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电磁波与光波

电磁波与光波

电磁波谱

概念

  • 可见光(visible light):λ=4000A−7000A\lambda = 4000A - 7000A

  • 红外线(infrared):λ=0.7μm−1mm\lambda = 0.7 \mu m - 1 mm

    • 用于夜视镜、辐射温度计、红外灯

  • 微波(microwave):λ=1mm−1m\lambda = 1mm - 1m

  • 无线电波(radio waves):λ>1m\lambda > 1m

    • 中波(MW):λ=3km−50m\lambda = 3km - 50m
    • 短波(SW):λ=50m−10m\lambda = 50m - 10m
    • 超短波(extra-SW):λ<1m\lambda < 1m

  • 紫外线(ultraviolet):λ=1nm−400nm\lambda = 1nm - 400 nm

  • X-rays:λ=0.01−10nm\lambda = 0.01 - 10nm

  • gamma rays:λ<10pm\lambda < 10 pm

电磁波的产生与发射

我们可以从麦克斯韦方程组推导出电磁波的性质

积分形式:

∬E⃗⋅dA⃗=q0ϵ0\iint \vec{E} \cdot d\vec{A} = \frac{q_{0}}{\epsilon_{0}} ∬B⃗⋅dA⃗=0\iint \vec{B} \cdot d \vec{A} = 0 ∮E⃗⋅dl⃗=−∬∂B⃗∂t⋅dA⃗\oint \vec{E} \cdot d \vec{l} = - \iint \frac{\partial \vec{B}}{\partial t} \cdot d \vec{A} ∮H⃗⋅dl⃗=i0+∬∂D⃗∂t⋅dA⃗\oint \vec{H} \cdot d \vec{l} = i_{0} + \iint \frac{\partial \vec{D}}{\partial t} \cdot d \vec{A}

微分形式(自由空间:ρ0=0,J0⃗=0\rho_{0} =0,\vec{J_{0}} = 0):

∇⋅E⃗=0\nabla \cdot \vec{E} = 0 ∇⋅B⃗=−∂B⃗∂t=−κmμ0∂H⃗∂t\nabla \cdot \vec{B} = - \frac{\partial \vec{B}}{\partial t} = - \kappa_{m} \mu_{0} \frac{\partial \vec{H}}{\partial t} ∇×H⃗=0\nabla \times \vec{H} = 0 ∇×H⃗=κeϵ0∂E⃗∂t\nabla \times \vec{H} = \kappa_{e} \epsilon_{0} \frac{\partial \vec{E}}{\partial t}

分量形式(把上式中的∇\nabla展开,得到如下式子):

∂Ex∂x+∂Ey∂y+∂Ez∂z=0\frac{\partial E_{x}}{\partial x} + \frac{\partial E_{y}}{\partial y} + \frac{\partial E_{z}}{\partial z} =0 ∣i^j^k^∂∂x∂∂y∂∂zEx⃗Ey⃗Ez⃗∣=−κmμ0(∂Hx⃗∂ti^+∂Hy⃗∂tj^+∂Hz⃗∂tk^)\begin{vmatrix} \widehat{i} && \widehat{j} && \widehat{k} \\ \frac{\partial}{\partial x} && \frac{\partial}{\partial y} && \frac{\partial}{\partial z} \\ \vec{E_{x}} && \vec{E_{y}} && \vec{E_{z}} \end{vmatrix} = -\kappa_{m} \mu_{0}(\frac{\partial \vec{H_{x}}}{\partial t} \widehat{i} + \frac{\partial \vec{H_{y}}}{\partial t} \widehat{j} + \frac{\partial \vec{H_{z}}}{\partial t} \widehat{k}) ∂Hx⃗∂x+∂Hy⃗∂y+∂Hz⃗∂z=0\frac{\partial \vec{H_{x}}}{\partial x} + \frac{\partial \vec{H_{y}}}{\partial y} + \frac{\partial \vec{H_{z}}}{\partial z} = 0 ∣i^j^k^∂∂x∂∂y∂∂zHx⃗Hy⃗Hz⃗∣=−κmμ0(∂Ex⃗∂ti^+∂Ey⃗∂tj^+∂Ez⃗∂tk^)\begin{vmatrix} \widehat{i} && \widehat{j} && \widehat{k} \\ \frac{\partial}{\partial x} && \frac{\partial}{\partial y} && \frac{\partial}{\partial z} \\ \vec{H_{x}} && \vec{H_{y}} && \vec{H_{z}} \end{vmatrix} = -\kappa_{m} \mu_{0}(\frac{\partial \vec{E_{x}}}{\partial t} \widehat{i} + \frac{\partial \vec{E_{y}}}{\partial t} \widehat{j} + \frac{\partial \vec{E_{z}}}{\partial t} \widehat{k})\notag

