# Graphically representing vectors with polar unit vectors without converting to Cartesian coordinates

## Short version :

How do you graphically represent a vector(without converting to Cartesian) given components in direction of $$\hat r$$ and $$\hat \theta$$ (unit vectors in polar coordinates)? and what is the correct way to calculate its magnitude? and why? (graphical explanation would be highly appreciated)
For example $$\vec r = 10\hat r + 30\hat \theta$$ .

## Long version :

If we -for example- want to represent the vector $$\vec r = 3\hat i + 2\hat j$$ in Cartesian coordinates we can do something like this : Simple vector representation in Cartesian plane (image).
As shown in the image we can imagine adding the $$\hat i$$ vectors one at a time and then adding the $$\hat j$$ vectors as these vectors have constant magnitude and direction relative to the axis.

With polar coordinates however, the first thing that comes to mind when representing vectors is the polar representation which uses a value for the magnitude of the vector ($$r$$) and a value for the angle which the vector makes with the polar axis ($$\theta$$) like so : Polar representation (image).

Based on that I thought that when representing a vector like this : $$\vec r = 10\hat r + 30\hat \theta$$ I would just use the value of the coefficients of the unit vectors as the value for $$r$$ and $$\theta$$.

According to these links (some might be a little off-topic), this is not the case :

And also it is stated that unit vectors in polar coordinates depend on the value of $$\theta$$ $$\hat{r} = cos\,(\theta)\,\hat{i} + sin\,(\theta)\,\hat{j}\ , \hat{\theta} = -sin\,(\theta)\,\hat{i} + cos\,(\theta)\,\hat{j}$$ Doesn't this mean that you can't represent the given vector ? as the unit vectors could be anything when not given $$\theta$$.

So is the representation going to look something like these (images below), or something different? :

And finally how is the magnitude calculated ? (as it is not the component in the direction of $$\hat r$$ according to the posts above)

$$\hat{\theta}$$ tells you the angle you need to be at from the positive $$x$$-axis, and $$\hat{r}$$ tells you how far you need to walk out from the origin. The magnitude is most certainly given by the $$\hat{r}$$ component. Perhaps your issue is that you're not adding/multiplying vectors in polar form properly. You cannot simply take, for example, $$v_1 = 1\hat{r} + \pi\hat{\theta}$$, and conclude that $$v_1 + v_1 = 2\hat{r} + 2\pi\hat{\theta}$$. Notice that the angle has been changed, which shouldn't happen for two vectors that are colinear.

It is better to represent vectors as $$re^{i\theta}$$. From which in our example we have, $$v_1 + v_1 = e^{i\pi} + e^{i\pi} = 2e^{i\pi}$$, which has an $$\hat{r}$$ of 2, and $$\hat{\theta}$$ of still $$\pi$$.

• Looking at the linked pages such as physicsforums.com/threads/…, that is not what $\hat\theta$ and $\hat r$ mean to the people writing those pages. I have never seen a reputable publication use $\hat\theta$ and $\hat r$ this way. Oct 8, 2018 at 21:48
• @DavidK Do you mean what the answer is saying is wrong or that the pages I linked are wrong ? Oct 9, 2018 at 9:23
• @MohamedAtef This answer is true if you use $(\hat r,\hat\theta)$ to signify a set of polar coordinates. In principle, if someone says that they are using the symbols that way we have to read them that way. In practice, people use conventions so they don't have to keep learning new symbology all the time, and as far as I can remember, polar coordinates would rather be written $(r,\theta)$ (or written with other letters such as $(\rho,\theta)$), while the symbols with "hats," if they are used, actually are vectors. Oct 9, 2018 at 10:10

There are two things that make this confusing. First, while you may be familiar with mathematics where there is one Cartesian coordinate system and one polar coordinate system, with well-known conversion formulas from one to the other, those are not the only plane coordinate systems people use. Second, people sometimes use the same letters to mean very different things, depending on things such as whether or not there is a "hat" on top of the letter.

