Verify $\nabla \times \textbf{f} = 0$ and Show Non-Existence of $\psi$ on $R_2$

In summary, the conversation discusses verifying that the vector field \textbf{f}(x,y,z) is equal to the curl of a scalar field on a given region. The first part, (A), involves finding a suitable scalar field \phi on a specific region R_1, and the speaker provides the solution \phi = arctan(x) + arccot(x) - arctan(y/x). For the second part, (B), the speaker is unsure how to proceed and eventually realizes that a potential function does not exist on region R_2 due to a contour around the pole (0,0) resulting in a non-zero path integral.
  • #1
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I wasn't quite sure how to do the second part of this question:

Given [tex]\textbf{f}(x,y,z) = (y/(x^2+y^2), -x/(x^2+y^2), 0)[/tex] where [tex](x,y) \neq (0,0)[/tex], verify that [tex]\nabla \times f = 0[/tex].

(A) Find a scalar field [tex]\phi[/tex] such that [tex]\textbf{f} = \nabla \phi[/tex] on [tex]R_1 = \{(x,y,z): y > 0\}[/tex].

(B) Show that there does NOT exist [tex]\psi[/tex] such that [tex]\textbf{f} = \nabla\psi[/tex] on [tex]R_2 = \{(x,y,z): (x,y) \neq (0,0)[/tex]For (A), I found [tex]\phi = [/tex] arctan(x) + arccot(x) - arctan(y/x).

I'm not sure how to do (B). In fact, I'm not even sure why it's true.
 
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  • #2
Never mind. It's pretty easy to show just by considering a contour going around the pole (x,y) = (0, 0). Considering the path integral around r(t) = (cos t, sin t, 0), the integral gives [tex]2\pi[/tex], whereas it should give 0 if a potential function did exist.
 

FAQ: Verify $\nabla \times \textbf{f} = 0$ and Show Non-Existence of $\psi$ on $R_2$

1. What does "Verify ∇ × f = 0" mean?

Verifying ∇ × f = 0 means checking if the curl (∇ × f) of a vector field f is equal to zero. This indicates that the vector field is irrotational, meaning that it has no rotational component. In other words, the vector field has a constant direction and does not rotate around any point.

2. Why is it important to verify ∇ × f = 0?

It is important to verify ∇ × f = 0 because it is a fundamental property of conservative vector fields. If a vector field is conservative, it means that it can be expressed as the gradient of a scalar function, known as the potential function. This has many applications in physics and engineering, such as in fluid dynamics and electromagnetics.

3. How can I verify ∇ × f = 0?

To verify ∇ × f = 0, you can use the vector calculus operation known as the curl. This involves taking the partial derivatives of the components of the vector field and checking if they satisfy the condition of being equal to zero. If all the partial derivatives are equal to zero, then the vector field is irrotational and ∇ × f = 0 is verified.

4. What does it mean to show the non-existence of ψ on R2?

Showing the non-existence of ψ on R2 means proving that a scalar function ψ does not exist on a two-dimensional space, known as the xy-plane. This means that there is no potential function for the vector field, and therefore the vector field is not conservative. In other words, the vector field has a non-zero curl (∇ × f ≠ 0).

5. How can I show the non-existence of ψ on R2?

To show the non-existence of ψ on R2, you can use the method of contradiction. Assume that a potential function ψ exists for the vector field, and then use the properties of the gradient to show that this leads to a contradiction. If a contradiction is reached, it means that the initial assumption was incorrect and therefore ψ does not exist on R2.

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