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Essential Electromagnetics Derivations

The goal here is not to reproduce a textbook. These are derivations worth remembering because each reveals a reusable physical structure.

1. Wave equation from Maxwell’s equations

In a homogeneous, source-free, linear medium,

\[\nabla\times\mathbf E=-\mu\frac{\partial\mathbf H}{\partial t}, \qquad \nabla\times\mathbf H=\epsilon\frac{\partial\mathbf E}{\partial t}.\]

Take the curl of Faraday’s law:

\[\nabla\times(\nabla\times\mathbf E) =-\mu\frac{\partial}{\partial t}(\nabla\times\mathbf H).\]

Use

\[\nabla\times(\nabla\times\mathbf E)=\nabla(\nabla\cdot\mathbf E)-\nabla^2\mathbf E.\]

With no free charge in a homogeneous medium, $\nabla\cdot\mathbf E=0$. Substitute Ampère–Maxwell:

\[-\nabla^2\mathbf E=-\mu\epsilon\frac{\partial^2\mathbf E}{\partial t^2}.\]

Therefore

\[\boxed{\nabla^2\mathbf E-\mu\epsilon\frac{\partial^2\mathbf E}{\partial t^2}=0}\]

with speed

\[\boxed{v=\frac1{\sqrt{\mu\epsilon}}}.\]
What the derivation teachesThe wave is not added as a separate postulate. It is already contained in the coupled curl structure of Maxwell's equations.

2. Poynting theorem

Start with

\[\nabla\times\mathbf H=\mathbf J+\frac{\partial\mathbf D}{\partial t}, \qquad \nabla\times\mathbf E=-\frac{\partial\mathbf B}{\partial t}.\]

Use the vector identity

\[\nabla\cdot(\mathbf E\times\mathbf H) =\mathbf H\cdot(\nabla\times\mathbf E)-\mathbf E\cdot(\nabla\times\mathbf H).\]

Substitution gives

\[\nabla\cdot(\mathbf E\times\mathbf H) =-\mathbf H\cdot\frac{\partial\mathbf B}{\partial t} -\mathbf E\cdot\mathbf J -\mathbf E\cdot\frac{\partial\mathbf D}{\partial t}.\]

For linear nondispersive media identify field-energy density

\[u=\frac12\mathbf E\cdot\mathbf D+\frac12\mathbf B\cdot\mathbf H.\]

Then

\[\boxed{ \nabla\cdot\mathbf S+ \frac{\partial u}{\partial t}+ \mathbf J\cdot\mathbf E=0, \qquad \mathbf S=\mathbf E\times\mathbf H. }\]

This is local electromagnetic energy conservation.

3. Reflection coefficient on a transmission line

At the load,

\[V=V^++V^-, \qquad I=\frac{V^+}{Z_0}-\frac{V^-}{Z_0}.\]

Enforce $Z_L=V/I$ and define $\Gamma=V^-/V^+$. Solving gives

\[\boxed{\Gamma_L=\frac{Z_L-Z_0}{Z_L+Z_0}}.\]

Special cases follow instantly:

4. Skin depth

In a good conductor, $\sigma\gg\omega\epsilon$. The propagation constant becomes approximately

\[\gamma\approx(1+j)\sqrt{\frac{\omega\mu\sigma}{2}}.\]

Thus attenuation constant

\[\alpha=\sqrt{\frac{\omega\mu\sigma}{2}}.\]

Define skin depth as the distance for amplitude to fall by $1/e$:

\[\boxed{\delta=\frac1\alpha=\sqrt{\frac{2}{\omega\mu\sigma}}}.\]

The key result is $\delta\propto f^{-1/2}$.

5. Friis scaling from power density and effective aperture

A transmitter with gain $G_t$ produces far-field power density

\[S=\frac{P_tG_t}{4\pi R^2}.\]

The receiving antenna captures

\[P_r=SA_e.\]

Using

\[A_e=\frac{G_r\lambda^2}{4\pi}\]

gives

\[\boxed{P_r=P_tG_tG_r\left(\frac{\lambda}{4\pi R}\right)^2}.\]
AssumptionsFar field, polarization alignment incorporated appropriately, impedance matching/losses treated consistently, unobstructed free-space propagation, and gains defined in the relevant directions.

6. Radar range resolution

For a signal bandwidth $B$, the characteristic compressed pulse/time resolution is roughly

\[\Delta t\sim\frac1B.\]

Radar round-trip delay is

\[\Delta t=\frac{2\Delta R}{c}.\]

Therefore

\[\boxed{\Delta R\approx\frac{c}{2B}}.\]

This makes clear why range resolution is fundamentally a bandwidth problem.

7. Larmor precession

A magnetic moment experiences torque

\[\boldsymbol\tau=\boldsymbol\mu\times\mathbf B.\]

For angular momentum $\mathbf F$ with $\boldsymbol\mu=\gamma\mathbf F$,

\[\frac{d\mathbf F}{dt}=\gamma\mathbf F\times\mathbf B.\]

The derivative is perpendicular to $\mathbf F$, so the magnitude stays approximately constant while the vector precesses. The angular frequency is

\[\boxed{\omega_L=|\gamma|B}.\]

8. Rabi frequency from the dipole Hamiltonian

For a classical oscillating electric field

\[\mathbf E(t)=\mathbf E_0\cos\omega t\]

and electric-dipole interaction

\[H_I=-\mathbf d\cdot\mathbf E(t),\]
the matrix element between $ g\rangle$ and $ e\rangle$ is proportional to
\[-\langle e|\mathbf d|g\rangle\cdot\mathbf E_0.\]

Under the usual resonant rotating-wave convention,

\[\boxed{\Omega=\frac{\mathbf d_{eg}\cdot\mathbf E_0}{\hbar}}\]

up to amplitude/convention definitions. Polarization enters through the projected vector matrix element.

9. AC Stark shift from dressed-state expansion

For a two-level system in the rotating frame, a common Hamiltonian is

\[H=\frac{\hbar}{2} \begin{pmatrix} 0 & \Omega\\ \Omega & -2\Delta \end{pmatrix}.\]

The eigenvalue separation involves

\[\sqrt{\Delta^2+\Omega^2}.\]
For $ \Omega/\Delta \ll1$,
\[\sqrt{\Delta^2+\Omega^2} \approx|\Delta|\left(1+\frac{\Omega^2}{2\Delta^2}\right).\]

The leading correction therefore scales as

\[\boxed{\delta\omega_{AC}\sim\frac{|\Omega|^2}{4\Delta}}\]

with sign and exact factor following the state/detuning convention.

What the derivation teachesThe far-detuned Stark shift is the perturbative shadow of full dressed-state level repulsion. Near resonance, use the complete dressed/Floquet model rather than forcing a quadratic-shift approximation.

10. Shockley–Ramo current from weighting potential

Define weighting field

\[\mathbf E_w=-\nabla\phi_w.\]

For charge $q$ moving with velocity $\mathbf v$,

\[\frac{d\phi_w}{dt}=\nabla\phi_w\cdot\mathbf v=-\mathbf E_w\cdot\mathbf v.\]

Thus, subject to electrode-current sign convention,

\[\boxed{i=q\mathbf v\cdot\mathbf E_w=-q\frac{d\phi_w}{dt}}.\]

This derivation makes the central point explicit: signal depends on motion through weighting potential, not only on charge collection at the electrode.

Derivation practice

For each derivation, try to identify:

That is more valuable than memorizing the algebra alone.