- #1
yungman
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I have a math question in this:http://en.wikipedia.org/wiki/Helmholtz_decomposition.
If you look at the proof, what is
[tex]\nabla_x\frac {1}{|\vec x-\vec x'|}\;and\; \nabla_x \times \frac {1}{|\vec x-\vec x'|}[/tex]
My thinking the first one is just the gradient of ##\frac {1}{|\vec x-\vec x'|}##. But the second one is puzzling. A curl has to perform on a vector...that is ##\nabla\times\vec A##, not ##\nabla\times K## where ##K## is a scalar function. Last I check, ##|\vec x-\vec x'|## is a scalar function.
Please help. Thanks
If you look at the proof, what is
[tex]\nabla_x\frac {1}{|\vec x-\vec x'|}\;and\; \nabla_x \times \frac {1}{|\vec x-\vec x'|}[/tex]
My thinking the first one is just the gradient of ##\frac {1}{|\vec x-\vec x'|}##. But the second one is puzzling. A curl has to perform on a vector...that is ##\nabla\times\vec A##, not ##\nabla\times K## where ##K## is a scalar function. Last I check, ##|\vec x-\vec x'|## is a scalar function.
Please help. Thanks