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Outer differentiation?

by Creator
Tags: differentiation, outer
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Creator
#1
Feb19-04, 10:08 AM
P: 559
Since I didn't get a response on the "Diff. Eqns." thread, I put it here.

Can someone give me a clear explaination of what is meant by "outer" differentiation.

As for example, when we differentiate the vector potential, dA/dx, in order to arrive at the magnetic field, B, why would this be called an 'outer' differentiation??


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turin
#2
Feb19-04, 10:03 PM
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Originally posted by Creator
... when we differentiate the vector potential, dA/dx, in order to arrive at the magnetic field, B, why would this be called an 'outer' differentiation??
Probably because d/dx is not exactly what you do. You take the cross product of the gradient operator with the vector potential to get the magnetic field. The cross product is sometimes called/related to the outer product, but that's tensor jargon with which I am unfamiliar.
Creator
#3
Feb20-04, 10:07 AM
P: 559
Thanks for the response, Turin;

and thanks for the correction. Oc course, you are accurate in saying it is the cross product of the gradient.
I guess then if someone says "dA/dt is the outer differentiation" I should interpret that to mean they are simply referring to it as being a cross product.??


turin
#4
Feb20-04, 11:07 AM
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P: 2,327
Outer differentiation?

Originally posted by Creator
... if someone says "dA/dt is the outer differentiation" I should interpret that to mean they are simply referring to it as being a cross product.??
I would ask them what they mean. I wouldn't know how to make sense of that.


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