Improper use of [nabla operator] in vector analysis

[Swapnil]

I am not too sure how you are using the distributive law of dot products to justify your claim. Could you please expand upon what you said.

 

[arildno]

Sure, I'll take a simple example using a STANDARD dot product.
Let us look at the 2-D case, but let our vectors be represented in DISTINCT bases:
\vec{v}=v_{1}\vec{i}_{1}+v_{2}\vec{i}_{2},\vec{w}=w_{1}\vec{j}_{1}+w_{2}\vec{j}_{2}
where the i's are orthogonal unit vectors, likewise the j's, but the i's are not necessarily perpendicular to the j's.
Thus, we have:
\vec{v}\cdot\vec{w}=\sum_{m=1}^{2}\sum_{n=1}^{2}(v_{m}\vec{i}_{m}\cdot{w_{n}\vec{j}_{n})
Knowing the values of the scalar products between the i's and j's then completes the picture.


In the case of an operator "vector", we just need to know the derivatives of unit vectors as well. Essentially, then, the operator acts as if it were written in a sort of different base than the regular vector (this is not surprising, since, strictly speaking, it isn't a vector in the first place).

 

[Swapnil]

Oh, I see. So going back to your previous post about divergence, you have

\nabla\cdot\vec{F}=(\vec{i}_{r}\frac{\partial}{\partial{r}}+\vec{i}_{\theta}\frac{\partial}{r\partial\theta}+\vec{k}\frac{\partial}{\partial{z}})\cdot (F_{r}\vec{i}_{r}+F_{\theta}\vec{i}_{\theta}+F_{z} \vec{k})

and then you using the definition of dot product you gave in the last post we get:

\nabla\cdot\vec{F}=\Big(\vec{i}_{r}\frac{\partial}{\partial{r}}\Big)\cdot \vec{F} +\Big(\vec{i}_{\theta}\frac{\partial}{r\partial\theta}\Big)\cdot \vec{F} + \Big(\vec{k}\frac{\partial}{\partial{z}}\Big)\cdot \vec{F}

Then you get:

\vec{i}_{r}\cdot\frac{\partial\vec{F}}{\partial{r} }+\vec{i}_{\theta}\cdot\frac{\partial\vec{F}}{r\partial\theta}+\vec{k}\cdot\frac{\partial\vec{F}}{\partial{z}}

The only "muddy" part is in the last step where you "move" the partial derivatives from the unit vectors to the vector function. Am I correct?

 

[arildno]

The only muddy part is that I switch the "dot" and the differential operator, and in that sense "moves" the diff. operator.
Apart from that switch, we can proceed as usual, for example:
\vec{i}_{\theta}\cdot\frac{\partial\vec{F}}{r\partial\theta}=\vec{i}_{\theta}\cdot(\frac{\partial{F}_{r}}{r\partial\theta}\vec{i}_{r}+\frac{F_{r}}{r}\vec{i}_{\theta}+\frac{1}{r}\frac{\partial{F}_{\theta}}{\partial\theta}\vec{i}_{\theta}-\frac{F_{\theta}}{r}\vec{i}_{r}++)=
\frac{F_{r}}{r}+\frac{1}{r}\frac{\partial{F}_{\theta}}{\partial\theta}
the other dot products cancelling by orthogonality relations.

I think that is what you meant by moving the diff. operator?


I would like to emphasize that this manipulative trick has nothing whatsoever to do with a proof of how the curl or div should look like when written in coordinate form.
Rather, it is a calculation device that brings out the correct formulae.

 

[Swapnil]

I meant that you treat the differential operator as a constant and move it to the vector function like the following:
\Big(\vec{i}_{r}\frac{\partial} {\partial{r}}\Big)\cdot \vec{F} = \vec{i}_{r}\cdot\Big(\frac{\partial\vec{F}}{\partial r}\Big)

But I think that's what you meant too by switching the dot and the differential operator.

 

[matt grime]

Obviously you have no idea what you are talking about. Let me show you...

Ooh, boy, now there's a red rag to a bull. You want to be careful saying things like that. At least this isn't sci.math where you would be ripped to shreds (and rightly so).

 

[Swapnil]

Obviously you have no idea what you are talking about. Let me show you.

OK, I am sorry arildno. I shouldn't have said that (I guess I am getting too cocky these days). I apologize.

 

[courtrigrad]

Ooh, boy, now there's a red rag to a bull. You want to be careful saying things like that. At least this isn't sci.math where you would be ripped to shreds (and rightly so).

What is sci.math?

 

[arildno]

OK, I am sorry arildno. I shouldn't have said that (I guess I am getting too cocky these days). I apologize.

No hard feelings.

 

[matt grime]

What is sci.math?

sci.math is an usenet news group, the original, if you will.

 

[Swapnil]

I would like to emphasize that this manipulative trick has nothing whatsoever to do with a proof of how the curl or div should look like when written in coordinate form.
Rather, it is a calculation device that brings out the correct formulae.

So does this manipulative trick works for any curvilinear coordinate system?

 

[matt grime]

Just do a change of variable argument in precisely the same way.