
#1
Jul2210, 07:09 PM

P: 250

According to various EM texts (Feynman, Griffiths, ...) Gauss’ law holds only in electrostatic situations. But using the point charge electric field solutions, I have found to date that it holds for a relativistically oscillating charge (wA = .99c) within at least 3 flux integration surfaces: (1) a sphere; (2) an ellipsoid; and (3) an egg. Is it possible that Gauss’ law is true for ALL source charge motions and ALL flux integration surfaces?




#2
Jul2210, 07:33 PM

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#3
Jul2210, 07:51 PM

P: 250

Thus far I've found that the law works for a relativistically oscillating particle (whose E field does not depend only on 1/r^2). Feynman flat out states that Gauss' law only holds in electrostatics. 



#4
Jul2210, 10:35 PM

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P: 11,255

Gauss' Law Universally True?Gauss's Law is one of Maxwell's Equations which (together with the Lorentz force law) define all of classical electromagnetism. Saying that Gauss's Law is not universally true is like saying that Maxwell's Equations aren't universally true (in the context of classical electrodynamics, that is). 



#5
Jul2310, 01:46 AM

Sci Advisor
P: 2,470

Seconded. Gauss Law always holds, but it is insufficient to determine the Efield unless you also know the curl of the field. In statics the later is always zero, so Gauss Law is all you need to solve a statics problem. That's probably what Feynman is talking about.




#6
Jul2310, 10:43 AM

P: 250

In general, the formula for the electric field of a moving point charge does NOT vary as 1/r^2, as Griffiths points out in Eq. 9.107 of his 2nd edition (see "Electric field of point charge, arbitrary motion" in the index, for other editions). I have found that Griffiths' general formula for E can always be COMPUTED, given a knowledge of the point charge's past motion (whatever that might be). That is how I found that the flux of a relativistically oscillating charge, through a few different shapes of enclosing surfaces, agreed with Gauss' law. 



#7
Jul2310, 11:00 AM

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P: 2,470

In that quote Feynman points out that the result is important in statics. It's still valid in dynamics. It just isn't useful, for reasons I described above.
And yes, the field from a moving charge is not a 1/r², but that's not because Gauss' Law stopped working. It's because you broke the symmetry which you relied on to use Gauss' Law to derive the field. 



#8
Jul2310, 11:26 AM

P: 250





#9
Jul2310, 10:47 PM

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PF Gold
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#10
Jul2610, 08:02 PM

P: 107

Consider a Gaussian beam of EM waves, propagating in the X direction. Consider a parallelopiped, whose edges are parallel to the X, Y and Z axes, the axis of the parallelopiped parallel to X not coincinding with X (best of all, shifted away from it by a distance comparable to the "width" of the beam, defined as so many standard deviations of the Gaussian function). Let the edges parallel to Y and Z be short compared to the "width"of the beam. Let the edge parallel to X be shrt compared to the wavelength of the EM wave considered. Is The Gauss law valid?




#12
Jul2710, 10:33 AM

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#13
Jul2710, 10:44 AM

P: 107

Since the intensity of the beam falls off rapidly as we move away from its axis, I do not see how the Gauss' law may be satisfied.




#14
Jul2710, 11:14 AM

P: 250





#15
Jul2710, 11:24 AM

P: 107

We can consider a beam whose intensity falls off rapidly as we move away from the X axis along the Y axis, but is independent of the Z coordinate.
It seems that the Gauss' law violation is clear in this case. 



#16
Jul2810, 02:13 AM

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#17
Jul2810, 05:33 AM

P: 107

Let us say that the E vector is parallel to the Y axis. I do not see why my last example does not violate the Gauss' law.




#18
Jul2810, 09:04 AM

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