電磁學(xué)亂七八糟的符號(hào)(一)

@(study)[DSP, markdown_study, LaTex_study] author:何偉寶

ψ = ∫ s F ? ? a ? n d S \psi = \int_s \vec F \bullet \vec a_n d S ψ=∫s?F ?a n?dS ψ = ∮ S F ? ? d S ? \psi = \oint_S \vec F \bullet d\vec S ψ=∮S?F ?dS

Γ = ∫ l F ? ? d l ? \Gamma=\int_l \vec F \bullet d\vec l Γ=∫l?F ?dl Γ = ∮ l F ? ? d l ? \Gamma=\oint_l \vec F \bullet d\vec l Γ=∮l?F ?dl

? = a ? x ? ? x + a ? y ? ? y + a ? z ? ? z \nabla =\vec a_x \frac{\partial }{\partial x}+\vec a_y \frac{\partial }{\partial y}+\vec a_z \frac{\partial }{\partial z} ?=a x??x??+a y??y??+a z??z??

? = a ? x ? 2 ? x 2 + a ? y ? 2 ? y 2 + a ? z ? 2 ? z 2 \nabla =\vec a_x \frac{\partial^2 }{\partial x^2}+\vec a_y \frac{\partial^2 }{\partial y^2}+\vec a_z \frac{\partial^2 }{\partial z^2} ?=a x??x2?2?+a y??y2?2?+a z??z2?2?

? × ( ? × F ? ) = ? ( ? ? F ? ) ? ? 2 F ? \nabla \times (\nabla \times \vec F) = \nabla(\nabla \bullet \vec F) -\nabla^2 \vec F ?×(?×F )=?(??F )??2F

g r a d u = a ? x ? u ? x + a ? y ? u ? y + a ? z ? u ? z grad u =\vec a_x \frac{\partial u}{\partial x}+\vec a_y \frac{\partial u}{\partial y}+\vec a_z \frac{\partial u}{\partial z} gradu=a x??x?u?+a y??y?u?+a z??z?u? g r a d u = ? u gradu=\nabla u gradu=?u

d i v F ? ? lim ? △ V → 0 ∮ S F ? d S ? △ V div \vec F \triangleq \lim_{\triangle V\to 0} \frac{\oint_S \vec F d \vec S}{\triangle V} divF ?△V→0lim?△V∮S?F dS ?

d i v F ? = ? F x ? x + ? F y ? y + ? F z ? z = ? ? F ? div \vec F=\frac{\partial F_x}{\partial x}+\frac{\partial F_y}{\partial y}+ \frac{\partial F_z}{\partial z} =\nabla \bullet \vec F divF =?x?Fx??+?y?Fy??+?z?Fz??=??F

∫ V ? ? F ? d V = ∮ l F ? d S ? \int_V \nabla \bullet \vec F d V =\oint_l \vec F d \vec S ∫V???F dV=∮l?F dS

γ n ? lim ? △ S → 0 ∮ l F ? d l ? △ S \gamma_n \triangleq \lim_{\triangle S\to 0} \frac{\oint_l \vec F d \vec l}{\triangle S} γn??△S→0lim?△S∮l?F dl ?

R ? m ? r o t F ? = a ? n ? lim ? △ S → 0 ∮ F ? d l ? △ S ? m a x \vec R_m \triangleq rot \vec F =\vec a_n \lgroup \lim_{\triangle S \to 0} \frac{\oint \vec F d \vec l }{\triangle S} \rgroup_{max} R m??rotF =a n??△S→0lim?△S∮F dl ??max?

r o t F ? = ? × F ? rot \vec F =\nabla \times \vec F rotF =?×F

∫ S ? × F ? ? d S ? = ∮ l F ? d l ? \int_S \nabla \times \vec F \bullet d \vec S = \oint_l \vec F d \vec l ∫S??×F ?dS =∮l?F dl

體電荷密度: ρ ( r ? ? ) = lim ? △ V → 0 △ q △ V ? = d q d V ? \rho (\vec r^{\bullet} ) = \lim_{\triangle V \to 0 } \frac{\triangle q}{\triangle V^\bullet} = \frac{d q}{d V^\bullet} ρ(r ?)=△V→0lim?△V?△q?=dV?dq? q = ∫ V ρ ( r ? ? ) d V ? q= \int_V \rho(\vec r^\bullet) d V^\bullet q=∫V?ρ(r ?)dV?