平面波

首先假设单一波源,在很远的自由空间中,在球面上取一弧面,可以近似为平面波;以其传播方向为zz轴,电场和磁场分别为xx和yy轴.

然后将上边的式子展开,可以得到以下八个方程

∂Ex∂x+∂Ey∂y+∂Ez∂z=0(1)\frac{\partial E_{x}}{\partial x} + \frac{\partial E_{y}}{\partial y} + \frac{\partial E_{z}}{\partial z} = 0 \tag{1} \\ ∂Ez∂y−∂Ey∂z=−κmμ0∂Hx∂t(2-1)\frac{\partial E_{z}}{\partial y} - \frac{\partial E_{y}}{\partial z} = -\kappa_{m} \mu_{0} \frac{\partial H_{x}}{\partial t} \tag{2-1} ∂Ex∂z−∂Ez∂x=−κmμ0∂Hy∂t(2-2)\frac{\partial E_{x}}{\partial z} - \frac{\partial E_{z}}{\partial x} = -\kappa_{m} \mu_{0} \frac{\partial H_{y}}{\partial t} \tag{2-2} ∂Ey∂x−∂Ex∂y=−κmμ0∂Hz∂t(2-3)\frac{\partial E_{y}}{\partial x} - \frac{\partial E_{x}}{\partial y} = -\kappa_{m} \mu_{0} \frac{\partial H_{z}}{\partial t} \tag{2-3} ∂Hx∂x+∂Hy∂y+∂Hz∂z=0(3)\frac{\partial H_{x}}{\partial x} + \frac{\partial H_{y}}{\partial y} + \frac{\partial H_{z}}{\partial z} =0 \tag{3} ∂Hz∂y−∂Hy∂z=κeϵ0∂Ex∂t(4-1)\frac{\partial H_{z}}{\partial y} - \frac{\partial H_{y}}{\partial z} =\kappa_{e} \epsilon_{0} \frac{\partial E_{x}}{\partial t} \tag{4-1} ∂Hx∂z−∂Hz∂x=κeϵ0∂Ey∂t(4-2)\frac{\partial H_{x}}{\partial z} - \frac{\partial H_{z}}{\partial x} =\kappa_{e} \epsilon_{0} \frac{\partial E_{y}}{\partial t} \tag{4-2} ∂Hy∂x−∂Hx∂y=κeϵ0∂Ez∂t(4-3)\frac{\partial H_{y}}{\partial x} - \frac{\partial H_{x}}{\partial y} =\kappa_{e} \epsilon_{0} \frac{\partial E_{z}}{\partial t} \tag{4-3}

横波

首先,横波在x和y方向的电场强度和磁场强度都是一样的,不会发生变化,所以

∂Ex∂x=∂Ey∂y=∂Hx∂x=∂Hy∂y=0\frac{\partial E_{x}}{\partial x} = \frac{\partial E_{y}}{\partial y} = \frac{\partial H_{x}}{\partial x} = \frac{\partial H_{y}}{\partial y} = 0 \notag

