# Definition of the position vector

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#1
In the cartesian coordinate system, the position vector is just a displacement vector from the origin to an arbitrary point, that lets you uniquely define a point. Simple enough.

But i have difficulty expanding this concept to polar coordinates, given that the basis vectors are not really defined at the origin. equallly I can't use the basis from the end of the vector, since we don't know where that point is until the position vector is defined.

Am i missing a trick here, or do I need a better definition for the position vector?
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6 years ago
#2
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6 years ago
#3
(Original post by Desk-Lamp)
In the cartesian coordinate system, the position vector is just a displacement vector from the origin to an arbitrary point, that lets you uniquely define a point. Simple enough.

But i have difficulty expanding this concept to polar coordinates, given that the basis vectors are not really defined at the origin. equallly I can't use the basis from the end of the vector, since we don't know where that point is until the position vector is defined.

Am i missing a trick here, or do I need a better definition for the position vector?
How do you quote other vectors in alternative coordinate systems?

I can't really visualise what a vector would be in a polar system because presumably it still has to resolve into orthogonal components so giving an angle and a modulus wouldn't seem to make any sense.

Edit: the wikipedia page on polar co-ordinates has a section entitled 'vector calculus' in it, the first thing it does is define the unit vector. I have literally no idea what is is on about, you need better than As maths to understand it.
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#4
(Original post by lerjj)
How do you quote other vectors in alternative coordinate systems?

I can't really visualise what a vector would be in a polar system because presumably it still has to resolve into orthogonal components so giving an angle and a modulus wouldn't seem to make any sense.

Edit: the wikipedia page on polar co-ordinates has a section entitled 'vector calculus' in it, the first thing it does is define the unit vector. I have literally no idea what is is on about, you need better than As maths to understand it.
In polar coordinates (and in fact most non-cartesian coordinate systems) the basis vectors are actually functions of position that vary from place to place. They don't have to be orthogonal, although they are in this case.

What bothers me is that the basis vectors don't seem well defined at the origin in the polar coordinate system.
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6 years ago
#5
The idea is exactly what you said in the first place however it can be extended to spaces with any coordinate system. Think of it this way:

Imagine a nice flat 2D surface (tabletop) and a large fishing net laid out across it. I am free to mould the fishing net into whatever shape I see fit (the net doesn't really work for polar coordinates unless you add extra parameters that you can cut up the net into continuous pieces and stick them together in a smooth manner, also we are somewhat restricted in the sense that the net is made of discrete pieces but you get the idea hopefully). Here the tabletop is like our space and the netting our chosen coordinate system. Now, we are able to choose any point of the net to be our origin, and so we stick a pin in the netting at that point (this freedom to pick the origin is a property of affine spaces). If we now choose another point on the surface on the table top, we define the position vector as the vector coming from our chosen origin and ending at the chosen point. Notice that this vector purely depends on the choice of origin we made and the point on the table top, not the shape the netting took i.e. the coordinate system. I am then able to choose a set of basis coordinates in which to decompose this vector, the most natural choice being that of the coordinate basis vectors. These are well defined at every point but the origin (in some cases). However, trivially the position vector of the origin is the zero vector (zero vector definition can be found in vector space definition).

Hopefully this helps and would be happy to answer more questions if you have them.

(Original post by Desk-Lamp)
In the cartesian coordinate system, the position vector is just a displacement vector from the origin to an arbitrary point, that lets you uniquely define a point. Simple enough.

But i have difficulty expanding this concept to polar coordinates, given that the basis vectors are not really defined at the origin. equallly I can't use the basis from the end of the vector, since we don't know where that point is until the position vector is defined.

Am i missing a trick here, or do I need a better definition for the position vector?
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#6
(Original post by WishingChaff)
I am then able to choose a set of basis coordinates in which to decompose this vector, the most natural choice being that of the coordinate basis vectors. These are well defined at every point but the origin (in some cases). However, trivially the position vector of the origin is the zero vector (zero vector definition can be found in vector space definition).
I feel like you might have said the answer but it still eludes me-
If the position vector is based at the origin, then isn't the lack of a basis at the origin a critical problem?