面電荷密度: ρ s ( r ? ? ) = lim ? △ S → 0 △ q △ S ? = d q d S ? \rho_s (\vec r^{\bullet} ) = \lim_{\triangle S \to 0 } \frac{\triangle q}{\triangle S^\bullet} = \frac{d q}{d S^\bullet} ρs?(r ?)=△S→0lim?△S?△q?=dS?dq? q = ∫ S ρ S ( r ? ? ) d S ? q= \int_S \rho_S(\vec r^\bullet) d S^\bullet q=∫S?ρS?(r ?)dS?

線電荷密度: ρ l ( r ? ? ) = lim ? △ l → 0 △ q △ l ? = d q d l ? \rho_l (\vec r^{\bullet} ) = \lim_{\triangle l \to 0 } \frac{\triangle q}{\triangle l^\bullet} = \frac{d q}{d l^\bullet} ρl?(r ?)=△l→0lim?△l?△q?=dl?dq? q = ∫ l ρ l ( r ? ? ) d l ? q= \int_l \rho_l(\vec r^\bullet) d l^\bullet q=∫l?ρl?(r ?)dl?

點(diǎn)電荷: q ( r ? ) = ∑ i = 1 N q i ( r ? i ) q(\vec r)= \sum_{i=1}^N q_i(\vec r_i) q(r )=i=1∑N?qi?(r i?)

電流: i = lim ? △ t → 0 △ q △ t = d q d t i = \lim_{\triangle t \to 0}\frac{\triangle q }{\triangle t}=\frac{d q}{d t} i=△t→0lim?△t△q?=dtdq?

體電流密度矢量:

J ? = a ? n lim ? △ S ? → 0 △ i △ S ? = a ? n d i d S ? \vec J= \vec a_n \lim_{\triangle S^\bullet \to 0} \frac{\triangle i}{\triangle S^\bullet}=\vec a_n \frac{di }{dS^\bullet} J =a n?△S?→0lim?△S?△i?=a n?dS?di?

i = ∫ s J ? ? d S ? i = \int_s \vec J \bullet d \vec S i=∫s?J ?dS

? ? J ? = ? ? ρ ? t \nabla \bullet \vec J=- \frac{\partial \rho}{\partial t} ??J =??t?ρ?

面電流密度:

J ? s = a ? n lim ? △ l ? → 0 △ i △ l ? = a ? n d i d l ? \vec J_s =\vec a_n \lim_{\triangle l^\bullet \to 0} \frac{\triangle i}{\triangle l^\bullet} = \vec a_n \frac{d i}{d l^\bullet} J s?=a n?△l?→0lim?△l?△i?=a n?dl?di?

i = ∫ l J ? s ? ( n ? × d l ? ? ) i = \int_l \vec J_s \bullet (\vec n \times d \vec l^\bullet) i=∫l?J s??(n ×dl ?)

由于靜態(tài)場(chǎng)的麥克斯韋方程組還沒有統(tǒng)一,這里就不寫了

E ? ? F ? q 0 \vec E \triangleq \frac{\vec F}{q_0} E ?q0?F ?

B ? ? μ 4 π ∮ l I d l ? × a R R 2 \vec B \triangleq \frac{\mu}{4\pi}\oint_l \frac{I d \vec l \times a_R}{R^2} B ?4πμ?∮l?R2Idl ×aR??

ε i n ? ? d ψ d t \varepsilon_{in} \triangleq -\frac{d \psi}{d t} εin???dtdψ? 其中 ψ \psi ψ為磁通量 ψ ? ∫ S B ? ? d S ? \psi \triangleq \int_S \vec B \bullet d \vec S ψ?∫S?B ?dS 所以: ε i n = ∫ s ? B ? ? t ? d S ? \varepsilon_{in} = \int_s \frac{\partial \vec B}{\partial t} \bullet d \vec S εin?=∫s??t?B ??dS

E ? ( r ? ) ? ? △ Φ ( r ? ) \vec E(\vec r) \triangleq -\triangle\Phi(\vec r) E (r )??△Φ(r )

Φ ( r ? ) = W q \Phi(\vec r)=\frac{W}{q} Φ(r )=qW?

電位的標(biāo)量泊松方程: ? 2 Φ ( r ? ) = ? ρ ( r ? ) ε 0 \nabla^2 \Phi(\vec r) = - \frac{\rho(\vec r)}{\varepsilon_0} ?2Φ(r )=?ε0?ρ(r )?

