Is the position a vector?

  • Thread starter Casco
  • Start date
  • #1
82
1

Main Question or Discussion Point

Is the position a vector???

Today I had a lecture on classical mechanics and the professor talked about vectors and how do we use them in physics to describe some physical quantities as velocity and acceleration. But he was very insistent about the idea that the position it is not a vector, that it just coincides with the case where we are describing positions in [itex]R^{n}[/itex] but when we go to places like a sphere, the position it is not longer a vector cause the position in the sphere, if we want to make a representation of it, has different unit vectors, dfferent of the case of [itex]R^{n}[/itex] cartesian coordinates. So the idea it is not clear for me now, what I understand now is that the position it is not a vector because in is only a location, doesn't make sense to give it a direction or magnitude, for example, if I assign two different cartesian coordinate systems to a particle, the position will be different in both systems, nevertheless the velocity and acceleration have to be the same in all coordinate systems because they have a defined direction and magnitud that only depends of how the particle is moving. That is the idea that I got about position, but I'd like to verify it, so that's why I'm here to receive commentaries about it. So it would be helpful if someone correct me or in other case confirm that my idea is correct.
 

Answers and Replies

  • #2
NascentOxygen
Staff Emeritus
Science Advisor
9,243
1,068
Last edited:
  • #3
185
3


your professor is being overly worrisome (but correct).

In normal, everyday, usage we describe particles by giving a 3-tupple of real numbers.
(x(t), y(t), z(t)). This is definitely a (Cartesian) vector. Or in Minkowskii-space for Special
Relativity a 4-tupple (t(τ), x(τ), y(τ), z(τ)). This is a (Minkowskii) vector.

However when we move to coordinates that might be curved, for instance, in general relativity,
or, perhaps more fitting to your class, generalized coordinates the position is not usually
a vector. (A vector being something that adds like arrows do).

Using the example of a particle on a sphere the angles [itex](\vartheta, \varphi)[/itex], certainly is a 2-tupple
describing the position. It's just not a vector. (There is no vector addition rule for the angles).
 
  • #4
vanhees71
Science Advisor
Insights Author
Gold Member
2019 Award
13,788
5,605


I would only clearly distinguish between vectors and components with respect to a basis. In affine space or even a Euclidean space, after choosing one point arbitrarily you have by definition a one-to-one map between points in this space and vectors pointing from the origin to any point of the affine manifold. Then, these vectors are mapped one to one to n-tupels of real numbers by choosing a basis and giving the decomposition of the vectors wrt. this basis.
 
  • #5
1,225
75


Hi. Subtracting the positions of two points make vector. We need two points, usually Origin and referred point, to generate vector. Regards.
 
  • #6
vanhees71
Science Advisor
Insights Author
Gold Member
2019 Award
13,788
5,605


That doesn't make sense! You cannot subtract two points. As I said you have to establish the structure of an affine space on the point manifold.

You have a set of points and a vector space (usually a finite dimensional one over the real numbers as scalars). Then you connect these two structures by certain axioms like that for any pair of points [itex]P_1,P_2[/itex] there exists one and only one vector [itex]\overrightarrow{P_1 P_2}[/itex] and for any point [itex]P_1[/itex] and any vector [itex]\vec{v}[/itex] there exists a point [itex]P_2[/itex] such that [itex]\vec{v}=\overrightarrow{P_1 P_2}[/itex]. For three points [itex]P_1,P_2,P_3[/itex] the identity [itex]\overrightarrow{P_1 P_2}+\overrightarrow{P_2 P_3}=\overrightarrow{P_1 P_3}[/itex] holds, etc.

If the vector space has additional structures like a norm or even a scalar product this structure in a natural way defines a norm over the affine manifold, i.e., a distance between two points or with a scalar product the point manifold inherits an Euclidean structure, with which you can define not only a distance but also angles etc.
 
  • #7
5,439
7


Actually, sweet springs does make sense although her post was really too short.

She is talking about the displacement vector, which is a proper vector.

I think one answer to the original question lies in the difference between free and bound vectors, rather than fancy mathematics, pitched way over the head of the casco, who is clearly just encountering this stuff.

I'm sorry my post is also too short but I don't have time to expand further this morning.
 
  • #8
vanhees71
Science Advisor
Insights Author
Gold Member
2019 Award
13,788
5,605


This is the normal way you teach vectors in the first semester, namely by drawing arrows between points, explain how to add those vectors, etc. Of course, here in the forum, I cannot easily draw these things, and then it sounds very abstract. It's however very important to make these things clear from the very beginning, particularly how to distinguish between vectors and components with respect to a basis. A force, e.g., is not a triple of real numbers but a very intuitive quantity. Of course, it's described with help of a triple of numbers after you've introduced a basis with respect to which you can write the components of the force!
 
  • #9
185
3


Actually, sweet springs does make sense although her post was really too short.

She is talking about the displacement vector, which is a proper vector.
it depends on the space you're working in.

in a euclidean space (by which i'll mean an affine space with a basis)
then position is a vector and displacement is a vector.

in an affine space then position is not a vector but displacement is. [see vanhees71' post]

in a space without affine structure (say a general manifold) neither position nor
displacement are vectors.
 
  • #10
5,439
7


It is interesting to note that qbert and vanhees are each referring to different concepts.

Personally I don't think it a good idea to confuse students at the outset with this.
 
  • #11
dextercioby
Science Advisor
Homework Helper
Insights Author
12,977
540


The position of a pointlike object is a physical concept which enters the the theory (mathematics) and the experiment by defining a coordinate system with respect to which the concept of <position> gets a meaning.

