# Doubt regarding polar vectors

## Homework Statement

What is the sum of position vectors of all points on a circle? Don't use Cartesian system.

## Homework Equations

Sum vector $$\vec s = \int_{\theta=0}^{\theta=2{\pi}}\int_{r=0}^{r=R} \, \,\vec P \, dr d\theta$$

where $$\vec P$$ is the position vector.

## The Attempt at a Solution

Using latex proving extremely difficult, so I'll just post a pic.

What I don't want is to write the unit vectors in cartesian form and solve.
I must have done something wrong, because the s vector should point to the origin.

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• pic.jpg
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Orodruin
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First of all:
Sum vector $$\vec s = \int_{\theta=0}^{\theta=2{\pi}}\int_{r=0}^{r=R} \, \,\vec P \, dr d\theta$$

where $$\vec P$$ is the position vector.
This is not the sum of the position vector on the circle. It is the sum of the position vector on the disk. The unit circle is the set of points such that ##x^2 + y^2 = 1##, not ##x^2 + y^2 \leq 1##.

Second, the position vector in polar coordinates is not ##\vec x = r\vec e_r + \theta\vec e_\theta##.

Third, the basis vectors ##\vec e_r## and ##\vec e_\theta## are not constant so you cannot integrate as if they were.

Finally, you should really not need to integrate at all. You could just use the symmetry properties of your integral. That would involve no coordinates at all.

Ok.
Firstly, sorry, I actually meant a disk(a filled circle! )

Secondly, then do you just write position as r*r_cap?

Thirdly, I knew I was doing something wrong about taking them as constants... So how do u evaluate them?

Finally, I actually want to solve the integral.
Is there no way to use the polar system only and then solve? Do people always first write in polar then convert to cartesian then solve things? Then what is the use of polar system?

Orodruin
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Secondly, then do you just write position as r*r_cap?
Yes.

Thirdly, I knew I was doing something wrong about taking them as constants... So how do u evaluate them?
You need to express your result in a particular basis. Since there is no particular point associated to your integral, it is unclear what polar basis this would be, the polar basis vectors are different at each point. This is an issue that will get worse once you get into curved spaces where the vectors do not even belong to the same vector space and it makes no sense to add them. If you want to do the integral, you first need to decide what basis you want to express the result in. This will involve choosing a Cartesian basis or, if you prefer, the polar basis associated to a particular point in the space. What you cannot do is to mix up the polar bases of different points when you add the vectors.

Finally, I actually want to solve the integral.
Is there no way to use the polar system only and then solve? Do people always first write in polar then convert to cartesian then solve things? Then what is the use of polar system?
Your big problem is that your result is a vector and that you are trying to use a non-constant vector basis. If your integral would be a scalar, you would not have this problem. Using different coordinate systems can be very helpful in order to take advantage of different symmetries your problem may display, in the case of polar coordinates, rotational symmetry in the plane.

Note that you can use polar coordinates while still using the Cartesian basis vectors to express your final result.

Do you mean system of unit vectors by basis?

You said there is no particular point associated with the integral. What does that mean?

I am trying to change the position vector smoothly so that it points to each and every point on the disk, and sum up all these vectors.
By symmetry, this sum should be a null vector. I could use Cartesian coordinates using Cartesian basis, or polar coordinates using Cartesian basis. But how do you evaluate using a polar basis?

Does this look alright to you?

Now, my question is:
What is the integral of this r hat from theta=0 to theta=some_other_angle?( say, from 0 to pi/2)
In the above picture, I used the symmetry to get the value 0.

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Ray Vickson
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Do you mean system of unit vectors by basis?

You said there is no particular point associated with the integral. What does that mean?

I am trying to change the position vector smoothly so that it points to each and every point on the disk, and sum up all these vectors.
By symmetry, this sum should be a null vector. I could use Cartesian coordinates using Cartesian basis, or polar coordinates using Cartesian basis. But how do you evaluate using a polar basis?

$$S = \int \! \! \int_D \left(x\, \mathbf{i} + y\, \mathbf{j} \right) \, dx \, dy,$$
where ##D## is a disk of the form ##D = \{ (x,y) : x^2 + y^2 \leq R^2 \}.## If you are not supposed to use cartesian coordinates to evaluate the integral, you can switch to polar coordinates, to get
$$S = \int_{r=0}^R \int_{\theta=0}^{2 \pi} \left( r \cos(\theta) \, \mathbf{i} + r \sin(\theta) \, \mathbf{j} \right) r \, dr \, d\theta.$$

$$S = \int \! \! \int_D \left(x\, \mathbf{i} + y\, \mathbf{j} \right) \, dx \, dy,$$
where ##D## is a disk of the form ##D = \{ (x,y) : x^2 + y^2 \leq R^2 \}.## If you are not supposed to use cartesian coordinates to evaluate the integral, you can switch to polar coordinates, to get
$$S = \int_{r=0}^R \int_{\theta=0}^{2 \pi} \left( r \cos(\theta) \, \mathbf{i} + r \sin(\theta) \, \mathbf{j} \right) r \, dr \, d\theta.$$

Is there any way to not use i or j ?
Like, forget that I know what are the x, y, and z axes. Now I don't know what is i.
Now, I know only about the polar system, with the position vector r and a unit vector pointing in its direction. Now I move that vector around, lengthen it...

