Prove the dipole potential is differentiable everywhere except at the surface

In summary: I have taken the limit of sum of continuous functions (superposition of continuous functions) to show that the partial derivative is continuous everywhere except at boundary ##S'##In general a rigorous proof ought to be built on perhaps some intermediate theorems. Doing it from first principles is probably asking too much. And, if I'm honest, of limited value.Finally, you need to decide whether the integral is Riemann or Lesbesgue.Is my proof correct? I cannot say for sure if your proof is correct because I am not familiar with the specific problem and its context. However, there are a few things that I would like to point out:1) In the beginning, you mention that you need to prove the
  • #1
Mike400
59
6
Moved from a technical forum, so homework template missing
The dipole potential is given by:

##\displaystyle\psi=\int_{V'} \dfrac{\rho}{|\mathbf{r}-\mathbf{r'}|} dV'
+\oint_{S'} \dfrac{\sigma}{|\mathbf{r}-\mathbf{r'}|} dS'##

I need to prove that ##\psi## is differentiable at points except at boundary ##S'## (where it is discontinuous)

I know if partial derivatives of ##\psi## "exist and are continuous" everywhere except at boundary ##S'##, then ##\psi## is differentiable everywhere except at boundary ##S'##.

How shall I proceed in showing partial derivatives of ##\psi## exist?

Thanks for any help in advance.
 
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  • #2
You need to show us how far you can get with this problem. My main question would be how rigorous you think you need to be?
 
  • #3
PeroK said:
You need to show us how far you can get with this problem. My main question would be how rigorous you think you need to be?
Let:
##\displaystyle\psi_{V}=\int_{V'} \dfrac{\rho}{|\mathbf{r}-\mathbf{r'}|} dV'##
##\displaystyle\psi_{S}= \oint_{S'} \dfrac{\sigma}{|\mathbf{r}-\mathbf{r'}|} dS'##

I need to prove the following two statements:

##(1)## Existence of partial derivatives of ##\psi## everywhere except at boundary ##S'##

My approach: I do not have any approach right now.
----

##(2)## Continuity of partial derivatives of ##\psi## everywhere except at boundary ##S'##

To prove ##(2)##, we need to prove the following:

(i) Continuity of partial derivatives of ##\displaystyle\psi_{V}## all over space

(ii) Continuity of partial derivatives of ##\displaystyle\psi_{S}## everywhere except at boundary ##S'##

----

My approach:

(i) I do not have a simple approach. However in this book (page 150) there is a rigorous proof.

(ii) If we fix ##\mathbf{r'}##, then ##\dfrac{x-x'}{|\mathbf{r}-\mathbf{r'}|^3}## is continuous in space except at ##\mathbf{r}=\mathbf{r'}##.

Therefore ##\dfrac{(\Delta{q_i}) x-x'}{|\mathbf{r}-\mathbf{r'}|^3}## is continuous in space except at ##\mathbf{r}=\mathbf{r'}##.

Since superposition of continuous functions is a continuous function:

##\displaystyle \lim \limits_{N \to \infty} \sum_{i=1}^N \dfrac{(\Delta{q_i}) x-x'}{|\mathbf{r}-\mathbf{r'}|^3}
=\int_{S'} \dfrac{(dq') x-x'}{|\mathbf{r}-\mathbf{r'}|^3}
=\int_{S'} \dfrac{(\sigma) x-x'}{|\mathbf{r}-\mathbf{r'}|^3}\ dS'=E_{S\ x}=\dfrac{\partial {\psi_S}}{\partial x}##

is continuous everywhere except at ##\mathbf{r}=\mathbf{r'}## points, i.e. boundary ##S'##
 
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  • #4
I'm not sure how much I can help. But, anyway:

1) If you take a charged spherical shell as an example. The electric field is discontinuous but the potential is continuous, but not differentiable at the surface.

2) You can find functions ##\rho## where this would not hold. You are going to have to make some assumptions about ##\rho## and ##\sigma## being "well-behaved".

3) The term in the denominator, technically, makes these improper integrals. Certainly for points within the volume or on the surface. That's another complexity.

