# Is a parameterized vector field still a vector field?

Suppose I have a vector field from $\mathbb{R}^3\rightarrow\mathbb{R}^3$,

$$\mathbf{A}(\mathbf{x})=A_1(\mathbf{x})\mathbf{\hat{e}}_1+A_2(\mathbf{x})\mathbf{\hat{e}}_2+A_3(\mathbf{x})\mathbf{\hat{e}}_3$$ where $\mathbf{x}=x_1\mathbf{\hat e}_1+x_2\mathbf{\hat e}_2+x_3\mathbf{\hat e}_3$.

Now suppose $\mathbf{x}$ is a function of, say $l$, $\mathbf{x}(l)=x_1(l)\mathbf{\hat e}_1+x_2(l)\mathbf{\hat e}_2+x_3(l)\mathbf{\hat e}_3$. The vector field is now $$\mathbf{A}(\mathbf{x}(l))=A_1(\mathbf{x}(l))\mathbf{\hat{e}}_1+A_2(\mathbf{x}(l))\mathbf{\hat{e}}_2+A_3(\mathbf{x}(l))\mathbf{\hat{e}}_3$$

Is this still a vector field $\mathbb{R}^3\rightarrow \mathbb{R}^3$?

If not, what is it called?

• It seems that you are restrictinf the given vector field to the image of the map $l\mapsto \mathbf x(l)$ – Hagen von Eitzen May 4 '17 at 14:54
• If you are setting up to take an integral of the vector field over a curve within $\mathbb R^3,$ you might start this way. – David K May 4 '17 at 14:55

When you later specify that the symbol $\x$ denotes a particular function $\R\to\R^3$ defined by $l \mapsto x_1(l)\e_1+x_2(l)\e_2+x_3(l)\e_3,$ you make it awkward to continue to use $\x$ as a dummy variable in the definitions of vector fields. But this does not "undefine" the vector field $\A,$ which can still be described using a different dummy variable.
So $\A$ still exists and is still a vector field $\R^3 \to \R^3.$ But in addition to that, $\x$ is a vector function $\R\to\R^3,$ and so is $\A\circ\x,$ the composition of the field $\A$ with the function $\x$; that is, $\mathbf F = \A\circ\x$ is a vector function $\mathbf F:\R\to\R^3$ defined by $l \mapsto \mathbf F(l) = \A(\x(l)).$