Divergence of ##\frac {1} {r^2} \hat r##

[ubergewehr273]

Basically a case where a positive charge q is placed in space which for convenience is taken as the origin. This electric field must have a large positive divergence but yet when evaluated mathematically we get 0. Also when we find divergence, we find it for a point right ? or is it possible to find it for a region as a whole ? If the former is true then at any point other than the origin, flux going in is greater than flux coming out and therefore the divergence of this vector field at this point should be negative. Where am I going wrong?

 

[kuruman]

... but yet when evaluated mathematically we get 0.

Be careful here. A term like $r/r$ is equal to one except at $r=0$ where it is undefined.

A divergence is a point function.

 

[ubergewehr273]

Be careful here. A term like $r/r$ is equal to one except at $r=0$ where it is undefined.

Lets just confine to finding divergence at a point other than the origin. Then shouldn't divergence be negative?

 

[kuruman]

For a point charge at the origin, the divergence is zero everywhere except at the origin. The appropriate Maxwell equation is $\vec{\nabla} \cdot \vec{E}=\rho(r)$ where in terms of the Dirac delta function the volume charge density is $\rho(r)=Q\delta(r)/(4\pi r^2)$.

 

[ubergewehr273]

Then shouldn't divergence be negative?

What's wrong with this?

 

[kuruman]

What's wrong is that you assert without proof that "flux going in is greater than flux coming out".

 

[ubergewehr273]

"flux going in is greater than flux coming out".

Since the field varies with $\frac {1} {r^2}$ isn't this true?

 

[kuruman]

Since the field varies with $\frac {1} {r^2}$ isn't this true?

Nope. The area of a sphere varies as $r^2$ so the flux through a sphere surrounding the charge will be the same regardless of the size of the sphere.

 

[Delta2]

If you do the math to calculate the divergence of $\frac{1}{r^2}\hat r$ you ll see that it is zero for any r except r=0 where it is not defined (because field becomes infinite at r=0). The intuitive meaning of divergence at a point is flux coming into that point minus flux coming out of that point (to be more precise flux regarding an infinitesimal area centered at that point), so it is always zero except if there is a source (or sink) at that point that generates flux in which case divergence becomes positive (source exists at that point) or negative (sink exists at that point).

 

[ubergewehr273]

If you do the math to calculate the divergence of $\frac{1}{r^2}\hat r$ you ll see that it is zero for any r except r=0 where it is not defined (because field becomes infinite at r=0). The intuitive meaning of divergence at a point is flux coming into that point minus flux coming out of that point (to be more precise flux regarding an infinitesimal area centered at that point), so it is always zero except if there is a source (or sink) at that point that generates flux in which case divergence becomes positive (source exists at that point) or negative (sink exists at that point).

So you're telling that for any vector field, when I calculate divergence at any point except a source or a sink, the result is zero?
Let's take an example where a field $\vec F=x^2 \hat i$ exists.
When I calculate divergence for this field I get $2x$ so the divergence is zero only at the origin. Does this mean that every other point in space is a source (positive x axis) or a sink(negative x axis) ?

 

[kuruman]

So you're telling that for any vector field, ...

Where did you see the wording "any vector field"?

Let's take an example where a field $\vec F=x^2 \hat{ i}$ exists. When I calculate divergence for this field I get $2x$ so the divergence is zero only at the origin.

Here we are talking about the divergence of a field due to a point charge, but since you brought it up, in this example the divergence is not zero only at the origin. It is zero in the entire yz plane.

 

[ubergewehr273]

Here we are talking about the divergence of a field due to a point charge, but since you brought it up, in this example the divergence is not zero only at the origin. It is zero in the entire yz plane.

Oh yes, sorry for the mistake.

Does this mean that every other point in space is a source (positive x axis) or a sink(negative x axis) ?

What about this?

 

[kuruman]

That's what it means.

 

[ubergewehr273]

I still have one doubt in mind. Could you help me in intuitively understanding the meaning of zero divergence for a field varying according to $\frac {1} {r^2} \hat r$, at any point but the origin.

 

[Delta2]

You just took some vector field which has divergence everywhere , however it is not the solution to some equation describing a setup in physical reality, you will not find this vector field as solution to Maxwell's equations or Navier-Stokes equations or some other equations that describe a vector field of a realistic setup.

 

[Delta2]

I still have one doubt in mind. Could you help me in intuitively understanding the meaning of zero divergence for a field varying according to $\frac {1} {r^2} \hat r$, at any point but the origin.

Since this field describes a situation where there are no sources except at the origin, hence the divergence at any point except at the origin will be zero.

 

[ubergewehr273]

Since this field describes a situation where there are no sources except at the origin, hence the divergence at any point except at the origin will be zero.

Thanks for the explanation

 

[kuruman]

Could you help me in intuitively understanding the meaning of zero divergence for a field varying according to $\frac {1} {r^2} \hat {r}$, at any point but the origin.

As an alternative explanation to the one offered by @Delta², here's one that makes use of the divergence theorem. Imagine a shell of arbitrary inner radius $R_1$ and outer radius $R_2$ centered at the source. In the limit $R_2 \rightarrow\infty$ and $R_1 \rightarrow 0$, "the flux through one shell must come out the outer shell because there are no sources inside the shell".

 

[ubergewehr273]

As an alternative explanation to the one offered by @Delta², here's one that makes use of the divergence theorem. Imagine a shell of arbitrary inner radius $R_1$ and outer radius $R_2$ centered at the source. In the limit $R_2 \rightarrow\infty$ and $R_1 \rightarrow 0$, "the flux through one shell must come out the outer shell because there are no sources inside the shell".

Thanks for the clarity.