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I'm looking at p. 16 of Fleisch's Student's Guide to Vectors & Tensors. He's talking about the relationship between the unit vector in 2D rectangular vs. polar coordinate systems. He gives these equations:

$\hat{r} = cos\,(\theta)\,\hat{i} + sin\,(\theta)\,\hat{j}\\ \hat{\theta} = -sin\,(\theta)\,\hat{i} + cos\,(\theta)\,\hat{j}$

I'm just not getting it. I understand how, in rectangular coordinates, $x = r \,cos\,(\theta)$, but the unit vectors are just not computing. Would appreciate a hint. Sorry, not a fantastic question. Thank you!

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Are you sure it's $\hat\theta$ and not, say, $\hat s$ meaning the other coordinate? –  Berci May 14 '13 at 23:53
hm, that's what it says here - a misprint? –  Peter May 15 '13 at 0:04
It's not a misprint. Fleisch likely knows what he is talking about. –  Brady Trainor May 15 '13 at 1:22
It's hard to help. The question is difficult to interpret.I don't think we have access to the text. One might say the polr coordinate vectors point in direction of increasing radius and angle each. One can become fluent in how they transform, and how they compare to the related differential forms. You can learn about conceps like contravariant versus covariant. It's not a small matter. I might recommend you simply work through any exampls and exercises you can find, for instance, hopfully your text has these. It's not a shallow topic.Perhaps someone quite fluent in these matters will comment. –  Brady Trainor May 15 '13 at 1:35
Thanks for the pointers to related concepts - I will look further into this. In context, the author is contrasting the Cartesian unit vectors $\hat{i}$ and $\hat{j}$ with the 2D Polar unit vectors $\hat{r}$ (radius) and $\hat{\theta}$ (degrees from x-axis). Immediately before the 2 equations he says "An important consequence [of the definition of polar unit vectors] is that the directions of $\hat{r}$ and $\hat{\theta}$ will be different at different locations...the dependence of the polar unit vectors on position can be seen in the following relations:" Thanks again for your help. –  Peter May 15 '13 at 2:06

3 Answers 3

up vote 1 down vote accepted

The symbols on the left side of those equations don't make any sense. If you wanted to change to a new pair of coordinates $(\hat{u}, \hat{v})$ by rotating through an angle $\theta$, then you would have $$ \left\{\begin{align} \hat{u} &= (\cos \theta) \hat{\imath} + (\sin \theta)\hat{\jmath} \\ \hat{v} &= (-\sin \theta) \hat{\imath} + (\cos \theta)\hat{\jmath}. \end{align}\right. $$

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The symbols in OP do make sense. –  Brady Trainor May 15 '13 at 1:17
Oh, I see what's going on. You are correct, @BradyTrainor. The OP is looking at an orthonormal frame at point $(r, \theta)$ in polar coordinates. This ends up being the standard frame $(\hat{\imath}, \hat{\jmath})$, rotated through angle $\theta$. –  Sammy Black May 15 '13 at 1:24
OK, this is a good clue. I'm getting the picture that if I move forward and learn about transformations between the coordinate systems (as Brady mentions above) I may be able to come back at that time and understand this. So thank you guys. –  Peter May 15 '13 at 2:17

The (endpoint of the) vector $(\cos\theta,\,\sin\theta)$ is on the unit circle, exactly at angle $\theta$ (if angles are measured from the $x$-axis, towards the $y$-axis). So, in polar form, we can say $r=1$ and $\theta=\theta$.

The other one, $(-\sin\theta,\,\cos\theta)$ is its rotated version, by $+90^\circ$. So, this has $\hat r=1$ and $\hat\theta=\theta+90^\circ$.

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this is not right. r-hat is not a scalar, OP has the correct formulas. As well for theta-hat. –  Brady Trainor May 15 '13 at 1:18

I too got stuck there, and found the answer in a YouTube video at: http://www.youtube.com/watch?v=WwQTTJdAJP8. The difficulty I had was confusing the geometry of polar coordinates, where x = rcos(theta) etc, with the vector situation, where the radius vector is resolved into its components in the x and y directions, so, applying the formulae for expressing a vector in terms of its components to the case of the polar unit radial vector, we get the equation quoted: r^=cos(θ)i^+sin(θ)j^. Attached is an image from the video that gives the geometry of this equation.

enter image description here

An image from the final moment of the video.

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