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math121b:02-10 [2020/02/09 23:55]
pzhou
math121b:02-10 [2020/02/22 18:03] (current)
pzhou
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 ====== 2020-02-10, Monday ====== ====== 2020-02-10, Monday ======
-\gdef\div{\text{div}} \gdef\vol{\text{Vol}}  +\gdef\div{\text{div}} \gdef\vol{\text{Vol}} \gdef\b{\mathbf} \gdef\d{\partial} 
-We first finish up the derivation of Laplacian last time. See the lecture note [[02-07]]. Then, we introduce two vector operationsdivergence and curlFinally, we give a summary of the mathematical treatment of tensor analysis+ 
 + 
 +We first finish up the derivation of Laplacian last time. See the lecture note [[02-07]]. Then, we review three concepts $df\nabla f, \nabla \cdot \b V( (\nabla \times \b Visabitspecialfor is a bit special for \R^3$) 
 + 
 + 
 +Then, we will follow Boas 10.8 and 10.9, to reconcilliate the math notation and physics notations 
  
-Then, we will follow Boas 10.9 and 10.10, to introduce the physic-engineer notations: the nabla operator \gdef\b{\mathbf} \b \nabla 
  
 ===== Differential of a function is a 1-form (covector field)===== ===== Differential of a function is a 1-form (covector field)=====
 In Cartesian coordinate, the differential of a function ff is  In Cartesian coordinate, the differential of a function ff is 
- df = \sum_i \frac{\df }{\d x_i} dx_i. + df = \sum_i \frac{\d f }{\d x_i} dx_i.
  
 In general coordinate (u1,,un)(u_1, \cdots, u_n), the differential of a function ff is  In general coordinate (u1,,un)(u_1, \cdots, u_n), the differential of a function ff is 
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 Note that gijg_{ij} and gijg^{ij} depends on the coordinate system.  Note that gijg_{ij} and gijg^{ij} depends on the coordinate system. 
 gij=g(ui,uj),gij=g(dui,duj). g_{ij} = g(\frac{\d}{\d u_i}, \frac{\d}{\d u_j}), \quad g^{ij} = g^*(d u_i, d u_j).  gij=g(ui,uj),gij=g(dui,duj). g_{ij} = g(\frac{\d}{\d u_i}, \frac{\d}{\d u_j}), \quad g^{ij} = g^*(d u_i, d u_j).
 +Beware that $\nabla u_i \neq \frac{\d}[\d u_i}$. 
  
 ** Notation ** $$ \nabla f = \grad f.$$  ** Notation ** $$ \nabla f = \grad f.$$ 
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 div(V)=i=1n1g(gVi)ui \div (\b V) = \sum_{i=1}^n \frac{1}{\sqrt{|g|}} \frac{\d (\sqrt{|g|} V^i)}{\d u_i}  div(V)=i=1n1g(gVi)ui \div (\b V) = \sum_{i=1}^n \frac{1}{\sqrt{|g|}} \frac{\d (\sqrt{|g|} V^i)}{\d u_i}
  
-Exercise: prove that for any compactly supported function φ\varphi ((a compactly supported function on Rn\R^n is a function that vanishes outside a sufficently large ball. )), we have  +The reason we have the above formula is that for any compactly supported function φ\varphi ((a compactly supported function on Rn\R^n is a function that vanishes outside a sufficently large ball. )), we have  
- \int_{\R^n} (\nabla \cdot \b V)(u)  f(u) \sqrt{|g|(u)} du_1\cdots d u_n = \int_{\R^n} g(\b V\nabla f(u)) \sqrt{|g|(u)} du_1\cdots d u_n + \int_{\R^n} (\nabla \cdot \b V)\,  \varphi\, \sqrt{|g|} du_1\cdots d u_n = \int_{\R^n} \b V \cdot (\nabla \varphi)\, \sqrt{|g|} du_1\cdots d u_n  
 + 
 +====== Back to Boas ====== 
 + 
 +===== Section 10.8 ===== 
 +For Cartesian coordinate, we have basis vectors i,j,k\b i, \b j, \b k.  
 + 
 +For spherical coordinate, we have **unit** basis vectors er,eθ,eϕ\b e_r, \b e_\theta, \b e_\phi, and corresponding coordinate basis vectors ar,aθ,aϕ\b a_r, \b a_\theta, \b a_\phi (not unit length). These an\b a_n corresponds to our coordinate basis tangent vectors:  
 +ar=r,aθ=θ, \b a_r = \frac{\d }{\d r}, \quad \b a_\theta = \frac{\d }{\d \theta}, \cdots  
 + 
 +See Example 2 on page 523, for how to consider a general curvilinear coordinate (x1,x2,x3)(x_1, x_2, x_3) on R3\R^3.  
 + 
 +The notation dsd \b s corresponds to  
 +i=1nxidxi(TpRn)(TpRn). \sum_{i=1}^n \frac{\d }{\d x_i} \otimes d x_i \in (T_p \R^n) \otimes (T_p \R^n)^*.  
 +An element TT in VVV \otimes V^* can be viewed as a linear operator VVV \to V, by inserting vVv \in V to the second slot of TT. In this sense dsd \b s is the identity operator on TpRnT_p \R^n. You might have seen in Quantum mechanics the bra-ket notation 1=nnn1 = \sum_n | n \rangle \otimes \langle n | (( \otimes sometimes omitted as usual in physics.)) It is the same thing, where nV| n \rangle \in V forms a basis and nV \langle n | \in V^* are the dual basis.   
 + 
 +** Orthogonal coordinate system (ortho-curvilinear coordinate) **, the matrix gijg_{ij} is diagonal, with entries hi2h_i^2 (not to be confused with our notation for dual basis). This is  
 +the case we will be considering mainly.  
 + 
 +===== Section 10.9 ===== 
 +Suppose we have orthogonal coordinate system (x1,x2,x3)(x_1, x_2, x_3), and **unit** basis vectors ei\b e_i, we have 
 +ai=xi=hiei. \b a_i = \frac{\d }{\d x_i} = h_i \b e_i.  
 + 
 +Given a vector field VV, we write its component in the basis of ei\b e_i (warning! this is not our usual notation, we usual write with basis xi\frac{\d }{\d x_i})  
 +V=iViei \b V = \sum_i V^i \b e_i .  
 + 
 +==== Divergence. ==== 
 +Try to do problem 1.  
 + 
 +An important property is the "Leibniz rule" 
 + \b \nabla \cdot (f \b V) = \b \nabla f \cdot \b V + f  \b \nabla  \cdot \b V. 
 + 
 +==== Curl ==== 
 +To compute the curl, we note the following rule 
 +×(fV)=(f)×V+f×V \b \nabla \times (f \b V) = \b \nabla(f) \times \b V + f \b \nabla \times \b V  
 +and  
 +×f=0 \b \nabla \times \nabla f = 0 .  
 + 
 +In the ortho-curvilinear coordinate, we can use the above rule to get a formula for the curl. I will not test on the curl operator in the orthocurvilinear case.  
 + 
 + 
 + 
 + 
  
  
  
math121b/02-10.1581321315.txt.gz · Last modified: 2020/02/09 23:55 by pzhou