Maxwell's equations

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  • #1
quasar987
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Here are some thoughts; if you have a different opinion, I'd be glad to hear it.

In the real world, on a small enough scale, linear charge distribution and surface charge distributions do not exist. It all comes down to [itex]\rho[/itex]. Same for the currents; there are only current densities [itex]\vec{J}[/itex]. So in lights of this, it is satifying to say that the Maxwell's equations are

[tex]\nabla \cdot E = \rho/\epsilon_0[/tex]

and

[tex]\nabla \times B = \mu_0 J + \mu_0 \epsilon_0 \partial E/ \partial t[/tex]

But in textbook exercices, as well as in real life approximations, we DO deal with linear and surface densities and currents. So in lights of this, aren't these two equations written in a more GENERAL form in their integral apearance:

[tex]\epsilon_0 \oint E \cdot da = Q_{enc}[/tex]

and

[tex]\oint B \cdot dl = \mu_0 I_{enc} + \mu_0 \epsilon_0 \int \frac{\partial E}{\partial t} \cdot da[/tex]

???
 

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  • #2
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[tex]\int \nabla \cdot E dV = \frac{1}{\epsilon_0} \int \rho dV[/tex]
[tex]\oint E dA = \frac{Q}{\epsilon_0}[/tex]

The same can be done with the others using Curl/Divergence Theorem.
 
  • #3
dextercioby
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The integral & the differential form are practically equivalent as it's almost impossible to produce integration domains which couldn't cope with Green's and Gauss-Ostrodradski's theorems which require certain analytic properties for such domains.

Daniel.
 
  • #4
quasar987
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Once again Daniel, you've lost me. What is your opinion about my statement "The two maxwell equ. in question are in a more general form written as integral as they are written as diff. equ. since in the former case they account for linear/surface charge/current"?
 
  • #5
Galileo
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quasar987 said:
Here are some thoughts; if you have a different opinion, I'd be glad to hear it.
In the real world, on a small enough scale, linear charge distribution and surface charge distributions do not exist. It all comes down to [itex]\rho[/itex]. Same for the currents; there are only current densities [itex]\vec{J}[/itex].
What do you mean with a linear charge distribution and a surface charge distribution?

Do you mean the difference between a collection of discrete point charge and a continuous charge distribution?
 
  • #6
Physics Monkey
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What Daniel is saying is that in ordinary situations, the integral form and differential form are completely equivalent. It is one line to go between them, a step that makes use of two theorems in mathematics, Stokes' Theorem and Gauss' Theorem.

In fact, many elementary texts introduce the integral form first because it is easier to use in the simplest symmetric cases and because most students aren't ready for the hardcore vector calculus yet. The two are equivalent however, and practical problems in electrostatics are almost invariably solved with potential methods using a differential equation. The trouble with taking the integral expresssions as a starting point is that it's not clear how to calculate the fields for charge/current distributions that are complicated and asymmetric. Also, the differential forms can handle singular charge/current distribtutions so there is no problem there.

Of course in the end, all this continuum stuff is approximate anyway. You've got to bust out electrons, protons, and so forth, and quantum mechanics and quantum electrodynamics if you really want to get the bottom of things.
 
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  • #7
Stingray
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Perhaps a more direct answer is that the differential equations can still be solved even if the charge and current densities are distributions. This is what is really being assumed when saying that you have surface or line charges. There is no real difference between the two formulations (as everyone else has said).
 
  • #8
jtbell
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You can account for point and line charges in the differential forms of Maxwell's equations, by representing them with Dirac delta functions. Actually, you need the delta functions in the integral formulation, too, in order to represent a point charge in an explicit integral over a continuous charge distribution.
 
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  • #9
As far as existence of surface and line charges is concerned I guess they do exist. Doesn't a conductor have surface charge density? it can not have volume charge density. the equations are equivalent mathematically but with different physical interpretations. But there is no loss of generality as we can always say that P(rho) in the 1st equation can be replaced by some equivalent surface/line charge density.
 
  • #10
quasar987
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kanupriya_iitd said:
As far as existence of surface and line charges is concerned I guess they do exist. Doesn't a conductor have surface charge density? it can not have volume charge density. the equations are equivalent mathematically but with different physical interpretations. But there is no loss of generality as we can always say that P(rho) in the 1st equation can be replaced by some equivalent surface/line charge density.
Can we restate the maxwell equations as such when we are dealing with surface charge and surface currents?

[tex]\nabla \cdot E = \sigma/\epsilon _0[/tex]

[tex]\nabla \times B = \mu_0 \vec{K} + \mu_0 \epsilon_0 \partial E/ \partial t[/tex]
 
  • #11
krab
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No. That's not even dimensionally correct. Imagine a metallic sphere, radius R, with a surface charge. Then [itex]\rho=\rho(r)=0[/itex] for all r not equal to R, and it is infinite at r=R. You can solve this with Gauss' law, and you will find E=0 for r<R, and jumps to E=Q/r^2 for r>R. You can solve it using the differential form, using limits, and you'll get the same answer. Just write the divergence in spherical coordinates and by spherical symmetry, the derivatives w.r.t. angle coordinates are zero, leaving only derivative w.r.t. r.
 

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