From: ted@rosencrantz.stcloudstate.edu Subject: Re: Question about Gauss's law Date: 12 Oct 2000 16:36:12 GMT Newsgroups: sci.physics.research Summary: [missing] In article <6nj8us8e589c5cf56trslamkp27d5jgm71@4ax.com>, wrote: >Let's say you have a positive charge distribution over an infinite >space. Intuitively, the field should be 0 at all points since their >shouldn't be a preferred direction. To actually compute this, I draw >a sphere of arbitrary size at an arbitrary point in this space. >Since space is uniform around the sphere, the field through it should >be uniform. This should mean that the field is the charge enclosed >divided by the area of the sphere times a constant. The same as if >all the enclosed charge was centered at the center of the sphere. >Unfortunatly, this doesn't make sense because I will calculate >different fields at the same point depending on the size and position >of my sphere which contradicts the intuitive answer of 0 everywhere. This is a famous paradox with a venerable history. I think Newton was aware of it (in the context of gravity, not electrostatics, but the math is precisely the same), and Laplace too. Congratulations on rediscovering it. The problem here is one of boundary conditions. Gauss's law has a unique solution for E, for a given density distribution, *if* you specify the boundary conditions that say what E is like at infinite distances. In the standard Gauss's-Law situations, like a sphere or a line of charge, you're implicitly assuming the boundary condition that E goes to zero far from the charge. But you can't make that assumption for charge that extends off to infinity. In fact, even the intuitive, symmetrical guess that E = 0 everywhere isn't a valid solution to Gauss's law. There just are no solutions to Gauss's law (with sensible boundary conditions) for a charge distribution that extends to infinity. Coulomb's law (or Newton's law for gravity) are ill-equipped to handle such distributions. -Ted ============================================================================== From: Aaron Chou Subject: Re: Question about Gauss's law Date: 17 Oct 2000 14:15:56 GMT Newsgroups: sci.physics.research Ted and Martin wrote: > In article <8s4p9r$itl4l$1@ID-28113.news.cis.dfn.de>, > wrote: > > >The problem here is one of boundary conditions. Gauss's law has a > >unique solution for E, for a given density distribution, *if* you > >specify the boundary conditions that say what E is like at infinite > >distances. > I would prefer to state it: Gauss's law has a unique solution for E as long as your Gaussian surface encloses all of the charge density. Otherwise, by the principle of superposition, any charges outside of your surface provide a background field which would lie underneath your calculated solution. I guess you could always argue that you could have a divergence free background E field which would then define preferred directions in space and/or give inhomogeneity and break Lorentz and Poincare symmetry. This would presumeably correspond to a charge distribution at infinity, so your Gaussian surface would have failed to enclose all of the charge. Then you would probably have to consider the topology of the universe before you could figure out what the allowed backgrounds are. > > I'm embarrassed to admit that I don't understand why this should be > (it's Monday morning, bear with me). Gauss's law is a consequence of > Maxwell's equations, which are local (div D = rho) and the divergence > theorem (surface integral D ds = volume integral div D dV). Where do > the boundary conditions appear? > This is Stokes Theorem in differential geometry. I don't think there are any relevant boundary conditions at infinity. We are integrating over only our finite size Gaussian surface, not over a surface at infinity as in quantum field theory. Of course, if you wanted to calculate the field at infinity, you could use an infinite size Gaussian surface and get infinite flux, but then you would have to divide by an infinite area. In the end you get E = 4/3 * rho * r as r goes to infinity. This gives the standard linear relationship between distance and redshift (Hubble's law). Cheers, Aaron ============================================================================== From: Aaron Chou Subject: Re: Question about Gauss's law Date: 13 Oct 2000 07:25:04 GMT Newsgroups: sci.physics.research mzoran@gte.net wrote: > I have a question about Gauss's law where the flux through a > Gausian surface is equal to the charge enclosed times a constant. > I've seen problems were they find the field in various rods using > Gauss's law and symetry to show that the field through the surface is > uniform so that only the enclosed charge affects the field. Well I > tried to use this on a more complex problem and I got an apparent > contradiction. > > Let's say you have a positive charge distribution over an infinite > space. Intuitively, the field should be 0 at all points since their > shouldn't be a preferred direction. To actually compute this, I draw > a sphere of arbitrary size at an arbitrary point in this space. > Since space is uniform around the sphere, the field through it should > be uniform. This should mean that the field is the charge enclosed > divided by the area of the sphere times a constant. The same as if > all the enclosed charge was centered at the center of the sphere. > Unfortunatly, this doesn't make sense because I will calculate > different fields at the same point depending on the size and position > of my sphere which contradicts the intuitive answer of 0 everywhere. Hah! You've just discovered the inherent instability of having a uniform charge density. The charges will naturally repel each other causing a net flux of charge to flow to infinity. To see this using your argument, consider a positive test charge in your constant positive charge density sea. Every sphere you choose will exert a force on your test charge, and by Newton's 3rd Law, the test charge will exert an equal and opposite force. This will be true for any test charge you decide to pick from the constant density sea. So the conclusion is that all charges move away from each other. This can only be true if the density is decreasing with time. Local charge conservation then tells us that charge is flowing to infinity. This is quite analogous to the Newtonian derivation of the expanding universe. In both cases, you can use Gauss's law to conclude that except for very special cases, a spatially constant density must be time-dependent. The only difference is that in the cosmology case, gravity is attractive, so the only explanation (assuming zero cosmological constant) for the presently observed expansion is that the universe was expanding in the past, and the kinetic energy of the expansion is keeping it going despite the attractive gravitational potential. PS. I think Maxime misunderstood your post. What he says is true, but of course you are only considering cases where the calculations are simplified by symmetries in the configuration. Cheers, Aaron