Speed of Light Derivation

The speed of light may seem like an arbitrary constant of nature but, in some sense, it is actually set by other properties of the world. These other properties are the strengths of the electric and magnetic fields which are defined by the constants that are used in determination of them, otherwise known as the permittivity constant ( $\epsilon_0=8.85 \times 10^{-12} \texttt{ }\textup{F/m}$ ) and permeability constant ( $\mu_0=4\pi \times 10^{-7}\texttt{ }\textup{H/m}$ ). Because light is simply an electromagnetic wave, one can derive its speed using these constants.Take a look at Maxwell’s equations below.

$\nabla \cdot E=\frac{\rho}{\epsilon_0}$
$\nabla \cdot B=0$
$\nabla \times E= -\frac{\partial B}{\partial t}$
$\nabla \times B= \mu_0 \epsilon_0 \frac{\partial E}{\partial t}+\mu_0 J$

Assume one defines an the electric field in space as $E=(0, 0, E_0\textup{sin}(y-vt))$ and the magnetic field in space as $B=(B_0\textup{sin}(y-vt), 0, 0)$ . This is basically a wave along the y-axis that travels with a speed of $v$ . It is visualized below.

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Assuming this happens in a vacuum with no charge or current density, these can be substituted into Maxwell’s equations.

Gauss’s Law (Electricity)

$\nabla \cdot E= \frac{\rho}{\epsilon _0}$
$\frac{\partial }{\partial z}\left ( E_0 \textup{sin}(y-vt) \right )= \frac{0}{\epsilon_0}$
$0=0$

Gauss’s Law (Magnetism)

$\nabla \cdot B=0$
$\frac{\partial }{\partial x}\left ( B_0 \textup{sin}(y-vt) \right )=0$

$0=0$

Faraday’s Law

$\nabla \times E= -\frac{\partial B}{\partial t}$
$\frac{\partial }{\partial y}\left ( E_0 \textup{sin}(y-vt) \right ) \widehat{i}=-\frac{\partial }{\partial t}\left ( B_0 \textup{sin}(y-vt)\widehat{i} \right )$
$E_0 \textup{cos}(y-vt)\widehat{i} =v B_0 \textup{cos}(y-vt)\widehat{i}$

Ampere’s Law

$\nabla \times B= \mu_0 \epsilon_0 \frac{\partial E}{\partial t}+\mu_0 J$
$-\frac{\partial }{\partial y}\left ( B_0\textup{sin}(y-vt) \right )\widehat{i}=\mu _0\epsilon _0 \frac{\partial }{\partial t} \left ( E_0 \textup{sin}(y-vt)\widehat{i} \right )+\mu _0 (0)$
$-B_0\textup{cos}(y-vt)\widehat{i}=-\mu _0\epsilon _0 vE_0 \textup{cos}(y-vt)\widehat{i}$

Faraday’s Law and Ampere’s Law seem only to work if $B_0 = \mu _0 \epsilon _0 vE_0$ and $E_0 = vB_0$ . This special case yields the following solution.

$E_0 = v(\mu_0 \epsilon_0 vE_0)$
$1= \mu_0 \epsilon_0v^2$
$\frac{1}{\sqrt{\mu_0 \epsilon_0}}=v$

From this, it is seen that if the wave is to follow Maxwell’s equations, its speed must be as it is defined above. This value evaluates to $\frac{1}{\sqrt{\mu_0 \epsilon_0}}=\frac{1}{\sqrt{(4\pi \times 10^{-7}) (8.85 \times 10^{-12})}}\approx2.99\times10^9 \frac{\textup{m}}{\textup{s}}$ which is simply the speed of light $c$ . The waves described represent light showing how light is simply an electromagnetic wave whose speed is dependent on its constituent parts: electricity and magnetism. This formulation is powerful in that it allows us to mathematically determine the speed of light which may yield less error than experimentation on the constant. It is important to note that this value can have some error due to the fact that the permeability and permittivity constants could also be slightly inaccurate.

In fact, even the magnitude of the electric field and magnetic field in an electromagnetic wave is not variable as rearranging the aforementioned formulas yield $E_0 = cB_0$ so there is an intricate proportional relation between them.

If you want to know more or see where I learned it from, read the book “Electricity and Magnetism” by Purcell and Morin. It should be in the “Books” section of this site.

aakashl

Speed of Light Derivation

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