Curl in spherical polar coordinates

[2019]

Hey, I've been stuck on this question for quite a while now:

Homework Statement

1a. Write down an expression for the position vector r in spherical polar coordinates.

1b. Show that for any function g(r) of r only, where r = |r|, the result \nabla x [g(r)r] = 0 is true. Why does this imply that there is a potential function associated with any vector field g(r)r?

Homework Equations

The Attempt at a Solution

So for (1a) I've written r = r\hat{e}_{r}
But for (1b) I really don't know what I'm doing, I know how to take the curl but not which function to use. So could anyone give me a clue as to where to start?
For the potential function bit I've written about the vector field being path-independent.

Thanks

 

[vela]

What is \nabla \times \mathbf{A} equal to when expressed in spherical coordinates?

 

[2019]

Well I've seen it a couple of times in determinant form, which I'm not sure how to type out on here! The bottom of this page for example:
http://hyperphysics.phy-astr.gsu.edu/hbase/curl.html

 

[vela]

OK, so just identify what the components of the vector field are equal to when the vector field is of the form g(r)r and plug it into that expression. You should find you can evaluate it without knowing the specific form of g(r).

 

[2019]

OK, so just identify what the components of the vector field are equal to when the vector field is of the form g(r)r and plug it into that expression. You should find you can evaluate it without knowing the specific form of g(r).

I don't really know how to do that. Am I supposed to use the answer from part (a)?

 

[vela]

In Cartesian coordinates, you have the three basis vectors (for R³), \hat{e}_x, \hat{e}_y, and \hat{e}_z, and the vector assigned to the point (x,y,z) can be expressed in terms of these basis vectors:
\mathbf{A} = A_x(x,y,z) \hat{e}_x + A_y(x,y,z) \hat{e}_y + A_z(x,y,z) \hat{e}_z
To calculate the curl in Cartesian coordinates, you need A_x, A_y, and A_z, which you simply identify as the coefficients of the basis vectors.

In spherical coordinates, it's convenient to use the three basis vectors \hat{e}_r, \hat{e}_\theta, and \hat{e}_\phi, and you can write
\mathbf{A} = A_r(r,\theta,\phi) \hat{e}_r + A_\theta(r,\theta,\phi) \hat{e}_\theta + A_\phi(r,\theta,\phi) \hat{e}_\phi
Here you do the same thing as before. To find A_r, for instance, you just find the coefficient of the basis vector \hat{e}_r.

In this problem, you're told the vector field is equal to \mathbf{A} = g(r)\mathbf{r}. Using your answer to part (a), you can express this in terms of the basis vectors so you can identify what the various components are.

 

[2019]

In Cartesian coordinates, you have the three basis vectors (for R³), \hat{e}_x, \hat{e}_y, and \hat{e}_z, and the vector assigned to the point (x,y,z) can be expressed in terms of these basis vectors:
\mathbf{A} = A_x(x,y,z) \hat{e}_x + A_y(x,y,z) \hat{e}_y + A_z(x,y,z) \hat{e}_z
To calculate the curl in Cartesian coordinates, you need A_x, A_y, and A_z, which you simply identify as the coefficients of the basis vectors.

In spherical coordinates, it's convenient to use the three basis vectors \hat{e}_r, \hat{e}_\theta, and \hat{e}_\phi, and you can write
\mathbf{A} = A_r(r,\theta,\phi) \hat{e}_r + A_\theta(r,\theta,\phi) \hat{e}_\theta + A_\phi(r,\theta,\phi) \hat{e}_\phi
Here you do the same thing as before. To find A_r, for instance, you just find the coefficient of the basis vector \hat{e}_r.

In this problem, you're told the vector field is equal to \mathbf{A} = g(r)\mathbf{r}. Using your answer to part (a), you can express this in terms of the basis vectors so you can identify what the various components are.

So, I get out that A_r is g(r)r, so just a function of r. Does this mean that there are no \theta or \phi terms, and when calculating the curl you never get \delta/\deltar of an r function, so it's zero?

 

[vela]

Yup, that's right.

 

[2019]

Yup, that's right.

Thanks a lot for your help
I wonder if you could help me with the next bit. It says:



I imagine that's completely wrong!

 

[vela]

Your expression for work is for a constant force. Here, the force varies with position, so you must use the general expressionW = \int_C \mathbf{F}\cdot d\mathbf{r} where C is the path the object follows.

 

[2019]

Your expression for work is for a constant force. Here, the force varies with position, so you must use the general expressionW = \int_C \mathbf{F}\cdot d\mathbf{r} where C is the path the object follows.

Ooh, okay, I thought there should be an integral somewhere! Thanks again