Maxwell equation in optics book vs in physics book, why?

constfang

I noticed that in many optics/photonics book, the Gauss law is expressed as:
[itex]\nabla.E = 0[/itex]
However, in wikipedia, it is
[itex]\nabla.E = \frac{\rho}{\epsilon_{0}}[/itex]
I assume that [itex]\rho[/itex] is zero in optics cases but why? I think that is because we assumed there is no electric charges on the way? then what if there is electric charges on the way? this lead to another question, can light be affected by going through a electric or magnetic field? I don't think so, but then why?

dextercioby

As far as I know the approximation of optics is that radiation is studied in the absence of sources, or in regions of space very far from the radiation sources. Thus the electric charges are discarded.

Light IS electric and magnetic field, so it can't be affected by itself...

clem

Neither of those equations are Gauss's law, but are one of Maxwell's equations as in your title.

ZapperZ

dextercioby is correct. In optics, you are often dealing with region of space that contains no charge. You're only interested in the field. So here, the charge density is zero, and that is the form of Gauss's law that you use.

And yes, those ARE Gauss's law, which is a part of Maxwell equations.

Zz.

constfang

Thank you, just a small follow up question, what if light is transmitted in a conductive medium that have positive electric charge? does that mean everything written in the optics textbook that based on the "no-charge" assumptions will not be applicable in this case?

ZapperZ

It depends on whether the charge interacts significantly with that light. If it causes something similar to "beam loading", or if it causes an absorption, then yes, you include such charge.

Zz.

dextercioby

You are right, he should have said: <the differential form of Gauss' law>.

constfang

I see, thank you.

ZapperZ

As opposed to the Integral form of Gauss's law?

I think they all qualified to be called "Gauss's law". We know enough mathematics to know that they are equivalent, and more importantly, can get from one to the other.

What is confusing about clem's comment is that it isn't a Gauss's law simply because it is part of Maxwell's equation. This is incorrect, because Maxwell's equation is a set of equations consisting of Gauss's law, Ampere's Law, and Faraday's law.

http://hyperphysics.phy-astr.gsu.edu...ric/maxeq.html

Zz.

vanhees71

One should, however, call Eq. (IV) on the hyperphysics page the "Ampere-Maxwell Law". Maxwell's "displacement current" was the crucial step, dinstinguishing Maxwell's theory from the other electromagnetic theories of his contemporaries (most notably Neumann's action-at-a-distance theory). This term lead Maxwell to the prediction of electromagnetic waves, and thus the discplacement current is one of Maxwell's greatest achievements. Finally Helmholtz, an early believer in Maxwell's concepts on the continent, initiated a research project to check the existence of em. waves, and this finally lead to their discovery by Heinrich Hertz.

RedX

I'm probably being nitpicky here, but optics textbook probably ought to have

[tex]\nabla \cdot D =0 [/tex]
since you will have charges on boundaries, although they're not "free" ( i.e., they're not put there by hand).

clem

It's a minor point, but there is a difference between 'law' and a derivation of that law.
The point of physics is often the derivation of a law or the derivation of an equation that follows from that law. Think of Snell's law or Murphy's law. Snell's law can be derlved from marching soldiers or wave propagation, but it IS neither.
Gauss's law is usually derived without reference or knowledge of the divergence operator. Gauss probably originally derived it that way. It is a remarkable achievement of theoretical physics to use the divergence theorem (sometimes misnamed 'Gauss's theorem') to show that the Maxwell equation div D=rho follows from Gauss's law. One can also go the other way and show that Gauss's law follows from the Maxwell equation. Showing that one thing can be derived from another does not make them the same thing.

ZapperZ

But other than a "historical significance", is there any issue with calling them Gauss's law? Is there an ambiguity here? Any student of physics would not be confused by such a label. After all, the integral form of Gauss's law is UNIQUELY CONNECTED to the differential form. So one can call it the integral form of gauss's law, or the differential form of gauss's law.

Do you also have a problem with the differential form of Ampere's and Faraday's law being called just Ampere's law and Faraday's law?

Zz.

clem

Faraday did not know any differential equations. Maxwell's achievement in showing that Faraday's descriptive law was derivable from a curl equation and v-v was remarkable.
I repeat, just because one equation is derivable from another does not make them the same thing. Also, Ampere never knew about the displacement current.
Calling two different things by the same name has confused some in this forum.

ZapperZ

But this name doesn't cause any ambiguity!

If you looked in some other thread, you'll see that I've argued about calling the process of electron liberation by photons in atoms as "photoelectric effect". The standard photoelectric effect is different than this photoionization effect, and calling them the same name causes confusion because they are not the same. So yes, I do know about the problem of calling different things by different names.

But in that case, there is an ambiguity there because the physics is different! So it isn't appropriate to call it by the same name. Is that the case here? The PHYSICS that both equations represent is the same thing! The mathematical form may be different, but it doesn't cause one to be ambiguous about what phenomenon or what physics description it is describing!

I see no physical distinction between one and the other.

Zz.