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The proof that the condition \(\nabla \times \mathbf{F}=0\) guarantees the path independence of the work \(\int_{1}^{2} \mathbf{F} \cdot d \mathbf{r}\) done by \(\mathbf{F}\) is unfortunately too lengthy to be included here. However, the following three exercises capture the main points: \(^{16}\) (a) Show that the path independence of \(\int_{1}^{2} \mathbf{F} \cdot d \mathbf{r}\) is equivalent to the statement that the integral \(\oint_{\mathrm{T}} \mathbf{F} \cdot d \mathbf{r}\) around any closed path \(\Gamma\) is zero. (By tradition, the symbol \(\oint\) is used for integrals around a closed path \(-\) a path that starts and stops at the same point.) [Hint: For any two points 1 and 2 and any two paths from 1 to 2 , consider the work done by \(\mathbf{F}\) going from 1 to 2 along the first path and then back to 1 along the second in the reverse direction. \((\) b) Stokes's theorem asserts that \(\oint_{\mathrm{T}} \mathbf{F} \cdot d \mathbf{r}=\int(\nabla \times \mathbf{F}) \cdot \hat{\mathbf{n}} d A,\) where the integral on the right is a surface integral over a surface for which the path \(\Gamma\) is the boundary, and \(\hat{\mathbf{n}}\) and \(d A\) are a unit normal to the surface and an element of area. Show that Stokes's theorem implies that if \(\nabla \times \mathbf{F}=0\) everywhere, then \(\oint_{\mathrm{T}} \mathbf{F} \cdot d \mathbf{r}=0 .\) (c) While the general proof of Stokes's theorem is beyond our scope here, the following special case is quite easy to prove (and is an important step toward the general proof): Let \(\Gamma\) denote a rectangular closed path lying in a plane perpendicular to the \(z\) direction and bounded by the lines \(x=B, x=B+b, y=C\) and \(y=C+c .\) For this simple path (traced counterclockwise as seen from above), prove Strokes's theorem that \(\oint_{\Gamma} \mathbf{F} \cdot d \mathbf{r}=\int(\mathbf{\nabla} \times \mathbf{F}) \cdot \hat{\mathbf{n}} d A\) where \(\hat{\mathbf{n}}=\hat{\mathbf{z}}\) and the integral on the right runs over the flat, rectangular area inside \(\Gamma\). [Hint: The integral on the left contains four terms, two of which are integrals over \(x\) and two over \(y\). If you pair them in this way, you can combine each pair into a single integral with an integrand of the form \(F_{x}(x, C+c, z)-F_{x}(x, C, z)\) (or a similar term with the roles of \(x\) and \(y\) exchanged). You can rewrite this integrand as an integral over \(y\) of \(\partial F_{x}(x, y, z) / \partial y\) (and similarly with the other term), and you're home.]

Short Answer

Expert verified
The condition \( \nabla \times \mathbf{F}=0 \) guarantees path independence via Stokes's theorem and closed loop integrals.

Step by step solution

01

Understanding Path Independence

Path independence means that the work done by vector field \( \mathbf{F} \) from point 1 to point 2 is the same irrespective of the path taken. Mathematically, this means the line integral \( \int_{1}^{2} \mathbf{F} \cdot d \mathbf{r} \) is unaffected by the choice of path.
02

Equivalence with Closed Path Integral

To show path independence is equivalent to \( \oint_{\Gamma} \mathbf{F} \cdot d \mathbf{r} = 0 \), consider two different paths from point 1 to point 2. If traveling along one path and returning on the other in reverse yields zero work (a closed loop), this implies path independence if \( \oint_{\Gamma} \mathbf{F} \cdot d \mathbf{r} = 0 \).
03

Applying Stokes's Theorem

Stokes's theorem states \( \oint_{\mathrm{T}} \mathbf{F} \cdot d \mathbf{r} = \int(abla \times \mathbf{F}) \cdot \hat{\mathbf{n}} \, dA \). If \( abla \times \mathbf{F} = 0 \) everywhere, then the surface integral of zero is zero, implying \( \oint_{\mathrm{T}} \mathbf{F} \cdot d \mathbf{r} = 0 \), guaranteeing path independence.
04

Proving Special Case of Stokes's Theorem

Consider a rectangular path \( \Gamma \) bounded by lines \( x = B, x = B+b, y = C, y = C+c \). Compute the line integral directly: it has four components (two pairs over \(x\) and \(y\)). Each pair has the form \( F_{x}(x, C+c, z) - F_{x}(x, C, z) \), allowing rewriting as a double integral over area, confirming Stokes's theorem for this path.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Line Integrals
Line integrals are a way to integrate along a curve or path in a vector field, represented by \( \int_{1}^{2} \mathbf{F} \cdot d \mathbf{r} \). Often used in physics and engineering, they help compute quantities like work done by a force field along a path. A line integral takes into account not just the value of the field along the path but also its direction.

