Physics에서의 Del 연산자
Explanation
In physics, an operator refers to a function that maps a function to another function. Among these, the del operator refers to a function that, given a function, results in a function that has the derivative of the given function as its function value. If the term operator is unfamiliar, you can simply understand it as a rule that computes a target. For example, if you insert $f$ into the function $\dfrac{d}{dx}$, the function value $f^{\prime}$ comes out.
$$ \dfrac{d }{dx} \left( f \right) = f^{\prime} $$
The del operator is usually introduced as follows.
$$ \nabla = \frac{ \partial }{ \partial x }\hat{\mathbf{x}}+\frac{ \partial }{ \partial y }\hat{\mathbf{y}}+\frac{ \partial }{ \partial z }\hat{\mathbf{z}} $$
As shown in the equation above, it is treated as if it were a vector, so it is sometimes denoted as $\vec{\nabla}$. By using this, we learn three operations on a scalar function $f$ and a vector function $\mathbf{A}=A_{x}\hat{\mathbf{x}}+A_{y}\hat{\mathbf{y}}+A_{z}\hat{\mathbf{z}}$. The three operations below are, from top to bottom, the Gradient of $f$, the Divergence of $\mathbf{A}$, and the Curl of $\mathbf{A}$.
$$ \begin{align*} \nabla f&=\frac{ \partial f }{ \partial x }\hat{\mathbf{x}}+\frac{ \partial f }{ \partial y }\hat{\mathbf{y}}+\frac{ \partial f }{ \partial z }\hat{\mathbf{z}} \\ \nabla \cdot \mathbf{A}&= \frac{ \partial A_{x} }{ \partial x }+\frac{ \partial A_{y} }{ \partial y }+\frac{ \partial A_{z} }{ \partial z } \\ \nabla\times \mathbf{A}&= \left( \frac{\partial A_{z}}{\partial y} - \frac{\partial A_{y}}{\partial z} \right)\hat{\mathbf{x}} + \left( \frac{\partial A_{x}}{\partial z} - \frac{ \partial A_{z}}{\partial x} \right)\hat{\mathbf{y}} + \left( \frac{\partial A_{y}}{\partial x}-\frac{\partial A_{x}}{\partial y} \right)\hat{\mathbf{z}} \end{align*} $$
As you can see, if you understand $\nabla$ as a vector, you can naturally accept the above calculation. However, understanding it this way is incorrect. When complex formulas appear, incorrect calculations can be made in many parts. Especially when the del operator is heavily involved in the formula, the calculation process and results may not be understood, leading to a waste of time.
The reason for writing the right-hand side value or vector the same way as the left-hand side is simply because it fits intuitively well; however, in reality, it is not the inner or outer product of $\nabla$ and $\mathbf{A}$. If you go into each document and look at the derivation process, you can understand it. Therefore, forget about the del operator and understand $\nabla f$, $\nabla \cdot \mathbf{A}$, and $\nabla \times \mathbf{A}$ as whole values or vectors.
(X) $\nabla f$ = Multiplication of the del operator and scalar function
(O) $\nabla f$ = A vector function that has as its components the derivatives of the given scalar function in three spatial coordinates, and has information about in which direction and how much $f$ increases.
Or
For the vector function $\mathbf{A} = (A_{x}, A_{y}, A_{z})$, $$ \frac{ d A_{x} }{ d x }+\frac{ d A_{y} }{ d y }+\frac{ dA_{z} }{ d z } $$ Mathematical expressions of such form frequently appear in physics, so instead of always writing them out in full, it’s agreed to simply express them as $$ \nabla \cdot \mathbf{A} $$ Conveniently defined as $\nabla = (\frac{ \partial }{ \partial x }, \frac{ \partial }{ \partial y }, \frac{ \partial }{ \partial z })$ because it’s intuitive and fits perfectly.
is the correct understanding.
Since the inner product of two vectors is commutable, if one thinks of $\nabla$ as a vector, one may think of $\nabla\cdot \mathbf{A}$ and $\mathbf{A}\cdot \nabla$ as the same. The two formulas are entirely different. Since $\nabla$ pertains to differentiation, the order is very important.
Understanding that the results of $x\left(\dfrac{df}{dx}\right)$ and $\dfrac{d(xf)}{dx}$ are not the same should make it easier. Therefore, $\mathbf{A}\cdot \nabla$ should not be understood as the inner product of two vectors but as a symbol created for easy expression because $A_{x}\frac{ \partial }{ \partial x }+A_{y}\frac{ \partial }{ \partial y }+A_{z}\frac{ \partial }{ \partial z }$ is too long. Another example is
$$ \nabla \times (\mathbf{A} \times \mathbf{B}) = (\mathbf{B} \cdot \nabla)\mathbf{A} - (\mathbf{A} \cdot \nabla)\mathbf{B} + \mathbf{A} (\nabla \cdot \mathbf{B}) - \mathbf{B} (\nabla \cdot \mathbf{A}) $$
which may lead one to mistakenly think that
$$ \nabla \times (\nabla \times \mathbf{A})=(\mathbf{A} \cdot \nabla)\nabla - (\nabla \cdot \nabla)\mathbf{A} + \nabla (\nabla \cdot \mathbf{A}) - \mathbf{A} (\nabla \cdot \nabla) $$
is valid when in reality
$$ \nabla \times (\nabla \times \mathbf{A})=\nabla(\nabla \cdot \mathbf{A})-\nabla ^{2} \mathbf{A} $$
is the correct formula. This example follows a similar context to the fact that the results of $\dfrac{d}{dx} (fg)=\dfrac{df}{dx}g+f\dfrac{dg}{dx}$ and $\dfrac{d}{dx} \left( \dfrac{df}{dx} \right) =\dfrac{d^2 f}{dx^2}$ are entirely different.