In polar coordinates, a position vector $$\vec r$$ might be written $$\vec r = (r, \theta).$$ Those coordinates cannot be manipulated like the Cartesian coordinates of a vector. The Cartesian coordinates $$(x,y)$$ correspond to a vector sum with coefficients $$x$$ and $$y,$$ namely $$x \hat \imath + y \hat \jmath$$ where $$\hat\imath$$ and $$\hat\jmath$$ are unit vectors in the $$x$$ and $$y$$ directions, but there is no general way to write $$\vec r$$ as a vector sum with coefficients $$r$$ and $$\theta.$$

Sometimes, however, people are interested in describing a point in polar coordinates, and they also want to answer certain kinds of questions about something that is happening at that point, such as the velocity or acceleration of a particle that is there at some moment in time. What they sometimes do then is to create a Cartesian coordinate system "custom-built" for that point in the plane: instead of using the usual unit vectors $$\hat\imath$$ and $$\hat\jmath$$ parallel to the $$x$$ and $$y$$ axes, they look at the vector $$\vec r$$ from the origin of their polar coordinates to the particular point of interest, and they make a unit vector $$\hat r$$ in the same direction as $$\vec r.$$ They then make another unit vector $$\hat \theta$$ perpendicular to $$\hat r,$$ usually pointing in the direction in which the polar coordinate $$\theta$$ would be increasing.

So the point in question, at distance and direction $$\vec r$$ from the origin, already has polar coordinates $$(r,\theta),$$ and we know how to get a second set of coordinates from this, namely the Cartesian coordinates $$x = r \cos\theta,$$ $$y = r \sin\theta$$; but now someone has introduced a third coordinate system different from either of these.

The new coordinate system is another Cartesian coordinate system, but in the general case it is not oriented the same way as the $$(x,y)$$ coordinate system (unless $$\theta$$ happens to be zero or some other whole multiple of $$2\pi$$), and we usually do not consider it to have the same origin as either the $$(x,y)$$ coordinates or the $$(r,\theta)$$ coordinates. If we think of its origin as being a point in the plane at all, we would most likely think of the point $$(r,\theta)$$ as the origin of this new system.

It would be very strange to want to write a position vector such as the position vector $$\vec r$$ in this new coordinate system; that's usually not what the new system is intended for. But it is very likely desirable to write a velocity vector or an acceleration vector in this new coordinate system. I would be surprised to see an equation such as $$\vec r = 10\hat r + 30\hat \theta$$ written in a book, because $$\vec r$$ is usually a position vector and the thing on the right side of the equation is not; but I would not be surprised at all to see a velocity vector written $$\vec v = 10\hat r + 30\hat \theta.$$

In summary, when you see an expression like $$10\hat r + 30\hat \theta,$$ you are not looking at a way of writing a vector using polar coordinates. You are looking at a set of Cartesian coordinates in a special Cartesian coordinate system. Since these coordinates are really Cartesian, you can use the usual Cartesian coordinate rules to add them (just adding the coordinates) or to find the magnitude of a vector (using the Pythagorean Theorem). And you should definitely not attempt to copy the polar coordinates of any point into this coordinate system; that is, in general $$(r = a,\theta = b) \neq a \hat r + b\theta$$ (the point whose polar coordinates are $$(r,\theta) = (a,b)$$ is not found by constructing the vector sum $$a \hat r + b\theta$$).

• This is a beautiful answer and I think I got it, so what you are saying is that $\hat r$ and $\hat \theta$ are just perpendicular unit vectors just like $\hat i$ and $\hat j$ but in a Cartesian coordinate system which is oriented differently depending on the coordinates of a certain point (origin of the new system) which also means that technically each point has it is own coordinate system so if a certain point changes its system will also change ($\hat r$ and $\hat \theta$), right ? Oct 9, 2018 at 9:52
• That is correct, the choice of which unit vectors to use for $\hat r$ and $\hat\theta$ depends on the point where you want to put the new origin. Oct 9, 2018 at 10:22