广义相对论-学习记录2-第二章-狭义相对论

第二章:狭义相对论

1、洛伦兹变换

Lorentz方程

  狭义相对性原理表示,自然规律对洛伦兹变换群是不变的。所谓的洛伦兹变换,是由一个时空坐标系 x α x^\alpha xα到另一个坐标系 x ′ α x'^\alpha x′α的变换(时空变换),这个线性变换具有如下形式:

x ′ α = Λ β α x β + a α x'^\alpha=\Lambda^\alpha_\beta x^\beta + a^\alpha x′α=Λβα​xβ+aα (1)

  其中, a α a^\alpha aα和 Λ β α \Lambda^\alpha_\beta Λβα​是常数, a α a^\alpha aα表示坐标, Λ β α \Lambda^\alpha_\beta Λβα​表示矩阵,它们满足条件:

Λ γ α Λ δ β η α β = η γ δ \Lambda^\alpha_\gamma\Lambda^\beta_\delta\eta_{\alpha\beta} =\eta_{\gamma\delta} Λγα​Λδβ​ηαβ​=ηγδ​ (2)

  上述关系式被称为Lorentz方程

关于Lorentz方程的几点说明

  (1)矩阵 η α β \eta_{\alpha\beta} ηαβ​具有如下形式:

η α β = { + 1 ,   α = β = 1 ,   o r   2 ,   o r   3 − 1 ,   α = β = 0 0 ,   α ≠ β \eta_{\alpha\beta}=\left\{\begin{aligned}&+1,\ \alpha=\beta=1,\ or\ 2,\ or\ 3\\ &-1,\ \alpha=\beta=0\\ &0,\ \alpha\neq \beta\end{aligned}\right. ηαβ​=⎩⎪⎨⎪⎧​​+1, α=β=1, or 2, or 3−1, α=β=00, α​=β​

  (2) x 1 x^1 x1, x 2 x^2 x2, x 3 x^3 x3是三个空间分量, x 0 = t x^0=t x0=t是时间分量

  (3)采用了爱因斯坦约定,所有的重复指标表示要求和(求和号被省略不写),例如,如果在式(1)中取 α = 0 \alpha=0 α=0,则(1)式为:

x ′ 0 = Λ 0 0 x 0 + Λ 1 0 x 1 + Λ 2 0 x 2 + Λ 3 0 x 3 + a 0 x'^0=\Lambda^0_0 x^0+\Lambda^0_1 x^1+\Lambda^0_2 x^2+\Lambda^0_3 x^3 + a^0 x′0=Λ00​x0+Λ10​x1+Λ20​x2+Λ30​x3+a0

  (4)Lorentz变换保持“原时” d τ d\tau dτ不变,即:

d τ 2 ≡ d t 2 − d x → 2 = − η α β d x α d x β d\tau^2\equiv dt^2-d\overrightarrow x^2=-\eta_{\alpha\beta}dx^\alpha dx^\beta dτ2≡dt2−dx 2=−ηαβ​dxαdxβ

  (关于第二个等号是如何变过去的: t = x 0 t=x^0 t=x0,而 x → \overrightarrow x x 对应的上角标分别是1,2,3,从右边出发可以推出左边)


证明:

  在坐标变换的过程中, x α → x ′ α x^\alpha\rightarrow x'^\alpha xα→x′α, d τ ′ 2 = − η α β d x ′ α d x ′ β d\tau'^2=-\eta_{\alpha\beta}dx'^\alpha dx'^\beta dτ′2=−ηαβ​dx′αdx′β。再对定义中的(1)式进行微分,得到:

d x ′ α = Λ γ α d x γ dx'^\alpha=\Lambda^\alpha_\gamma dx^\gamma dx′α=Λγα​dxγ

  由此,分别将: Λ δ α d x δ \Lambda^\alpha_\delta dx^\delta Λδα​dxδ和 Λ γ β d x γ \Lambda^\beta_\gamma dx^\gamma Λγβ​dxγ代入到 d τ ′ 2 d\tau'^2 dτ′2右端的两个 d x ′ dx' dx′中:

