relationship of polar unit vectors to rectangular
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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$\begingroup$To understand the formula:
1) draw a Cartesian coordinates with a unit circle (centred at origin with a radius of 1).
2) draw a vector pointing towards 2-oclock (so we have $\theta = 30$, this is arbitrary but I prefer this 30 degree angle as I tend to confuse the sin and cos on a 45 degree angle) with a length of 1, so this vector ($\vec{V}$) ends on the unit circle.
3) The key point to remember is that we are converting the unit vectors of the Cartesian system ($\hat{i}$ and $\hat{j}$) to those of the polar system ($\hat{r}$ and $\hat{\theta}$). Unit vector by definition is length 1.
4) To get to the radial unit vector $\hat{r}$: move $1\times cos(\theta)$ units along the x direction ($cos(\theta) \hat{i}$), then move $1\times sin(\theta)$ units along the y direction ($sin(\theta) \hat{j}$). That is the 1st equation: $\hat{r} = cos(\theta) \hat{i} + sin(\theta) \hat{j}$. Note that this how vector additions work. Draw it out on your paper and you will figure it out immediately.
5) Now to get the tangential unit vector ($\hat{\theta}$): that is, by definition, right-angle to $\hat{r}$, with a length of 1. So you will know that all you need to do is switch the positions of $cos(\theta)$ and $sin(\theta)$ in your 1st equation, and add a minus sign to one of them (so the dot product of these 2 resultant vectors is 0, equivalent to perpendicular). As we define anti-clock wise as the positive direction, the minus sign goes to the $\hat{i}$ direction. So here it is the 2nd equation: $\hat{\theta} = -sin(\theta) \hat{i} + cos(\theta） \hat{j}$.
6) If you dislike the step (5) way. Draw that $\hat{\theta}$ out on your unit circle: starting from origin, pointing towards 11-oclock (right angle to $\hat{r}$) and ending on the unit circle (length=1). To get to that end point by moving only along x- and y- directions, first move $-1\times sin(\theta)$ units on x-, then $1\times cos(\theta)$ units on y-. You get the same formula.
To reverse the conversion: from polar to Cartesian, you could simply do some pure algebra on the 2 equations we just derived, or use geometry to "move to" the target point you wish to derive.
$\endgroup$ $\begingroup$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 $$ \left\{\begin{align} \hat{u} &= (\cos \theta) \hat{\imath} + (\sin \theta)\hat{\jmath} \\ \hat{v} &= (-\sin \theta) \hat{\imath} + (\cos \theta)\hat{\jmath}. \end{align}\right. $$
$\endgroup$ 3 $\begingroup$I too got stuck there, and found the answer in a YouTube video at: . 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.
An image from the final moment of the video.
$\endgroup$ $\begingroup$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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