# Clarification on product rule for vectors in non-Cartesian coordinates

Consider position vectors $\vec{r}_1(t)$ and $\vec{r}_2(t)$ that depend on time. At least in Cartesian coordinates, the derivative of the sum is given by:

$$\frac{d}{dt}(\vec{r}_1 + \vec{r}_2) = \frac{d\vec{r}_1}{dt} + \frac{d\vec{r}_2}{dt}$$

Does this hold when the vectors are expressed in any coordinate system? My guess/hope is yes. I'm trying to think of this in a coordinate-free way, especially since a vector's existence could care less if there is a coordinate system or not. If you have two arrows and add them together, should the rate of change of the sum vector equal the sum of the rates of change of the individual vectors? If this is true geometrically, then it should be true no matter what coordinate system $\vec{r}_1$ and $\vec{r}_2$ are expressed in. Is this correct? My book doesn't prove this geometrically. It only proves the above equation by supposing that $\vec{r}_1$ and $\vec{r}_2$ are expressed in cartesian coordinates. In 2D polar $\vec{r}_1 = r_1\hat{r}(\phi_1)$ and $\vec{r}_2 = r_2\hat{r}(\phi_2)$. Only when $\phi_1$ = $\phi_2$ can you just add and say that $\vec{r} = (r_1 + r_2)\hat{r}(\phi_1)$. I'm just getting a little confused on which rules/definitions are coordinate-free and which are coordinate dependent.

Likewise, do the product rules for dot/cross products hold in any coordinate system? In Cartesian coordinates at least, the equations are:

$$\frac{d}{dt}(\vec{r}_1 \cdot \vec{r}_2) = \frac{d\vec{r}_1}{dt} \cdot \vec{r}_2+ \vec{r}_1 \cdot \frac{d\vec{r}_2}{dt} \\ \frac{d}{dt}(\vec{r}_1 \times \vec{r}_2) = \frac{d\vec{r}_1}{dt} \times \vec{r}_2 + \vec{r}_1 \times \frac{d\vec{r}_2}{dt}$$

Again my hope/guess is that this is true in any coordinate system (by any coordinate system, I specifically mean spherical/cylindrical/2D polar coordinates). All the coordinate systems that I just stated are at any given instant, equivalent to a Cartesian coordinate system since the unit vectors (in the order that we write them) form a right-handed coordinate system and are mutually perpendicular (they therefore set up a Cartesian coordinate system at that instant). If the unit vectors are right-handed and mutually perpendicular then I think the dot/cross product between vectors expressed in these other systems have exactly the same form as the dot/cross product in Cartesian coordinates (for the dot product of 2 position vectors - we have the same problem as in the above paragraph - but for the dot product between say a position vector and a force expressed in the same unit vectors $\hat{r}(\phi_1)$ and $\hat{\phi}(\phi_1)$ it's exactly the same as Cartesian coordinates I think). Anyways, if these 3 equations hold geometrically, then my guess is that they will hold for any coordinate system. Do these 3 equations work when vectors are expressed in other coordinate systems? Is there an easy proof? The first equation can probably be derived with the limit definition. My book still does everything in Cartesian coordinates though.

## 1 Answer

Yes, the "sum rule", $\frac{d(f+ g)}{du}= \frac{df}{du}+ \frac{df}{du}$, and the "product rule", $\frac{d(fg)}{du}= f(u)\frac{df}{du}+ \frac{df}{du}g(u)$, hold in any coordinate system.