d τ ′ 2 = − η α β Λ δ α Λ γ β d x δ d x γ = − η δ γ d x δ d x γ = − η α β d x α d x β = d τ 2 \begin{aligned}d\tau'^2&=-\eta_{\alpha\beta}\Lambda^\alpha_\delta\Lambda_\gamma^\beta dx^\delta dx^\gamma\\ &=-\eta_{\delta\gamma}dx^{\delta}dx^{\gamma}\\ &=-\eta_{\alpha\beta}dx^\alpha dx^\beta\\ &=d\tau^2\end{aligned} dτ′2​=−ηαβ​Λδα​Λγβ​dxδdxγ=−ηδγ​dxδdxγ=−ηαβ​dxαdxβ=dτ2​

QED.


  (5)Lorentz变换是保持 d τ d\tau dτ不变的仅有的非奇异变换 x → x ′ x\rightarrow x' x→x′

非奇异的含义: x ′ ( x ) x'(x) x′(x)和 x ( x ′ ) x(x') x(x′)都是正规的可微分函数,并使矩阵 ∂ x ′ α ∂ x β \dfrac{\partial x'^\alpha}{\partial x^\beta} ∂xβ∂x′α​有确定的逆矩阵 ∂ x β ∂ x ′ α \dfrac{\partial x^\beta}{\partial x'^\alpha} ∂x′α∂xβ​


证明:

  存在性已在说明(4)中证明,接下来只需要证明唯一性,即先考虑最一般的坐标变换,在加上各种条件后最后只剩下这一种变换

  首先需要证明是线性变换,随后需要证明满足Lorentz方程

  先考虑最一般的坐标变换 x α → x ′ α x^\alpha\rightarrow x'^\alpha xα→x′α,一般地把 d τ → d τ ′ d\tau\rightarrow d\tau' dτ→dτ′

d τ ′ 2 = − η α β d x ′ α d x ′ β d\tau'^2=-\eta_{\alpha\beta}dx'^\alpha dx'^\beta dτ′2=−ηαβ​dx′αdx′β

  目标: d τ ′ 2 = d τ 2 = − η γ δ d x γ d x δ d\tau'^2=d\tau^2=-\eta_{\gamma\delta} dx^\gamma dx^\delta dτ′2=dτ2=−ηγδ​dxγdxδ

  将 ∂ x ′ α ∂ x γ d x γ \dfrac{\partial x'^\alpha}{\partial x^\gamma}dx^\gamma ∂xγ∂x′α​dxγ代入到 d x ′ α dx'^\alpha dx′α,将 ∂ x ′ β ∂ x δ d x δ \dfrac{\partial x'^\beta}{\partial x^\delta}dx^\delta ∂xδ∂x′β​dxδ代入到 d x ′ β dx'^\beta dx′β:

d τ ′ 2 = − η α β ∂ x ′ α ∂ x γ ∂ x ′ β ∂ x δ d x γ d x δ d\tau'^2=-\eta_{\alpha\beta}\dfrac{\partial x'^\alpha}{\partial x^\gamma}\dfrac{\partial x'^\beta}{\partial x^\delta}dx^\gamma dx^\delta dτ′2=−ηαβ​∂xγ∂x′α​∂xδ∂x′β​dxγdxδ

  与目标结合对比,两边相消,得到:

η γ δ = η α β ∂ x ′ α ∂ x γ ∂ x ′ β ∂ x δ \eta_{\gamma\delta}=\eta_{\alpha\beta}\dfrac{\partial x'^\alpha}{\partial x^\gamma}\dfrac{\partial x'^\beta}{\partial x^\delta} ηγδ​=ηαβ​∂xγ∂x′α​∂xδ∂x′β​

  上式两端对 x ϵ x^\epsilon xϵ求导,得:

