Inverse of the curl


by ikaal
Tags: curl, inverse
ikaal
ikaal is offline
#1
Aug14-10, 09:02 PM
P: 1
I want to express A as a function of B in the following equation:

curl{A}=B

So I need the inverse of the curl operator, but I don't know if it exist.

Thanks.
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Curl
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#2
Aug14-10, 09:48 PM
P: 751
It's called vector potential
Redbelly98
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#3
Aug14-10, 11:14 PM
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There is no unique solution for A. You can always add a vector field of zero curl to one solution and get another solution.

HallsofIvy
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#4
Aug15-10, 05:42 AM
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Thanks
PF Gold
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Inverse of the curl


If A= f(x,y,z)i+ g(x,y,z)j+ h(x,y,z)k then curl A = (h_y- g_z)i+ (f_z- h_x)j+ (g_x- f_y)k.

If you are given that curl A= B= p(x,y,z)i+ q(x,y,z)j+ r(x, y, z)k then you must solve the system of equation h_y- g_z= p, f_z- h_x= q, g_x- f_y= r.

Since those are partial differential equations, the "constants of integration" will be functions of x, y, z. That is why, as RedBelly98 says, "You can always add a vector field of zero curl".
arildno
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#5
Aug15-10, 06:20 AM
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Note that this non-uniqueness is not something that is a strange facet by the vector potential only.

You know of it from before, as the so-called "constant of integration".
When doing partials, functions of the other variables will be "constants" with respect to that variable you have differentiated with respect to.


A far more important question, though, is, not uniqueness vs. non-uniqueness, but existence vs. non-existence of the vector potential(s)!

Do you know, given B, how to be certain that at least one "A" exists?
Anthony
Anthony is offline
#6
Aug16-10, 03:54 PM
P: 83
Quote Quote by ikaal View Post
I want to express A as a function of B in the following equation:

curl{A}=B

So I need the inverse of the curl operator, but I don't know if it exist.

Thanks.
Consider the vector field defined by:

[tex] \mathbf{A}(\mathbf{x}) = \int_0^1 \mathbf{B}(\lambda \mathbf{x}) \wedge (\lambda\mathbf{x})\, \mathrm{d}\lambda[/tex].

You might like to show that if [tex]\nabla\cdot\mathbf{B}=0[/tex], then [tex]\nabla \wedge \mathbf{A} = \mathbf{B}[/tex]. Obviously this [tex]\mathbf{A}[/tex] is not unique.


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