则由(1)(1)式,我们有

∂Ex∂x=0\frac{\partial E_{x}}{\partial x} = 0 \notag

由(2−3)(2-3)式,有

∂Hz∂t=0\frac{\partial H_{z}}{\partial t} = 0 \notag

同样地,由(3)(3)式可得

∂Hz∂z=0\frac{\partial H_{z}}{\partial z} = 0 \notag

由(4−3)(4-3),有

∂Ez∂t=0\frac{\partial E_{z}}{\partial t} = 0 \notag

所以电场和磁场在zz轴的分量与时间和z轴都无关,可以设为constantconstant

E⊥k,H⊥kE \perp k ,H \perp k \notag

电场垂直磁场

运用Ez=Hz=0E_{z} = H_{z} = 0,我们 有

(2−1)(2-1)式

∂Ey∂z=κmμ0∂Hx∂t(2-1)\frac{\partial E_{y}}{\partial z} =\kappa_{m} \mu_{0} \frac{\partial H_{x}}{\partial t} \tag{2-1} \notag

(2−2′)(2-2')式

∂Ex∂z=−κmμ0∂Hy∂t(2-2’)\frac{\partial E_{x}}{\partial z} =-\kappa_{m} \mu_{0} \frac{\partial H_{y}}{\partial t} \tag{2-2'}

(4−1′)(4-1')式

∂Hy∂z=−κeϵ0∂Ex∂t(4-1’)\frac{\partial H_{y}}{\partial z} =-\kappa_{e} \epsilon_{0} \frac{\partial E_{x}}{\partial t} \tag{4-1'}

(4−2)(4-2)式

∂Hx∂z=κeϵ0∂Ey∂t\frac{\partial H_{x}}{\partial z} =\kappa_{e} \epsilon_{0} \frac{\partial E_{y}}{\partial t} \notag

上边四个式子只包含Ey,Ex,Hy,HXE_{y},E_{x},H_{y},H_{X},说明电场,磁场只在x,yx,y方向有分量.

由于x,yx,y的方向是任意的,那么我们取xx的方向为电场方向,就有

∂Hx∂z=0=∂Hx∂t\frac{\partial H_{x}}{\partial z} = 0 = \frac{\partial H_{x}}{\partial t} \notag

所以磁场强度方向与电场强度方向垂直,我们就证明了E⃗⊥H⃗\vec{E} \perp \vec{H}

!!! note 其实就是把Ey=0E_{y} = 0带入(2−1)(2-1)和(4−2)(4-2),就得到的上边的结论.

波动方程

麦克斯韦:原来光就是电磁波

对(2−2′)(2-2')式同时对tt求偏导

∂2Ex∂z2=−κmμ0∂∂t∂Hy∂z=κmμ0Keϵ0∂2Ex∂t2\frac{\partial^{2} E_{x}}{\partial z^{2}} = - \kappa_{m} \mu_{0}\frac{\partial}{\partial t}\frac{\partial H_{y}}{\partial z} = \kappa_{m}\mu_{0}K_{e}\epsilon_{0} \frac{\partial^{2} E_{x}}{\partial t^{2}} \notag

同理对(4−1′)(4-1')操作,得到如下方程

∂2Ex∂z2−κeμ0Kmϵ0∂2Ex∂t2=0\frac{\partial^{2} E_{x}}{\partial z^{2}} - \kappa_{e}\mu_{0}K_{m}\epsilon_{0} \frac{\partial^{2} E_{x}}{\partial t^{2}} = 0 \notag ∂2Hy∂z2−κeμ0Kmϵ0∂2Hy∂t2=0\frac{\partial^{2} H_{y}}{\partial z^{2}} - \kappa_{e}\mu_{0}K_{m}\epsilon_{0} \frac{\partial^{2} H_{y}}{\partial t^{2}} = 0 \notag

猜根,有:

