Understanding Polar Vectors on a Circle

In summary: This is the same as$$S = \int \! \! \int_0^D \left(x\, \mathbf{i} + y\, \mathbf{j} \right) \, dx \, dy$$
  • #1
Kaguro
221
57

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.
pic.jpg


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. Please help.
 

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  • #2
First of all:
Kaguro said:
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.
 
  • #3
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?
 
  • #4
Kaguro said:
Secondly, then do you just write position as r*r_cap?
Yes.

Kaguro said:
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.

Kaguro said:
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.
 
  • #5
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?
 
  • #6
Does this look alright to you?
The_pic.jpg


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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  • #7
Kaguro said:
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?

Your sum is
$$S = \int \! \! \int_D \left(x\, \mathbf{i} + y\, \mathbf{j} \right) \, dx \, dy,$$
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
$$S = \int_{r=0}^R \int_{\theta=0}^{2 \pi} \left( r \cos(\theta) \, \mathbf{i} + r \sin(\theta) \, \mathbf{j} \right) r \, dr \, d\theta.$$
 
  • #8
Ray Vickson said:
Your sum is
$$S = \int \! \! \int_D \left(x\, \mathbf{i} + y\, \mathbf{j} \right) \, dx \, dy,$$
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)
$$\int_0^{2\pi}\hat r \, d{\theta}$$
 
  • #9
Kaguro said:
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.

Kaguro said:
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.

Kaguro said:
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.
 
  • #10
Kaguro said:
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.
 
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  • #11
Orodruin said:
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.
 
  • #12
Orodruin said:
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.
 
  • #13
Now, can you guys tell me, what is the purpose of using ##\hat r ## and ## \hat {\theta}\, ## ?
Are they defined for just "fun"?
 
  • #14
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.
 
  • #15
Orodruin said:
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.
I'll follow your footsteps.
 
  • #16
Kaguro said:
And after writing the first line, you will destroy everything else and write in terms of i and j.
I do not follow your logic here. Where exactly do you think anything is being "destroyed"? You are simply looking at it from a bad perspective when you require integration. As I already told you, integration that should yield a vector is not a good example.
 
  • #17
Orodruin said:
I do not follow your logic here. Where exactly do you think anything is being "destroyed"? You are simply looking at it from a bad perspective when you require integration. As I already told you, integration that should yield a vector is not a good example.
OK. I think what I wished to do, has no mathematical meaning...

Fine, I'll stick to things that work, for now.

Thank you very much for making me understand.
 

1. What is a polar vector?

A polar vector is a type of vector that is defined by its magnitude and direction. It is typically represented by an arrow pointing in the direction of the vector, with the length of the arrow representing the magnitude. In other words, a polar vector is a quantity that has both size and direction.

2. What is the difference between a polar vector and a Cartesian vector?

The main difference between a polar vector and a Cartesian vector is the way in which they are represented. While a polar vector is represented by an arrow with magnitude and direction, a Cartesian vector is represented by its components along the x, y, and z axes. Additionally, polar vectors are typically used in two-dimensional systems, while Cartesian vectors are used in three-dimensional systems.

3. How do I convert a polar vector to a Cartesian vector?

To convert a polar vector to a Cartesian vector, you can use the following formulas: x = r * cos(theta) and y = r * sin(theta), where r is the magnitude of the vector and theta is the angle between the vector and the positive x-axis. Once you have the x and y components, you can combine them to form a Cartesian vector.

4. Can a polar vector have negative magnitude?

No, a polar vector cannot have negative magnitude. Magnitude is a measure of size and cannot be negative. However, the direction of a polar vector can be negative if it is pointing in the opposite direction of the positive axis.

5. How do I add and subtract polar vectors?

To add or subtract polar vectors, you can use the parallelogram rule or the head-to-tail method. Draw the vectors with their tails at the same point and connect the arrows to form a parallelogram. The diagonal of the parallelogram represents the resultant vector, which is the sum or difference of the original vectors. Alternatively, you can also use the formulas: x = x1 + x2 and y = y1 + y2, where x1 and y1 are the components of the first vector and x2 and y2 are the components of the second vector.

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