Is a position vector with an argument or not? $\mathbf{r}$ vs $\mathbf{r}(t)$

Question

I thought a position vector was a vector without an argument: $$ \mathbf{r}=(x,y,z)=x\mathbf{\hat e}_x+y\mathbf{\hat e}_y+z\mathbf{\hat e}_z$$ So for a vector field $\mathbf{E}(\mathbf{r})=\mathbf{E}(x,y,z)$.

However, wikipedia define a position vector with an argument, a vector function: $$ \mathbf{r}(t)=(x(t),y(t),z(t))=x(t)\mathbf{\hat e}_x+y(t)\mathbf{\hat e}_y+z(t)\mathbf{\hat e}_z$$ But then a vector field is $\mathbf{E}(\mathbf{r}(t))=\mathbf{E}(x(t),y(t),z(t))$?

Which is correct? And which is correct as an argument for a vector field?

EDIT: I didn't mean $t$ necessarily is the time, $t$ just could be any variable of your choice.

Answer 1

Well the answer is whether it's time dependent or not? If it is, there's an argument $t$ if it isn't there isn't.

For example supposing you solved $F(\mathbf{r})=m\ddot{\mathbf{r}}$. Then generally $r$ will depend on $t$ as the particle traces out the curve of its motion in a force field $F$. I've hidden the implicit $t$. There is of course the special case where $\mathrm{r}$ is constant in time.

Now $F$ is vector field and in this case we could have $F(t,\mathbf{r}(t),\dot{\mathbf{r}}(t), \ldots)$ .

Answer 2

A position vector of a point is the vector pointing from the origin to that point. If the point is fixed, then the position vector has no argument. However, if the point is moving with time, then the position vector depends on a time parameter. In this case we have a parametrization $\textbf{r}(t)$.

A vector field $\textbf{F}$ is typically defined on the entire plan $\mathbb{R}^2$ or $3$-space $\mathbb{R}^3$, so would be of the form $\textbf{F}(x,y)$ or $\textbf{F}(x,y,z)$.

If you have a parametrization $\textbf{r}(t)$ that outputs a vector for every time $t$, then (provided the number of coordinates matches), you could plug that in to a vector field. That is, if $\textbf{r}(t)=(x(t),y(t),z(t))$ and you have a vector field $\textbf{F}(x,y,z)$, then $\textbf{F}(\textbf{r}(t))=F(x(t),y(t),z(t))$. In this case you're only looking at the vector field along the path $\textbf{r}(t)$.