Divergence & Curl -- Is multiplication by a partial derivative operator allowed?

[Fascheue]

Divergence & curl are written as the dot/cross product of a gradient.

If we take the dot product or cross product of a gradient, we have to multiply a function by a partial derivative operator.

is multiplication by a partial derivative operator allowed? Or is this just an abuse of notation

 

[Delta2]

I wouldn't say abuse of notation, rather a convention we make that the multiplication by the operator is the same as to apply the operator (to the function).

 

[Fascheue]

I wouldn't say abuse of notation, rather a convention we make that the multiplication by the operator is the same as to apply the operator (to the function).

like this?

$\begin{equation*} \frac{\partial}{\partial x} * y = \frac{\partial y}{\partial x} \end{equation*} ?$

Couldn’t we then do this?$\begin{equation*} \frac{\partial(x^2)}{\partial x} * y = \frac{\partial }{\partial x} * x^2*y = \frac{\partial(x^2y)}{\partial x} \end{equation*}$

Which is not correct.

 

[etotheipi]

No, $\frac{\partial}{\partial x}$ is an operator, whilst $\frac{\partial(x^2)}{\partial x}$ is a number.

 

[Delta2]

We have two different multiplications here which you consider to be the same kind but they are not and there is no reason to assume that the associative property holds (because they are different kind of multiplications) and indeed it doesn't hold.
Multiplication 1 ($*$): Multiplication of an operator with a function. This (*) is also called in some books as an external operation because it is between elements from different sets (the set of differential operators and the set of functions)
Multiplication 2 ($\cdot$): Multiplication of two functions, this is an internal operation because it is between functions, i.e elements of the same set the set of all functions.

We have no reason to assume that the associative property holds between these two different operations that is that $(a*b)\cdot c=a*(b\cdot c)$ and indeed it does not hold in this case.($a=\frac{\partial}{\partial x}, b=x^2, c=y$).

You got one point here though, that we use the same symbol for these two different operations, so we have a little abuse of notation (or symbols) here indeed.

 

[Fascheue]

We have two different multiplications here which you consider to be the same kind but they are not and there is no reason to assume that the associative property holds (because they are different kind of multiplications) and indeed it doesn't hold.
Multiplication 1 ($*$): Multiplication of an operator with a function. This (*) is also called in some books as an external operation because it is between elements from different sets (the set of differential operators and the set of functions)
Multiplication 2 ($\cdot$): Multiplication of two functions, this is an internal operation because it is between functions, i.e elements of the same set the set of all functions.

We have no reason to assume that the associative property holds between these two different operations that is that $(a*b)\cdot c=a*(b\cdot c)$ and indeed it does not hold in this case.($a=\frac{\partial}{\partial x}, b=x^2, c=y$).

You got one point here though, that we use the same symbol for these two different operations, so we have a little abuse of notation (or symbols) here indeed.

Are there also two kinds of dot products then? One involving multiplication 1 and the other involving multiplication 2?

 

[Delta2]

Are there also two kinds of dot products then? One involving multiplication 1 and the other involving multiplication 2?

΅Well yes this dot product $\nabla\cdot \vec{E}$ is different than this $\vec{E}\cdot\vec{E}$ though we use the same symbol.