Uniqueness Theorem's for Vector Fields

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pmb

Main Question or Discussion Point

I recall an prof of mine (from a grad EM course) telling me that once the divergence and curl are of a vector field is specified then the vector field is determined up to an additive constant.

Helmhotz's Theorem states that any vector field whose divergence and curl vanish at infinity can be written as the sum of may be written as the sum of an irrotational part and a solenoidal part - I.e. as the sum of the curl of a vector field and the grad of a scalar field.

Not exactly the same but kind of close


What is that theorem? Does anyone know this? Is it true that if I define the divergence and the curl that the vector fields are unique to an additive constant? If so then I would like to see a proof.

Thanks

Pete
 

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  • #2
selfAdjoint
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I think your prof should have said "up to an arbitrary gradient. Because curl grad (anything) = 0 and div grad (anything) = 0. This is the source of gauge invariance.
 
  • #3
pmb
Originally posted by selfAdjoint
I think your prof should have said "up to an arbitrary gradient. Because curl grad (anything) = 0 and div grad (anything) = 0. This is the source of gauge invariance.
If so then the gradient can't be arbitrary. Suppose the field in question is A. Then the div and curl are given by assumption

div A = Q
curl A = H

where Q is a scalar function and H is a vector function.


Now if we let A' = A + grad S

where S is another arbitrary scalar function. Taking the divergence of A' we have

div A' = div A + del^2 S = Q + del^2 S

But we've stated that the field in question must have the divergence given and therefore

div A' = Q + del^2 S = Q

therefore

del^2 = Q

So the function S must satisfy Laplace's equation and hence is not arbitrary.

Pmb
 
  • #4
Hurkyl
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So the function S must satisfy Laplace's equation and hence is not arbitrary.
Therein lies the proof you seek. Just do the steps the other way around! IIRC any solution of Laplace's equation has zero divergence and zero curl, right? (It's been a while since I've looked into that stuff a lot) Then you can add any solution of Laplace's equation to any solution to div A = Q & curl A = H to obtain a new solution, and all solutions are of this form.
 
  • #5
pmb
Originally posted by Hurkyl
Therein lies the proof you seek. Just do the steps the other way around! IIRC any solution of Laplace's equation has zero divergence and zero curl, right? (It's been a while since I've looked into that stuff a lot) Then you can add any solution of Laplace's equation to any solution to div A = Q & curl A = H to obtain a new solution, and all solutions are of this form.
What do you mean by "Any solution of Laplace's equation has zero divergance"??

Solutions Laplace's equation are scalar fields - not vector fields

Pete
 
  • #6
Hurkyl
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Er, brain fart. :smile:

I meant the homogeneous wave equation.
 
  • #7
pmb
Originally posted by Hurkyl
Er, brain fart. :smile:

I meant the homogeneous wave equation.
I'm sorry but I still don't understand. What wave equation? There's no wave equation here? In fact the particular example that I'm interesting in eventually solving is a static field
 
  • #8
Hurkyl
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Er, ignore me, I was working a different problem!
 
  • #9
pmb
Originally posted by Hurkyl
Er, ignore me, I was working a different problem!
Okay. I contacted a friend of mine. Turns out that this is a theorem proved in Panofsky and Phillips. I should get it in the mail by the weekend.

Thanks for your help

Pete
 
  • #10
Hurkyl
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Ok, I think I know what I'm talking about this time!

Suppose we have div A = Q and curl A = H

Suppose V is a degree of freedom for A, so that:

div (A + V) = Q and curl (A + V) = H

the latter implies curl V = 0, and via Stokes' Theorem and the Fundamental Theorem, V = grad &phi for some scalar &phi. Therefore, any degree of freedom of the original equation must be the gradient of a scalar field &phi.

If &phi is any solution to Laplace's equation, then

div grad &phi = 0 and curl grad &phi = 0


So the degrees of freedom in the original equation is precisely the set of gradients of solutions to Laplace's equation.


As to the proof that any solution exists, I'm gonna cheat and say look at Wolfram for an explicit formula.

http://mathworld.wolfram.com/HelmholtzsTheorem.html


Ok I'm only gonna partially cheat.

If you're given that div V = Q and curl V = 0, then you can prove some V exists as follows:

Via Stokes' theorem, all contour integrals of V are zero, and thus line integrals of V are path independant, and the fundamental theorem for line integrals yields a scalar function &phi whose gradient is V.


So to solve the original equation, you just need to find a vector field W with curl W = H, and then apply the above reasoning to adust the divergence properly. For this part, I'm tempted to say that you simply apply green's theorem:

&int&int&int curl F dV = &int&int F * dA

integrated over a spherical ball/shell, and take the limit as the radius goes to 0, but I can't rigorously prove it off hand.
 
  • #11
pmb
Here's the correct statement of the theorem. From page 1 of Panofsky and Phillips

"All vector fields are uniquely defined if their circulation densities and source densities are given functions of the coordinates at all points in space, and if the totality of sources, as well as the source density, is zero at infinity."


Pete
 
  • #12
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Originally posted by pmb
I recall an prof of mine (from a grad EM course) telling me that once the divergence and curl are of a vector field is specified then the vector field is determined up to an additive constant.
I think you mean that the potential field is determined up to an additive constant.

The proof can be found in Reitz, Milford & Christy.
 
  • #13
pmb
Re: Re: Uniqueness Theorem's for Vector Fields

Originally posted by asvani
I think you mean that the potential field is determined up to an additive constant.

The proof can be found in Reitz, Milford & Christy.
No. I don't mean that. If you have a vector potential "A" (e.g the magnetic vector potential) and you are given the divergence and curl of A then you can add a constant vector C and still get the same divergence and curl. Specifically speaking


In the 1st edition of an EM text by Panofsky and Phillips the authors write on page 1
--------------------------------------------------------------
"All vector fields are uniquely defined if their circulation densities and source densities are given functions of the coordinates at all points in space, and if the totality of sources, as well as the source density, is zero at infinity."
--------------------------------------------------------------

Let the field be called V and let the given div and curl be

div V = s
curl V = c

They give a proof of this statement and summarize:
--------------------------------------------------------------
(a) If the source density "s" and the circulation density "c" of a vector field V are given for a finite region of space and there are no sources at infinity, then V is uniquely defined.
(b) If V has sources "s" but no circulation "c," V is derivable from a scalar potential Phi.
(c) At a point in space where "s" and "c" vanish V is derivable from a scalar potential Phi for which div (grad Phi) = 0, or from a vector potential A for which curl (curl A) = 0. We may add that at such points the field is said to be 'harmonic."
--------------------------------------------------------------

Does anyone think the above is wrong?

Pete
 

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