Can't find potential of vector field

[Addez123]

Homework Statement

There's two exercises where I'm suppose to find if there's a potential and if so determine it.
$$1. A = 3r^2sin v e_r + r^3cosv e_v$$
$$2. A = 3r^2 sin v e_r + r^2 cos v e_v$$

Relevant Equations

None

1. To find the solution simply integrate the e_r section by dr.
$$\nabla g = A$$
$$g = \int 3r^2sin v dr = r^3sinv + f(v)$$
Then integrate the e_v section similarly:
$$g = \int r^3cosv dv = r^3sinv + f(r)$$

From these we can see that $g = r^3sinv + C$
But the answer is apparently that there is no solution since ∇ × A does not equal zero.
Wtf? Take the derivative of $r^3sinv$ with respect to r and v and you'll literally get A!
So how does it "not exist"?

Same problem with number 2.
$$g = \int 3r^2 sin v dr = r^3 sin v + f(v)$$
$$g = \int r^2 cos v dv = r^2 sin v + f(r)$$
From this you can CLEARLY see there is NO solution available.
Yet this exercise has the audacity to claim that
$$g = r^3 cos v$$
would be a solution!
Have they even tried taking the derivative of that with respect to literally ANYTHING?!
$d/dr (r^3 cos v) = 3r^2 COS v$, which is not $A_r$.

Are they trying to make me go mad??

 

[Orodruin]

Take the derivative of r3sinv with respect to r and v and you'll literally get A!

That's not how the gradient works in curvilinear coordinates.

Please see https://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates#Del_formula

 

[Addez123]

That's not how the gradient works in curvilinear coordinates.

My textbook does not cover potentials except in cartesian coordinates.

What's wrong with my approach?
I need to add a jacobian somewhere or something?

 

[pasmith]

In plane polars, <br /> \nabla g = \frac{\partial g}{\partial r} e_r + \frac 1r \frac{\partial g}{\partial v} e_v.

 

[topsquark]

My textbook does not cover potentials except in cartesian coordinates.

What's wrong with my approach?
I need to add a jacobian somewhere or something?

You would need to convert r, $e_r$, v, and $e_v$ to Cartesian coordinates. It's technically possible but it's going to be a real mess.

The concepts are identical if you use polar coordinates and it will give you some good experience to work it that way. Coordinate transformations are a good technique to learn when solving these types of problems, and many others.

-Dan

 

[Addez123]

Ahh ok I get it!
So when solving for $e_v$ in 1.
I should have done
$$g = \int r^3cosv * 1/r dv = r^2sinv + f(r)$$
which then has no solution.
Is that correct?

 

[Orodruin]

My textbook does not cover potentials except in cartesian coordinates.

But does it cover the derivative operators in curvilinear coordinates? That should be enough since the potential concept in itself is not coordinate dependent.

Edit: … and if it doesn’t, then consider another textbook that does (nudge, nudge, wink, wink)

 

[pasmith]

Ahh ok I get it!
So when solving for $e_v$ in 1.
I should have done
$$g = \int r^3cosv * 1/r dv = r^2sinv + f(r)$$
which then has no solution.
Is that correct?

No, you have <br /> \begin{split}<br /> \frac{\partial g}{\partial r} &amp;= A_r \\<br /> \frac 1r \frac{\partial g}{\partial v} &amp;= A_v \end{split} so if g exists then <br /> \int A_r\,dr + B(v) = \int rA_v\,dv + C(r).

 

[Orodruin]

No, you have <br /> \begin{split}<br /> \frac{\partial g}{\partial r} &amp;= A_r \\<br /> \frac 1r \frac{\partial g}{\partial v} &amp;= A_v \end{split} so if g exists then <br /> \int A_r\,dr + B(v) = \int rA_v\,dv + C(r).

I would write it slightly differently. If a potential $\phi$ exists, then
$$
\phi = \int^{\vec x} \vec A\cdot d\vec x.
$$
This is coordinate independent. In polar coordinates however
$$
d\vec x = \vec e_r dr + \vec e_v r \, dv
$$
and so
$$
\phi = \int^{\vec x} (A_r dr + rA_v dv).
$$
Integrate along any curve to find the result.

The result is of course the same.

 

[WWGD]

Don't we need the condition $U_x=V_y$ , in Cartesian , as a necessary condition?

 

[vanhees71]

First calculate the curl (I'd extend it to 3D to use the standard formulae for cylinder coordinates ;-)). If the curl doesn't vanish there's no scalar potential. Otherwise, I'd use the hint in #8.