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Cross product as a gradient?

by Mentia
Tags: cross, gradient, product
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Mentia
#1
Feb23-12, 03:45 PM
P: 10
Is it possible to nontrivially represent the cross product of a vector field [itex]\vec{f}(x,y,z)[/itex] with its conjugate as the gradient of some scalar field [itex]\phi(x,y,z)[/itex]?

In other words, can the PDE

[itex]\vec{\nabla}\phi(x,y,z) = \vec{f}(x,y,z)\times\vec{f}^\ast(x,y,z)[/itex]

be nontrivially (no constant field [itex]\vec{f}[/itex]) solved?

If not, why? If so, can you give an example of such a scalar field? This problem has popped up in my research and I'm afraid my PDE skills are lacking.
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meldraft
#2
Feb24-12, 06:54 AM
P: 280
If I remember correctly, you cannot take the cross product of two vector fields, you need the external product. It is difficult to otherwise define which vectors you are operating on.

Does your question stem from Laplace's equation?
Mentia
#3
Feb24-12, 10:05 AM
P: 10
Let [itex]\vec{f}(x,y,z)=f_x(x,y,z)\hat{x}+f_y(x,y,z)\hat{y}+f_z(x,y,z)\hat{z}[/itex]. Then [itex]\vec{f}\times\vec{f}^\ast =(f_yf_z^\ast -f_y^\ast f_z)\hat{x}+(f_zf_x^\ast -f_z^\ast f_x)\hat{y}+(f_xf_y^\ast -f_x^\ast f_y)\hat{z}=2i\left [ \text{Im}(f_yf_z^\ast)\hat{x} + \text{Im}(f_zf_x^\ast)\hat{y} + \text{Im}(f_xf_y^\ast)\hat{z} \right ][/itex].

Does there exist a non-constant and necessarily complex [itex]\vec{f}[/itex] for which there exists a [itex]\phi[/itex] that satisfies [itex]\vec{\nabla}\phi=\vec{f}\times\vec{f}^\ast[/itex]?

meldraft
#4
Feb27-12, 08:17 AM
P: 280
Cross product as a gradient?

Hmmm, complex vector spaces, nice

Here's some tips that may help you: In order for a potential function to exist, it must satisfy Laplace's equation. Try calculating the divergence of the formula you derived and see whether it suggests something meaningful.

As for it being necessarily complex, the condition is that:

[tex]f\neq f^*[/tex]

Try starting with the condition [itex]f=f^*[/itex] in the divergence of [itex]\nabla φ[/itex]. It should lead you to a non-true statement.


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