lec04

# lec04 - Scott Hughes 10 February 2005 Massachusetts...

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Unformatted text preview: Scott Hughes 10 February 2005 Massachusetts Institute of Technology Department of Physics 8.022 Spring 2005 Lecture 4: Wrap up of vector calculus: Poisson & Laplace equations; curl 4.1 Summary: Vector calculus so far We have learned several mathematical operations which fall into the category of vector calculus. In Cartesian coordinates, these operations can be written in very compact form using the following operator: vector ∇ ≡ ˆ x ∂ ∂x + ˆ y ∂ ∂y + ˆ z ∂ ∂z . The first vector calculus operation we learned is the gradient . It converts the electric potential into the electric field : vector E =- grad φ =- vector ∇ φ . The second operation is the divergence , which relates the electric field to the charge density: div vector E = 4 πρ . Via Gauss’s theorem (also known as the divergence theorem ), we can relate the flux of any vector field vector F through a closed surface S to the integral of the divergence of vector F over the volume enclosed by S : contintegraldisplay S vector F · d vector A = integraldisplay V div vector F dV . This follows from the definition of the divergence: div vector F = lim V → contintegraltext S vector F · d vector A V . 34 4.2 Divergence in Cartesian coordinates So far, we’ve only defined the divergence as a particular limit. We now want to develop a concrete calculation showing its value. To do so, consider an infinitesimal cube with sides ∆ x , ∆ y , and ∆ z , centered on the coordinate x , y , z : (x,y,z) x y z ∆ ∆ z ∆ y x [The large dot in the center of this cube is only there to mark the coordinate ( x, y, z ) — it doesn’t mean anything else.] We take this region to be filled with a vector field vector F . To begin, we want to compute the flux of vector F over this cube. Let’s look at the flux over just the z faces (the top and bottom): ∆Φ z = integraldisplay top & bottom vector F · d vector A similarequal ∆ x ∆ y [ F z ( x, y, z + ∆ z/ 2)- F z ( x, y, z- ∆ z/ 2)] . Since the cube is taken to be very small (indeed, we will soon take the limit ∆ x, ∆ y, ∆ z → 0), we approximate the integral. Notice that the contribution from the bottom face ( z- ∆ z/ 2) enters with a minus sign, consistent with that face pointing down. Now, to evaluate the function under the integrand, we make a Taylor expansion about z : ∆Φ z similarequal ∆ x ∆ y bracketleftBigg F z ( x, y, z ) + ∆ z 2 ∂F z ∂z bracketrightBigg- ∆ x ∆ y bracketleftBigg F z ( x, y, z )- ∆ z 2 ∂F z ∂z bracketrightBigg similarequal ∆ x ∆ y ∆ z ∂F z ∂z . Repeating this exercise to find the fluxes through the other four sides, we find ∆Φ x similarequal ∆ x ∆ y ∆ z ∂F x ∂x ∆Φ y similarequal ∆ x ∆ y ∆ z ∂F y ∂y ....
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lec04 - Scott Hughes 10 February 2005 Massachusetts...

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