Exercises — Speed of light c = 1 - √(ε₀ μ₀)

$c=\frac{1}{\sqrt{{\epsilon }_0{\mu }_0}}$

• ${\epsilon }_0$ = permittivity of free space — how readily empty space permits (supports) an electric field.
• ${\mu }_0$ = permeability of free space — how readily empty space supports a magnetic field. Both are "stiffnesses" of vacuum. Bigger stiffness → slower relay → smaller $c$, which is exactly why they sit in the denominator.

The product ${\epsilon }_0{\mu }_0$ has units ${\text{s}}^2/{\text{m}}^2$ (this is $(\text{time}/\text{distance}{)}^2$, i.e. a slowness squared).

• $\sqrt{{\epsilon }_0{\mu }_0}$ → units $\text{s}/\text{m}$ = a slowness, not a speed. ✗
• $1/\sqrt{{\epsilon }_0{\mu }_0}$ → units $\text{m}/\text{s}$ = a speed. ✓ So the reciprocal form is the only one that can be a speed.

Step 1 — multiply the constants. WHY: the formula needs the product ${\epsilon }_0{\mu }_0$ first. ${\mu }_0=4\pi \times 10^{-7}=1.2566\times 10^{-6}{\text{T m A}}^{-1}$ ${\epsilon }_0{\mu }_0=(8.85\times 10^{-12})(1.2566\times 10^{-6})=1.112\times 10^{-17}{\text{s}}^2{\text{m}}^{-2}$ Step 2 — square root. WHY: the formula has  ​ in the denominator. $\sqrt{1.112\times 10^{-17}}=3.335\times 10^{-9}{\text{s m}}^{-1}$ Step 3 — invert. WHY: $c$ is one over that slowness. $c=\frac{1}{3.335\times 10^{-9}}=2.998\times 10^8{\text{m s}}^{-1}\approx 3.00\times 10^8{\text{m s}}^{-1}$

Step 1 — index. WHY: $n=\sqrt{{\epsilon }_r{\mu }_r}$ comes from running the same derivation with material constants (see Refractive Index and n = c/v). $n=\sqrt{2.25\times 1}=1.5$ Step 2 — speed. WHY: $v=c/n$. $v=\frac{3.00\times 10^8}{1.5}=2.00\times 10^8{\text{m s}}^{-1}$

WHY this tool: for a plane EM wave the same Maxwell equations force $E=cB$ at every instant (see Electromagnetic Waves). $E_0=c{B}_0=(3.00\times 10^8)(1.0\times 10^{-8})=3.0{\text{V m}}^{-1}$ The electric amplitude is numerically about $3\times 10^8$ times the magnetic one — that huge factor is just $c$.

The flaw is using the pre-Maxwell Ampère law $\nabla \times \vec{B}={\mu }_0\vec{J}$. Maxwell added the displacement current term: $\nabla \times \vec{B}={\mu }_0\vec{J}+{\mu }_0{\epsilon }_0\frac{\partial \vec{E}}{\partial t}.$ In vacuum $\vec{J}=0$, but the second term survives whenever $\vec{E}$ changes in time. That term is exactly what appears in Step 3 of the wave-equation derivation; without it the equation ${\nabla }^2\vec{E}={\mu }_0{\epsilon }_0{\partial }^2\vec{E}/\partial t^2$ never forms and there is no light. See Displacement Current and the Ampère–Maxwell Law.

The standard 1-D Wave Equation is $\frac{{\partial }^{2}f}{\partial x^2}=\frac{1}{v^2}\frac{{\partial }^{2}f}{\partial t^2}.$ Matching term-by-term with ${\partial }^{2}E/\partial x^2={\mu }_0{\epsilon }_0{\partial }^{2}E/\partial t^2$ gives $\frac{1}{v^2}={\mu }_0{\epsilon }_0\Rightarrow v=\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}=c.$ Units: ${\epsilon }_0{\mu }_0$ carries ${\text{s}}^2{\text{m}}^{-2}$, so $1/\sqrt{{\epsilon }_0{\mu }_0}$ carries $\sqrt{{\text{m}}^2{\text{s}}^{-2}}={\text{m s}}^{-1}$. ✓ The picture below shows the two curves ${\partial }^2/\partial x^2$ and ${\partial }^2/\partial t^2$ being forced into proportion.

First material: $n=\sqrt{4\cdot 1}=2$, so $v=c/2$ — it slows by a factor 2. Second material: $n=\sqrt{1\cdot 4}=2$, so also $v=c/2$ — the same slowdown. Reveal: ${\epsilon }_r$ and ${\mu }_r$ enter symmetrically through their product under the square root. Only the product ${\epsilon }_r{\mu }_r$ controls the speed; a given slowdown can be produced by extra electric "give" or extra magnetic "give" equally.

${\epsilon }_0$ is fixed by Coulomb's law (force between charges); ${\mu }_0$ is fixed by the force between current-carrying wires. Neither experiment involves light. Combining: $c=\frac{1}{\sqrt{(8.85\times 10^{-12})(4\pi \times 10^{-7})}}=3.00\times 10^8{\text{m s}}^{-1}.$ This matches the optically measured speed of light. Since the two constants came purely from electricity and magnetism, the match is a proof that light is an electromagnetic phenomenon — a unification of three previously separate subjects.