{Ex=Ex0ei(ωt−kz)Hy=Hy0ei(ωt−kz)\left\{ \begin{matrix} E_{x} = E_{x0}e^{i(\omega t - kz)} \\ H_{y} = H_{y0}e^{i(\omega t - kz)} \end{matrix} \right. \notag

ω=2πT\omega = \frac{2\pi}{T}是角频率,k=2πλk = \frac{2\pi}{\lambda}是波矢,也叫波数

带回方程,得到

k2=κeϵ0κmμ0ω2⇒k=ωκeϵ0κmμ0k^{2} = \kappa_{e}\epsilon_{0}\kappa_m\mu_0\omega^2 \Rightarrow k = \omega\sqrt{\kappa_e\epsilon_0\kappa_m\mu_0} \notag

又因为

v=ωk=1κeϵ0κmμ0v = \frac{\omega}{k} = \frac{1}{\sqrt{\kappa_e\epsilon_0\kappa_m\mu_0}} \notag

!!! note 而真空中,磁导率κm\kappa_m和介电常数κe\kappa_e都为1,所以代入计算得到v=c=3.0×108m/sv = c = 3.0 \times 10^8 m/s,我们就计算出了光速

我们定义$\sqrt{\kappa_e\kappa_m} = n$,就是折射率,所以可以推导出光学中的

$$
v= \frac{c}{n} \notag
$$

电场和磁场

由(2−2′)(2-2')式,将我们猜根得到的Ex,HyE_{x},H_{y}代入,得到如下式子

−ikEx0ei(ωt−kx)=−κmμ0iωHy0ei(ωt−kx)kEx0=κmμ0ωHy0Ex0=κmμ0ωkHy0=κmμ0vHy0=κmμ01κeϵ0κmμ0Hy0κeϵ0Ex0=κmμ0Hy0κeϵ0Ex0eiϕE=κmμ0Hy0eiϕH\begin{align*} -ikE_{x0}e^{i(\omega t - kx)} &= -\kappa_m\mu_0i\omega H_{y0}e^{i(\omega t - kx)} \\ kE_{x0} &= \kappa_m\mu_0\omega H_{y0} \\ E_{x0} &= \kappa_m\mu_0\frac{\omega}{k}H_{y0} = \kappa_m\mu_0v H_{y0} \\ &= \kappa_m\mu_0\frac{1}{\sqrt{\kappa_e\epsilon_0\kappa_m\mu_0}}H_{y0} \\ \sqrt{\kappa_e\epsilon_0}E_{x0} &= \sqrt{\kappa_m\mu_0}H_{y0} \\ \sqrt{\kappa_e\epsilon_0}E_{x0}e^{i\phi_E} &= \sqrt{\kappa_m\mu_0}H_{y0}e^{i\phi_H} \end{align*} \notag

通过上式我们可得以下两个方程

{κeϵ0E0=κmμ0H0(振幅相等)ϕE=ϕH(相位相同)\left\{ \begin{matrix} \sqrt{\kappa_e\epsilon_0}E_{0} = \sqrt{\kappa_m\mu_0}H_{0} (振幅相等) \\ \phi_{E} = \phi_{H} (相位相同) \end{matrix} \right. \notag

真空中,κe=κm=1\kappa_e = \kappa_m = 1

所以

ϵ0E0=μ0H0⇒E0=μ0H0ϵ0μ0=cB0(c为光速)\sqrt{\epsilon_0}E_{0} = \sqrt{\mu_0}H_0 \\ \Rightarrow E_{0} = \frac{\mu_0H_0}{\sqrt{\epsilon_0\mu_0}} = cB_{0}(c为光速) \notag

!!! note “电场强度与磁感应强度” 我们发现E0=cB0E_{0} = cB_{0},电场强度和磁感应强度之间只差了一个常数

电磁波的能流密度和动量

单位体积内电磁波的能量包括电场和磁场两部分

  • 电场能量:UE=12ϵ0E2U_E = \frac{1}{2}\epsilon_0E^2
  • 磁场能量:UB=12B2μ0U_B = \frac{1}{2}\frac{B^2}{\mu_0}

