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Gauss' Law Universally True?

by GRDixon
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GRDixon
#1
Jul22-10, 07:09 PM
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According to various EM texts (Feynman, Griffiths, ...) Gaussí 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 Gaussí law is true for ALL source charge motions and ALL flux integration surfaces?
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jtbell
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Jul22-10, 07:33 PM
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Quote Quote by GRDixon View Post
According to various EM texts (Feynman, Griffiths, ...) Gaussí law holds only in electrostatic situations.
Where does Griffiths say this? Third edition preferably, because that's the one I have next to me right now.
GRDixon
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Jul22-10, 07:51 PM
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Quote Quote by jtbell View Post
Where does Griffiths say this? Third edition preferably, because that's the one I have next to me right now.
I have the 2nd Edition. Just following Eq. 2.10 in that edition Griffiths says, "Notice that it all hinges on the 1/r^2 character of Coulomb's law; without that the crucial cancellation in (2.9) would not take place, and the flux of E would depend on the surface chosen, not merely on the total charge enclosed."

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.

jtbell
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Jul22-10, 10:35 PM
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Gauss' Law Universally True?

Quote Quote by GRDixon View Post
I have the 2nd Edition. Just following Eq. 2.10 in that edition
OK, I found it... it's after Eq. 2.13 in the 3rd edition. You're reading more into that statement than is warranted. At that point, Griffiths is dealing only with electrostatics, in which Gauss's Law and Coulomb's Law are basically equivalent, and electrodynamics hasn't even entered into the picture yet.

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).

Feynman flat out states that Gauss' law only holds in electrostatics.
Without a specific citation, I can only surmise that you're misinterpreting Feynman in a similar way.
K^2
#5
Jul23-10, 01:46 AM
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Seconded. Gauss Law always holds, but it is insufficient to determine the E-field 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.
GRDixon
#6
Jul23-10, 10:43 AM
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Quote Quote by jtbell View Post

Without a specific citation, I can only surmise that you're misinterpreting Feynman in a similar way.
Perhaps I am. re a specific citation: in Sect. 4-6 of "TFLonP" V2, Feynman states "Our result is an important law of the electrostatic field, called Gauss' law."

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.
K^2
#7
Jul23-10, 11:00 AM
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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/r≤, 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.
GRDixon
#8
Jul23-10, 11:26 AM
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Quote Quote by K^2 View Post
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/r≤, 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.
Thanks to all. If either Feynman or Griffiths discusses the more general applicability of Gauss' law (outside of electrostatics), I haven't read about it. That's what got me to wondering in the first place. In any case, we all seem to be in agreement: Gauss' law holds for all source charge motions, and for all surfaces of flux integration. What a gem!
Born2bwire
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Jul23-10, 10:47 PM
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Quote Quote by GRDixon View Post
Thanks to all. If either Feynman or Griffiths discusses the more general applicability of Gauss' law (outside of electrostatics), I haven't read about it. That's what got me to wondering in the first place. In any case, we all seem to be in agreement: Gauss' law holds for all source charge motions, and for all surfaces of flux integration. What a gem!
Often in electrostatics or magnetostatics we can derive relationships using only one of the Maxwell's equations. For example we can get Coulomb's law from Gauss' law and the magnetic field of an infinite wire from Ampere's Law. However, when we move to dynamics, the electric and magnetic fields become coupled and we generally lose symmetry. This requires us to use all of Maxwell's equations to properly define a system. So Gauss' law is still used in dynamics, however it is used in combination with the other equations.
htg
#10
Jul26-10, 08:02 PM
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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?
jtbell
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Jul26-10, 08:19 PM
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Yes.
GRDixon
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Jul27-10, 10:33 AM
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Quote Quote by htg View Post
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?
Since divE=0 at all points in the beam, I would think Gauss' law must be valid. That is, flux in through one surface of the parallelopipid must equal flux out through the opposing surface. More generally, I believe that Gauss' law is universally true. All of my own personal attempts to find a violation have been fruitless.
htg
#13
Jul27-10, 10:44 AM
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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.
GRDixon
#14
Jul27-10, 11:14 AM
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Quote Quote by htg View Post
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.
It may be true that the intensity falls off as we move away from the x axis. But it increases perpendicular to that axis as the light energy disperses. The divergence entails derivatives in all 3 directions. Even the most collimated laser beams spread with distance. Laser beams pointed at the moon illuminate lunar surface areas much larger than the beam's cross section at its origination point. For what it's worth, I am sympathetic to your skepticism, perhaps because texts don't discuss the applicability/validity of Gauss' law in electrodynamic (especially relativistic) situations. But ultimately divE=rho/eps0 appears to apply in all cases, even though one of its ramifications (Gauss' law) is mostly invoked in electrostatics.
htg
#15
Jul27-10, 11:24 AM
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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.
Born2bwire
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Jul28-10, 02:13 AM
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Quote Quote by htg View Post
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.
Electromagnetic waves in an isotropic source-free region are divergence free. Intensity is not enough, the polarization of the wave matters as flux is dependent upon the dot product of the surface's normal with the vector field.
htg
#17
Jul28-10, 05:33 AM
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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.
Born2bwire
#18
Jul28-10, 09:04 AM
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Quote Quote by htg View Post
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.
How could it? Since the electric field is polarized along the Y axis, then the k-vector must be in the z direction. If you were to imagine a rectangular solid as your Gaussian surface then the flux at the lower x-z side will cancel out the flux at the upper x-z side.


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