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Maxwell's equations predict that the speed of light is constant

The aim of this proof is to transform the Maxwell's equations into an equation that describes electromagnetic waves (the one-dimensional wave equation):

[math]\nabla^2\vec E=\frac{1}{\color{red}{v}^2} \frac{{\partial}^2\vec E}{{\partial}t^2}[/math]

This equation allow us to calculate the speed of an electric wave ($\color{red}{v}$) which is equal to the speed of light. With a mathematical development from Maxwell's equations, we will see that the speed of light only depends on constants. This idea of a constant speed of light is historically very important for modern physics since it gave Albert Einstein the idea to develop the theory of relativity.

Prerequisite information[edit]

  • Be able to use these four vector operators:
  1. Curl: $\nabla\times$
  2. Divergence: $\nabla\cdot$
  3. Gradient: $\nabla$
  4. Laplacian: $\nabla^2$

The proof[edit]

Here are the four Maxwell's equations for an electric field $\vec E$ and a magnetic field $\vec B$ in a linear medium:

  • $\nabla\cdot\vec E=\frac{\rho }{\varepsilon }$
  • ${\nabla\times\vec E}=\frac{-\partial\vec{B}}{{\partial}t}$
  • $\nabla\cdot\vec B=0$
  • $\nabla\times\vec B=\mu\vec j+\varepsilon\mu \frac{{\partial}\vec E}{{\partial}t}$

We will calculate the speed of light in the vacuum, this is why we replace $\mu$ (permeability) and $\varepsilon$ (permittivity) by $\mu_0$ (vacuum permeability ≈ 1.256×10−6) and $\varepsilon_0$ (vacuum permittivity ≈ 8.854× 10−12). In order to get the result we must consider that, because we are in the vacuum, there is no charge. This is why the current density ($\vec j$) and the charge density ($\rho$) are equal to 0. After making those modifications, we get:

  1. $\nabla\cdot\vec E=0$
  2. ${\nabla\times\vec E}=\frac{-{\partial}{\vec B}}{{\partial}t}$
  3. $\nabla\cdot\vec B=0$
  4. $\nabla\times\vec B=\varepsilon _0\mu _0\frac{{\partial}\vec E}{{\partial}t}$

In order to get the wave equation, we first have to isolate the electric field $\vec E$ [math]\nabla^2\color{red}{\vec E}=\frac{1}{v^2} \frac{{\partial}^2\color{red}{\vec E}}{{\partial}t^2}[/math].. To do this, we have to take the curl of the equation $\textbf{(2)}$ and differentiate the equation $\textbf{(4)}$ with respect to time.

  1. $\nabla\times(\nabla\times\vec E)=\nabla\times(\frac{-{\partial}\vec B}{{\partial}t})$
  2. $\frac{\partial}{\partial t}\nabla\times\vec B=\varepsilon _0\mu _0\frac{{\partial}^2\vec E}{{\partial}t^2}$

We can see that in the right-hand side of $\textbf{(5)}$, there is a derivation with respect to time and then a derivation with $\nabla\times$. In the left-hand side of $\textbf{(6)}$ it is the opposite case. Because the derivative is a commutative operation, we have can equalize $\textbf{-(5)}$ and $\textbf{(6)}$:

[math]-\frac{{\partial}}{{\partial}t}\nabla\times\vec B=-\nabla\times(\frac{{\partial}\vec B}{{\partial}t})[/math]

We substitute the right hand-side of the equation $\textbf{(5)}$ by right-hand side of the equation $\textbf{(6)}$:

  1. $-\varepsilon _0\mu _0\frac{\partial^2\vec E}{{\partial}t^2}=\nabla\times(\nabla\times\vec E)$

The electric field $\vec E$ is now isolated. However, there is no Laplacian $\nabla^2$, this is why it does not look like the wave equation of the electric field The laplacian in the wave equation : [math]\color{red}{\nabla^2}\vec E=\frac{1}{v^2} \frac{{\partial}^2\vec E}{{\partial}t^2}[/math].. In order to get the Laplacian in the equation $\textbf{(7)}$, we have to prove this equality:

  1. $\nabla\times(\nabla\times\vec E)=\nabla(\nabla\cdot\vec E)-\nabla^2\vec E$

By developing both sides of the equation with the vector notation, we get:

