1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Gauss' Law Universally True?

  1. Jul 22, 2010 #1
    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. jcsd
  3. Jul 22, 2010 #2

    jtbell

    User Avatar

    Staff: Mentor

    Where does Griffiths say this? Third edition preferably, because that's the one I have next to me right now.
     
  4. Jul 22, 2010 #3
    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.
     
  5. Jul 22, 2010 #4

    jtbell

    User Avatar

    Staff: Mentor

    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).

    Without a specific citation, I can only surmise that you're misinterpreting Feynman in a similar way.
     
  6. Jul 23, 2010 #5

    K^2

    User Avatar
    Science Advisor

    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.
     
  7. Jul 23, 2010 #6
    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.
     
  8. Jul 23, 2010 #7

    K^2

    User Avatar
    Science Advisor

    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.
     
  9. Jul 23, 2010 #8
    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!
     
  10. Jul 23, 2010 #9

    Born2bwire

    User Avatar
    Science Advisor
    Gold Member

    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.
     
  11. Jul 26, 2010 #10

    htg

    User Avatar

    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. Jul 26, 2010 #11

    jtbell

    User Avatar

    Staff: Mentor

    Yes.
     
  13. Jul 27, 2010 #12
    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.
     
  14. Jul 27, 2010 #13

    htg

    User Avatar

    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.
     
  15. Jul 27, 2010 #14
    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.
     
  16. Jul 27, 2010 #15

    htg

    User Avatar

    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.
     
  17. Jul 28, 2010 #16

    Born2bwire

    User Avatar
    Science Advisor
    Gold Member

    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.
     
  18. Jul 28, 2010 #17

    htg

    User Avatar

    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.
     
    Last edited: Jul 28, 2010
  19. Jul 28, 2010 #18

    Born2bwire

    User Avatar
    Science Advisor
    Gold Member

    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.
     
  20. Jul 28, 2010 #19

    htg

    User Avatar

    If the intensity of the beam fades quickly as we move away from the X axis (I said the energy would propagate in the X direction), then I do not see how the fluxes through parallel surfaces of my cuboid would cancel each other.
     
  21. Jul 28, 2010 #20

    Born2bwire

    User Avatar
    Science Advisor
    Gold Member

    A better question is how would you create such a field? The problem with your question is simply that it goes against the foundations of electromagnetics. Maxwell's Equations and the Lorentz Force are the five equations that govern all of classical electromagnetics. What you are attempting here is to come up with a unphysical problem. You can describe any kind of electromagnetic wave but that doesn't mean that it is physically feasible unless it satisfies Maxwell's equations.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Gauss' Law Universally True?
  1. Gauss's Law (Replies: 2)

  2. Gauss's Law (Replies: 6)

Loading...