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## Main Question or Discussion Point

I have searched the forums and the internet to see various discussions about the divergence of an electric field, or more directly, the divergence of

If I write the radial vector in cartesian coordinates, something of which I am more familiar with, I can by the definition of divergence calculate ∇(

We then turn to the divergence theorem. We compute a volume integral of the divergence of

This is where I have seen the "spike at r = 0" idea come in to explain things, to render the previous not a contradiction.

My view, however, is that the volume integral simply does not make sense when it includes r = 0. It seems undefined. The surface integral is well defined because there is no division by zero.

Rather than saying there is a spike at r = 0 or the divergence theorem is wrong, couldn't we say that because the volume integral is not well defined and the surface integral is, that we

I haven't looked at a good proof of the divergence theorem, so I ask, need it hold for such cases as this? I am sure there are properties that vector fields must hold in order for the divergence theorem to hold on them. For example, if it holds for continuous vector fields by not necessarily discontinuous vector fields, the above would not necessarily be a contradiction. We could then construct an new object to fill in what is undefined. In this case, we use the Dirac delta function.

Nevertheless, I just want to make sure that we are defining the volume integral containing r = 0 to be 4∏ because it doesn't seem to give rise to contradictions, not actually saying that the summation of volume elements over ∇(

**r**/r^3. I still don't understand this "spike at r = 0" idea, and really don't believe it. It simply seems to be an idea that fills in the gaps of what really should be a plain definition.If I write the radial vector in cartesian coordinates, something of which I am more familiar with, I can by the definition of divergence calculate ∇(

**r**/r^3) for r ≠ 0. It is clear that no matter how small we let r become, the divergence is zero. We cannot say anything about the divergence for r = 0. I understand this.We then turn to the divergence theorem. We compute a volume integral of the divergence of

**r**/r^3 where this volume contains r = 0, by computing a closed surface integral where this surface is the volume boundary. We get 4∏ for which certainly is not zero.This is where I have seen the "spike at r = 0" idea come in to explain things, to render the previous not a contradiction.

My view, however, is that the volume integral simply does not make sense when it includes r = 0. It seems undefined. The surface integral is well defined because there is no division by zero.

Rather than saying there is a spike at r = 0 or the divergence theorem is wrong, couldn't we say that because the volume integral is not well defined and the surface integral is, that we

*define*the the divergence of**r**/r^3 to be 4∏δ(**r**)? That is, we leave no room for misleading ideas?I haven't looked at a good proof of the divergence theorem, so I ask, need it hold for such cases as this? I am sure there are properties that vector fields must hold in order for the divergence theorem to hold on them. For example, if it holds for continuous vector fields by not necessarily discontinuous vector fields, the above would not necessarily be a contradiction. We could then construct an new object to fill in what is undefined. In this case, we use the Dirac delta function.

Nevertheless, I just want to make sure that we are defining the volume integral containing r = 0 to be 4∏ because it doesn't seem to give rise to contradictions, not actually saying that the summation of volume elements over ∇(

**r**/r^3) could ever be 4∏. That seems absurd to me.