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Curl and Divergence

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Curl and Divergence As a mnemonic device, one can think of the curl of F~ as the symbolic cross product: ~ ~ δ δ δ curl(F ) = r × F = ( δx ; δy ; δz ) × (F1; F2; F3). Curl and Divergence Definition Let F~ = (F1; F2; F3) be a vector field. The curl of F~ is the vector field defined by δF δF δF δF δF δF curl(F~ ) = 3 − 2 ; 1 − 3 ; 2 − 1 . δy δz δz δx δx δy Curl and Divergence Definition Let F~ = (F1; F2; F3) be a vector field. The curl of F~ is the vector field defined by δF δF δF δF δF δF curl(F~ ) = 3 − 2 ; 1 − 3 ; 2 − 1 . δy δz δz δx δx δy As a mnemonic device, one can think of the curl of F~ as the symbolic cross product: ~ ~ δ δ δ curl(F ) = r × F = ( δx ; δy ; δz ) × (F1; F2; F3). For a mnemonic device, we can think of the divergence as the symbolic dot product: ~ ~ δ δ δ div(F ) = r · F = ( δx ; δy ; δz ) · (F1; F2; F3). Curl and Divergence, contd. Definition Again let F~ = (F1; F2; F3) be a vector field. The divergence of F~ is the real-valued function in three variables defined by δF δF δF div(F~ ) = 1 + 2 + 3 . δx δy δz Curl and Divergence, contd. Definition Again let F~ = (F1; F2; F3) be a vector field. The divergence of F~ is the real-valued function in three variables defined by δF δF δF div(F~ ) = 1 + 2 + 3 . δx δy δz For a mnemonic device, we can think of the divergence as the symbolic dot product: ~ ~ δ δ δ div(F ) = r · F = ( δx ; δy ; δz ) · (F1; F2; F3). Imagine taking a paddle-wheel (which can spin in any direction) and fixing it at a point (x; y; z). Then the curl vector of F~ at (x; y; z) may be imagined as the axis on which the fluid makes the wheel spin according to the right-hand rule: that is, if you stick your right thumb up in the direction of the curl, the wheel will spin in the direction that your fingers curl. The magnitude of the curl vector is how fast the wheel rotates. The divergence of F~ represents the expansion/compression of the fluid at a given point (x; y; z). A positive divergence corresponds to fluid expansion, i.e. the fluid is generally moving away from the point, while a negative divergence corresponds to fluid compression, i.e. the fluid is generally moving toward the point. Physical Significance The physical applications of the notions of curl and divergence of a vector field are impossible to fully capture within the scope of this class (and this slide!). However, we can give some terse indications in the context of fluid dynamics. Think of F~ as representing the velocity field of a three-dimensional body of liquid in motion. The divergence of F~ represents the expansion/compression of the fluid at a given point (x; y; z). A positive divergence corresponds to fluid expansion, i.e. the fluid is generally moving away from the point, while a negative divergence corresponds to fluid compression, i.e. the fluid is generally moving toward the point. Physical Significance The physical applications of the notions of curl and divergence of a vector field are impossible to fully capture within the scope of this class (and this slide!). However, we can give some terse indications in the context of fluid dynamics. Think of F~ as representing the velocity field of a three-dimensional body of liquid in motion. Imagine taking a paddle-wheel (which can spin in any direction) and fixing it at a point (x; y; z). Then the curl vector of F~ at (x; y; z) may be imagined as the axis on which the fluid makes the wheel spin according to the right-hand rule: that is, if you stick your right thumb up in the direction of the curl, the wheel will spin in the direction that your fingers curl. The magnitude of the curl vector is how fast the wheel rotates. Physical Significance The physical applications of the notions of curl and divergence of a vector field are impossible to fully capture within the scope of this class (and this slide!). However, we can give some terse indications in the context of fluid dynamics. Think of F~ as representing the velocity field of a three-dimensional body of liquid in motion. Imagine taking a paddle-wheel (which can spin in any direction) and fixing it at a point (x; y; z). Then the curl vector of F~ at (x; y; z) may