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| Let's write down Maxwell's equations for a homogeneous medium without free charges and currents: | | Let's write down Maxwell's equations for a homogeneous medium without free charges and currents: |
| $$\text{div} \vec{E} = 0$$ | | $$\text{div} \vec{E} = 0$$ |
| $$\text{div} \vec{B} = 0$$ | | $$\text{div} \vec{B} = 0$$ |
| $$\text{rot} \vec{E} = -\frac{\partial \vec{B}}{\partial t}$$ | | $$\text{rot} \vec{E} = -\frac{\partial \vec{B}}{\partial t}$$ |
| $$\text{rot} \vec{B} = \mu\mu_0\varepsilon\varepsilon_0 \frac{\partial \vec{E}}{\partial t}$$ | | $$\text{rot} \vec{B} = \mu\mu_0\varepsilon\varepsilon_0 \frac{\partial \vec{E}}{\partial t}$$ |
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| Let a plane electromagnetic wave propagate along the $z$-axis. We direct the vector $\vec{E}$ along the $x$-axis ($E_x$) and the vector $\vec{B}$ along the $y$-axis ($B_y$). Then, in projections, the curl equations will take the form: | | Let a plane electromagnetic wave propagate along the $z$-axis. We direct the vector $\vec{E}$ along the $x$-axis ($E_x$) and the vector $\vec{B}$ along the $y$-axis ($B_y$). Then, in projections, the curl equations will take the form: |
| $$\frac{\partial E_x}{\partial z} = -\frac{\partial B_y}{\partial t} \quad (1)$$ | | $$\frac{\partial E_x}{\partial z} = -\frac{\partial B_y}{\partial t} \quad (1)$$ |
| $$-\frac{\partial B_y}{\partial z} = \mu\mu_0\varepsilon\varepsilon_0 \frac{\partial E_x}{\partial t} \implies \frac{\partial B_y}{\partial z} = -\mu\mu_0\varepsilon\varepsilon_0 \frac{\partial E_x}{\partial t} \quad (2)$$ | | $$-\frac{\partial B_y}{\partial z} = \mu\mu_0\varepsilon\varepsilon_0 \frac{\partial E_x}{\partial t} \implies \frac{\partial B_y}{\partial z} = -\mu\mu_0\varepsilon\varepsilon_0 \frac{\partial E_x}{\partial t} \quad (2)$$ |
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| Let's differentiate equation (1) with respect to the coordinate $z$: | | Let's differentiate equation (1) with respect to the coordinate $z$: |
| $$\frac{\partial^2 E_x}{\partial z^2} = -\frac{\partial}{\partial z} \left( \frac{\partial B_y}{\partial t} \right) = -\frac{\partial}{\partial t} \left( \frac{\partial B_y}{\partial z} \right)$$ | | $$\frac{\partial^2 E_x}{\partial z^2} = -\frac{\partial}{\partial z} \left( \frac{\partial B_y}{\partial t} \right) = -\frac{\partial}{\partial t} \left( \frac{\partial B_y}{\partial z} \right)$$ |
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| Substitute equation (2) into this expression: | | Substitute equation (2) into this expression: |
| $$\frac{\partial^2 E_x}{\partial z^2} = -\frac{\partial}{\partial t} \left( -\mu\mu_0\varepsilon\varepsilon_0 \frac{\partial E_x}{\partial t} \right) = \mu\mu_0\varepsilon\varepsilon_0 \frac{\partial^2 E_x}{\partial t^2}$$ | | $$\frac{\partial^2 E_x}{\partial z^2} = -\frac{\partial}{\partial t} \left( -\mu\mu_0\varepsilon\varepsilon_0 \frac{\partial E_x}{\partial t} \right) = \mu\mu_0\varepsilon\varepsilon_0 \frac{\partial^2 E_x}{\partial t^2}$$ |
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| We have obtained the classical wave equation, which in its general form is written as: | | We have obtained the classical wave equation, which in its general form is written as: |
| $$\frac{\partial^2 E_x}{\partial z^2} = \frac{1}{v^2} \frac{\partial^2 E_x}{\partial t^2}$$ | | $$\frac{\partial^2 E_x}{\partial z^2} = \frac{1}{v^2} \frac{\partial^2 E_x}{\partial t^2}$$ |
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| Comparing the coefficients, we find the wave propagation speed $v$: | | Comparing the coefficients, we find the wave propagation speed $v$: |
| $$\frac{1}{v^2} = \mu\mu_0\varepsilon\varepsilon_0 \implies v = \frac{1}{\sqrt{\mu\mu_0\varepsilon\varepsilon_0}}$$ | | $$\frac{1}{v^2} = \mu\mu_0\varepsilon\varepsilon_0 \implies v = \frac{1}{\sqrt{\mu\mu_0\varepsilon\varepsilon_0}}$$ |
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| Given that the speed of light in a vacuum is $c = \frac{1}{\sqrt{\mu_0\varepsilon_0}}$, we obtain the required relationship: | | Given that the speed of light in a vacuum is $c = \frac{1}{\sqrt{\mu_0\varepsilon_0}}$, we obtain the required relationship: |
| $$v = \frac{c}{\sqrt{\mu\varepsilon}}$$ | | $$v = \frac{c}{\sqrt{\mu\varepsilon}}$$ |
