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The proof that the condition ×F=0 guarantees the path independence of the work 12Fdr done by 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 12Fdr is equivalent to the statement that the integral TFdr around any closed path Γ is zero. (By tradition, the symbol 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 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 TFdr=(×F)n^dA, where the integral on the right is a surface integral over a surface for which the path Γ is the boundary, and n^ and dA are a unit normal to the surface and an element of area. Show that Stokes's theorem implies that if ×F=0 everywhere, then TFdr=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 Γ 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 ΓFdr=(×F)n^dA where n^=z^ and the integral on the right runs over the flat, rectangular area inside Γ. [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 Fx(x,C+c,z)Fx(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 Fx(x,y,z)/y (and similarly with the other term), and you're home.]

Short Answer

Expert verified
The condition ×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 F from point 1 to point 2 is the same irrespective of the path taken. Mathematically, this means the line integral 12Fdr is unaffected by the choice of path.
02

Equivalence with Closed Path Integral

To show path independence is equivalent to ΓFdr=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 ΓFdr=0.
03

Applying Stokes's Theorem

Stokes's theorem states TFdr=(abla×F)n^dA. If abla×F=0 everywhere, then the surface integral of zero is zero, implying TFdr=0, guaranteeing path independence.
04

Proving Special Case of Stokes's Theorem

Consider a rectangular path Γ 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 Fx(x,C+c,z)Fx(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 12Fdr. 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 TFdr=(abla×F)n^dAThis 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×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×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

For a system of N particles subject to a uniform gravitational field g acting vertically down, prove that the total gravitational potential energy is the same as if all the mass were concentrated at the center of mass of the system; that is, U=αUα=MgY where M=mα is the total mass and R=(X,Y,Z) is the position of the CM, with the y coordinate measured vertically up. [Hint: We know from Problem 4.5 that Uα=mαgyα.]

(a) Consider an electron (charge e and mass m ) in a circular orbit of radius r around a fixed proton (charge +e ). Remembering that the inward Coulomb force ke2/r2 is what gives the electron its centripetal acceleration, prove that the electron's KE is equal to 12 times its PE; that is, T=12U and hence E=12U. (This result is a consequence of the so-called virial theorem. See Problem 4.41.) Now consider the following inelastic collision of an electron with a hydrogen atom: Electron number 1 is in a circular orbit of radius r around a fixed proton. (This is the hydrogen atom.) Electron 2 approaches from afar with kinetic energy T2. When the second electron hits the atom, the first electron is knocked free, and the second is captured in a circular orbit of radius r. (b) Write down an expression for the total energy of the three-particle system in general. (Your answer should contain five terms, three PEs but only two KEs, since the proton is considered fixed.) (c) Identify the values of all five terms and the total energy E long before the collision occurs, and again long after it is all over. What is the KE of the outgoing electron 1 once it is far away? Give your answers in terms of the variables T2,r, and r.

If a particle's potential energy is U(r)=k(x2+y2+z2), where k is a constant, what is the force on the particle?

Near to the point where I am standing on the surface of Planet X, the gravitational force on a mass m is vertically down but has magnitude mγy2 where γ is a constant and y is the mass's height above the horizontal ground. (a) Find the work done by gravity on a mass m moving from r1 to r2, and use your answer to show that gravity on Planet X, although most unusual, is still conservative. Find the corresponding potential energy. (b) Still on the same planet, I thread a bead on a curved, frictionless, rigid wire, which extends from ground level to a height h above the ground. Show clearly in a picture the forces on the bead when it is somewhere on the wire. (Just name the forces so it's clear what they are; don't worry about their magnitude.) Which of the forces are conservative and which are not? (c) If I release the bead from rest at a height h, how fast will it be going when it reaches the ground?

Verify that the gravitational force GMmr^/r2 on a point mass m at r, due to a fixed point mass M at the origin, is conservative and calculate the corresponding potential energy.

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