则单位体积内电磁波的能量:

U=∭(12ϵ0E2+12B2μ0)dvU = \iiint (\frac{1}{2}\epsilon_0E^2 + \frac{1}{2}\frac{B^2}{\mu_0})dv \notag

更一般的,我们知道D⃗=κeϵ0E⃗,B⃗=κmμ0H⃗\vec{D} = \kappa_e\epsilon_0\vec{E},\vec{B} = \kappa_m\mu_0\vec{H},那么

U=UE+UB=∭(12D⃗⋅E⃗+12B⃗⋅H⃗)dvdUdt=ddt∭(12D⃗⋅E⃗+12B⃗⋅H⃗)dv=12∭∂∂t(D⃗⋅E⃗+B⃗⋅H⃗)dv\begin{align*} U = U_E + U_B &= \iiint(\frac{1}{2}\vec{D}\cdot \vec{E} + \frac{1}{2}\vec{B}\cdot\vec{H})dv \\ \frac{dU}{dt} &= \frac{d}{dt}\iiint(\frac{1}{2}\vec{D}\cdot \vec{E} + \frac{1}{2}\vec{B}\cdot\vec{H})dv \\ &= \frac{1}{2}\iiint \frac{\partial}{\partial t}(\vec{D}\cdot\vec{E} + \vec{B} \cdot\vec{H})dv \end{align*}

对积分内部展开:

∂∂t(D⃗⋅E⃗+B⃗⋅H⃗)=κeϵ0∂∂t(E⃗⋅E⃗)+κmμ0∂∂t(H⃗⋅H⃗)=2κeϵ0E⃗⋅∂E⃗∂t+2κmμ0H⃗⋅∂H⃗∂t=2E⃗⋅∂D⃗∂t+2H⃗⋅∂B⃗∂t\begin{align*} \frac{\partial}{\partial t}(\vec{D}\cdot\vec{E} + \vec{B} \cdot\vec{H}) &= \kappa_e\epsilon_0\frac{\partial}{\partial t}(\vec{E} \cdot \vec{E}) + \kappa_m\mu_0\frac{\partial}{\partial t}(\vec{H}\cdot\vec{H}) \notag \\ &= 2\kappa_e\epsilon_0\vec{E}\cdot\frac{\partial \vec{E}}{\partial t} + 2\kappa_m\mu_0\vec{H}\cdot\frac{\partial \vec{H}}{\partial t} \\ &= 2\vec{E}\cdot \frac{\partial \vec{D}}{\partial t} + 2\vec{H} \cdot \frac{\partial \vec{B}}{\partial t} \end{align*}

在麦克斯韦方程中:

∂D⃗∂t=∇×H⃗−J0⃗\frac{\partial \vec{D}}{\partial t} = \nabla \times \vec{H} - \vec{J_{0}} ∂B⃗∂t=−∇×E⃗\frac{\partial \vec{B}}{\partial t} = -\nabla \times \vec{E} \notag

代入上式可得:

=2E⃗⋅(∇×H⃗−J0⃗)−2H⃗⋅(∇×E⃗)=2[E⃗⋅(∇×H⃗)−H⃗⋅(∇×E⃗)−J0⃗⋅E⃗]=−2∇⋅(E⃗×H⃗)−2J0⃗⋅E⃗\begin{align*} &= 2\vec{E} \cdot (\nabla \times \vec{H} - \vec{J_{0}}) - 2\vec{H} \cdot (\nabla \times \vec{E}) \\ &= 2[\vec{E} \cdot (\nabla \times \vec{H}) - \vec{H} \cdot (\nabla \times \vec{E}) - \vec{J_{0}} \cdot \vec{E}] \\ &= -2\nabla \cdot (\vec{E} \times \vec{H}) - 2\vec{J_{0}}\cdot \vec{E} \notag \end{align*}