η α β ∂ 2 x ′ α ∂ x γ ∂ x ϵ ∂ x ′ β ∂ x δ + η α β ∂ x ′ α ∂ x γ ∂ 2 x ′ β ∂ x δ ∂ x ϵ = 0 \eta_{\alpha\beta}\dfrac{\partial^2 x'^\alpha}{\partial x^\gamma\partial x^\epsilon}\dfrac{\partial x'^\beta}{\partial x^\delta}+\eta_{\alpha\beta}\dfrac{\partial x'^\alpha}{\partial x^\gamma}\dfrac{\partial^2x'^\beta}{\partial x^\delta \partial x^\epsilon}=0 ηαβ​∂xγ∂xϵ∂2x′α​∂xδ∂x′β​+ηαβ​∂xγ∂x′α​∂xδ∂xϵ∂2x′β​=0(2)

  (2)式中 γ \gamma γ与 ϵ \epsilon ϵ互换,得:

η α β ∂ 2 x ′ α ∂ x ϵ ∂ x γ ∂ x ′ β ∂ x δ + η α β ∂ x ′ α ∂ x ϵ ∂ 2 x ′ β ∂ x δ ∂ x γ = 0 \eta_{\alpha\beta}\dfrac{\partial^2 x'^\alpha}{\partial x^\epsilon\partial x^\gamma}\dfrac{\partial x'^\beta}{\partial x^\delta}+\eta_{\alpha\beta}\dfrac{\partial x'^\alpha}{\partial x^\epsilon}\dfrac{\partial^2x'^\beta}{\partial x^\delta \partial x^\gamma}=0 ηαβ​∂xϵ∂xγ∂2x′α​∂xδ∂x′β​+ηαβ​∂xϵ∂x′α​∂xδ∂xγ∂2x′β​=0(3)

  继续(2)式中 δ \delta δ与 ϵ \epsilon ϵ互换,得:

η α β ∂ 2 x ′ α ∂ x γ ∂ x δ ∂ x ′ β ∂ x ϵ + η α β ∂ x ′ α ∂ x γ ∂ 2 x ′ β ∂ x δ ∂ x ϵ = 0 \eta_{\alpha\beta}\dfrac{\partial^2 x'^\alpha}{\partial x^\gamma\partial x^\delta}\dfrac{\partial x'^\beta}{\partial x^\epsilon}+\eta_{\alpha\beta}\dfrac{\partial x'^\alpha}{\partial x^\gamma}\dfrac{\partial^2x'^\beta}{\partial x^\delta \partial x^\epsilon}=0 ηαβ​∂xγ∂xδ∂2x′α​∂xϵ∂x′β​+ηαβ​∂xγ∂x′α​∂xδ∂xϵ∂2x′β​=0(4)

  对(4)式的第一项,进行 α \alpha α和 β \beta β的互换,得:

η α β ∂ 2 x ′ β ∂ x γ ∂ x δ ∂ x ′ α ∂ x ϵ + η α β ∂ x ′ α ∂ x γ ∂ 2 x ′ β ∂ x δ ∂ x ϵ = 0 \eta_{\alpha\beta}\dfrac{\partial^2 x'^\beta}{\partial x^\gamma\partial x^\delta}\dfrac{\partial x'^\alpha}{\partial x^\epsilon}+\eta_{\alpha\beta}\dfrac{\partial x'^\alpha}{\partial x^\gamma}\dfrac{\partial^2x'^\beta}{\partial x^\delta \partial x^\epsilon}=0 ηαβ​∂xγ∂xδ∂2x′β​∂xϵ∂x′α​+ηαβ​∂xγ∂x′α​∂xδ∂xϵ∂2x′β​=0(4’)

  (2)+(3)-(4’),最终得到:

η α β ∂ 2 x ′ α ∂ x γ ∂ x ϵ ∂ x ′ β ∂ x δ = 0 ∂ 2 x ′ α ∂ x γ ∂ x ϵ = 0 \begin{aligned}\eta_{\alpha\beta}\dfrac{\partial^2 x'^\alpha}{\partial x^\gamma\partial x^\epsilon}\dfrac{\partial x'^\beta}{\partial x^\delta}&=0\\ \dfrac{\partial^2 x'^\alpha}{\partial x^\gamma \partial x^\epsilon}&=0\end{aligned} ηαβ​∂xγ∂xϵ∂2x′α​∂xδ∂x′β​∂xγ∂xϵ∂2x′α​​=0=0​

