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.

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