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Inverse of the curl 
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#1
Aug1410, 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. 


#2
Aug1410, 09:48 PM

P: 757

It's called vector potential



#3
Aug1410, 11:14 PM

Mentor
P: 12,070

There is no unique solution for A. You can always add a vector field of zero curl to one solution and get another solution.



#4
Aug1510, 05:42 AM

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PF Gold
P: 39,345

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". 


#5
Aug1510, 06:20 AM

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PF Gold
P: 12,016

Note that this nonuniqueness is not something that is a strange facet by the vector potential only.
You know of it from before, as the socalled "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. nonuniqueness, but existence vs. nonexistence of the vector potential(s)! Do you know, given B, how to be certain that at least one "A" exists? 


#6
Aug1610, 03:54 PM

P: 83

[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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