Say for instance i wanted to express the vector of a point 3 units directly above my chosen origin; in 2D cartesian coords I just look what the basis is at the origin and write the vector in terms of that basis i.e. 3i + 0j. As far as I can see, this approach just breaks down in Polar coords since i look for the basis vectors at the origin and don't find any.
0
6 years ago
#7
(Original post by Desk-Lamp)
I feel like you might have said the answer but it still eludes me-
If the position vector is based at the origin, then isn't the lack of a basis at the origin a critical problem?

Say for instance i wanted to express the vector of a point 3 units directly above my chosen origin; in 2D cartesian coords I just look what the basis is at the origin and write the vector in terms of that basis i.e. 3i + 0j. As far as I can see, this approach just breaks down in Polar coords since i look for the basis vectors at the origin and don't find any.
In your example here, your confusion lies in evaluating the basis vectors at the origin, as opposed to the point chosen in .

So, in your example consider the 2D surface with polar coordinates across it. You define an origin and move what you define as 3 units in your cartesian coordinate system. Now, the vector coming from the origin and ending at the point (0,3) is what we define to be the position vector. Notice that the vector in cylindrical co-ordinates will be the same as in cartesian co-ordinates. The problem you had lies in choosing a suitable basis to evaluate this vector. It happened to be the case that the coordinate basis created at the origin (in cartesian coordinates) is exactly the same as the covariant basis generated from any point in . In the case of polar coordinates, I now have different coordinate basis vectors. In fact, the basis vectors depend the choice of coordinate on the flat space. So the position vector in polar coordinates is simply given by:

where,

The point to take from this is that the position vector exists independent of the coordinate system (up to a choice of origin). We then need to choose a suitable basis to evaluate the vector in. Here we normally choose the coordinate basis i.e. the basis vectors generated by the chosen coordinate system put across the surface.
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#8
What you are saying about the geometric nature of the vectors makes sense, but it doesn't really answer the question I intended, perhaps I ought to rephrase it a little.

What I was querying was the non unique nature of position vectors in polar coordinates. For instance, look at your equation
.

The way I see it, the two sides of that equation can't be equivalent. The leftg hand side is a geometric object independent of choice of coordinates, but the left hand side could describe an infinite number of different vectors, only one of which is a correct representation of R.
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6 years ago
#9
The way to look at is as follows. The quantity on the left is a purely geometric quantity and we can evaluate it in anyway we see fit. Your original post was regarding "What is the position vector"? The simple answer is that it is the thing we have on the left.

(Original post by Desk-Lamp)
What you are saying about the geometric nature of the vectors makes sense, but it doesn't really answer the question I intended, perhaps I ought to rephrase it a little.

What I was querying was the non unique nature of position vectors in polar coordinates. For instance, look at your equation
.

The way I see it, the two sides of that equation can't be equivalent. The leftg hand side is a geometric object independent of choice of coordinates, but the left hand side could describe an infinite number of different vectors, only one of which is a correct representation of R.
With regards to this question, I think you are misunderstanding what looks like. The representation of in this manner is perfectly fine. Here, corresponds to exactly one vector and not a field of vectors.

I am struggling to see what your problem is here, so if you could elaborate more that would help me to answer your question. Thanks.
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#10
if corresponds to only one vector, then which one does it correspond to, in cartesian basis vectors?
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6 years ago
#11
We already know the transformations between cartesian and polar coordinates. So,

You now need to simply evaluate the basis vector in the cartesian basis. Do you know how to do that?