[math]\left(\begin{matrix}\frac{{\partial}}{{\partial}y}[\frac{{\partial}E_y}{{\partial}x}-\frac{{\partial}E_x}{{\partial}y}]-\frac{{\partial}}{{\partial}z}[\frac{{\partial}E_x}{{\partial}z}-\frac{{\partial}E_z}{{\partial}x}]\\\frac{{\partial}}{{\partial}z}[\frac{{\partial}E_z}{{\partial}y}-\frac{{\partial}E_y}{{\partial}z}]-\frac{{\partial}}{{\partial}x}[\frac{{\partial}E_y}{{\partial}x}-\frac{{\partial}E_x}{{\partial}y}]\\\frac{{\partial}}{{\partial}x}[\frac{{\partial}E_x}{{\partial}z}-\frac{\mathit{dE}_z}{{\partial}x}]-\frac{{\partial}}{{\partial}y}[\frac{{\partial}E_z}{{\partial}y}-\frac{{\partial}E_y}{{\partial}z}]\end{matrix}\right)= \left(\begin{matrix}\frac{{\partial}}{{\partial}x}[\frac{{\partial}E_x}{{\partial}x}+\frac{{\partial}E_y}{{\partial}y}+\frac{{\partial}E_z}{{\partial}z}]\\\frac{{\partial}}{{\partial}y}[\frac{{\partial}E_x}{{\partial}x}+\frac{{\partial}E_y}{{\partial}y}+\frac{{\partial}E_z}{{\partial}z}]\\\frac{{\partial}}{{\partial}z}[\frac{{\partial}E_x}{{\partial}x}+\frac{{\partial}E_y}{{\partial}y}+\frac{{\partial}E_z}{{\partial}z}]\end{matrix}\right)-\left(\begin{matrix}\frac{{\partial}^2E_x}{{\partial}x^2}+\frac{{\partial}^2E_x}{{\partial}y^2}+\frac{{\partial}^2E_x}{{\partial}z^2}\\\frac{{\partial}^2E_y}{{\partial}x^2}+\frac{{\partial}^2E_y}{{\partial}y^2}+\frac{{\partial}^2E_y}{{\partial}z^2}\\\frac{{\partial}^2E_z}{{\partial}x^2}+\frac{{\partial}^2E_z}{{\partial}y^2}+\frac{{\partial}^2E_z}{{\partial}z^2}\end{matrix}\right) [/math]

By using the properties of the derivative, we prove that the equality $\textbf{(8)}$ is correct:

[math]\left(\begin{matrix}\frac{{\partial}^2E_y}{{\partial}y{\partial}x}-\frac{{\partial}^2E_x}{{\partial}y^2}-\frac{{\partial}^2E_x}{{\partial}z^2}+\frac{{\partial}^2E_z}{{\partial}z{\partial}x}\\\frac{{\partial}^2E_z}{{\partial}z{\partial}y}-\frac{{\partial}^2E_y}{{\partial}z^2}-\frac{{\partial}^2E_y}{{\partial}x^2}+\frac{{\partial}^2E_x}{{\partial}x{\partial}y}\\\frac{{\partial}^2E_x}{{\partial}x{\partial}z}-\frac{{\partial}^2E_z}{{\partial}x^2}-\frac{{\partial}^2E_z}{{\partial}y^2}+\frac{{\partial}^2E_y}{{\partial}y{\partial}z}\end{matrix}\right) =\left(\begin{matrix}\frac{{\partial}^2E_y}{{\partial}y{\partial}x}-\frac{{\partial}^2E_x}{{\partial}y^2}-\frac{{\partial}^2E_x}{{\partial}z^2}+\frac{{\partial}^2E_z}{{\partial}z{\partial}x}\\\frac{{\partial}^2E_z}{{\partial}z{\partial}y}-\frac{{\partial}^2E_y}{{\partial}z^2}-\frac{{\partial}^2E_y}{{\partial}x^2}+\frac{{\partial}^2E_x}{{\partial}x{\partial}y}\\\frac{{\partial}^2E_x}{{\partial}x{\partial}z}-\frac{{\partial}^2E_z}{{\partial}x^2}-\frac{{\partial}^2E_z}{{\partial}y^2}+\frac{{\partial}^2E_y}{{\partial}y{\partial}z}\end{matrix}\right)[/math]

Now, we know that both $\textbf{(7)}$ and $\textbf{(8)}$ are correct. We combine them in one single equation:

[math]-\varepsilon _0\mu _0\frac{{\partial}^2\vec E}{{\partial}t^2}=\nabla(\nabla\cdot\vec E)-\nabla^2\vec E[/math]

As stated at the beginning of the proof, $\nabla\cdot\vec E=0$ because we calculate the speed of light in vacuum (see equation $\textbf{(1)}$). This is why we get:

  1. $\nabla^2\vec E=\varepsilon _0\mu _0\frac{{\partial}^2\vec E}{{\partial}t^2}$

If we compare the equation $\textbf{(9)}$ with the one-dimensional wave equation [math]\nabla^2\vec E=\frac{1}{\color{red}{v}^2} \frac{{\partial}^2\vec E}{{\partial}t^2}[/math], we can deduce that:

[math]v=\frac{1}{\sqrt{\epsilon_{0}\mu_{0}}}[/math]

Where $\epsilon_{0}$ Vacuum permittivity and $\mu_{0}$ Vacuum permeability are both constant. This is why the speed of light is constant in the vacuum. It is possible to extend this principle to other references, because Maxwell's equations are Lorentz invariant $\,\,\#$

Note[edit]

Note that there are two way to interpret this result:

  1. The speed of light was constant with respect to the aether. This theory turned out to be false. Read more about the Luminiferous aether.
  2. The theory of Relativity