# Vector addition for differentials in the context of electric potential

Recently, my professor drew the following diagrams to explain $$\vec{ds}$$ (in the context of electric potential, where $$V=-\int\vec{E}\cdot\vec{ds}$$ He showed us the following diagrams and accompanying conclusions:

I'm just generally confused as to how he arrived at these conclusions. The first one I completely understand (it's just simply $$dx$$, $$dy$$, and $$dz$$ in their respective directions), the second one I half-understand (for example, shouldn't the second term be $$dr d\phi\hat{\phi}$$ instead of $$r d\phi\hat{\phi}$$?), and the third one I'm just lost on. Any help would be greatly appreciated.

I think that your professor is showing the differential vector for infinestimal change in each coordinate component.

The diagrams correspond to cartesian, cylindrical and spherical coordinate systems from top to bottom order.

In cartesian coordinate system, the coordinate components or axes are independent. So, infinestimal change $$dx$$ in x-axis doesn't make a difference in other axes (y-axis and z-axis). So, if we make a change $$dx$$ in x-axis we simply get $$dx$$, for y-axis $$dy$$ and for z-axis $$dz$$. These changes are in the direction of unit vectors of each axis. So the resulting infinestimal change vector for cartesian coordinates:

$$d\vec{s}=dx\hat{x}+dy\hat{y}+dz\hat{z}$$

For cylindrical coordinates, axes are also independent, r-axis (a.k.a. axial-distance or radial axis) and z-axis (a.k.a axial coordinate or height) are as same as in the cartesian coordinates, so for infinestimal change in r-axis corresponds to change $$dr$$ in the direction of $$\hat{r}$$ unit vector and infinestimal change in z-axis corresponds to change $$dz$$ in the direction of $$\hat{z}$$. But phi-axis (a.k.a. azimuthal coordinate) is an angle rather than being a distance or lenght, so its unit vector $$\hat{\phi}$$ is in the direction of the change of the $$\phi$$ angle. Imagine a point described in cylindrical coordinates $$(r, \phi, z)$$; $$r$$ describes the horizontal distance to the origin of the point, $$\phi$$ describes the angle between the point and the +r-axis. If we change $$\phi$$ angle, we scan area on polar plane ($$r-\phi$$ plane) which looks similar to pizza slice. And lastly $$z$$ describes the vertical distance of the point to the polar plane. By changing $$\phi$$ by an infinestimal angle $$d\phi$$; we, again, scan an area over the polar plane which looks like pizza slice. This area's both sides are the same length $$r$$, the angle between the sides is $$d\phi$$, so the arc length between the tip of the sides can be calculated by multiplying side lenght and the angle between sides: $$rd\phi$$ and the change in angle is in the direction of the $$\hat{\phi}$$. So the infinestimal change vector of the point can be given as:

$$d\vec{s}=dr\hat{r}+rd\phi\hat{\phi}+dz\hat{z}$$

Lastly for the spherical coordinate system we have r-axis (radial coordinate) which is also length similar to the axes of cartesian coordinates. So the infinestimal change $$dr$$ in r-axis correspondes to change $$dr$$ in the direction of $$\hat{r}$$. For a given point in spherical coordinates $$(r, \theta, \phi)$$; $$r$$ is the distance to the origin. $$\theta$$ is an polar angle between the point and vertical axis (or polar axis). $$\phi$$ is the angle between the point and horizontal axis. Change in $$\theta$$ is corresponds to the same change in spherical coordinates with the same reasoning: $$rd\theta\hat{\theta}$$. Calculation of change in $$\phi$$ on the other hand is a bit different. Change $$d\phi$$ in $$\phi$$ also scans an area over the horizontal plane, but the length of sides of the pizza shape are projection length of the point, the projection length of the point on the horizontal plane is $$r\sin(\theta)$$, if we multiply this side length with the angle between sides $$d\phi$$ we get the corresponding change $$r\sin(\theta)d\phi\hat{\phi}$$ and this change is in the direction of the change in angle $$\hat{\phi}$$. The infinestimal change vector for spherical coordinates is as follows:

$$d\vec{s}=dr\hat{r}+rd\theta\hat{\theta}+r\sin(\theta)d\phi\hat{\phi}$$

• Thank you!!! This makes so much more sense now knowing that $\phi$ is the angle from the horizontal axis. Now I completely understand why the cylindrical and spherical coordinates are the way they are :) Commented May 21, 2023 at 2:42
• @JBatswani I'm glad I could help you. The terms I used in the answer may not match the terms actually used in terminology like "infinestimal change vector" etc. But I hope you got an insight of the analogy of these calculations. Have a great day! Commented May 21, 2023 at 11:34

The infinitesimal displacement $$d\vec{s}$$ is derived from $$\vec{s}$$.

In cartesian coordinates : $$\vec{s}=x\vec{i}+y\vec{j}+z\vec{k}$$ $$d\vec{s}=dx\vec{i}+xd\vec{i}+dy\vec{j}+yd\vec{j}+dz\vec{k}+zd\vec{k}=dx\vec{i}+dy\vec{j}+dz\vec{k}$$

Because $$d\vec{i}=d\vec{j}=d\vec{k}=0$$

Your issue comes from the fact that the infinitesimal displacement is different in polar and spherical coordinates:

$$\vec{s}=r\vec{r}+\vec{z}$$

$$d\vec{s}=dr\vec{r}+r\frac{d\vec{r}}{d\phi}d\phi + zdz= dr\vec{r}+rd\phi\hat{\phi} + dz$$

Because $$\frac{d\vec{r}}{d\phi}=\hat{\phi}$$. In spherical coordinates, we obtain slightly the same expression.

$$\vec{s}=r\vec{r}$$

$$d\vec{s}=dr\vec{r}+r\frac{d\vec{r}}{d\theta}d\theta+rsin\theta \frac{d\vec{r}}{d\phi}d\phi=dr\vec{r}+rd\theta\hat{\theta}+rsin\theta d\phi\hat{\phi}$$

The diagrams are slightly misleading, because the infinitesimal changes in angular quantities are shown as being quite large. It may be more helpful to draw separate diagrams showing how $$\vec{s}$$ changes under a small change in each of the cylindrical and spherical coordinates.

In any case, the terms in the expressions for $$d \vec{s}$$ should not contain more than one infinitesimal change $$d a$$. Such terms are higher order and become negligible in the limit where the changes in quantities approach $$0$$, compared with those containing a single change.