Time Derivative of Unit Vectors

[PhysicsKid0123]

Quick question (a little rusty on this): Why don't unit vectors in Cartesian Coordinates not change with time? For example, suppose [tex] \mathbf{r} (t) = x(t) \mathbf{x} + y(t) \mathbf{y} + z(t) \mathbf{z}[/tex] 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: [tex] \frac{d }{dt}\mathbf{x} = 0[/tex]

 

[vancouver_water]

They can change with time. [itex] \mathbb{r(t)} = \sin{t}\mathbb{x} + \cos{t}\mathbb{y} [/itex] 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.

 

[PhysicsKid0123]

They can change with time. [itex] \mathbb{r(t)} = \sin{t}\mathbb{x} + \cos{t}\mathbb{y} [/itex] 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 [tex] \mathbf{e}_1 = \mathbf{x}, \mathbf{e}_2 = \mathbf{y}, \mathbf{e}_3 = \mathbf{z}[/tex]

 

[vancouver_water]

##\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.

 

[PhysicsKid0123]

##\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.