Two common difficul:es with HW 2: Problem 1c: v = r n ˆr

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1 Two common difficul:es with HW : Problem 1c: For what values of n does the divergence of v = r n ˆr diverge at the origin? In this context, diverge means becomes arbitrarily large ( goes to infinity ). We have v = ( n + )r n 1, except at n = -, where that expression gives 0/0. So, it diverges (nega:ve exponent) for all n < 1, except, possibly, n = -. What about n = -? That s the situa:on that gives the Dirac delta func:on at the origin. So, the answer is, the divergence diverges (!) for all n < - 1. Problem c: What is the field at the center of a regular n- sided polygon, when n is odd? It is not sufficient to simply say, It s zero by symmetry. More detail is needed. I suggest either of two arguments: Suppose the field is non- zero. Then, it must point in some direc:on. Now, rotate the problem by 360 /n. We have the same configura:on of charges, but the field vector has rotated by 360 /n. This is only possible if the field = 0. Draw a line through one vertex and the center. The problem has mirror symmetry about this line. So, the field cannot have a non- zero component perpendicular to that line. This is true for any similar line (through a different vertex), so the field must have two dis:nct zero components. Therefore the field = 0. Physics

2 Laplace s Equation (3.1) Let s go through a more systema:c treatment of the problem. In general, V = ρ. Let s start with a simpler problem, a charge- free region. ε 0 In that case, we have Laplace s equa:on, V = 0. In Cartesian coordinates, this is: V x + V y + V z = 0 Start with the 1- dimensional case (no y or z dependence): This describes a parallel- plate capacitor, or similar geometry. Solu:on: V(x) is a linear func:on: How do we determine the constants, a and b? We need to specify two boundary condi:ons. For example: V x 1 V x ( ) = V 1 and V ( x ) = V Alternatively: V ( x 1 ) = V 1 and dv dx x1 ( ) = ax + b = a d V dx = 0 Nomenclature: Specifica:on of V is called a Dirichlet boundary condi:on. Specifica:on of the deriva:ve is called a Neumann boundary condi:on. This is mixed boundary condi:ons. x x 1 x y Physics

3 1D in spherical coordinates: Suppose there is no angular (θ or φ) dependence. Then: 1 r d dr ( ) r dv r dr ( ) = 0 r dv r dr = C 1 V ( r ) = C 1 r + C We have either the poten:al due to a point charge, or a constant (which has no physical significance). That s the only spherically symmetric possibility. 1D in cylindrical coordinates: Suppose there is no angular (φ or z) dependence. Then: 1 s ( ) d ds s dv s ds ( ) = 0 s dv s ds In G s nota:on, = C 1 V ( s) = C 1 ln( s) + C s is the radial We have either the poten:al due to an infinite line charge, or a constant. coordinate. Note: The only possible 1D angular dependence is V(θ) or V(φ) = C 1. Try it. Physics

4 Now, - dimensions: Things are not so simple in - D. We now have a par:al differen:al equa:on: V x + V y = 0 One important difference between ordinary and par:al differen:al equa:ons: To obtain a unique solu:on (to determine all the arbitrary constants) to a p.d.e., one must specify V(x,y) or dv/dn on the boundary of the region of interest (perhaps at ), or some equivalent amount of informa:on. We ll defer examples un:l we learn how to solve Laplace s equa:on by separa:on of variables. The near term plan: Mean Value theorem (briefly) Uniqueness theorems Solu:on techniques in D and 3D. It s useful to understand some of the general features before plunging into specific problems. Physics

5 A Short Diversion (Feynman 7- ) This is an adver:sement for Math 446 (Complex Variables) : Consider func:ons of complex numbers: z = F w ( ) = F( x + iy) w = x + iy Complex numbers Real numbers Here s an example: We can write F as the sum of two func:ons, one for the real part and one for the imaginary part. F( w) = U ( w) + iv ( w), where U = x - y and V = xy are both real func:ons of x and y. We now steal a theorem from Math 446: For any differen:able (holomorphic) func:on, F(w), U and V sa:sfy: Well, you might say, so what? F( w) w = ( x + iy) = x y + ixy U x = + V y V x = U y Cauchy- Riemann equa:ons Physics

