15Oct/100

## Speed of light in a vacuum

Electric fields travel at the speed of light - a constant 299,792,458 meters per second regardless of wavelength.  However, this fact is only true for the ideal case where the waves are traveling through vacuum away from any other charge sources.  In non-ideal situations, where an electric field is traveling through another medium, the speed of the field is slowed.  But how slow?

In order to understand how different mediums effect an electric field's velocity, we first need to understand why the speed of light travels at the velocity it does in that idealized case.  The first part of this multi-part series is going to discuss the general form of the wave equation followed by a derivation of the speed of light in vacuum.

The Wave Equation:
The wave equation, is a second order partial differential equation in both time and space describing a traveling plane wave.  In the one-dimensional case, the equation is:

$\frac{\partial^2\psi\left(x,t\right)}{\partial x^2}=\frac{1}{v^2}\cdot\frac{\partial^2\psi\left(x,t\right)}{\partial t^2}$

Where $\psi$ is a scalar function representing the plane wave in both time and space, x is the direction in which the wave is traveling, v is the phase velocity of the wave, and t is time.  In the two-dimensional case, we add an additional spatial term to the equation:

$\frac{\partial^2\psi\left(x,y,t\right)}{\partial x^2}+\frac{\partial^2\psi\left(x,y,t\right)}{\partial y^2}=\frac{1}{v^2}\cdot\frac{\partial^2\psi\left(x,y,t\right)}{\partial t^2}$

Where the wave $\psi$  is now traveling in both the x and y directions.  This equation can be simplified by using the spacial Laplacian operator defined as:

$\displaystyle\nabla^2\vec{f}=\sum\limits_{i=1}^n\frac{\partial^2\vec{f}}{\partial x_i^2}$

Yielding the general form of the wave equation as:

$\nabla^2\vec{\psi}=\frac{1}{v^2}\cdot\frac{\partial^2\vec{\psi}}{\partial t^2}$

For our purposes of the electric field case, the variables in the general form of the wave equation are modified to:

$\nabla^2\vec{E}=\frac{1}{c^2}\cdot\frac{\partial^2\vec{E}}{\partial t^2}$

The Speed of Light in Vacuum
Now that we have the wave equation for an electric field, we can move onto deriving the speed of light.  To do this, we start with Maxwell's equations in differential form:

$\begin{matrix} \nabla\times\vec{E}&=&-\frac{\partial\vec{B}}{\partial t}&&&\;\left(1\right)\\\nabla\bullet\vec{E}&=&\frac{\rho}{\varepsilon_0}&&&\;\left(2\right)\\\nabla\times\vec{B}&=&\mu_0\cdot\varepsilon_0\cdot\frac{\partial\vec{E}}{\partial t}&+&\mu_0\cdot\vec{J}&\;\left(3\right)\\\nabla\bullet\vec{B}&=&0&&&\;\left(4\right) \end{matrix}$

We want to calculate the field's velocity in the ideal case of a vacuum where there are no electric charges, therefore it is safe to assume that the total charge density $\rho=0$, and the total current density $\vec{J}=0$.  Therefore eq.2 & eq.3 from above become:

$\begin{matrix} \nabla\bullet\vec{E}&=&0&\;\left(5\right)\\\nabla\times\vec{B}&=&\mu_0\cdot\varepsilon_0\cdot\frac{\partial\vec{E}}{\partial t}&\;\left(6\right) \end{matrix}$

In order to arrive at the wave equation, we need to find the Laplacian of $\vec{E}$.  In order to get it into this form, we first take the curl on both sides of eq.1:

$\nabla\times\nabla\times\vec{E}=-\frac{\partial}{\partial t}\left(\nabla\times\vec{B}\right)\;\left(7\right)$

Plugging eq.6 into eq.7 we get:

$\begin{matrix} \nabla\times\nabla\times\vec{E}&=&-\frac{\partial}{\partial t}\left(\mu_0\cdot\varepsilon_0\cdot\frac{\partial\vec{E}}{\partial t}\right)\\&=&-\mu_0\cdot\varepsilon_0\cdot\frac{\partial^2\vec{E}}{\partial t^2}\end{matrix}\;(8)$

We then leverage the vector identities:

$\begin{matrix} \nabla\times\nabla\times\vec{f}=\nabla\left(\nabla\bullet\vec{f}\right)-\nabla^2\vec{f}\\\nabla 0=0\end{matrix}$

Applying the first identity to eq.8 we get:

$\nabla\left(\nabla\bullet\vec{E}\right)-\nabla^2\vec{E}=-\mu_0\cdot\varepsilon_0\cdot\frac{\partial^2\vec{E}}{\partial t^2}\;(9)$

Then using eq.5 with the second identity on eq.9:

$\nabla^2\vec{E}=\mu_0\cdot\varepsilon_0\cdot\frac{\partial^2\vec{E}}{\partial t^2}$

Arriving at a form resembling the generalized wave equation from above where $\frac{1}{v^2}=\mu_0\cdot\varepsilon_0$, therefore the phase velocity can be represented as:

$v=\frac{1}{\sqrt{\mu_0\cdot\varepsilon_0}}=c$