) per unit length and compare with the standard inductance formula \(W_B=\frac12LI^2\), \(L=\mu_0n^2\pi a^2\) per unit length.

8.4 Exercise 4: Poynting’s Theorem for a Current-Carrying Resistor (Essential)

A cylindrical resistor of radius \(a\) and length \(L\) carries a steady current \(I\) along its axis, with a uniform axial field \(E=V/L\) on and inside its surface, where \(V=IR\) is the voltage drop along its length.

a) Using Ampère’s law, find \(\mathbf B\) on the lateral surface \(r=a\).

b) Compute \(\mathbf S=\mathbf E\times\mathbf B/\mu_0\) on that surface and show that it points radially inward.

c) Compute the total inward flux of \(\mathbf S\) through the lateral surface and show that it equals \(IV=I^2R\) — the power dissipated as Joule heating, entering the resistor entirely through the fields at its surface rather than “along the wire.”

Feel free to create issues, ask questions, or suggest improvements in the GitHub repository.