電位的標(biāo)量拉普拉斯方程: ? 2 Φ ( r ? ) = 0 \nabla^2 \Phi(\vec r) = 0 ?2Φ(r )=0

B ? ( r ? ) ? ? × A ? ( r ? ) \vec B (\vec r )\triangleq \nabla \times \vec A(\vec r) B (r )??×A (r )

庫(kù)侖規(guī)范: ? ? A ? = 0 \nabla \bullet \vec A = 0 ??A =0

磁矢位的矢量泊松方程: ? 2 A ? ( r ? ) = ? μ 0 J ? ( r ? ) \nabla^2 \vec A (\vec r )=- \mu_0 \vec J (\vec r) ?2A (r )=?μ0?J (r )

磁矢位的矢量拉普拉斯方程 ? 2 A ? ( r ? ) = 0 \nabla^2 \vec A (\vec r )=0 ?2A (r )=0

磁矩m: m ? ? I ? S ? \vec m \triangleq \vec I \vec S m ?I S

P ? ( r ? ) = lim ? △ V → 0 ∑ i p ? i △ V \vec P(\vec r)=\lim_{\triangle V \to 0} \frac{\sum_i \vec p_i}{\triangle V} P (r )=△V→0lim?△V∑i?p ?i?? P ? = χ e ε 0 E ? \vec P = \chi_e \varepsilon_0 \vec E P =χe?ε0?E 其中 χ e \chi_e χe?為電極化率

D ? ( r ? ) ? ε 0 E ? ( r ? ) + P ? ( r ? ) \vec D(\vec r) \triangleq \varepsilon_0 \vec E(\vec r)+\vec P(\vec r) D (r )?ε0?E (r )+P (r ) 所以有:

∫ s D ? ( r ? ) ? d S ? = q \int_s \vec D(\vec r) \bullet d \vec S =q ∫s?D (r )?dS =q

? ? D ? ( r ? ) = ρ ( r ? ) \nabla \bullet \vec D(\vec r) = \rho(\vec r) ??D (r )=ρ(r )

D ? = ε E ? \vec D = \varepsilon \vec E D =εE

M ? ( r ? ) = lim ? △ V → 0 ∑ i m ? i △ V \vec M(\vec r)=\lim_{\triangle V \to 0} \frac{\sum_i \vec m_i}{\triangle V} M (r )=△V→0lim?△V∑i?m i?? M ? = χ m H \vec M = \chi_m H M =χm?H 其中 χ m \chi_m χm?為磁化率

H ? ( r ? ) = B ? ( r ? ) μ 0 ? M ? ( r ? ) \vec H(\vec r)=\frac{\vec B(\vec r)}{\mu_0}-\vec M(\vec r) H (r )=μ0?B (r )??M (r ) ∮ l H ? ? d l ? = I \oint_l \vec H\bullet d\vec l=I ∮l?H ?dl =I ? × H ? ( r ? ) = J ? ( r ? ) \nabla \times \vec H (\vec r )=\vec J(\vec r) ?×H (r )=J (r ) B ? = μ H ? \vec B=\mu \vec H B =μH

J ? ( r ? ) = σ E ? ( r ? ) \vec J(\vec r)=\sigma \vec E(\vec r) J (r )=σE (r ) 其中 σ \sigma σ為電導(dǎo)率

p ( r ? ) = J ? ( r ? ) ? E ? ( r ? ) = σ E 2 ( r ? ) p(\vec r)=\vec J(\vec r)\bullet \vec E(\vec r)=\sigma E^2(\vec r) p(r )=J (r )?E (r )=σE2(r )

a ? n × ( E ? 1 ? E ? 2 ) = 0 , E 1 t = E 2 t \vec a_n \times (\vec E_1 -\vec E_2)=0,\quad \quad E_{1t}=E_{2t} a n?×(E 1??E 2?)=0,E1t?=E2t? a ? n × ( H ? 1 ? H ? 2 ) = J ? s , H 1 t = H 2 t \vec a_n \times (\vec H_1 -\vec H_2)=\vec J_s,\quad \quad H_{1t}=H_{2t} a n?×(H 1??H 2?)=J s?,H1t?=H2t? a ? n ? ( D ? 1 ? D ? 2 ) = ρ s , D 1 n ? D 2 n = ρ s \vec a_n \bullet (\vec D_1 -\vec D_2) =\rho_s, \quad D_{1n}-D_{2n}=\rho_s a n??(D 1??D 2?)=ρs?,D1n??D2n?=ρs? a ? n ? ( B ? 1 ? B ? 2 ) = 0 , B 1 n = B 2 n \vec a_n \bullet (\vec B_1 - \vec B_2)=0,\quad \quad B_{1n}=B_{2n} a n??(B 1??B 2?)=0,B1n?=B2n?