4) Possibly a counterexample, like the one above, is enough to show that in general the potential is not differentiable across a surface charge. In fact, if ##\sigma## is zero on some area, then the potential may well be differentiable at a point in that region. So, I'd just get an example of a non-differentiable potential.

5) Apart from that you'll just have to use the definition of a partial derivative. It's enough to do it for one variable.

To do this rigorously would be a nightmare IMHO.

Sorry I can't be of more help.

Maybe @fresh_42 has some thoughts?
 
  • #5
I don't know how much this helps, but it seems you need to use the FTC in 2-dimensions, which is Green's theorem, though I don't have a full understanding of how to do this. Is this what you are aiming for?
 
  • #6
I was told by moderators to show my work. I have worked it out myself in the best possible way I could. It just needs to be checked (if there are any mistakes).
242337

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continued below
 
  • #7
242342

242343


Are there any mistakes or anything missing?
 

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  • #8
Can anyone point out the flaws in my work if there are any.
 
  • #9
I had a look at it. In general the result won't be true. It will depend on ##\rho## being well behaved in some way.

Any rigorous proof should identify the required properties of ##\rho##.

The first thing you did was take a limit inside an integral. That in general is not possible.

On the main body you seemed to prove continuity outside the region of charge, but there was nothing to justify continuity at the singularity. A rigorous proof would have to take the limit of proper integrals in this case.

In general a rigorous proof ought to be built on perhaps some intermediate theorems. Doing it from first principles is probably asking too much. And, if I'm honest, of limited value.

Finally, you need to decide whether the integral is Riemann or Lesbesgue.
 
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  • #10
PeroK said:
Any rigorous proof should identify the required properties of ##\rho##.
##\rho## is continuous, bounded and its domain ##V'## is finite. Any more properties required to be added??

PeroK said:
The first thing you did was take a limit inside an integral. That in general is not possible.
I was also not convinced myself with part I and part II in the first place. The convergence theorems seem to be too esoteric to me. Do you know another simpler method to show the existence of partial derivatives of ##\psi^V## and ##\psi^S##?

PeroK said:
On the main body you seemed to prove continuity outside the region of charge, but there was nothing to justify continuity at the singularity. A rigorous proof would have to take the limit of proper integrals in this case.
I think I have done. Please have a look at part V

PeroK said:
Finally, you need to decide whether the integral is Riemann or Lebesgue.
I am using Riemann integration. (Haven't studied Lebesgue yet).

Note: I am a physics student, not a math student. So I am not familiar with the convergence theorems and Lebesgue integration. However all of the physics books I have never discuss the differentiability of
potentials and fields. This dissatisfies me and so I am trying to get out of this difficulty. Can you please suggest whether I really need to go through the proofs of differentiability in the first place, or can simply assume them to be continuously differentiable as physicists do. If I am to adopt the latter, how can the assumption be justified?
 

1. What is the dipole potential?

The dipole potential is a concept in physics that describes the electric potential created by a pair of opposite charges, known as a dipole. It represents the potential energy per unit charge at a point in space due to the presence of the dipole.

2. Why is it important to prove that the dipole potential is differentiable everywhere except at the surface?

Proving that the dipole potential is differentiable everywhere except at the surface is important because it helps us understand the behavior of electric fields and how they interact with dipoles. It also allows us to make more accurate predictions and calculations in various scientific and engineering applications.

3. What does it mean for the dipole potential to be differentiable everywhere except at the surface?

A function is considered differentiable at a point if it has a well-defined tangent at that point. In the case of the dipole potential, this means that the potential function is smooth and continuous everywhere except at the surface of the dipole, where it is not defined due to the singularity of the charges.

4. How is the differentiability of the dipole potential related to the behavior of electric fields?

The differentiability of the dipole potential is directly related to the behavior of electric fields. The electric field is the negative gradient of the potential function, so if the potential is differentiable, the electric field will also be continuous and well-behaved. This allows us to accurately predict and analyze the behavior of electric fields around dipoles.

5. What evidence supports the claim that the dipole potential is differentiable everywhere except at the surface?

There are several mathematical proofs that demonstrate the differentiability of the dipole potential. These include using the definition of differentiability, the Cauchy-Riemann equations, and the continuity of the electric field. Additionally, experimental evidence from various studies and applications also supports this claim.

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