How to prove that electric field propagates like a wave?

[Adesh]

I was studying Feynman Lectures on Physics Volume 1 chapter 29. In there he proves that electric field propagates like a wave. Here is my attempt (in image), please tell me my mistake.
Thank you

 

[stevendaryl]

I think you're getting off track. There are two types of electromagnetic fields: static and propagating. If I have a charge $Q$, it has an associated static electric field: $E = \frac{Q}{r^2}$. It does NOT propagate like a wave. It's constant. Similarly, a constant current density such as a current-carrying wire will produce a static magnetic field. That doesn't propagate as a wave, either.

But in empty space, far from any charges, you can still have an electromagnetic field. A changing electric field acts a "source" for the magnetic field. Similarly, a changing magnetic field acts a source for the electric field. So if both the electric field and magnetic field are changing, they can create a self-sustaining wave.

You can't derive that fact from Coulomb's law for static fields (or at least, I don't know how, maybe Feynman did).

 

[BvU]

Hi Adesh, ,

I looked at the lecture and see that Feynman concentrates on $(t-r/c)$ to reason the disturbance propagates outward with speed $c$. As you calculate, the amplitude diminishes with $r$, so you get a similarity and not an exact equality.

The $1/r$ has to do with the energy radiated, that has to be distributed over an area that increases with $r$. As Feynman explains much more clearly in 29.2

So: kudos for your efffort , point taken, but: read on !

 

[Adesh]

I think you're getting off track. There are two types of electromagnetic fields: static and propagating. If I have a charge $Q$, it has an associated static electric field: $E = \frac{Q}{r^2}$. It does NOT propagate like a wave. It's constant. Similarly, a constant current density such as a current-carrying wire will produce a static magnetic field. That doesn't propagate as a wave, either.

But in empty space, far from any charges, you can still have an electromagnetic field. A changing electric field acts a "source" for the magnetic field. Similarly, a changing magnetic field acts a source for the electric field. So if both the electric field and magnetic field are changing, they can create a self-sustaining wave.

You can't derive that fact from Coulomb's law for static fields (or at least, I don't know how, maybe Feynman did).

Thank you for answering me, you can see here the way Feynman did http://www.feynmanlectures.caltech.edu/I_29.html

 

[Adesh]

Hi Adesh, ,

I looked at the lecture and see that Feynman concentrates on $(t-r/c)$ to reason the disturbance propagates outward with speed $c$. As you calculate, the amplitude diminishes with $r$, so you get a similarity and not an exact equality.

The $1/r$ has to do with the energy radiated, that has to be distributed over an area that increases with $r$. As Feynman explains much more clearly in 29.2

So: kudos for your efffort , point taken, but: read on !

Thank you for being so kind. I really like your answer.

 

[hilbert2]

The equations of electromagnetic field in vacuum are

$\displaystyle\frac{\partial^2 \vec{E}}{\partial t^2} - c^2 \vec{\nabla}^2 \vec{E} = 0$

$\displaystyle\frac{\partial^2 \vec{B}}{\partial t^2} - c^2 \vec{\nabla}^2 \vec{B} = 0$

The fact that it's the second time derivative $\frac{\partial^2}{\partial t^2}$ in the equations is what makes these wave equations. Put a partial derivative of any other order in there, and conservation of energy is violated.