靜電場(chǎng)能量密度: ω e = 1 2 ε E 2 \omega_e = \frac 12 \varepsilon E^2 ωe?=21?εE2 ω e = 1 2 D ? ( r ? ) ? E ? ( r ? ) \omega_e = \frac 12 \vec D(\vec r )\bullet \vec E(\vec r) ωe?=21?D (r )?E (r ) 靜磁場(chǎng)能量密度: ω m = 1 2 μ H 2 \omega_m = \frac 12 \mu H^2 ωm?=21?μH2 ω m = 1 2 H ? ( r ? ) ? B ? ( r ? ) \omega_m = \frac 12 \vec H(\vec r )\bullet \vec B(\vec r) ωm?=21?H (r )?B (r )

{ ∮ l E ? ( r ? , t ) ? d l ? = ? ∫ S ? B ? ( r ? , t ) ? t ? d S ? , ? × E ? ( r ? , t ) = ? ? B ? ( r ? , t ) ? t ∮ l H ? ( r ? , t ) ? d l ? = ∫ S ( J ? ( r ? , t ) + ? D ? ( r ? , t ) ? t ) , ? × H ? ( r ? , t ) = J ? ( r ? , t ) + ? D ? ( r ? , t ) ? t ∮ S D ? ( r ? , t ) ? d S ? = ∫ V ρ ( r ? , t ) d V , ? ? D ? ( r ? , t ) = ρ ( r ? , t ) ∮ S B ? ( r ? , t ) ? d S ? = 0 , ? ? B ? ( r ? , t ) = 0 \begin{cases} \oint_l \vec E(\vec r,t)\bullet d \vec l = -\int_S \frac{\partial \vec B(\vec r,t)}{\partial t} \bullet d \vec S , \quad\quad \nabla \times \vec E(\vec r,t) = - \frac{\partial \vec B(\vec r,t)}{\partial t} \\ \oint_l \vec H(\vec r,t)\bullet d\vec l = \int_S (\vec J(\vec r,t)+\frac{\partial \vec D(\vec r,t)}{\partial t}),\quad \nabla \times \vec H(\vec r,t)=\vec J(\vec r,t)+\frac{\partial \vec D(\vec r,t)}{\partial t}\\ \oint_S \vec D(\vec r,t)\bullet d \vec S = \int_V \rho(\vec r,t)dV,\quad\quad\quad\quad \nabla \bullet \vec D(\vec r,t)=\rho(\vec r,t)\\ \oint_S \vec B(\vec r ,t)\bullet d \vec S =0 ,\quad \quad\quad\quad\quad\quad\quad\quad\nabla \bullet \vec B(\vec r,t)=0 \end{cases} ????????????∮l?E (r ,t)?dl =?∫S??t?B (r ,t)??dS ,?×E (r ,t)=??t?B (r ,t)?∮l?H (r ,t)?dl =∫S?(J (r ,t)+?t?D (r ,t)?),?×H (r ,t)=J (r ,t)+?t?D (r ,t)?∮S?D (r ,t)?dS =∫V?ρ(r ,t)dV,??D (r ,t)=ρ(r ,t)∮S?B (r ,t)?dS =0,??B (r ,t)=0?

E ? = ? ? Φ ? ? A ? ? t \vec E=-\nabla\Phi -\frac{\partial \vec A}{\partial t} E =??Φ??t?A ?

洛倫茲條件(洛倫茲規(guī)范): ? ? A ? = ? μ ε ? Φ ? t \nabla \bullet \vec A=-\mu \varepsilon \frac{\partial \Phi}{\partial t} ??A =?με?t?Φ? 非齊次波動(dòng)方程(動(dòng)態(tài)退化可以得到其他規(guī)范): ? 2 Φ ( r ? , t ) ? μ ε ? 2 Φ ( r ? , t ) ? t 2 = ? ρ ( r ? , t ) ε \nabla^2 \Phi(\vec r,t)-\mu\varepsilon\frac{\partial^2\Phi(\vec r,t)}{\partial t^2}=- \frac{\rho(\vec r,t)}{\varepsilon} ?2Φ(r ,t)?με?t2?2Φ(r ,t)?=?ερ(r ,t)?

