# Proof of speed of light from Maxwell's equations in integral form

Is it possible to prove that

$$\displaystyle c = \frac 1 {\sqrt{\epsilon_0 \mu_0}}$$

using Maxwell's equation in integral form? Recently, I saw this kind of proof by Professor Walter Lewin in one of his 8.02x lectures. But I could not understand the derivation in the video lecture.

Here is the link to the lecture (Derivation part starts from 13:24) https://youtu.be/D3tnZzhSISo?feature=shared

• Why did you ask “Is it possible to prove…?” when you really want to know “How can I prove…?”. Commented Jan 24 at 18:41
• In the Giorgi MKS system $\mu_0$ is defined to be $4\pi 10^{-7}$ $[\frac{Vsec}{Am}]$ and $\epsilon_0$ is defined as $\epsilon_0 = \frac{1}{c^2 \mu_0}$ where $c$ is the speed of light in vacuum. Whether the speed of the Hertzian electromagnetic wave derived from Helmholtz's equation is really the same $c$ as that of light is an experimental fact and is not something to be proved. That the two are really in fact the same is the regular daily experience of every microwave and/or radar engineer. Commented Jan 24 at 21:41
• The proof is straightforward using Maxwell’s equations in differential form, which as you probably know are equivalent to Maxwell’s equations in integral form. So what is the point in using only the equations in integral form to derive $c$? Commented Jan 24 at 23:58
• @Ghoster I am an high school student and did not know anything about divergence and curl. That's why I have asked whether it is possible to prove using integral form of Maxwell's equations. Commented Jan 25 at 11:55
• This question, the comments, and answer demonstrate the confusion caused by SI units. Commented Jan 25 at 20:55

Admittedly the derivation in the lecture was quite terse, So it may have been hard to follow.

The lecture makes this approach for a plane EM wave travelling in $$+z$$ direction $$\vec{E}(z,t)=E_o\hat{x}\cos(kz-\omega t) \tag{1}$$ $$\vec{B}(z,t)=B_o\hat{y}\cos(kz-\omega t) \tag{2}$$ with some constants ($$E_0,B_0,k,\omega$$).

Like every EM field this EM wave needs to satisfy the Ampere-Maxwell law $$\oint\vec{B}\cdot d\vec{l} = \mu_0\epsilon_0\frac{d}{dt}\underbrace{\int\vec{E}\cdot d\vec{A}}_{\Phi_E}. \tag{3}$$ We can use this law to deduce a relation between the constants in (1) and (2). For applying the Ampere-Maxwell law we choose as the closed loop this rectangle in the $$yz$$ plane.

Then we can calculate the change rate of the electric flux $$\Phi_E$$ appearing on the right side of (3) with the $$\vec{E}$$ field from (1). \begin{align} \frac{d\Phi_E(t)}{dt} &=\frac{d}{dt}\int\vec{E}(z,t)\cdot d\vec{A} \\ &=\frac{d}{dt}\int_0^l dy \int_0^{\lambda/4}dz\ E_0\cos(kz-\omega t) \\ &=\int_0^l dy \int_0^{\lambda/4}dz\ E_0\frac{d}{dt}\cos(kz-\omega t) \\ &=\int_0^l dy \int_0^{\lambda/4}dz\ E_0\omega\sin(kz-\omega t) \end{align} For simplicity we consider this especially for $$t=0$$: \begin{align} \frac{d\Phi_E(t=0)}{dt} &=\int_0^l dy \int_0^{\lambda/4}dz E_0\omega\sin(kz) \\ &=lE_0\omega\left[-\frac{1}{k}\cos(kz)\right]_{z=0}^{z=\lambda/4} \\ &=lE_0\frac{\omega}{k}\left[-\cos(kz)\right]_{z=0}^{z=\lambda/4} &\text{remember }k=\frac{2\pi}{\lambda} \\ &=lE_0\frac{\omega}{k}\left(-\cos\left(\frac{\pi}{2}\right)+\cos(0)\right) \\ &=lE_0\frac{\omega}{k}(0+1) &\text{remember }c=\frac{\omega}{k} \\ &=lE_0c \tag{4} \end{align}

The loop integral on the left side of (3) is easier to calculate. Again for simplicity we calculate it only for $$t=0$$. Then only the left edge (at $$z=0$$) of the rectangular loop contributes to the integral. \begin{align} &\oint\vec{B}(z,t=0)\cdot d\vec{l} \\ &=\int_0^l dy\ B_0 \cos(kz-\omega t) \\ &=\int_0^l dy\ B_0 \\ &=lB_0 \tag{5} \end{align} Inserting the results (4) and (5) into the Ampere-Maxwell law (3) we get $$lB_0=\mu_0\epsilon_0 lE_0c$$ or $$B_0=\mu_0\epsilon_0 E_0c \tag{6}$$

The EM wave also needs to satisfy Faraday's law. $$\oint\vec{E}\cdot d\vec{l} = -\frac{d}{dt}\underbrace{\int\vec{B}\cdot d\vec{A}}_{\Phi_B}. \tag{7}$$ We can use this law to deduce one more relation between the constants in (1) and (2). For applying Faraday's law we choose as the closed loop this rectangle in the $$xz$$ plane.

Then calculations very similar to the above yield $$\frac{d\Phi_B(t=0)}{dt}=-lB_0c \tag{8}$$ and $$\oint\vec{E}(z,t=0)\cdot d\vec{l}=lE_0 \tag{9}$$ Inserting the results (8) and (9) into Faraday's law (7) we get $$lE_0=lB_0c$$ or $$E_0=B_0c \tag{10}$$

Now we can combine (6) and (10) and find $$c=\frac{1}{\sqrt{\epsilon_0\mu_0}}$$

• Thank you so much! Commented Jan 25 at 11:57