Curl in spherical polar coordinates

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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 [itex]\nabla[/itex] 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[itex]\hat{e}[/itex][itex]_{r}[/itex]
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
 
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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).
 
vela said:
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)?
 
In Cartesian coordinates, you have the three basis vectors (for R3), [itex]\hat{e}_x[/itex], [itex]\hat{e}_y[/itex], and [itex]\hat{e}_z[/itex], and the vector assigned to the point (x,y,z) can be expressed in terms of these basis vectors:
[tex]\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[/tex]
To calculate the curl in Cartesian coordinates, you need Ax, Ay, and Az, which you simply identify as the coefficients of the basis vectors.

In spherical coordinates, it's convenient to use the three basis vectors [itex]\hat{e}_r[/itex], [itex]\hat{e}_\theta[/itex], and [itex]\hat{e}_\phi[/itex], and you can write
[tex]\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[/tex]
Here you do the same thing as before. To find Ar, for instance, you just find the coefficient of the basis vector [itex]\hat{e}_r[/itex].

In this problem, you're told the vector field is equal to [itex]\mathbf{A} = g(r)\mathbf{r}[/itex]. 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.
 
vela said:
In Cartesian coordinates, you have the three basis vectors (for R3), [itex]\hat{e}_x[/itex], [itex]\hat{e}_y[/itex], and [itex]\hat{e}_z[/itex], and the vector assigned to the point (x,y,z) can be expressed in terms of these basis vectors:
[tex]\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[/tex]
To calculate the curl in Cartesian coordinates, you need Ax, Ay, and Az, which you simply identify as the coefficients of the basis vectors.

In spherical coordinates, it's convenient to use the three basis vectors [itex]\hat{e}_r[/itex], [itex]\hat{e}_\theta[/itex], and [itex]\hat{e}_\phi[/itex], and you can write
[tex]\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[/tex]
Here you do the same thing as before. To find Ar, for instance, you just find the coefficient of the basis vector [itex]\hat{e}_r[/itex].

In this problem, you're told the vector field is equal to [itex]\mathbf{A} = g(r)\mathbf{r}[/itex]. 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 Ar is g(r)r, so just a function of r. Does this mean that there are no [itex]\theta[/itex] or [itex]\phi[/itex] terms, and when calculating the curl you never get [itex]\delta[/itex]/[itex]\delta[/itex]r of an r function, so it's zero?
 
vela said:
Yup, that's right.

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



I imagine that's completely wrong!
 
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vela said:
Your expression for work is for a constant force. Here, the force varies with position, so you must use the general expression[tex]W = \int_C \mathbf{F}\cdot d\mathbf{r}[/tex] where C is the path the object follows.

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