If you imagine moving through a force field, the work done is calculated by summing up contributions from each infinitesimal segment. This involves the component of the vector field in the direction of movement and can depend on the path taken.

Path independence is an important concept related to line integrals. It means the integral has the same value regardless of the path taken between any two points. In mathematical terms, if a vector field is conservative, or in other words, if it can be expressed as the gradient of a scalar potential, the work done will only depend on the initial and final points, leading to path independence.
Stokes's Theorem
Stokes's Theorem is a key result in vector calculus that connects line integrals over a closed path with surface integrals over a surface bounded by the path. The theorem is summarized by the equation \[ \oint_{\mathrm{T}} \mathbf{F} \cdot d \mathbf{r} = \int (abla \times \mathbf{F}) \cdot \hat{\mathbf{n}} \, dA \]This relation is powerful because it allows us to transform a difficult line integral around a loop into a potentially simpler surface integral over the area enclosed by the loop.

The surface integral involves the curl of the vector field, \(abla \times \mathbf{F}\), which can be thought of as a measure of rotation or swirling strength of the field. If the curl is zero everywhere over a surface, then the original line integral around the boundary of the surface becomes zero.
  • Stokes's Theorem provides the foundation for showing that a zero curl implies path independence over any surface with the same boundary.
  • Understanding this theorem helps in solving complex integrals by converting them from one-dimensional paths to two-dimensional surfaces.
Curl of a Vector Field
The curl of a vector field, denoted \( abla \times \mathbf{F} \), is a vector that describes the rotation at a point in the field. Think of it as capturing how much and in what direction the field twists or circulates around that point. It is especially useful in assessing the behavior of flow fields, like air or water currents.

When the curl is zero, the field is considered irrotational, indicating two key aspects:
  • No local rotation occurs anywhere within the field.
  • The vector field is conservative, and line integrals within the field are path-independent.
This absence of circulation means we can transform line integrals depending on the path into surface integrals, evaluating conditions over surfaces instead.

This concept is central to proving path independence in vector calculus. It is directly tied to Stokes's theorem. A zero curl across any chosen surface verifies the path independence of line integrals over its boundary, emphasizing consistency in approaches across different paths.

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Most popular questions from this chapter

In one dimension, it is obvious that a force obeying Hooke's law is conservative (since \(F=-k x\) depends only on the position \(x,\) and this is sufficient to guarantee that \(F\) is conservative in one dimension). Consider instead a spring that obeys Hooke's law and has one end fixed at the origin, but whose other end is free to move in all three dimensions. (The spring could be fastened to a point in the ceiling and be supporting a bouncing mass \(m\) at its other end, for instance.) Write down the force \(\mathbf{F}(\mathbf{r})\) exerted by the spring in terms of its length \(r\) and its equilibrium length \(r_{\mathrm{o}} .\) Prove that this force is conservative. [Hints: Is the force central? Assume that the spring does not bend.]

A particle of mass \(m_{1}\) and speed \(v_{1}\) collides with a second particle of mass \(m_{2}\) at rest. If the collision is perfectly inelastic (the two particles lock together and move off as one) what fraction of the kinetic energy is lost in the collision? Comment on your answer for the cases that \(m_{1} \ll m_{2}\) and that \(m_{2} \ll m_{1}\).

A charge \(q\) in a uniform electric field \(\mathbf{E}_{0}\) experiences a constant force \(\mathbf{F}=q \mathbf{E}_{0}\). (a) Show that this force is conservative and verify that the potential energy of the charge at position \(\mathbf{r}\) is \(U(\mathbf{r})=-q \mathbf{E}_{0} \cdot \mathbf{r}\). (b) By doing the necessary derivatives, check that \(\mathbf{F}=-\nabla U\).

A mass \(m\) is in a uniform gravitational field, which exerts the usual force \(F=m g\) vertically down, but with \(g\) varying with time, \(g=g(t) .\) Choosing axes with \(y\) measured vertically up and defining \(U=m g y\) as usual, show that \(\mathbf{F}=-\nabla U\) as usual, but, by differentiating \(E=\frac{1}{2} m v^{2}+U\) with respect to \(t,\) show that \(E\) is not conserved.

(a) The force exerted by a one-dimensional spring, fixed at one end, is \(F=-k x,\) where \(x\) is the displacement of the other end from its equilibrium position. Assuming that this force is conservative (which it is) show that the corresponding potential energy is \(U=\frac{1}{2} k x^{2},\) if we choose \(U\) to be zero at the equilibrium position. (b) Suppose that this spring is hung vertically from the ceiling with a mass \(m\) suspended from the other end and constrained to move in the vertical direction only. Find the extension \(x_{\mathrm{o}}\) of the new equilibrium position with the suspended mass. Show that the total potential energy (spring plus gravity) has the same form \(\frac{1}{2} k y^{2}\) if we use the coordinate \(y\) equal to the displacement measured from the new equilibrium position at \(x=x_{\mathrm{o}}\) (and redefine our reference point so that \(U=0\) at \(y=0\) ).

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