Four dimensional version of Gauss' law

[vin300]

I'm having trouble interpreting the four dimensional version of Gauss' law. In the original version, a vector would be integrated around a closed 2D surface and this would be equal to the integral of divergence over the enclosed volume. In the newer version, the vector(or tensor) is integrated over a 3D hypersurface and this is equal to the integral of the covariant derivative over the 4D volume. Now, how could the integral over a bounded 3D volume be equal to the integral over an unbounded 4D volume? I mean, how could the integral over a fixed 3D volume be equal to the integral over 4D volume that is carried on forever in time?

 

[Dale]

I'm having trouble interpreting the four dimensional version of Gauss' law. In the original version, a vector would be integrated around a closed 2D surface and this would be equal to the integral of divergence over the enclosed volume. In the newer version, the vector(or tensor) is integrated over a 3D hypersurface and this is equal to the integral of the covariant derivative over the 4D volume. Now, how could the integral over a bounded 3D volume be equal to the integral over an unbounded 4D volume? I mean, how could the integral over a fixed 3D volume be equal to the integral over 4D volume that is carried on forever in time?

What specific equation do you mean by a 4D version of Gauss' law? Usually, in classical electrodynamics, Ampere's law and Gauss' law are combined into a single 4D (tensor) law. I don't think that I have ever seen Gauss' law on its own in a 4D version.

 

[Nugatory]

I'm having trouble interpreting the four dimensional version of Gauss' law. In the original version, a vector would be integrated around a closed 2D surface and this would be equal to the integral of divergence over the enclosed volume. In the newer version, the vector(or tensor) is integrated over a 3D hypersurface and this is equal to the integral of the covariant derivative over the 4D volume. Now, how could the integral over a bounded 3D volume be equal to the integral over an unbounded 4D volume? I mean, how could the integral over a fixed 3D volume be equal to the integral over 4D volume that is carried on forever in time?

Just as the 3D version works by integrating over a closed 2D surface that encloses (because it is closed!) a finite 3D volume... The 4D analogue works by integrating over a closed 3D hypersurface that encloses a finite 4D volume.

A 3D hypersurface that allows time to go on forever is not closed in four-dimensional spacetime. Trying to apply Gauss's theorem to it would be like trying to apply the traditional three-dimensional version of Gauss's theorem when the surface is defined by x²+y²=1 with z unconstrained - that is, a cylinder with infinite extent in both directions on the z axis.

 

[DrGreg]

What specific equation do you mean by a 4D version of Gauss' law? Usually, in classical electrodynamics, Ampere's law and Gauss' law are combined into a single 4D (tensor) law. I don't think that I have ever seen Gauss' law on its own in a 4D version.

I think vin300 means Gauss's theorem a.k.a the divergence theorem.

 

[Muphrid]

I think it might help if you think about the relationship between the N-dimensional region and its N-1-dimensional boundary. In this way, you realize that for the boundary to be finite, the region also has to be finite. This should resolve your conundrum.

I also would not call this "Gauss's law" in 4D. It's somewhat common to hear this called the generalized Stokes theorem. The divergence theorem and the (usual) Stokes theorem are all special cases of the generalized Stokes theorem. I prefer to think of it more as an extension of the fundamental theorem of calculus.

 

[vin300]

I get it now. Thanks. Had to do some thinking. Also, I think it would be impossible to create a four dimensional analogue of Stokes' theorem (the one in which the curl is integrated over an open surface).

 

[Muphrid]

Not at all. All the lower-dimensional versions of the fundamental theorem still apply in a higher-dimensional space.

 

[ApplePion]

"Four dimensional version of Gauss' law"

First let's put Gauss' Law in a convenient form in 3 dimensional Cartesian coordinates.

It will be better to use an algebraic approach, rather than the geometric approach used in Purcell and other places.

(Note, all derivatives are intended to be partial derivatives. Ex means E with a subscript of x, etc.)

Consider Integral of ( dEx/dx + dEy/dy + dEz/dz) dx dy dz

This contains three terms. Let's look at the first termIntegral of (dEx/dx) dx dy dz.Since the integral of the derivative of a function is the difference of the function at the boundaries, Integral (dEx/dx) dx is [Ex (at one boundary) - Ex (at the other boundary)] . From this you can see that Integral of (dEx/dx) dx dy dz is going to be Integral of [Ex(at one boundary) - Ex (at the other boundary)] dy dz. If you do this with all three terms you can see that the volume integral of the divergence of a quantity is the surface integral of the flux of that quantity.

The procedure I just showed you can straightforwardly be extended to more than three dimensions.