Why Don't Unit Vectors in Cartesian Coordinates Change with Time?

Click For Summary

Discussion Overview

The discussion revolves around the behavior of unit vectors in Cartesian coordinates and whether they change with time. Participants explore the implications of time-dependent position vectors and the nature of unit vectors in a Cartesian framework.

Discussion Character

  • Debate/contested
  • Conceptual clarification

Main Points Raised

  • One participant questions why unit vectors in Cartesian coordinates do not change with time, specifically referencing the expression \(\frac{d}{dt}\mathbf{x} = 0\).
  • Another participant argues that unit vectors can change with time, providing an example of a time-dependent vector \(\mathbb{r(t)} = \sin{t}\mathbb{x} + \cos{t}\mathbb{y}\) which indicates that if components depend on time, the derivative will not be zero.
  • A participant clarifies that the discussion is about the unit vectors themselves, such as \(\mathbf{e}_1 = \mathbf{x}, \mathbf{e}_2 = \mathbf{y}, \mathbf{e}_3 = \mathbf{z}\), rather than the position vectors.
  • Another participant states that the unit vector \(\mathbb{x}\) can be defined as the vector from the origin to the point \((1,0,0)\), concluding that since these points do not change with time, the vector remains constant.

Areas of Agreement / Disagreement

Participants express disagreement regarding whether unit vectors in Cartesian coordinates can change with time. Some argue that they remain constant, while others provide examples suggesting they can vary depending on the context.

Contextual Notes

The discussion highlights the dependency of unit vectors on the definitions used and the conditions under which they are considered. There is an unresolved aspect regarding the implications of time dependence in different contexts.

PhysicsKid0123
Messages
95
Reaction score
1
Quick question (a little rusty on this): Why don't unit vectors in Cartesian Coordinates not change with time? For example, suppose \mathbf{r} (t) = x(t) \mathbf{x} + y(t) \mathbf{y} + z(t) \mathbf{z} How exactly do we know that the unit vectors don't change with time?

Or in other words, what is the argument that justifies this expression: \frac{d }{dt}\mathbf{x} = 0
 
Last edited:
Physics news on Phys.org
They can change with time. \mathbb{r(t)} = \sin{t}\mathbb{x} + \cos{t}\mathbb{y} is a unit vector that changes with time. If none of the components depend on time, the derivative will be 0. Otherwise the derivative will be another vector.
 
vancouver_water said:
They can change with time. \mathbb{r(t)} = \sin{t}\mathbb{x} + \cos{t}\mathbb{y} is a unit vector that changes with time. If none of the components depend on time, the derivative will be 0. Otherwise the derivative will be another vector.
I'm talking about the unit vectors in Cartesian coordinates themselves \mathbf{e}_1 = \mathbf{x}, \mathbf{e}_2 = \mathbf{y}, \mathbf{e}_3 = \mathbf{z}
 
##\mathbb{x}## for example can be defined as the vector from the origin to the point ## (1,0,0) ##. Since the two points are not changing with time, the vector won't change with time either.
 
vancouver_water said:
##\mathbb{x}## for example can be defined as the vector from the origin to the point ## (1,0,0) ##. Since the two points are not changing with time, the vector won't change with time either.
okay, that's true, now I remember. Thanks.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
996
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K