Spoiler:
Show

(Original post by Desk-Lamp)
if corresponds to only one vector, then which one does it correspond to, in cartesian basis vectors?
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#12
Yes, but you can't evaluate these polar basis vectors unless you know where they are in the space, which completely defies the point.
What you've given me is a position vector that I can't use to identify a position unless i already know the position it's pointing to
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6 years ago
#13
I already know the position vector as it is given. Your picture of the vector appears to be wrong. This, as I said before, corresponds to exactly 1 vector. Do you see why this is?

(Original post by Desk-Lamp)
Yes, but you can't evaluate these polar basis vectors unless you know where they are in the space, which completely defies the point.
What you've given me is a position vector that I can't use to identify a position unless i already know the position it's pointing to
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#14
(Original post by WishingChaff)
I already know the position vector as it is given. Your picture of the vector appears to be wrong. This, as I said before, corresponds to exactly 1 vector. Do you see why this is?
If it corresponds to a distinct vector, then why can't you express it's components explicitly in the cartesian coordinate system?
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6 years ago
#15
We appear to be going in circles here as I have already defined the transformation laws between polar and cartesian coordinates.

Think of it this way. I have the grid across the 2D plane. Now, draw a line from the origin to a point say 1m away at an angle of . This will be our position vector.

If we evaluate this vector in the polar basis we get:

where is a unit vector pointing in the direction of .

Now this same vector in the cartesian basis will be:

Does this example help?

(Original post by Desk-Lamp)
If it corresponds to a distinct vector, then why can't you express it's components explicitly in the cartesian coordinate system?
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#16
(Original post by WishingChaff)
If we evaluate this vector in the polar basis we get:

How?

The basis vectors are one-one functions of position, so at which at which position are you taking this basis? the end of the vector?
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6 years ago
#17
Ah I see your confusion. You are thinking that the basis vector comes from a point in space and thus 2 times that basis vector moves you 2 times a long from that point along that direction. This is not true.

Indeed we evaluate basis vectors at a point but that does not mean they stay fixed at that particular point.

In fact, you can think of these vectors pointing in the direction defined at the point coming from the origin and not the point chosen. So, we evaluate the basis vector at a point in the space and then move it back to the origin. Then, when we say "move 9 times along that vector" we add the vector up 9 times from the origin, NOT the point.

This can be seen by considering the following. We know that (property of vector space) for and vector . If the vector was coming from the chosen point then we would have a contradiction. As we would have 2 origins defined on the space. Thus, the vector must be coming from the origin.

Has this cleared it up?

(Original post by Desk-Lamp)
How?

The basis vectors are one-one functions of position, so at which at which position are you taking this basis? the end of the vector?
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#18

It still doesn't answer the original question (which i probably didn't express clearly enough) which is how do you uniquely define a point in the space with a position vector?

To give an example, if I define vectors in cartesian coordinates A = 2i + 0j and B = 0i + 2j then following your reasoning both vectors be written A = 2r, B = 2r in the polar coordinate system. Now I know that those two rs are different vectors, but if all I gave you were the position vectors you would have no way of knowing.
0
6 years ago
#19
However, this is purely a notational problem not a physical/mathematical one. You acknowledge that in your example the two r's are in fact different vectors and thus need to be distinguished by adding a prime on one. With this, no problem arises.

(Original post by Desk-Lamp)

It still doesn't answer the original question (which i probably didn't express clearly enough) which is how do you uniquely define a point in the space with a position vector?

To give an example, if I define vectors in cartesian coordinates A = 2i + 0j and B = 0i + 2j then following your reasoning both vectors be written A = 2r, B = 2r in the polar coordinate system. Now I know that those two rs are different vectors, but if all I gave you were the position vectors you would have no way of knowing.
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#20
(Original post by WishingChaff)
However, this is purely a notational problem not a physical/mathematical one. You acknowledge that in your example the two r's are in fact different vectors and thus need to be distinguished by adding a prime on one. With this, no problem arises.
Your definition of problem and mine clearly differ .
Granted, there are no paradoxes / contradictions, but I have a position vector that doesn't label a point. In my book that is something of a problem.
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