? 2 A ( r ? , t ) ? μ ε ? 2 A ( r ? , t ) ? t 2 = ? μ J ? ( r ? , t ) \nabla^2 A(\vec r,t)-\mu\varepsilon\frac{\partial^2 A(\vec r,t)}{\partial t^2}= -\mu \vec J(\vec r,t) ?2A(r ,t)?με?t2?2A(r ,t)?=?μJ (r ,t)

S ? ( r ? , t ) ? E ? ( r ? , t ) × H ? ( r ? , t ) \vec S (\vec r,t) \triangleq \vec E(\vec r,t)\times \vec H(\vec r,t) S (r ,t)?E (r ,t)×H (r ,t)

? ? ? S ? = ? ω ? t + p -\nabla \bullet \vec S=\frac{\partial\omega}{\partial t}+p ???S =?t?ω?+p

? ∮ S S ? ( r ? , t ) ? d S ? = ? ? t ∫ V ω ( r ? , t ) d V + ∫ V p ( r ? , t ) d V -\oint_S \vec S(\vec r,t)\bullet d \vec S=\frac{\partial}{\partial t}\int_V \omega(\vec r,t)d V+\int_Vp(\vec r,t)dV ?∮S?S (r ,t)?dS =?t??∫V?ω(r ,t)dV+∫V?p(r ,t)dV

u ( z , t ) = R e { [ U 0 ( z ) e j ? ] e j ω t } = R e { U ˙ ( z ) e j ω t } u(z,t)=Re\{ [U_0(z)e^{j\phi}]e^{j\omega t} \} = Re \{ \dot{U}(z) e^{j\omega t} \} u(z,t)=Re{[U0?(z)ej?]ejωt}=Re{U˙(z)ejωt} U ˙ ( z ) = U 0 ( z ) e j ? \dot{U}(z)=U_0(z)e^{j\phi} U˙(z)=U0?(z)ej?

? × E ? = j ω B ? \nabla \times \vec E=j\omega \vec B ?×E =jωB ? × H ? = J ? + j ω D ? \nabla \times \vec H =\vec J + j \omega \vec D ?×H =J +jωD E ? ˙ = a ? x E x ˙ ( r ? ) + a ? y E y ˙ ( r ? ) + a ? z E z ˙ ( r ? ) \dot{\vec E}=\vec a_x\dot{E_x}(\vec r)+\vec a_y\dot{E_y}(\vec r)+\vec a_z\dot{E_z}(\vec r) E ˙=a x?Ex?˙?(r )+a y?Ey?˙?(r )+a z?Ez?˙?(r )

? ? A ? ( r ? ) = ? j ω μ ε Φ ( r ? ) \nabla \bullet \vec A(\vec r) = -j\omega \mu\varepsilon \Phi(\vec r) ??A (r )=?jωμεΦ(r )

? 2 Φ ( r ? ) + ω 2 μ ε Φ ( r ? ) = ? ρ ( r ? ) ε \nabla^2\Phi(\vec r)+\omega^2\mu\varepsilon\Phi(\vec r)=-\frac{\rho(\vec r)}{\varepsilon} ?2Φ(r )+ω2μεΦ(r )=?ερ(r )? ? 2 A ? ( r ? ) + ω 2 μ ε A ? ( r ? ) = ? μ J ? ( r ? ) \nabla^2 \vec A(\vec r)+\omega^2\mu\varepsilon \vec A(\vec r)=-\mu \vec J(\vec r) ?2A (r )+ω2μεA (r )=?μJ (r ) 令 k 2 = ω 2 μ ε k^2=\omega^2\mu\varepsilon k2=ω2με有: 非齊次亥姆霍茲方程: ? 2 Φ ( r ? ) + k 2 Φ ( r ? ) = ? ρ ( r ? ) ε \nabla^2\Phi(\vec r)+k^2\Phi(\vec r)=-\frac{\rho(\vec r)}{\varepsilon} ?2Φ(r )+k2Φ(r )=?ερ(r )? ? 2 A ? ( r ? ) + k 2 A ? ( r ? ) = ? μ J ? ( r ? ) \nabla^2 \vec A(\vec r)+k^2 \vec A(\vec r)=-\mu \vec J(\vec r) ?2A (r )+k2A (r )=?μJ (r ) 齊次亥姆霍茲方程: ? 2 Φ ( r ? ) + k 2 Φ ( r ? ) = 0 \nabla^2\Phi(\vec r)+k^2\Phi(\vec r)=0 ?2Φ(r )+k2Φ(r )=0 ? 2 A ? ( r ? ) + k 2 A ? ( r ? ) = 0 \nabla^2 \vec A(\vec r)+k^2 \vec A(\vec r)=0 ?2A (r )+k2A (r )=0

η 0 = μ 0 ε 0 \eta_0=\sqrt{\frac{\mu_0}{\varepsilon_0}} η0?=ε0?μ0?? ?

