Infinitesimal element of length in the $ \hat{\theta} $ direction

Question

Why author defines infinitesimal element of length in the $ \hat{\theta} $ direction exactly as $ dl_\theta = rd\theta $ ?

Why we cannot define the infinitesimal element of length in the $ \hat{\phi} $ direction directly as $ dl_\phi = rd\phi $ ? If I draw circle like this :

Answer 1

Here's an intuitive argument. If you imagine being on a circle or radius $r$, then if your angular position changes by $\Delta \theta$ you will have moved a distance $r\Delta \theta $. This suggests the same relationship should hold at an 'infinitesimal' level.

Answer 2

$dl_\theta $ is dependent on $r $: bigger for larger $r $, smaller for smaller. Like on a record, the farther away from the center you are, the more ground you cover ($l_\theta =r\theta $) for an angle $\theta $. Similarly for $d\theta $.

If you have seen the infinitesimal unit of length done in calculus, it is similar ($ds=(dx^2+dy^2)^{\frac12} $)...

Answer 3

To get the formula for the length of the circle of radius $r$ you have to integrate $r\,d\theta$ rather than $d\theta$ (otherwise you get the wrong answer).

The latitude that the point is on is a circle of radius $r\sin\theta$ in this notation. That's the explanation for the factor of $\sin\theta$.

Answer 4

The confusion here is because $\mathrm{d}r$, $\mathrm{d}\mathbf{l}$, and other similar expressions are not infinitesimal displacements.

The closest sense in which such quantities relate to infinitesimal displacements is that we can treat $\mathrm{d}r$ as giving one of the coordinates of a displacement — it's the same ideas as how we use $r$ to express one of the coordinates of a point in spherical coordinates.

Coordinates are relative to a basis; given the selection of coordinate functions $(r , \theta, \phi)$, there are standard basis vectors $(\Delta r, \Delta \theta, \Delta \phi)$. The vector $\Delta r$, for example, points in the direction in which $\theta$ and $\phi$ remain constant and $r$ increases, and the magnitude of $\Delta r$ is the inverse of the rate of $r$'s growth.

(i.e. since $r$'s rate of growth were constant, the magnitude of $\Delta r$ is how far you need to go in order for $r$ go increase by one unit)

Remark: The vector $\Delta r$ depends on all three choices of coordinate; it's rather unfortunate that the notation makes it look like it depends on $r$ alone. Be careful to remember this when working with multiple coordinate systems! For example, in polar coordinates, $\Delta r$ would refer to something else entirely.

Remark: $\mathrm{d}r$ doesn't suffer this problem; it means the same thing in both spherical coordinates and polar coordinates. But only if you use the notation to refer to the correct thing: a differential form.

For an arbitrary displacement $\mathbf{v}$, we have the identity

$$ \mathbf{v} = \mathrm{d}r(\mathbf{v}) \Delta r + \mathrm{d}\theta(\mathbf{v}) \Delta \theta + \mathrm{d}\phi(\mathbf{v}) \Delta \phi $$

where I've explicitly written $\mathrm{d}r(\mathbf{v})$ to express that we want the value of $\mathrm{d}r$ at the vector $v$. This is the same idea as using $r(P)$ to denote the $r$-coordinate of the point $P$.

The problem with this expression is that the author wants to use unit vectors rather than the standard basis vectors. The normalization to unit vectors are (using your text's notation for the unit vectors)

• $\Delta r = \hat{r}$
• $\Delta \theta = r \hat{\theta} $
• $\Delta \phi = r \sin(\theta) \hat{\phi} $

So, if we make these substitutions, for the displacement $\mathbf{v}$ we get

$$ \mathbf{v} = \mathrm{d}r(\mathbf{v}) \hat{r} + r \mathrm{d}\theta(\mathbf{v}) \hat{\theta} + r \sin(\theta) \mathrm{d}\phi(\mathbf{v}) \hat\phi $$