- #1

Addez123

- 199

- 21

- 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??

$$\nabla g = A$$

$$g = \int 3r^2sin v dr = r^3sinv + f(v)$$

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??