# Why would one write a vector field as a derivative?

The motivation goes like this.

- When we define things mathematically, we want to use as few separate objects as possible. We don't want to define a new object independently if it can be defined in terms of existing things.
- Suppose a particle moves so that when it is at position $\mathbf{r}$, its velocity is $\mathbf{v}(\mathbf{r})$, where $\mathbf{v}$ is a vector field. Then if there is some function $f(\mathbf{r})$, then the particle sees $$\frac{df}{dt} = v^i \frac{\partial f}{\partial x^i}$$ by the chain rule. That is, if we interpret a vector field as a velocity field, and give it a function $f$, then we can compute another function $df/dt$, which is the rate of change of $f$ seen by a particle following the flow of the vector field, as if it were a velocity field.
- By glancing at the chain rule, you see that if you know $df/dt$ for every $f$, then you know what the vector field is.
- Hence, when we work in the more general setting of a manifold, where it's not immediately clear how to define a vector field in the usual way ("an arrow at every point"), we can use this in reverse to define what a vector field is. That is, a vector field $v$
**is**a map of functions $f \mapsto v(f) = df/dt$ which obeys certain properties. - Note that not every vector field should be
*physically*regarded as a velocity field. We're just making mathematical definitions here. The definitions are chosen to make the formalism as clean and simple as possible, possibly at the expense of intuition. - Translating between the two is very simple. For example, the field $$\mathbf{E}(\mathbf{r}) = x \hat{i} + xy \hat{j}$$ translates to $$\mathbf{E}(\mathbf{r}) = x \frac{\partial}{\partial x} + x y \frac{\partial}{\partial y}.$$ That is, whenever you see $\partial/\partial x$ you can just imagine it as a unit vector pointing in the $x$ direction. The intuition is the same, because the two definitions obey the same properties.

There are many examples of this in mathematics. For example, you might think $\log(x)$ is defined as "the number of times you have to multiple by $e$ to get to $x$", but it's unclear how to rigorously define that. Then, through semi-rigorous manipulation you can show that
$$\log(x) = \int_1^x \frac{dx'}{x'}.$$
Now the mathematician sees this and chooses to *define* $\log(x)$ as this integral. This is simpler, because it automatically works for any real $x$ and uses only the notion of an integral, which we already know. Then one can derive the "intuitive" properties of the logarithm, such as $\log(xy) = \log(x) + \log(y)$. By using a less intuitive definition, the formalism becomes simpler. And once you show that this definition is equivalent to the intuitive one, you are "allowed" to just keep on using the same intuition you started with, so you get the best of both worlds!

**Edit:** the OP asks for examples where this definition of a vector field is more practically useful. I can think of two off the top of my head. First, how do basis vectors transform when you change coordinates from $x^i$ to $y^j$? In the usual formalism you may have to memorize a formula, but with the derivatives it follows from the chain rule,
$$\frac{\partial}{\partial x^i} = \frac{\partial y^j}{\partial x^i} \frac{\partial}{\partial y^j} \equiv J^j_i \frac{\partial}{\partial y^j}$$
where $J$ is the Jacobian matrix. Next, suppose the vector field actually is a velocity field, and you want to calculate $f(x(t))$ given $x(0)$, i.e. you want to know where you'll be if you follow the flow for time $t$. In this formalism, that's a one-liner. It's just
$$f(x(t)) = (e^{t v} f(x))|_{x = x(0)}.$$
To prove this, expand the exponential in a Taylor series.

Of course, the great thing is that once you prove two formalisms are equivalent, you can use the intuition from either one interchangably, because you know they're both equally valid. So you gain intuition for a few new cases, without losing any intuition you had before. You can always change back and forth, much like the same software program can run on different hardware.

A one-line motivation is as follows:

You can identify a vector (field) with the "directional derivative" along that vector (field).

Given a point and a vector at that point, you can (try to) differentiate a function at that point in that direction.

In coordinates, the relation between your $X$ and your $\vec{A}=\sum_{i=1}^n A_i \vec{e}_i$ is $$ X= \vec{A} \cdot \nabla = \sum_{i=1}^n A_i \frac{\partial}{\partial x_i}.$$

A vector (field) is a "dynamical" object, a way of "transforming space": take the ODE, associated with $\vec{A}$, i.e. $\boldsymbol{r}'= \vec{A}(\boldsymbol{r})$, and look at the flow. Once you have the flow, you can differentiate your functions along it.