6 It s easy to show that both U and V sa:sfy Laplace s equa:on: U = 0 and V = 0 So, from one func:on, F, we obtain two solu:ons of Laplace s equa:on Caveat: This only works in - D. Example: Consider V = xy (from previous slide). Here s a graph of curves of constant V: Because V = 0 if x = 0 or y = 0, this func:on is the solu:on for the poten:al when the situa:on is two grounded metal plates on the x and y axes. Rela:on between U and V: Curves of constant U are perpendicular to curves of constant V. E is everywhere perpendicular to constant V. (why?) So, the U curves show us the field lines. Note: The roles of U and V can be reversed. Look at the Cauchy- Riemann equa:ons. Remember, deriva:ves are slopes See Feynman for more graphs Physics

7 Proof of the Mean Value Theorem in 3D: V ( r c ) = V av = 1 4π R Surface V da Prove it for a single point charge outside the sphere. The general situa:on follows by superposi:on. Put the center of the sphere at the origin and the charge at z 0. Do the integral in spherical coordinates: V av = 1 4π R q π π R sinθdθdϕ 4πε 0 z 0 + R 1 = 0 0 ( z = r 0 Rcosθ ) V(r c ) R q = V ( r 4πε 0 z c ) 0 For z 0 > R. G does the details. The theorem also works in 1D and D (if the geometry really is 1D or D): z q z 0 r θ R Average of V around the circumference = Value of V at the center x The most important consequence is Earnshaw s Theorem: V cannot have a maximum or minimum inside a charge- free region. You cannot confine a charge with electrosta:c fields only. For z 0 < R, the answer is q 4πε 0 R (HW). Comment: G says that the MVT suggests the method of relaxa-on as a numerical solu:on technique. This technique is very inefficient, approaching the correct solu:on logarithmically. Physics

8 Uniqueness of Solutions ( ) You may recall: I claimed that uniqueness of solu:ons requires specifying V or dv/dn (or some combina:on) at the boundaries of the region of interest. Let s prove it. Region of interest Comments: In D, the boundaries are lines. In 3D, they are surfaces. The outer boundary might be at. Set V = 0 there. I ll prove that these condi:ons are sufficient, not necessary. Why are we bothering with this?? Don t be proud If you can find a solu:on by any nefarious means (e.g., images), you re done. There aren t any other solu:ons. There are three theorems: When V is specified, When dv/dn (= E perp ) is specified, and When charge (on conductors) is specified. G proves the first and third. I ll do the first and second. Boundaries Physics

9 Proofs by contradic:on: 1: Suppose there are two dis:nct solu:ons, V 1 and V, sa:sfying the same boundary condi:ons. Then, because the equa:ons are linear: V 3 = V 1 V is also a solu:on to Laplace s equa:on, with different boundary condi:ons: V 3 = 0 on all boundaries. However, the mean value theorem tells us that V 3 = 0 everywhere in the region. Therefore, V 1 = V. Note that this proof also applies to Poisson s equa:on, where ρ 0 in the region : The boundary condi:on is on E perp, so compare two dis:nct solu:ons, E 1 and E. As above, consider E 3 = E 1 E. E 3 = 0 everywhere in the region of interest, even if ρ 0. Also, E 3perp = 0 on every surface. Use G s vector iden:ty #5: f ( A) = f ( A ) + A ( f ). Let f be V 3 and A be E 3. Then, ( V E3 3 ) = V ( 3 E3 ) + E 3 ( V ) 3 = 0 E 3 Now, integrate over the region of interest and use the divergence theorem: ( V E3 3 )dvol = volume E 3 dvol volume Therefore, E 3 = 0 everywhere. = V E3 3 ( ˆn )da = 0 surfaces End 9/13/13 Physics