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Chapter 14: Problem 4

Why does a conservative vector field produce zero circulation around a closed curve?

Short Answer

Expert verified
Answer: Yes, a conservative vector field has zero circulation around a closed curve. This is because the curl of a gradient is always zero, and according to Stokes' theorem, the circulation around a closed curve is equal to the surface integral of the curl of the vector field. Since the curl is zero for a conservative vector field, the circulation around a closed curve is also zero.

Step by step solution

01

Define a conservative vector field

A vector field F is called conservative if it is the gradient of a scalar potential function, denoted as F = 鈭噁, where f is a smooth scalar function and 鈭� is the gradient operator.
02

Introduce the closed curve

Let C be a closed curve in the domain of the conservative vector field F. The circulation around the closed curve C is given by the line integral of the vector field along the curve C: Circulation = 鈭甠C F 鈰� dr
03

Apply Stokes' theorem

To relate the circulation of F around curve C to the curl of F, we can utilize Stokes' theorem. Stokes' theorem states that the circulation around a closed curve C is equal to the surface integral of the curl of F over a surface S bounded by the curve C: 鈭甠C F 鈰� dr = 鈭琠S curl F 鈰� dS
04

Find curl of F

Since F is a conservative vector field (F = 鈭噁), we have: curl F = curl(鈭噁) However, it is a known result that the curl of a gradient is always zero: curl(鈭噁) = 0
05

Show zero circulation

Now, using the result we obtained in Step 4, we can substitute it back into the expression from Step 3: 鈭甠C F 鈰� dr = 鈭琠S 0 鈰� dS = 0 Therefore, the circulation of a conservative vector field around a closed curve C is always zero. This implies that a conservative vector field does not produce any circulation around a closed curve.

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Most popular questions from this chapter

Evaluate each line integral using a method of your choice. $$\begin{aligned} &\oint_{C} \mathbf{F} \cdot d \mathbf{r}, \text { where } \mathbf{F}=\left\langle 2 x y+z^{2}, x^{2}, 2 x z\right\rangle \text { and } C \text { is the circle }\\\ &\mathbf{r}(t)=\langle 3 \cos t, 4 \cos t, 5 \sin t\rangle, \text { for } 0 \leq t \leq 2 \pi \end{aligned}$$

a. Let \(\mathbf{a}=\langle 0,1,0\rangle, \mathbf{r}=\langle x, y, z\rangle,\) and consider the rotation field \(\mathbf{F}=\mathbf{a} \times \mathbf{r} .\) Use the right-hand rule for cross products to find the direction of \(\mathbf{F}\) at the points (0,1,1),(1,1,0),(0,1,-1), and (-1,1,0). b. With \(\mathbf{a}=\langle 0,1,0\rangle,\) explain why the rotation field \(\mathbf{F}=\mathbf{a} \times \mathbf{r}\) circles the \(y\) -axis in the counterclockwise direction looking along a from head to tail (that is, in the negative \(y\) -direction).

The French physicist Andr茅-Marie Amp猫re \((1775-1836)\) discovered that an electrical current \(I\) in a wire produces a magnetic field \(\mathbf{B} .\) A special case of Amp猫re's Law relates the current to the magnetic field through the equation \(\oint_{C} \mathbf{B} \cdot d \mathbf{r}=\mu I,\) where \(C\) is any closed curve through which the wire passes and \(\mu\) is a physical constant. Assume that the current \(I\) is given in terms of the current density \(\mathbf{J}\) as \(I=\iint_{S} \mathbf{J} \cdot \mathbf{n} d S\) where \(S\) is an oriented surface with \(C\) as a boundary. Use Stokes' Theorem to show that an equivalent form of Amp猫re's Law is \(\nabla \times \mathbf{B}=\mu \mathbf{J}\)

Prove that for a real number \(p,\) with \(\mathbf{r}=\langle x, y, z\rangle, \nabla\left(\frac{1}{|\mathbf{r}|^{p}}\right)=\frac{-p \mathbf{r}}{|\mathbf{r}|^{p+2}}.\)

The rotation of a three-dimensional velocity field \(\mathbf{V}=\langle u, v, w\rangle\) is measured by the vorticity \(\omega=\nabla \times \mathbf{V} .\) If \(\omega=\mathbf{0}\) at all points in the domain, the flow is irrotational. a. Which of the following velocity fields is irrotational: \(\mathbf{V}=\langle 2,-3 y, 5 z\rangle\) or \(\mathbf{V}=\langle y, x-z,-y\rangle ?\) b. Recall that for a two-dimensional source-free flow \(\mathbf{V}=(u, v, 0),\) a stream function \(\psi(x, y)\) may be defined such that \(u=\psi_{y}\) and \(v=-\psi_{x} .\) For such a two-dimensional flow, let \(\zeta=\mathbf{k} \cdot \nabla \times \mathbf{V}\) be the \(\mathbf{k}\) -component of the vorticity. Show that \(\nabla^{2} \psi=\nabla \cdot \nabla \psi=-\zeta\). c. Consider the stream function \(\psi(x, y)=\sin x \sin y\) on the square region \(R=\\{(x, y): 0 \leq x \leq \pi, 0 \leq y \leq \pi\\}\). Find the velocity components \(u\) and \(v\); then sketch the velocity field. d. For the stream function in part (c), find the vorticity function \(\zeta\) as defined in part (b). Plot several level curves of the vorticity function. Where on \(R\) is it a maximum? A minimum?
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