The divergence of 1/r^2 fields

[Aikon]

Hi people,

It is my first post here :)

It is not a homework problem because i solved it (and at least i think i am right...).
The question is as follows: In problem 1.16 (Intro. to eletrodynamics, Grifitths, 3rd edition) he asks to calculate the divergence of the function v=1/r²r (bold is for vector and italic for versor).

That divergent results zero! And i want to know what this zero means mathematicaly, physicaly, ...?

 

[Pythagorean]

zero distance between the particles?

The resulting field is infinite, which isn't really physically meaningful, and certainly not observed, but in approximations to zero you can say it's really big.

 

[newflyer]

The trick is to calculate it by "brute force" and then using the divergence theorem, putting those two together gives you the definition of the so called Dirac delta function, which is what the divergence really is.

 

[Dale]

In problem 1.16 (Intro. to eletrodynamics, Grifitths, 3rd edition) he asks to calculate the divergence of the function v=1/r²r (bold is for vector and italic for versor).

That divergent results zero! And i want to know what this zero means mathematicaly, physicaly, ...?

Mathematically it means that the net flux across the surface of a differential volume is zero.

Physically it depends on what v is. If for example v is the E-field then it means there is no source of the E-field (charge) within the differential volume.

 

[Pythagorean]

Oh, the divergence of 1/r^2r. Whoops.

 

[yungman]

The form

\vec A = A\frac {\hat r}{r^2}

is an inverse square field. It is proven the divergence of an inverse square field is always equal zero.

If vector A is an inverse square field, it is a conservative field. and

\vec A = \nabla \Phi \hbox { where } \Phi \hbox { is some scalar function.}

Work done on a particle by A is path independent and equal to zero if it is a closed path...yada yada.

Google " conservative field " to get more detail.

Electro static field E is a conservative field.

Watch out, divergence of a vector equal zero do not imply the vector is a conservative field. It only say the divergence of a conservative field is zero.

 

[JaWiB]

The form

\vec A = A\frac {\hat r}{r^2}

is an inverse square field. It is proven the divergence of an inverse square field is always equal zero.

Except at r=0 of course

 

[yungman]

Except at r=0 of course

Forgot, conservative is not defined at r=0...I guess that falls into the yada yada part!

See, the yada yada is a very good way to CYA!

 

[dgOnPhys]

The form

\vec A = A\frac {\hat r}{r^2}

is an inverse square field. It is proven the divergence of an inverse square field is always equal zero.

If vector A is an inverse square field, it is a conservative field. and

\vec A = \nabla \Phi \hbox { where } \Phi \hbox { is some scalar function.}

Work done on a particle by A is path independent and equal to zero if it is a closed path...yada yada.

Google " conservative field " to get more detail.

Electro static field E is a conservative field.

Watch out, divergence of a vector equal zero do not imply the vector is a conservative field. It only say the divergence of a conservative field is zero.

This is somewhat correct:
1) the divergence of
\vec A = A\frac {\hat r}{r^2}
is zero NEARLY everywhere and the 'nearly' is very important because the integral of this divergence over a volume including the origin is not zero. Using distribution theory you would write
div( A\frac {\hat r}{r^2})=4A\pi\delta(\vec r)
The divergence basically describe the density of sources of a field. The basic example of this field is the electrostatic field generated by a point-like charge

2) the divergence of a conservative field is far from being always zero. A field with zero divergence is usually called solenoidal, a field with zero curl is called conservative. Any general field can be decomposed in the sum of a soleinodal and a conservative field (Helmoltz theorem).

 

[yungman]

This is somewhat correct:
1) the divergence of
\vec A = A\frac {\hat r}{r^2}
is zero NEARLY everywhere and the 'nearly' is very important because the integral of this divergence over a volume including the origin is not zero. Using distribution theory you would write
div( A\frac {\hat r}{r^2})=4A\pi\delta(\vec r)
The divergence basically describe the density of sources of a field. The basic example of this field is the electrostatic field generated by a point-like charge

2) the divergence of a conservative field is far from being always zero. A field with zero divergence is usually called solenoidal, a field with zero curl is called conservative. Any general field can be decomposed in the sum of a soleinodal and a conservative field (Helmoltz theorem).

There should be a whole section in a lot of calculus books, not going to list all the characteristics here. I suggested OP to search for the info on web. All the remain belong to the yada yada.

Yes the origin is not defined, but if you expand the divergence of the given conservative field, it is equal to zero for all points that the field is defined...that is excluding origin or |r|=0. Isn't it?

 

[jtbell]

divergence of the function v=1/r²r (bold is for vector and italic for versor).

Congratulations! You have taught me a new word.

In 25 years of teaching physics and 15 years of studying it before that, I don't think I have ever before seen the word "versor." But a Google search took me to a Wikipedia page which indicates that "versor" is indeed sometimes used to refer to what is usually called a "unit vector" in English. Maybe in some other languages this word (or rather its equivalent) is used more.

It's a nice word, and I wish I could use it in my classes, but I'm afraid that it would confuse many of my students because none of our textbooks use it.