Position Vectors: A Geometric Explanation

In summary, the conversation discusses the concept of position vectors in two different reference frames and how they are not invariant under coordinate transformations. The idea of a position vector as a geometric object independent of any chosen frame of reference is also questioned. The conclusion is that a position vector is not a misnomer, but rather a vector whose origin may change depending on the reference frame.
  • #1
ehrenfest
2,020
1
[SOLVED] position vectors

Homework Statement


Say you have two reference frames R_1 and R_2, where the origin of R_2 is far away from the origin of R_1. Then say you have a point in space P. The position vector of P in R_1 is r_1 and the position vector of P in R_2 is r_2.
What I am confused about is that vectors are supposed to be tensors which are supposed to be geometric objects that are invariant under coordinate transformations, right? Only their components change. But when you look at r_1 it is not the same as r_2. I don't understand that?

This is from Wikipedia: "The intuition underlying the tensor concept is inherently geometrical: as an object in and of itself, a tensor is independent of any chosen frame of reference."

What am I missing?

Homework Equations





The Attempt at a Solution

 
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  • #2
The notion of position vector is not coordinate independent. (When you talk about a 'position vector' you're talking about a vector from the origin to some point.)

If we talk about vectors in terms of the vector's origin and destination, you will find that
[tex]<d_x-o_x,d_y-o_y,d_z-o_z>[/tex]
where the d's and o's are the coordinates of the vector's origin and destination, you will find that the terms are invariant over translation.
 
  • #3
Please confirm the following statement:

a position vector is a misnomer; it is not really a vector since it is not really a tensor since it is not really a geometric object that is independent of the IRF
 
  • #4
ehrenfest said:
Please confirm the following statement:

a position vector is a misnomer; it is not really a vector since it is not really a tensor since it is not really a geometric object that is independent of the IRF

http://mathworld.wolfram.com/Vector.html

No, that's incorrect. Rather, I meant that if [itex]\vec{v}[/itex] is a position vector in one reference frame let's call that reference frame [itex]A[/itex], that doesn't guarantee that [itex]\vec{v}[/itex] is a position vector in some other reference frame [itex]B[/itex].

Let's say we have [itex]\vec{v}=\vec{AB}[/itex]. Now, if [itex]A[/itex] happens to be the origin, then we can call [itex]\vec{v}[/itex] a position vector, but when we change reference frames, [itex]A[/itex] might no longer be the origin, and the [tex]\vec{v}[/itex], while still a vector, is not a position vector anymore.

N.B.: I can't recall seeing any formal definition of position vector, but in my experience it refers to 'vector with it's tail/base/start at the origin.'
 
  • #5
i see. thanks.
 

FAQ: Position Vectors: A Geometric Explanation

What is a position vector?

A position vector is a vector that describes the position of a point in space relative to an origin point. It is typically denoted as r and expressed in terms of its components, rx, ry, and rz, for a three-dimensional space.

How is the magnitude of a position vector calculated?

The magnitude or length of a position vector is calculated using the Pythagorean theorem. It is the square root of the sum of the squares of its components, r = √(rx2 + ry2 + rz2).

What is the difference between a position vector and a displacement vector?

A position vector describes the location of a point in space relative to an origin point, while a displacement vector describes the change in position of an object from one point to another. The magnitude and direction of a displacement vector depends on the initial and final positions, while a position vector remains constant regardless of the object's movement.

How do you find the angle between two position vectors?

The angle between two position vectors, r1 and r2, can be found using the dot product formula, cosθ = (r1 ⋅ r2) / (∥r1∥ ∥r2∥), where ∥r1∥ and ∥r2∥ are the magnitudes of the two vectors. The angle can then be determined by taking the inverse cosine of the resulting value.

Can a position vector be negative?

Yes, a position vector can have negative components depending on the coordinate system being used. For example, in a Cartesian coordinate system, a vector with a negative x-component would point in the opposite direction of the positive x-axis. However, the magnitude of a position vector cannot be negative as it represents a distance or length.

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