Can one say that $\int_{r_{a}}^{r_{b}} \mathrm{d}\vec{r}$ is the infinite sum of infinitesimal magnitudes, $\mathrm{d}r$, with direction, $\hat{r}$?

Question

Say, for example, that you had a situation where you needed to some up an infinite number of infinitesimally small portions of some vector, $\vec{r}$. This vector, $\vec{r}$, can be then defined as

\begin{equation} \vec{r} = \int_{r_{a}}^{r_{b}} \mathrm{d}\vec{r} \end{equation}

which states that the infinite summation of these infinitesimally small vector contributions, $\mathrm{d}r$, will total to the vector, $\vec{r}$.

We can say that this vector, $\vec{r}$, can be split into its components of some direction, $\hat{r}$, and magnitude, $r$, given by the following relation

\begin{equation} \hat{r} = \dfrac{\vec{r}}{r} \end{equation}

Using this, we could say that the vector, $\vec{r}$, can be expressed by the following \begin{equation} \vec{r} = r\hat{r} \end{equation}

From the integral which was stated above, we know that, $\vec{r}$, can be expressed as the infinite sum of the infinitesimally small vector contributions, $\mathrm{d}\vec{r}$, so perhaps we can express this differential as

\begin{equation} \mathrm{d}\vec{r} = \hat{r}\mathrm{d}r \end{equation}

If this is true, then I could rewrite the previous integral as

\begin{equation} \vec{r} = \int_{r_{a}}^{r_{b}} \hat{r}\mathrm{d}r \end{equation}

This makes more sense to me geometrically. I can picture this as going along some function interval which describes the direction of the vector at any point, and providing it with a magnitude at that direction. Summing up all these magnitudes pointing in some, $\hat{r}$, direction should, in turn, provide me with the total vector, $\vec{r}$.

Answer 1

Yes, you can. In physics documentation, it is very common to see vectors as a magnitude with an associated direction $r\hat r$ since having constant "track" over the direction of where the calculation is taking place is useful at times.

Answer 2

You have $\mathrm d\vec r = \hat r\,\mathrm dr + r\,\mathrm d\hat r$, as you would expect from the product rule. It doesn't make sense for the limits of integration of $r$ to depend on $r$, but we can say that $\vec r(b) - \vec r(a) = \int_a^b \hat r\,\mathrm dr + \int_a^b r\,\mathrm d\hat r$, because the chain rule shows that, since $\hat r = \frac1 r\vec r$, we have $\mathrm d\hat r = \frac1 r\mathrm d\vec r - \frac1 r\hat r\,\mathrm dr$.

You already noted that the formula $\mathrm d\vec r = \hat r\,\mathrm dr$ works in a special case; you said “all portions of $\mathrm dr$ are pointing in the direction of $\vec r$”, which doesn't directly make sense to me ($\mathrm dr$ is scalar, so doesn't have ‘portions’ in an obvious sense), but perhaps you meant “if $\mathrm d\vec r$ is always parallel to $\vec r$”. We describe this by saying that all the acceleration of $\vec r$ is tangential. In this case $\hat r$ is constant, so $r\,\mathrm d\hat r$ is $0$, and we recover your formula.

At the opposite extreme, if $\vec r$ has constant length $r$ (i.e., it moves around a circle centred at the origin), then $\mathrm dr$ is $0$. We describe this situation by saying that the acceleration of $\vec r$ is normal, and we have the formula $\vec r(b) - \vec r(a) = \int_a^b r\,\mathrm d\vec r$.

In general, the acceleration of $\vec r$ has some tangential component and some normal component. You have basically discovered the 2D part of the Frenet–Serret frame! Congratulations!