Del operator

Henje

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Dec 16, 2014
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i'm going through some tricky vector product arrangements and rearrangements. some involve expressions like phi del (i mean the symbol phi followed by the operator del) - no dot or cross product involved. what exactly does this mean? is it the same as del phi? simply the gradient of phi? i can't find any websites which clarify this. any help appreciated
 
Without a "\(\displaystyle \times\)" or \(\displaystyle \cdot\) after del, \(\displaystyle \nabla\), when then applied to a function, we have the gradient rather than a curl or divergence. It must be applied to a scalar function and the result, after multiplying by the scalar function, \(\displaystyle \phi(x, y, z)\) is a vector function:

For any scalar function, f(x), \(\displaystyle \phi\nabla\) applied to f is the vector function \(\displaystyle \phi\nabla f= \phi(x,y,z)\frac{\partial f}{\partial x}\vec{i}+ \phi(x,y,z)\frac{\partial f}{\partial x}\vec{j}+ \phi(x,y,z)\frac{\partial f}{\partial z}\vec{k}\).

Of course, there are many variations on that. If \(\displaystyle \vec{\phi}= \phi_1\vec{i}+ \phi_2\vec{j}+ \phi_3\vec{k}\) were a vector function, you could have \(\displaystyle \vec{\phi}\cdot\nabla\) which, applied to scalar function f, would give \(\displaystyle \phi_1\frac{\partial f}{\partial x}+ \phi_2\frac{\partial f}{\partial y}+ \phi_3\frac{\partial f}{\partial z}\) and similarly for \(\displaystyle \vec{\phi}\times\nabla \).

Or we could have \(\displaystyle \phi\nabla\cdot\), \(\displaystyle \phi\nabla\times\), or \(\displaystyle \vec{\phi}\times\nabla\times\) which would have to be applied to a vector function.
 
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