be imagined as the axis on which the fluid makes the wheel spin according to the right-hand rule: that is, if you stick your right thumb up in the direction of the curl, the wheel will spin in the direction that your fingers curl. The magnitude of the curl vector is how fast the wheel rotates. The divergence of F~ represents the expansion/compression of the fluid at a given point (x; y; z). A positive divergence corresponds to fluid expansion, i.e. the fluid is generally moving away from the point, while a negative divergence corresponds to fluid compression, i.e. the fluid is generally moving toward the point. Linearity of Curl and Divergence If F~ , G~ are any two vectors fields then curl(F~ + G~ ) = curl(F~ ) + curl(G~ ) and div(F~ + G~ ) = div(F~ ) + div(G~ ), and if c is any constant then curl(cF~ ) = c curl(F~ ) and div(cF~ ) = c div(F~ ). In other words curl and div are linear transformations. Boundary Orientations Let G(u; v) be a smooth one-to-one parametrization with domain 3 D of a two-dimensional surface S in R . Define the boundary of S, denoted δS, to be the image of δD under G. We informally fix a boundary orientation on S as follows: if you are a normal vector standing on the surface S walking along the boundary with the correct orientation, then the surface is on your left and the void is on your right. (See picture.) Stokes' Theorem Theorem (Stokes' Theorem) Let S be a surface parametrized by a smooth one-to-one function G(u; v) with domain D, where δD is comprised of simple closed curves. Then I ZZ F~ · ds = curl(F~ ) · dS. δS S Example Verify Stokes' theorem for the field F~ = (−y; 2x; x + z) and the upper unit hemisphere S, with outward-pointing normal vectors. Now compute: I Z 2π F~ · ds = F~ (~c(t)) · ~c 0(t)dt δS 0 Z 2π = (− sin t; 2 cos t; cos t) · (− sin t; cos t; 0)dt 0 Z 2π = (sin2 t + 2 cos2 t)dt 0 Z 2π = (1 + cos2 t)dt 0 Z 2π 3 1 = + cos 2t dt 0 2 2 3 1 2π = t + sin 2t = 3π: 2 4 0 Solution Known: F~ = (−y; 2x; x + z) First we will just compute the line integral about the boundary δS, which is the unit circle in the xy-plane oriented counterclockwise. We can parametrize δS in the usual way: ~c(t) = (cos t; sin t; 0) on the domain [0; 2π]. Solution Known: F~ = (−y; 2x; x + z) First we will just compute the line integral about the boundary δS, which is the unit circle in the xy-plane oriented counterclockwise. We can parametrize δS in the usual way: ~c(t) = (cos t; sin t; 0) on the domain [0; 2π]. Now compute: I Z 2π F~ · ds = F~ (~c(t)) · ~c 0(t)dt δS 0 Z 2π = (− sin t; 2 cos t; cos t) · (− sin t; cos t; 0)dt 0 Z 2π = (sin2 t + 2 cos2 t)dt 0 Z 2π = (1 + cos2 t)dt 0 Z 2π 3 1 = + cos 2t dt 0 2 2 3 1 2π = t + sin 2t = 3π: 2 4 0 Now compute the normal: ~ ~ Tθ = (− sin θ sin φ, cos θ sin φ, 0) and Tφ = (cos θ cos φ, sin θ cos φ, − sin φ); ~ ~ 2 2 ~n = Tθ × Tφ = (cos θ sin φ, sin θ sin φ, sin φ cos φ). At this moment we verify mentally or graphically that we have chosen the correct orientation for our normal vector ~n. (If we had accidentally parametrized S in such a way that ~n faced the wrong direction, then we could fix the problem by just flipping the sign on ~n.) Solution, contd. Known: F~ = (−y; 2x; x + z) H ~ δS F · ds = 3π Now we will compute the flux integral over S. First we parametrize the upper unit hemisphere in the usual way: π G(θ; φ) = (cos θ sin φ, sin θ sin φ, cos φ) on the domain [0; 2π] × [0; 2 ]. ~ ~ 2 2 ~n = Tθ × Tφ = (cos θ sin φ, sin θ sin φ, sin φ cos φ). At this moment we verify mentally or graphically that we have chosen the correct orientation for our normal vector ~n. (If we had accidentally parametrized S in such a way that ~n faced the wrong direction, then we could fix the problem by just flipping the sign on ~n.) Solution, contd. Known: F~ = (−y; 2x; x + z) H ~ δS F · ds = 3π Now we will compute the flux integral over S. First we parametrize the upper unit hemisphere in the usual way: π G(θ; φ) = (cos θ sin φ, sin θ sin φ, cos φ) on the domain [0; 2π] × [0; 2 ]. Now compute the normal: ~ ~ Tθ = (− sin θ sin φ, cos θ sin φ, 0) and Tφ = (cos θ cos φ, sin θ cos φ, − sin φ); At this moment we verify mentally or graphically that we have chosen the correct orientation for our normal vector ~n.
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