(a) From $E=cB$: $B_0=E_0/c=(1.0\times 10^3)/(3.00\times 10^8)=3.33\times 10^{-6}\text{T}$. (b) $E_0/B_0$ has units $\frac{\text{V/m}}{\text{T}}$. Since $1\text{T}=1{\text{V⋅s⋅m}}^{-2}$, $\frac{\text{V/m}}{{\text{V⋅s⋅m}}^{-2}}=\frac{\text{V}}{\text{m}}\cdot \frac{{\text{m}}^2}{\text{V⋅s}}=\frac{\text{m}}{\text{s}}.$ ✓ So $E_0/B_0=c=3.00\times 10^8{\text{m s}}^{-1}$, consistent with (a).

WHY: $c\propto 1/\sqrt{{\epsilon }_0{\mu }_0}$, so multiplying ${\epsilon }_0$ by 10 multiplies $c$ by $1/\sqrt{10}$. $c^{\prime }=\frac{c}{\sqrt{10}}=0.316c\approx 9.49\times 10^7{\text{m s}}^{-1}.$ A "stiffer" (more permitting) vacuum lets the field-relay proceed only about a third as fast.

Source-free Maxwell equations (see Maxwell's Equations): $\nabla \cdot \vec{E}=0,\nabla \cdot \vec{B}=0,\nabla \times \vec{E}=-\frac{\partial \vec{B}}{\partial t},\nabla \times \vec{B}={\mu }_0{\epsilon }_0\frac{\partial \vec{E}}{\partial t}.$ Step 1 — curl the Ampère–Maxwell law. WHAT: apply $\nabla \times$ to $\nabla \times \vec{B}={\mu }_0{\epsilon }_0\partial \vec{E}/\partial t$. WHY: to isolate a single equation in $\vec{B}$ by later substituting the other equation for $\nabla \times \vec{E}$. $\nabla \times (\nabla \times \vec{B})={\mu }_0{\epsilon }_0\frac{\partial }{\partial t}(\nabla \times \vec{E}).$ Step 2 — vector identity. WHAT: $\nabla \times (\nabla \times \vec{B})=\nabla (\nabla \cdot \vec{B})-{\nabla }^2\vec{B}$. WHY: converts the double curl into a Laplacian. Since $\nabla \cdot \vec{B}=0$, the first term dies: $-{\nabla }^2\vec{B}={\mu }_0{\epsilon }_0\frac{\partial }{\partial t}(\nabla \times \vec{E}).$ Step 3 — substitute Faraday $\nabla \times \vec{E}=-\partial \vec{B}/\partial t$. WHY: couples the equations and brings in the time derivative: $-{\nabla }^2\vec{B}={\mu }_0{\epsilon }_0\frac{\partial }{\partial t}\left(-\frac{\partial \vec{B}}{\partial t}\right)=-{\mu }_0{\epsilon }_0\frac{{\partial }^2\vec{B}}{\partial t^2}.$ Step 4 — clean up: ${\nabla }^2\vec{B}={\mu }_0{\epsilon }_0\frac{{\partial }^2\vec{B}}{\partial t^2}.$ Identical in form to the $\vec{E}$ equation ⇒ same speed $v=1/\sqrt{{\mu }_0{\epsilon }_0}=c$. So $\vec{E}$ and $\vec{B}$ travel locked together at one speed. The figure shows the relay: each field's change sources the other.

Step 1 — differentiate. WHY: Faraday in 1-D relates the space slope of $E$ to the time rate of $B$. $\frac{\partial E}{\partial x}=E_{0}k\cos (kx-\omega t),-\frac{\partial B}{\partial t}=+B_0\omega \cos (kx-\omega t).$ Step 2 — equate coefficients of the common $\cos (kx-\omega t)$: $E_{0}k=B_0\omega \Rightarrow \frac{E_0}{B_0}=\frac{\omega }{k}.$ Step 3 — identify the phase speed. WHY: for $\sin (kx-\omega t)$ the wave speed is $\omega /k=c$. Hence $\frac{E_0}{B_0}=c\Rightarrow E_0=c{B}_0.■$

(a) $E_0=c{B}_0=(3.00\times 10^8)(2.0\times 10^{-6})=6.0\times 10^2{\text{V m}}^{-1}=600{\text{V m}}^{-1}$. (b) $n=\sqrt{2.25}=1.5$, so $v=c/n=(3.00\times 10^8)/1.5=2.00\times 10^8{\text{m s}}^{-1}$. (c) Frequency $f$ is fixed at a boundary; $v=f\lambda$, so $\lambda \propto v$: $\frac{{\lambda }_{\text{medium}}}{{\lambda }_{\text{vacuum}}}=\frac{v}{c}=\frac{1}{n}=\frac{1}{1.5}=\mathrm{0.667.}$ The wavelength shrinks by the same factor the speed drops.

• L2.1: $c\approx 3.00\times 10^8{\text{m s}}^{-1}$
• L2.2: $n=1.5$, $v=2.00\times 10^8{\text{m s}}^{-1}$
• L2.3: $E_0=3.0{\text{V m}}^{-1}$
• L3.3: both slow by factor 2 ($n=2$)
• L4.2: $B_0=3.33\times 10^{-6}\text{T}$
• L4.3: $c^{\prime }=0.316c\approx 9.49\times 10^7{\text{m s}}^{-1}$
• L5.3: $E_0=600{\text{V m}}^{-1}$; $v=2.00\times 10^8{\text{m s}}^{-1}$; $\lambda$ ratio $=0.667$