1
$\begingroup$

I'm a undergraduate and I was reading about the polar coordinate system specifically this paper. I don't understand the term: $$\frac{de_r}{d\theta} = e_\theta \text{, and } \frac{de_\theta}{d\theta} = -e_r$$
I don't see how you can have a the derivative of $de_r$ over $d\theta$ since they are not the same variable, if that makes any sence.

$\endgroup$
1

2 Answers 2

2
$\begingroup$

In reference to the diagram, ask yourself what happens to the unit vector $\hat{e}_r$ as you increase the angle $\varphi$ (as measured counterclockwise from the rightward pointing horizontal axis) by an infinitesimal amount $d\varphi$. Note that they use $\varphi$ in this diagram instead of $\theta$.

In words, the tail of the vector $\hat{e}_r$ moves counterclockwise along the circumference of the circle, with its tip always pointing radially outwards from the centre.

The difference between the transformed $\hat{e}_r$ (call it $\hat{e}_{r1}$), and the original $\hat{e}_r$ (call it $\hat{e}_{r0}$), is another vector (call it $d\hat{e}_r$), which points from the tip of $\hat{e}_{r0}$ to the tip of $\hat{e}_{r1}.$ That is, $d\hat{e}_r=\hat{e}_{r1}-\hat{e}_{r0}.$

In the limit as $d\varphi\rightarrow 0,$ that is, as $\hat{e}_{r1}$ and $\hat{e}_{r0}$ become closer and closer together, $d\hat{e}_r$ becomes closer and closer to a vector pointing perpendicular to $\hat{e}_{r0}$, i.e. it becomes close and closer to the vector $\hat{e}_\varphi$. Draw some pictures to convince yourself of this

In other words, as $d\varphi\rightarrow 0,$ $d\hat{e}_r\rightarrow \hat{e}_\varphi,$ which is to say that $$\frac{d\hat{e}_r}{d\varphi} = \hat{e}_\varphi.$$

polar coordinates

Reference for the diagram

Edit:

To address @R.W.Bird's comment. I'll try to show that the magnitude of $\frac{d\hat{e}_r}{d\varphi} = \hat{e}_\varphi$ indeed is 1.

The magnitude of $d\hat{e}_r$ is given by

$$ \left|{d\hat{e}_r}\right|=\left|\hat{e}_{r1}-\hat{e}_{r0}\right|=\sqrt{\hat{e}_{r1}^2-2\hat{e}_{r1}\cdot\hat{e}_{r0}+\hat{e}_{r0}^2}.$$

If we use the fact that by definition $\left|\hat{e}_{r}\right|=1$, then

$$ \left|{d\hat{e}_r}\right|=\sqrt{1-2\cos(d\varphi)+1}=\sqrt{2-2\cos(d\varphi)},$$

since the angle between $\hat{e}_{r0}$ and $\hat{e}_{r1}$ is $d\varphi$. Now,

$$ \left|\frac{{\left|d\hat{e}_r\right|}}{d\varphi}\right|=\left|\frac{d}{d\varphi}\sqrt{2-2\cos(d\varphi)}\right|=\left|\frac{\sin(d\varphi)}{\sqrt{2-2\cos(d\varphi)}}\right|.$$

As $d\varphi\rightarrow 0,$

$$\lim_{d\varphi\rightarrow 0}\left|\frac{{\left|d\hat{e}_r\right|}}{d\varphi}\right|=\lim_{d\varphi\rightarrow 0}\left|\frac{\sin(d\varphi)}{\sqrt{2-2\cos(d\varphi)}}\right|$$

$$=\lim_{d\varphi\rightarrow 0}\left|\frac{d}{d\varphi}\frac{\sin(d\varphi)}{\sqrt{2-2\cos(d\varphi)}}\right|$$

$$=\lim_{d\varphi\rightarrow 0}\left|\frac{\cos (d\varphi)}{\sqrt{2-2 \cos (d\varphi)}}-\frac{\sin ^{2}(d\varphi)}{(2-2 \cos (d\varphi))^{3 / 2}}\right|$$

$$=1$$

where from the first to second line we have used l'Hopital's rule. This argument is a little hand-wavy, but it gets us the right answer... sorry about that.

$\endgroup$
2
  • $\begingroup$ Your explanations of the directions are clear enough. It is not so obvious that the magnitude of the derivative should be 1. $\endgroup$
    – R.W. Bird
    Oct 31, 2019 at 17:43
  • 1
    $\begingroup$ Thanks alot, I understand alot better now, Thanks! $\endgroup$
    – LAMagix
    Nov 1, 2019 at 16:34
0
$\begingroup$

$\hat{e}_r$ is a radial unit vector. It obviously points in different directions for different values of $\theta$. So it has a derivative expressing that rate of change.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.