# Confirm the characteristic of Position Vector.

1. Mar 22, 2011

### yungman

I want to confirm this, for any position vector.

1) They are radial vector that start from the origin to a point in space.

2) If A is a position vector:

$$\nabla X \vec A \;=\; 0 \;\hbox { and }\; \nabla \cdot \vec A \;\hbox { not equal to zero. }$$

3) Any position vector in spherical coordinates contain only $\hat R$ term:

$$\vec A _{(x,y,z)} = \vec A_{(R,\theta,\phi)} = \hat R A_R \;\hbox { with }\; A_R=|\vec A|$$

Am I correct?

Last edited: Mar 22, 2011
2. Mar 22, 2011

### I like Serena

All true except for the curl.
The curl is only 0 if A_r is independent of theta and independent of phi.

You can see how this works on http://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates

[EDIT]Hold on, a little more specific is that the curl is 0 for a position vector.
But for a position vector A_r is equal to r.[/EDIT]

3. Mar 22, 2011

### yungman

But $\vec A_R$ by definition is a vector field defining A by a radius vector ( position vector) r. Position vector is originated at the origin and the arrow head at some (x,y,z).

So in this sense, curl of a position vector has to be zero.

Am I correct?

Thanks

BTW, I am reading up in the calculus book too, the components of a position vector are constant respect to the coordinates but can be a function of time t so the vector can trace out a curve with time.

4. Mar 22, 2011

### I like Serena

A vector field is usually defined by a local base of unit vectors.

For instance, in polar coordinates the following holds:

A_r is the magnitude in the direction of the position vector.
Furthermore we have A_phi and A_theta that are magnitudes perpendicular to the direction vector (in the directions that phi resp. theta would go).
A_r, A_phi, and A_theta are functions of r, theta, and phi.

Note that this local base of unit vectors changes direction when moving to another position vector.

One of the simplest vector fields is the one where A_r=r and A_theta=A_phi=0
Another one is A_r=1 and A_theta=A_phi=0.

These vector fields have curl A=0 and div A!=0

5. Mar 22, 2011

### yungman

Sorry, I mis-read. I thought you meant

$$\vec A _{(\vec r)}$$

I read back, you meant the R component of A in spherical coordinate. By definition of position vector, the A_r has to be constant because it start at origin and end at the point. So the curl has to be zero.