Q & A.

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1 Q & A

2 §1 Sep 21 何斯慧

3 Answer: As shown in Fig 1.4 (p17), n is the normal of the boundary surface, t is the normal of the loop and is tangent to the surface. So t x n is the direction of the long arm. Using the rule of mixed product, (t x n)* E =(n x E)* t, we get vector forms of boundary relations for E and H.

4 del operator

5 2) let me use the vector formula to verify,

6 Taylor expansion -see (4.22) (p150)

7 3) p39, Green function is a solution of point source, it will be used to solve boundary value problem. Neumann problem usually is a exterior problem, where surface charge density is given (electric field or normal derivative of the potential). So total surface is inner part plus infinity.

8 4) P 42, Green’s 1st identity

9 5) p44, variational approach
Derive EoS Provide approximate but accurate solutions Systems in equilibrium have minimal energy. energy-like functional has a stationary minimum if \psi satisfy equation of motion (Poisson equation, Schrodinger eq, Dirac eq)

10 Consider the change of the functional
Provided , the change vanishes if \psi satisfies Poisson eq, where g is a source function.

11 Recipe to find an approximate solution
Choose a flexible trial function \phi( a, b, c,…) Calculate the functional I( a, b, c…) Seek an extremum by setting the partial derivatives with respect to the parameters - Then solve coupled algebraic linear equations With the optimun parameters, the trial solution is the best possible approximation with the particular functional (!) form chosen. See p46 Fig 1.9

12 6) p103 The point x is on the z axis, so \gamma =0,
P_l(1)=1, so eq reduces to (3.37).

13 §2 Sep 24, 刘冶华 e.a. For 1D light wave propagating through an interface, it can be shown that E+E”=E’ and the equality of their derivative w.r.t. x. (like \psi in QM). E is tangent to surface, so the 1st equation is actually n x (E_2 - E_1)=0. The relation of derivative can be derived from continuity of H, which is proportional to curl E. Above results work for normal incidence as well.

14 混合0502（光电0504） 徐博

15 §3 1、第179页式5.21是由上一等式经分部积分（integration by part应该是分部积分吧）所得 ，但关于三重积分的分布积分法则我们在数学分析中没有学过，所以您能不能在上课的时候把 这一方面的数学内容将一下，在书上有多处都用到了分部积分。 Treatment in the book is not clear, see my notes

16 2、第32页上D（x)的定义中两表面S与S‘无限接近的同时面电荷密度趋于无穷大，但d与σ的成 绩为什么可以取有限定值，这是不是一个实验结论？
3、第34页中使用了立体角，但立体角的内容我在解析几何书上没有找到，老师能不能简单的 讲一下立体角的积分法则？ 4、第213页的式5.146中，积分第二项为零是为什么？ Using Gauss’s law, the surface integral vanishes.

17 §4 何斯慧 第41页中，式（1.54）怎样由上面的式子得到？
第45页中，式（1.72）上面的一段话“The reason is that, if the source density g is well behaved and finite at the origin, Gauss’s law shows that ψhas a maximum or minimum there with vanishing slope.”没看懂这一解释。 第181页中，Sect5.5的上面一段“If △ψ=0 holds in all space, ψ must be at most a constant provided there are no sources at infinity.”为什么？

18 第181页中，（5.33）怎样从Fig.5.5得到？ 第185页中，（5。52）怎样得到？ 第185页倒数第三行， 为什么？
The trick works also for charge density of the ring (p123)

19 7．第102页中，power series 怎样算出来？
还有（3.38）怎么得到？ 8.第189页中， （5.67）上面，为什么x → Bk(0), 而不是x→0 ? 还有（5。69）下面的那一段话是什么意思？ 9.第197页中，为什么（5.100）比（5.97）多了后面一项？ 10.想问一下下面的结论是否正确？ 一组粒子，既可以描述成带电荷量q i’和磁荷量g i’,也可描述成只带电荷q i 其中 μg i’ g j’/4πr2 + q i’ q j’/4πr2ε = q i q j/4πr2ε 这样，可以把磁场（B= μg i’ /4πr2 ），磁荷转化成电场，电荷。于是，在电磁场中运动的电荷，磁荷所受的力，就等于对应的静止的电荷在电场中所受的力，只要用库仑定律就可以求出总的力了。

20 为常矢量 例: 求 其中 为常矢量 例: 求 解：

21 Formulae for Del operator
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.