S ? a v ( r ? ) = 1 T ∫ 0 T S ? ( r ? , t ) d t = 1 2 [ E ? 0 ( r ? ) × H ? 0 ( r ? ) ] c o s ( ? e ? ? n ) \vec S_av(\vec r)=\frac 1T\int_0^T\vec S(\vec r,t)dt=\frac 12 [\vec E_0(\vec r)\times \vec H_0(\vec r)]cos(\phi_e-\phi_n) S a?v(r )=T1?∫0T?S (r ,t)dt=21?[E 0?(r )×H 0?(r )]cos(?e???n?)

S ˙ ( r ? ) = 1 2 E ? ( r ? ) × H ? ? ( r ? ) = 1 2 E ? 0 ( r ? ) e ? j ? e × H ? 0 ( r ? ) e j ? n = 1 2 [ E ? 0 ( r ? ) × H ? 0 ( r ? ) ] e ? e ? ? n \dot{S}(\vec r)=\frac 12 \vec E(\vec r) \times \vec H^*(\vec r)=\frac 12 \vec E_0(\vec r)e^{-j\phi_e}\times \vec H_0(\vec r )e^{j\phi_n}=\frac 12[\vec E_0(\vec r)\times \vec H_0(\vec r)]e^{\phi_e-\phi_n} S˙(r )=21?E (r )×H ?(r )=21?E 0?(r )e?j?e?×H 0?(r )ej?n?=21?[E 0?(r )×H 0?(r )]e?e???n?

其中: S ? a v ( r ? ) = R e { S ˙ ( r ? ) } \vec S_av(\vec r)=Re\{ \dot{S}(\vec r) \} S a?v(r )=Re{S˙(r )}

? ∮ s S ˙ ( r ? ) ? d S ˙ = j 2 ω ∫ V [ ω m ? a v ( r ? ) ? ω e ? a v ( r ? ) ] d V + ∫ V p a v ( r ? ) d V -\oint_s \dot{S}(\vec r)\bullet d \dot{S} =j2\omega \int_V[\omega_{m-av}(\vec r)-\omega_{e-av}(\vec r)]dV +\int_V p_{av}(\vec r)dV ?∮s?S˙(r )?dS˙=j2ω∫V?[ωm?av?(r )?ωe?av?(r )]dV+∫V?pav?(r )dV

其中: ω a v ( r ? ) = 1 4 [ E ? ( r ? ) ? D ? ? ( r ? ) + B ? ( r ? ) ? H ? ? ( r ? ) ] = 1 4 [ ε ∣ E ? ( r ? ) ∣ 2 + μ ∣ H ? ( r ? ) ∣ 2 ] = R e ω ( r ? ) \omega_av (\vec r)=\frac 14[\vec E(\vec r)\bullet \vec D^*(\vec r)+\vec B(\vec r)\bullet \vec H^*(\vec r)]=\frac 14[\varepsilon|\vec E(\vec r)|^2 + \mu|\vec H(\vec r)|^2 ]=Re\omega(\vec r) ωa?v(r )=41?[E (r )?D ?(r )+B (r )?H ?(r )]=41?[ε∣E (r )∣2+μ∣H (r )∣2]=Reω(r ) p a v ( r ? ) = 1 2 E ? ( r ? ) ? J ? ? ( r ? ) = 1 2 σ ∣ E ? ( r ? ) ∣ 2 = R e p ( r ? ) p_{av}(\vec r)=\frac 12 \vec E(\vec r )\bullet \vec J^*(\vec r) =\frac 12 \sigma |\vec E(\vec r)|^2 =Rep(\vec r) pav?(r )=21?E (r )?J ?(r )=21?σ∣E (r )∣2=Rep(r )

天書雖然可怕,但,他還是你爸爸 也就,100條公式而已,前四章

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