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I am trying to get the gradient of function f using the following coordinate-free del operator.

$\nabla = \lim_{\Delta v \to 0} \frac{\iint_S dS}{\Delta v} $

For infinitesimal volume from $ (r, \phi , z) $ to $(r+ \Delta, \phi + \Delta \phi, z+\Delta z) $

here,

$$ \Delta v = r \Delta r \Delta \phi \Delta z $$

Through r-direction we can calculate the following

$$ \iint_{S_r} f dS = f(r+\Delta r, \phi, z)(r+\Delta r)\Delta \phi \Delta z - f(r,\phi, z)r\Delta \phi \Delta z $$ Then, (when $ \Delta v \to 0 $) $$ \frac{\iint_{S_r} f dS}{\Delta v} = \frac{\partial f}{\partial r} + \frac{1}{r}f = \frac{1}{r}\frac{\partial}{\partial r} (rf)\ \cdots \ (1) $$

But, as you know, the r component of the gradient is $ \frac{\partial f}{\partial r} $

Surprisingly, the calculation is correct in getting divergence!

Can somebody tell me what's wrong with the calculation in gradient?

Allen Math
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1 Answers1

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That’s only a definition for del as a divergence operator. The definition for del as a gradient operator is different.

The only reason we use the same symbol is because you can think of it as a vector of derivatives in coordinates.

Bendy
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  • From "Vector Analysis" by Homer Newell Jr, it says (1) is not the correct expression for the gradient and the author wants the readers to show where the error occurred and correct it. Indeed, I'd ask him if possible. – Allen Math Dec 16 '19 at 21:03
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    Don't really appreciate the downvote as someone that's trying to help you. As I said in my answer, the definition $\nabla = \lim_{\Delta v \to 0} \frac{\iint_S dS}{\Delta v}$ that you used is for del as a divergence; you can't calculate a gradient using this. In fact they're completely different operators, despite the fact that we use the same symbols for them. You can see that your definition of del doesn't make sense for a gradient, since it returns a scalar and a gradient should be a vector. – Bendy Dec 17 '19 at 09:38