最后运用高斯定理化简:

dUdt=−∭∇⋅(E⃗×H⃗)dv−∭(J0⋅E⃗)dv=−∬(E⃗×H⃗)⋅dA−∭(J0⋅E⃗)dv\begin{align*} \frac{dU}{dt} &= - \iiint\nabla \cdot (\vec{E} \times \vec{H})dv - \iiint (J_{0} \cdot \vec{E})dv \notag \\ &= -\iint(\vec{E} \times \vec{H})\cdot dA - \iiint (J_{0} \cdot \vec{E})dv \end{align*}

我们现在关注第二项到底是什么意思

在欧姆定律中,我们有以下公式,E⃗\vec{E}为电场,K⃗\vec{K}为非静电力

J0⃗=σ(E⃗+K⃗)⇒E⃗=1σJ0⃗−K⃗=ρJ0−K⃗\vec{J_{0}} = \sigma(\vec{E} + \vec{K}) \notag \\ \Rightarrow \vec{E} = \frac{1}{\sigma}\vec{J_{0}} - \vec{K} = \rho J_{0} - \vec{K}

把这个积分放到均匀的圆筒里边,那么我们对vv积分其实就是乘以ΔA⋅Δl\Delta A \cdot \Delta l,从而进行如下变换

∭(J0⋅E⃗)dv=(J0⋅E⃗)ΔA⋅Δl=J0⋅(ρJ0−K⃗)ΔA⋅Δl=ρJ02ΔA−J0⋅K⃗ΔA⋅Δl=ρΔlΔA(J0ΔA)2−(J0ΔA)(K⃗⋅Δl)=Ri02−I0Δϵ(Δϵ是电动势)=Q−P\begin{align*} \iiint(J_{0} \cdot \vec{E})dv &= (J_{0} \cdot \vec{E})\Delta A \cdot \Delta l \\ &= J_{0} \cdot (\rho J_{0} - \vec{K})\Delta A \cdot \Delta l \\ &= \rho J_{0}^{2}\Delta A - J_{0}\cdot \vec{K}\Delta A \cdot \Delta l \\ &= \rho \frac{\Delta l}{\Delta A}(J_{0}\Delta A)^{2} - (J_{0}\Delta A)(\vec{K}\cdot \Delta l) \\ &= Ri_{0}^{2} - I_{0}\Delta \epsilon(\Delta \epsilon是电动势) \\ &= Q - P \end{align*} \notag

所以我们记S⃗=E⃗×H⃗\vec{S} = \vec{E} \times \vec{H},为Poynting Vertor(玻印廷矢量)Poynting \ Vertor(玻印廷矢量),那么

dUdt=−∬S⃗⋅dA⃗−Q+P\frac{dU}{dt} = - \iint \vec{S}\cdot d\vec{A} - Q + P \notag

!!! note “理解上式” dUdt\frac{dU}{dt}是单位时间内电场能量与磁场能量之和的变化; −∬S⃗⋅dA⃗- \iint \vec{S}\cdot d\vec{A}是通过表面向外辐射的能量; QQ是产生的热量

Poynting Vertor

定义单位时间,单位面积内的能量流动

S⃗=E⃗×H⃗=E⃗×B⃗μ0=E⃗2μ0c\vec{S} = \vec{E} \times \vec{H} = \frac{\vec{E} \times \vec{B}}{\mu_0} = \frac{\vec{E}^{2}}{{\mu_0}c} \notag

定义Z0=μ0c=377ΩZ_{0} = \mu_0c = 377 \Omega

从而S=E2377ΩS = \frac{E^2}{377\Omega}

!!! tip “电磁波的强度” 电磁波的强度II实际上就是SS的平均值

$$
I = \langle S\rangle = \frac{\langle E^2\rangle}{Z_0} = \frac{E_{max}^2}{377\Omega}\langle sin^2(kz-wt)\rangle = \frac{1}{2}\frac{E_{max}^2}{377\Omega}
$$

