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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 :

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    $\begingroup$ 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). $\endgroup$ – Mikhail Katz Oct 19 '17 at 9:02
  • $\begingroup$ @MikhailKatz To get length of infinitesimal arc we multiplying $ r $ on $ d\theta $ which in its turn coincides with chord because those are very small. Right ? $\endgroup$ – Quark Oct 19 '17 at 9:45
  • $\begingroup$ If the angle increment $d\theta$ is very small then the corresponding chord is almost the same as the arc, correct. But the arc is exactly $r\,d\theta$. $\endgroup$ – Mikhail Katz Oct 19 '17 at 9:50
  • $\begingroup$ @MikhailKatz Why only arc is $ rd\theta $ not both arc and chord ? But $ dl_\theta $ is vector not arc. $\endgroup$ – Quark Oct 19 '17 at 10:02
  • $\begingroup$ The answer depends on whether you are studying physics or mathematics. If you are studying physics, the answer is "they are so close that for all practical purposes there is no difference between them". If you are studying mathematics you may have to learn about limits and/or infinitesimals. $\endgroup$ – Mikhail Katz Oct 19 '17 at 10:18
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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.

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$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} $)...

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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 $\sin\theta$.

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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

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

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  • $\begingroup$ Judging from the login name, the OP is a physics student. Therefore they are indeed infinitesimal displacements. $\endgroup$ – Mikhail Katz Oct 20 '17 at 7:23
  • $\begingroup$ @MikhailKatz: Even in the way the textbook uses them shows they can't be infinitesimal displacements; they aren't used like vectors, they are used like scalars. $\endgroup$ – user14972 Oct 20 '17 at 8:13
  • $\begingroup$ Can we settle for calling them infinitesimal displacements in a certain direction? That's what the book says after all. I don't think it would be accurate to claim that the book describes dr as a differential form. $\endgroup$ – Mikhail Katz Oct 20 '17 at 8:20

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