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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?

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  • $\begingroup$ It seems that you are restrictinf the given vector field to the image of the map $l\mapsto \mathbf x(l)$ $\endgroup$ – Hagen von Eitzen May 4 '17 at 14:54
  • $\begingroup$ 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. $\endgroup$ – David K May 4 '17 at 14:55
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Here's my take on this.

When you say that $\newcommand{A}{{\mathbf A}}\newcommand{x}{{\mathbf x}} \newcommand{e}{\mathbf{\hat{e}}} \A(\x)=A_1(\x)\e_1+A_2(\x)\e_2+A_3(\x)\e_3$ is a vector field $\newcommand{R}{\mathbb{R}}\R^3 \to \R^3,$ you mean that $\A: \R^3\to\R^3$ is the function defined by $\x \mapsto A_1(\x)\e_1+A_2(\x)\e_2+A_3(\x)\e_3.$ In that definition, $\x$ is a dummy variable; it would be equally valid to say that $\A$ is defined by ${\mathbf y} \mapsto A_1({\mathbf y})\e_1+A_2({\mathbf y})\e_2+A_3({\mathbf y})\e_3$ or ${\mathbf m} \mapsto A_1({\mathbf m})\e_1+A_2({\mathbf m})\e_2+A_3({\mathbf m})\e_3.$

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 $l \mapsto \mathbf F(l) = \A(\x(l)).$

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