Now I want to sum all these vectors to get a final vector s.

I don't know anything about cartesian coordinates.

Now, is there any way to do this?(Except by pointing out symmetry)
$$\int_0^{2\pi}\hat r \, d{\theta}$$

Orodruin
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You said there is no particular point associated with the integral. What does that mean?
You cannot expect to express your answer in terms of the polar basis vectors because your integral is the sum of vectors at different points and the polar basis depends on the point. If you choose the polar basis of a particular point that would be no different from choosing to use a set of Cartesian basis vectors. This is your major problem and you will not advance with this problem until you understand this point.

Like, forget that I know what are the x, y, and z axes.
You do not "know" what these axes are. You choose what these axes are by choosing what Cartesian system to use. Due to the symmetry of your problem, it does not matter what Cartesian basis you choose.

Now, I know only about the polar system, with the position vector r and a unit vector pointing in its direction. Now I move that vector around, lengthen it...
There is a way, but until you understand the point above about the basis being position dependent. It involves describing the basis and its dependence on position in the form of Christoffel symbols ##\Gamma_{ab}^c## such that ##\partial_a \vec e_b = \Gamma_{ab}^c \vec e_c##. You then have to parallel transport each vector to a single point where you can add them up, using the basis of that point. The parallel transport is unique since you are looking at a flat Euclidean space. This belongs to more advanced concepts than what you are likely looking at at the moment.

Ray Vickson
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Is there any way to not use i or j ?
Like, forget that I know what are the x, y, and z axes. Now I don't know what is i.
Now, I know only about the polar system, with the position vector r and a unit vector pointing in its direction. Now I move that vector around, lengthen it...

Now I want to sum all these vectors to get a final vector s.

I don't know anything about cartesian coordinates.

Now, is there any way to do this?(Except by pointing out symmetry)
$$\int_0^{2\pi}\hat r \, d{\theta}$$

You have described what you would like to do in getting the "sum". However, that is essentially meaningless; you really do need to define what you mean by the sum, maybe through some limiting procedure for finite sums, or something similar. Then you need to worry about whether the thing you are trying to define actually exists, and is unique. These are no small tasks.

The definition I gave in post #7 (and hinted at in #4) seems to get at what you want, and the fact that it refers to a cartesian coordinate system is no problem at all. You are dealing with some 2-dimensional vector space based on geometry, so we understand what it means for two vectors to have an angle of ##\theta## between them, and we understand what it means for two vectors to be perpendicular to each other. In such a space you can always introduce two orthonormal vectors as a basis---calling them ##\mathbf{i}## and ##\mathbf{j}##---and then write any other vector in the form ##x\, \mathbf{i} + y\, \mathbf{j}.## If we are not given such a system, we can just go ahead and build it, because of basic properties of vectors and vector spaces.

Of course, we could, instead, choose two non-orthogonal vectors as a basis, but computations would become more tedious in such a system.

Last edited:
You cannot expect to express your answer in terms of the polar basis vectors because your integral is the sum of vectors at different points and the polar basis depends on the point.

Ok. I understand this much.

It involves describing the basis and its dependence on position in the form of Christoffel symbols Γcab\Gamma_{ab}^c such that ∂a→eb=Γcab→ec\partial_a \vec e_b = \Gamma_{ab}^c \vec e_c. You then have to parallel transport each vector to a single point where you can add them up, using the basis of that point. The parallel transport is unique since you are looking at a flat Euclidean space.

Well...
I think I'm better off using the good old Cartesian basis.

Now, can you guys tell me, what is the purpose of using ##\hat r ## and ## \hat {\theta}\, ## ?
Are they defined for just "fun"?

Orodruin
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Why have you chosen to work with an integral in searching for a meaning to use a curvilinear coordinate basis? As I have told you several times already, using a non-constant basis is not the best for that as it is not clear in the basis of which point you wish to express your answer. In more general curved spaces there also will not exist a Cartesian basis and the tangent spaces of different points will really be different vector spaces and it will make no sense whatsoever to add vectors from different points. The use of different coordinate bases is most convenient when you are looking at vector fields that display certain symmetries, such as the field from a point charge.

The use of different coordinate bases is most convenient when you are looking at vector fields that display certain symmetries, such as the field from a point charge.

And after writing the first line, you will destroy everything else and write in terms of i and j.
Nice.

Orodruin
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