# I Gauss' Law for electromagnetic radiation?

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1. Dec 4, 2017

### Elmer Correa

For the proof I've read that verifies transverse electromagnetic waves are consistent with Gauss' Law, there seems to be the suggestion that the magnetic and electric field at a given small length c(dt), along which the waves travel, propagate infinitely backwards and forwards in their respective axis. In this way, the same electric or magnetic flux enters one side of the surface as exits an adjacent parallel surface. I assume then that there is no way to set up a Gaussian surface so that a "source" either wave is inside the surface, meaning that the waves have to infinitely propagate on a certain axis at any given instant, right? Otherwise I don't think I totally understand the proof. While I'm on the subject, I also am unsure if EM waves leave a trail going from there point of propagation forward, back to their source. Or is this only the case if the source is continually emitting the waves? Thanks in advance for any clarification.

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Last edited: Dec 4, 2017
2. Dec 4, 2017

### Staff: Mentor

They are assuming plane waves, which by definition have that property. You can do the proof for any EM wave, but it isn’t as trivial.

However, I wonder why bother with such a proof. An EM wave is defined as a vacuum solution to Maxwell’s equations. So by definition it satisfies all of Maxwell’s equations, which obviously includes Gauss’ Law

3. Dec 4, 2017

### Chandra Prayaga

The picture shown in the attached file is misleading. I assume that since we are talking of a transverse wave, the directions of the electric and magnetic fields shown mean that the wave is propagating in the x direction. Then, at a given time, in any plane perpendicular to the x axis, at any given point, say x = x0, the electric field is the same. It has the same magnitude and direction. So does the magnetic field. At that same instant of time, the electric field in a plane at a different value of x is different, unless this new value of x is exactly one wavelength away from x0. The picture shows the same value for the electric field at different values of x, which is not correct.
The wave is propagating along the x axis. There is no propagation in either the y or the z direction.
As far as the proof goes, once you draw a transverse wave, you are already assuming Gauss's law.

4. Dec 4, 2017

### Elmer Correa

Could you elaborate on what exactly about a transverse wave assumes Gauss’s Law?

5. Dec 4, 2017

### Staff: Mentor

A transverse wave (or any other EM field) is a solution to Maxwell’s equations, so Gauss’ law is assumed, along with Faraday’s law and all of the other laws contained in Maxwell’s equations.

It is like if you are given the unit circle then you don’t have to prove whether $x^2+y^2=1$. You know that it does because the unit circle is defined as the set of points that satisfy that relation. Similarly, any EM field must satisfy Gauss’ law or it wouldn’t be an EM field.

6. Dec 4, 2017

### Elmer Correa

Hmmm ok. Can you also say whether my interpretation that the electric and magnetic waves propagate infinitely in either direction along their respective axis is correct for transverse waves?

7. Dec 4, 2017

### Staff: Mentor

A plane wave traveling in the x direction is a function of x and t only. It is the same at all y and z locations, off to infinity.

8. Dec 4, 2017

### ZapperZ

Staff Emeritus
Often times, it is so much easier to just simply show it. Look at how they used the source-free form of Maxwell equation for Gauss's Law and no-monopole to show that for plane wave solution, E and B are perpendicular to k.

http://farside.ph.utexas.edu/teaching/em/lectures/node48.html

Zz.