Derive light speed from Maxwell equation set

Hi there. It has been half a year since I updated this blog. I was doing my projects and staying unemployed at home. Recently, I started my Uni life and became busy with the assignment. The hazy life style makes me want to make study notes to keep up with the progress. However, instead of making them on paper or Microsoft Word, it is a good idea to make them publicly available so it also helps other.

So, here’s my first study note. It is about the derivation of light speed from the Maxwell equation set, as I think it is fundamental to understand electromagnetism. Without further ado, here is it.

Maxwell equation

$$
\begin{align}
\nabla\cdot\overrightarrow{\boldsymbol{D}} &= \rho_v && Gauss’s\ law\ of\ Electric\ Field \\
\nabla\cdot\overrightarrow{\boldsymbol{B}} &= 0 && Gauss’s\ law\ of\ Magnetic\ Field \\
\nabla\times\overrightarrow{\boldsymbol{E}} &= - \cfrac{d\overrightarrow{\boldsymbol{B}}}{dt} && Faraday’s\ Law \\
\nabla\times\overrightarrow{\boldsymbol{H}} &= \boldsymbol{J} + \cfrac{d\overrightarrow{\boldsymbol{D}}}{dt} && Ampere’s\ law
\end{align}
$$

First of all, let’s start with the Maxwell equation. Light is a time-varying electromagnetic wave, so we take the last two equations, which are time related, Faraday’s Law and Ampere’s Law. The principle of how light waves keep oscillating and move forward is by letting the electric field generate the magnetic field while the magnetic field generates the electric field. Similar to a pendulum, constantly exchanging energy between kinetic energy and gravitational potential energy.

Inter-relate $\overrightarrow{\boldsymbol{A}}$ and $\boldsymbol{V}$

However, it is essential to convert the fields, $\overrightarrow{\boldsymbol{E}}$ and $\overrightarrow{\boldsymbol{H}}$, into potential variable, $\overrightarrow{\boldsymbol{A}}$ and $\boldsymbol{V}$, as it gives more fundamental view.

$$
\begin{align}
-\nabla\boldsymbol{V}+\cfrac{d\overrightarrow{\boldsymbol{A}}}{dt} &= \overrightarrow{\boldsymbol{E}} && Electric\ potential\\
-\nabla\boldsymbol{V} &= \overrightarrow{\boldsymbol{E}} && Electric\ potential\\
\nabla\times\overrightarrow{\boldsymbol{A}} &= \overrightarrow{\boldsymbol{B}} && Magnetic\ potential\ vector\\
\end{align}
$$

Before we use the potential variable, we need to inter-relate them algebraly. To do this, we can use Faraday’s law

$$
\begin{align}
\nabla\times\overrightarrow{\boldsymbol{E}} &= - \cfrac{d\overrightarrow{\boldsymbol{B}}}{dt} && Faraday’s\ Law \\
\nabla\times-\nabla\boldsymbol{V} &= - \cfrac{d\left(\nabla\times\overrightarrow{\boldsymbol{A}}\right)}{dt} \\
\nabla\times-\nabla\boldsymbol{V} &= \nabla\times-\cfrac{d\overrightarrow{\boldsymbol{A}}}{dt} \\
\nabla\times\left(-\nabla\boldsymbol{V}-\cfrac{d\overrightarrow{\boldsymbol{A}}}{dt}\right) &= 0 \\
\nabla\times\boldsymbol{E} &= 0 \\
\end{align}
$$
Since the $\cfrac{d\overrightarrow{\boldsymbol{A}}}{dt}$ is not violate the conservative field of $\overrightarrow{\boldsymbol{E}}$, it is safe to place it into the $\overrightarrow{\boldsymbol{E}}$. So,
$$
\begin{align}
-\nabla\boldsymbol{V}-\cfrac{d\overrightarrow{\boldsymbol{A}}}{dt} &= \overrightarrow{\boldsymbol{E}}
\end{align}
$$

Derive wave equation of light waves

Next, we need to use another equation, which is Ampere-Maxwell’s Law.