!!! note “电场能量密度和磁场能量密度的关系” 由于μE=12ϵ0E2,μB=12B2μ0\mu_{E} = \frac{1}{2}\epsilon_{0}E^2,\mu_B = \frac{1}{2}\frac{B^2}{\mu_0}

而$B = \frac{E}{c}$,那么$\mu_B = \frac{1}{2}\frac{E^2}{C^2\mu_0} = \frac{1}{2}\epsilon_0E^2 = \mu_E$ 

所以二者能量各占一半

电磁波的能量密度可以表示为:$\mu = \mu_E + \mu_B = \epsilon_0E^2$

!!! note “电磁波的强度” I=c⟨μ⟩=cϵ0⟨E2⟩=cϵ0Emax22=12Emax2μ0c=Emax22⋅377Ω=Erms2377Ω I = c\langle\mu\rangle = c\epsilon_0\langle E^2\rangle = c\epsilon_0\frac{E_{max}^2}{2} = \frac{1}{2}\frac{E_{max}^{2}}{\mu_0c} = \frac{E_{max}^2}{2 \cdot 377 \Omega} = \frac{E_{rms}^2}{377\Omega}

电路中的能量传输

如上图所示电路,考虑与电源正极相连的导线,导线内部存在一个电场,那么由于∮E⃗⋅dl⃗=−dΦBdt=0\oint \vec{E} \cdot d\vec{l} = -\frac{d\Phi_B}{dt} = 0,导体外部一定存在一个方向相同的电场.

再加上一个垂直导线的电场,根据S⃗=E⃗×H⃗\vec{S} = \vec{E} \times \vec{H},我们可以得到能量流动的方向,一方面流向电阻,另一方面被导线消耗.

与电源负极相连的导线也是类似的

电磁波的动量

假设一个有一个力 ΔF⃗\Delta \vec{F},这个力会让电荷做功,即ΔW=ΔF⋅Δl\Delta W = \Delta F \cdot \Delta l;而这部分功就是这个物体吸收的净能量;

ΔF⃗⋅cΔt=(Sin⃗−Sout⃗)⋅ΔAΔt\Delta \vec{F} \cdot c \Delta t = (\vec{S_{in}} - \vec{S_{out}}) \cdot \Delta A \Delta t \notag

故

ΔF⃗=1c(Sin⃗−Sout⃗)⋅ΔA\Delta \vec{F} = \frac{1}{c}(\vec{S_{in}} - \vec{S_{out}}) \cdot \Delta A \notag

矢量减

光压

单位面积上的力

P⃗=1c(Sin⃗−Sout⃗)\vec{P} = \frac{1}{c}(\vec{S_{in}} - \vec{S_{out}}) \notag

动量密度

单位体积内的动量

Δg=F⋅ΔtΔAcΔt=FcΔA=1c2(Sin⃗−Sout⃗)\Delta g = \frac{F \cdot \Delta t}{\Delta Ac\Delta t} = \frac{F}{c\Delta A} = \frac{1}{c^2}(\vec{S_{in}} - \vec{S_{out}}) \notag

!!! note “动量密度” gin=1c2Sin⃗g_{in} = \frac{1}{c^2}\vec{S_{in}}为入射光的动量密度,gout=1c2Sout⃗g_{out} = \frac{1}{c^2}\vec{S_{out}}为反射光的动量密度

!!! tip “光压” 对于白体,Sin=SoutS_{in} = S_{out},故gin=goutg_{in} = g_{out}

$$
P = \frac{2}{c}\vec{S_{in}} \notag
$$

对于黑体,$S_{out} = 0$,故$g_{in} = \frac{1}{c^2}\vec{S_{in}}$

$$
P = \frac{1}{c}\vec{S_{in}}\notag
$$