# How do we get a expression for speed of light $\frac{1}{\sqrt{\mu\epsilon}}$ using Maxwell's equation? [duplicate]

I am a $$12^{th}$$ class student in India. I am quite new to these interesting concepts. And, I think I've learnt about electrostatics, magnetism, Maxwell's equations very thoroughly. But, my books doesn't convey satisfactory information about how $$c$$ is calculated. It just simply that: using Maxwell's equations, $$c=\frac{1}{\sqrt{\mu\epsilon}}$$.

Also, I've been reading many questions asked about light and photons on this site, stating even the electric wave and magnetic wave are out of phase (which sometimes seems obvious to me), but I lack systemetic resources to learn about them.

I just need some help like how the velocity was calculated or even some resources.

Probably, this part is more opinion based but can someone please also suggest some good resources to learn these things (free). I heavily rely on sites Khanacademy

[Clarification after two answers used the differential form of Maxwell’s equations:]

The main problem is the equations taught to me are different which I find elsewhere:

A nice answer to this question thnx to @Elio Fabri

• Have you checked, for example, en.wikipedia.org/wiki/…? – Dvij D.C. Mar 5 '19 at 17:25
• @Dvij Mankad I have tried that, but I think all the things I have been taught are quite different. – PranshuKhandal Mar 5 '19 at 17:31
• @Dvij Mankad Plz look at edit, the problem i face is that things taught to me are not compatible with what i find.. – PranshuKhandal Mar 5 '19 at 17:48
• I rolled it back because it ignored the fact that there were two answers which answered the question as originally asked. With your edit, the answers make no sense. – G. Smith Mar 5 '19 at 18:06
• I'm amazed in seeing that apparently nobody is understanding what the OP is asking for. I'm going to prepare a totally different answer. – Elio Fabri Mar 10 '19 at 14:22

You can start from Maxwell's equations with no source in free space (a vacuum).

$$\begin{array}{r}{\vec{\nabla} \cdot \vec{E}=0} \\ {\vec{\nabla} \cdot \vec{B}=0} \\ {\vec{\nabla} \times \vec{E}=-\frac{\partial \vec{B}}{\partial t}} \\ {\vec{\nabla} \times \vec{B}=\mu_{0} \epsilon_{0} \frac{\partial E}{\partial t}}\end{array}$$

Taking the curl of the third and forth equations give us 2 equations that show that Maxwell's equations in free space generate $$\vec{E}$$ and $$\vec{B}$$ fields that obey the wave equation. Comparing to the wave equation,

$$\frac{\partial^{2} \Psi}{\partial x^{2}}=\frac{1}{v^{2}} \frac{\partial^{2} \Psi}{\partial t^{2}}$$

we can see that

$$\frac{1}{v^2} = \mu_0 \epsilon_0$$

so that finally

$$v = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$$

• Hi! Please check the guidelines for how to answer homework-like questions: physics.meta.stackexchange.com/questions/714/… – Dvij D.C. Mar 5 '19 at 17:27
• Hi! Sorry - new to here. Is this more like it? – BritChick Mar 5 '19 at 17:38
• thnx for help :) but i need a small favor, the maxwell's equations taught to me are a bit different, see in the edit, cab you just plz help me relating them with real equations, thank you for your help – PranshuKhandal Mar 5 '19 at 17:48
• You are looking at the equations in their integral form. It is common to convert and use them in a differential form (I find them easier to memorise in this form!) - see WikiHow – BritChick Mar 5 '19 at 17:52
• "hmwphysics" sounds like "h[o]m[ework]physics" ;) – Yashas Mar 5 '19 at 17:56

You find the speed of light by using Maxwell's equations to derive wave equations for the electric and magnetic fields in which you discover that their speed of propagation is $$1/\sqrt{\mu_0\epsilon_0}$$. The details are below.

Start with Maxwell's equations in the absence of any charge density or current density:

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

You see from the fourth and second equations how a changing electric field induces a magnetic field, and a changing magnetic field induces an electric field, so it seems plausible that they can "keep each other going", with a wave as a result.

To see this, first take the curl of the second and fourth equations:

$$\nabla\times(\nabla\times\mathbf{E})=-\frac{\partial}{\partial t}\nabla\times\mathbf{B}$$

$$\nabla\times(\nabla\times\mathbf{B})=\mu_0\epsilon_0\frac{\partial}{\partial t}\nabla\times\mathbf{E}$$

Now use the second and fourth equations to replace the curls on the right-hand-sides:

$$\nabla\times(\nabla\times\mathbf{E})=-\mu_0\epsilon_0\frac{\partial^2\mathbf{E}}{\partial t^2}$$

$$\nabla\times(\nabla\times\mathbf{B})=-\mu_0\epsilon_0\frac{\partial^2\mathbf{B}}{\partial t^2}$$

You see that we now have one equation with just $$\mathbf{E}$$ and one with just$$\mathbf{B}$$.

There is a mathematical identity for the curl of a curl of any vector field $$\mathbf{V}$$:

$$\nabla\times(\nabla\times\mathbf{V})=\nabla(\nabla\cdot\mathbf{V})-\nabla^2\mathbf{V}$$

Using this, and the first and third Maxwell equations to get rid of the divergence term, we get

$$\nabla^2\mathbf{E}=\mu_0\epsilon_0\frac{\partial^2\mathbf{E}}{\partial t^2}$$

$$\nabla^2\mathbf{B}=\mu_0\epsilon_0\frac{\partial^2\mathbf{B}}{\partial t^2}$$

These are three-dimensional wave equations for each component of the two fields, with the wave velocity equal to $$1/\sqrt{\mu_0\epsilon_0}$$.

• thnx for help :) but i need a small favor, the maxwell's equations taught to me are a bit different, see in the edit, cab you just plz help me relating them with real equations, thank you for your help – PranshuKhandal Mar 5 '19 at 17:47
• That is not a small favor. Learning how the differential form of Maxwell's equations relates to the integral form that you've learned is not something one can cover in a comment. – G. Smith Mar 5 '19 at 17:50