$$
\begin{align}
\nabla\times\overrightarrow{\boldsymbol{H}} &= \sigma\boldsymbol{J} + \sigma\cfrac{d\overrightarrow{\boldsymbol{D}}}{dt} && Ampere-Maxwell’s\ law \\
\nabla\times\left(\nabla\times\overrightarrow{\boldsymbol{A}}\right) &= \sigma_{0}\boldsymbol{J} + \sigma_{0}\epsilon_{0}\cfrac{d\left(-\nabla\boldsymbol{V}-\cfrac{d\overrightarrow{\boldsymbol{A}}}{dt}\right)}{dt} && Subtitue\ with\ \overrightarrow{\boldsymbol{A}}\ and\ \boldsymbol{V}
\end{align}
$$

Since we are deriving in free space, there is zero conductivity and the $\boldsymbol{J}$ can be cancelled. Also, let’s use the vector identity to expand the double cross product.

$$
\begin{align}
\nabla\left(\nabla\cdot\overrightarrow{\boldsymbol{A}}\right)-\nabla^2\overrightarrow{\boldsymbol{A}} &= \sigma_{0}\epsilon_{0}\cfrac{d\left(-\nabla\boldsymbol{V}-\cfrac{d\overrightarrow{\boldsymbol{A}}}{dt}\right)}{dt} \\
\nabla\left(\nabla\cdot\overrightarrow{\boldsymbol{A}}\right)-\nabla^2\overrightarrow{\boldsymbol{A}} &= -\nabla\left(\sigma_{0}\epsilon_{0}\cfrac{d\boldsymbol{V}}{dt}\right)-\sigma_{0}\epsilon_{0}\cfrac{d^2\overrightarrow{\boldsymbol{A}}}{dt^2} \\
\nabla\left(\nabla\cdot\overrightarrow{\boldsymbol{A}}\right)+\nabla\left(\sigma_{0}\epsilon_{0}\cfrac{d\boldsymbol{V}}{dt}\right)-\nabla^2\overrightarrow{\boldsymbol{A}} &= -\sigma_{0}\epsilon_{0}\cfrac{d^2\overrightarrow{\boldsymbol{A}}}{dt^2}
\end{align}
$$

Now, we can use Lorentz gauge to cancel out the divergence part. We will cover the Lorentz gauge in next few blog.

$$
\begin{align}
\nabla\cdot\overrightarrow{\boldsymbol{A}}+\sigma_{0}\epsilon_{0}\cfrac{d\boldsymbol{V}}{dt} &= 0 && Lorentz’s\ Gauge \\
\nabla\left(\nabla\cdot\overrightarrow{\boldsymbol{A}}+\sigma_{0}\epsilon_{0}\cfrac{d\boldsymbol{V}}{dt}\right)-\nabla^2\overrightarrow{\boldsymbol{A}} &= -\sigma_{0}\epsilon_{0}\cfrac{d^2\overrightarrow{\boldsymbol{A}}}{dt^2} \\
-\nabla^2\overrightarrow{\boldsymbol{A}} &= -\sigma_{0}\epsilon_{0}\cfrac{d^2\overrightarrow{\boldsymbol{A}}}{dt^2}\\
\cfrac{d^2\overrightarrow{\boldsymbol{A}}}{dt^2} &= \cfrac{1}{\sigma_{0}\epsilon_{0}}\nabla^2\overrightarrow{\boldsymbol{A}}\\
\cfrac{d^2\boldsymbol{U}}{dt^2} &= c^2\cfrac{d^2\boldsymbol{U}}{dx^2} && Wave\ Equation \\
c^2&=\cfrac{1}{\sigma_{0}\epsilon_{0}}\\
c&=\sqrt{\cfrac{1}{\sigma_{0}\epsilon_{0}}}\\
\end{align}
$$

So, here is it. We derive the Ampere-Maxwell equation into a homogeneous wave equation. Let’s see whether the light-speed formula is correct.

Conclusion

$$
\begin{align}
c&=\sqrt{\cfrac{1}{\sigma_{0}\epsilon_{0}}}\\
\sigma_{0} &= 8.854\times10^{−12} F^{-1}m^{-1}\\
\epsilon_{0} &= 4\pi\times10^{-7} NA^{-2}\\
c&=\sqrt{\cfrac{1}{8.854\times10^{−12}\times4\pi\times10^{-7}}}\\
&=299795637.7ms^{-1}
\end{align}
$$