Mathematically, one makes the "position of a point particle'' as an element P of the underlying set of a (finite-dimensional) real manifold and the coordinate system is just a local chart mapping the open sets containing P onto an open subset of R^n (n finite).

So by "position of a point particle" we will mean an n-tuple of real numbers. Measuring (or calculating) the "position of a point particle" means measuring or calculating n real numbers.

(I hope I remember this right)
 
  • #12
5,439
7


The position of a pointlike object is a physical concept which enters the the theory (mathematics) and the experiment by defining a coordinate system with respect to which the concept of <position> gets a meaning.

Mathematically, one makes the "position of a point particle'' as an element P of the underlying set of a (finite-dimensional) real manifold and the coordinate system is just a local chart mapping the open sets containing P onto an open subset of R^n (n finite).

So by "position of a point particle" we will mean an n-tuple of real numbers. Measuring (or calculating) the "position of a point particle" means measuring or calculating n real numbers.
Gosh, how does this help someone just moving on (in Physics) from a definition such as "a vector is an object with magnitude and direction?
 
  • #13
251
10


The position of a pointlike object is a physical concept which enters the the theory (mathematics) and the experiment by defining a coordinate system with respect to which the concept of <position> gets a meaning.

Mathematically, one makes the "position of a point particle'' as an element P of the underlying set of a (finite-dimensional) real manifold and the coordinate system is just a local chart mapping the open sets containing P onto an open subset of R^n (n finite).

So by "position of a point particle" we will mean an n-tuple of real numbers. Measuring (or calculating) the "position of a point particle" means measuring or calculating n real numbers.

(I hope I remember this right)
Long time ago, when I was a teenager, my Physics teacher explained to me what a vector is resorting to the position as a simple example. Now I realize I don't understand what a vector is.
 
  • #14
867
60


A vector is very simply a geometric object with direction and magnitude.

Tell your professor there is a briefcase full of diamonds 1037 meters south by southwest of his current location but he'll never be able to find it because you haven't told him its position, just given him a vector.
 
  • #15
5,439
7


A vector is very simply a geometric object with direction and magnitude.
All vectors have a magnitude, not all vectors have a direction.
 
  • #16
Nugatory
Mentor
12,500
4,995


But he was very insistent about the idea that the position it is not a vector, that it just coincides with the case where we are describing positions in [itex]R^{n}[/itex]
The problem here is that there are multiple definitions of "vector" kicking around.

In the most widely used and non-specialized definition, something is a vector if you can add and subtract it using the rules of vector arithmetic - and by that definition the position vector is a perfectly good vector. It's also a very useful vector for solving many real and important problems; for example, if you're doing planetary orbits or any other central-force problem, you'll want to put the origin at the center and use position vectors relative to that origin in your calculations.

But there is a more rigorous definition of a vector, one that says that a vector maintains its logical identity and physical meaning no matter what coordinate system you're using. The numerical values of the components will change as you shift from one coordinate system to another, of course, but they change according to particular rules that preserve the physical meaning of the vector. By this definition, the position vector is not a "true vector" (also called a "rank 1 contravariant tensor"); it's a "pseudo-vector" because it is defined relative to the origin of a coordinate system so is only meaningful in that coordinate system.

I suspect that your professor is trying to get you to pay attention to this second definition and not explaining very clearly. In practice, the distinction between pseudo-vectors and true vectors matters only if you mistakenly try to transform a position vector from one coordinate system to another - transform the coordinates of the endpoint and construct a new position vector relative to the new endpoint and you'll be fine.
 
  • #17
185
3


All vectors have a magnitude, not all vectors have a direction.
What is your definition of a vector?

Yes, I do understand that to 0 vector can't be said to have a direction. But one of the points
that seems to have come up is that you might not be comfortable with non-arrowy vectors.
And that is precisely what the OP was asking about. vanhees71 and I were both answering
the original question: why (or when) is position not a vector.
 
  • #18
81
0


A position is a vector from some origin which defaults to (0,0,0) in 3 dimensions.
 
  • #19
185
3


The problem here is that there are multiple definitions of "vector" kicking around.

In the most widely used and non-specialized definition, something is a vector if you can add and subtract it using the rules of vector arithmetic - and by that definition the position vector is a perfectly good vector.
That's precisely the issue. The position of an object isn't necessarily a vector. Specifically in analytic
mechanics it becomes convenient to divorce ourselves from the notion of the position being specified
by cartesian coordinates and instead specify the position with generalized coordinates which don't add
vectorially. Or, another example, the coordinates of points in general relativity. Or,for that matter,
the state vector in QM might be considered a position. Or, the thermodynamic state, .....

But there is a more rigorous definition of a vector, one that says that a vector maintains its logical identity and physical meaning no matter what coordinate system you're using. The numerical values of the components will change as you shift from one coordinate system to another, of course, but they change according to particular rules that preserve the physical meaning of the vector. By this definition, the position vector is not a "true vector" (also called a "rank 1 contravariant tensor"); it's a "pseudo-vector" because it is defined relative to the origin of a coordinate system so is only meaningful in that coordinate system.
This is just wrong - position is never a pseudo-vector, even when position is a vector.

I suspect that your professor is trying to get you to pay attention [ ... ] and not explaining very clearly.
This i think is true. (edit mine)
 

Related Threads for: Is the position a vector?

  • Last Post
Replies
1
Views
484
Replies
8
Views
2K
Replies
0
Views
914
Replies
8
Views
11K
Replies
8
Views
4K
Replies
1
Views
3K
  • Last Post
Replies
16
Views
20K
Top