  上式的通解即为线性函数-定义中的(1)式,且注意到当定义中的(1)式代入到下式时:

η γ δ = η α β ∂ x ′ α ∂ x γ ∂ x ′ β ∂ x δ \eta_{\gamma\delta}=\eta_{\alpha\beta}\dfrac{\partial x'^\alpha}{\partial x^\gamma}\dfrac{\partial x'^\beta}{\partial x^\delta} ηγδ​=ηαβ​∂xγ∂x′α​∂xδ∂x′β​

   Λ β α \Lambda^\alpha_\beta Λβα​必然符合定义中的(2)式,由此证明是Lorentz变换(通解满足定义1式,而 Λ β α \Lambda^\alpha_\beta Λβα​又满足了定义2式,由此看来,这满足了Lorentz的定义,就是Lorentz变换了,同时也证明了唯一性)

QED.


  (6)对于光速运动的粒子:

d τ 2 = d t 2 [ 1 − ( d x → d t ) 2 ] = d t 2 ( 1 − 1 ) = 0 d\tau^2=dt^2\left[1-\left(\dfrac{d\overrightarrow x}{dt}\right)^2\right]=dt^2(1-1)=0 dτ2=dt2⎣⎡​1−(dtdx ​)2⎦⎤​=dt2(1−1)=0

  即 d τ = 0 d\tau=0 dτ=0

  (7)对满足Lorentz方程的所有Lorentz变换集合,组成非齐次的Lorentz群(或称为Poincare群)。若取 a α = 0 a^\alpha=0 aα=0,则称齐次的Lorentz群,简称Lorentz群

  (8)如果是正的Lorentz,则需要加上附加条件(之后如无特殊情况,讨论的将会都是正Lorentz群):

Λ 0 0 ≥ 1 ,   ∣ Λ ∣ = + 1 \Lambda^0_0\ge 1,\ |\Lambda|=+1 Λ00​≥1, ∣Λ∣=+1

  【1】:一般的 Λ 0 0 \Lambda^0_0 Λ00​

  Lorentz方程 γ = δ = 0 \gamma=\delta=0 γ=δ=0

   Λ 0 α Λ 0 β η α β = η 00 = − 1 \Lambda^\alpha_0\Lambda^\beta_0\eta_{\alpha\beta}=\eta_{00}=-1 Λ0α​Λ0β​ηαβ​=η00​=−1

   ( Λ 0 0 ) 2 = 1 + ∑ i = 1 , 2 , 3 ( Λ 0 i ) 2 ≥ 1 (\Lambda^0_0)^2=1+\sum\limits_{i=1,2,3}(\Lambda^i_0)^2\ge 1 (Λ00​)2=1+i=1,2,3∑​(Λ0i​)2≥1

  由上述不等式,有:

   { Λ 0 0 ≥ 1 Λ 0 0 ≤ − 1 \left\{\begin{aligned}&\Lambda^0_0\ge 1\\ &\Lambda_0^0\le -1\end{aligned}\right. {​Λ00​≥1Λ00​≤−1​

  只取第一种情况

  【2】:一般的 Λ \Lambda Λ的行列式

  把Lorentz方程写成矩阵形式

∣ η ∣ = ∣ Λ T η Λ ∣ |\eta|=|\Lambda^T\eta\Lambda| ∣η∣=∣ΛTηΛ∣

  得到: ∣ Λ ∣ 2 = 1 |\Lambda|^2=1 ∣Λ∣2=1, ∣ Λ ∣ = ± 1 |\Lambda|=\pm 1 ∣Λ∣=±1,只取 ∣ Λ ∣ = + 1 |\Lambda|=+1 ∣Λ∣=+1,此即正Lorentz群

  【3】正Lorentz群 Λ β α \Lambda^\alpha_\beta Λβα​可以通过连续参数变换,变换到单位元素 δ β α \delta^\alpha_\beta δβα​,反之亦然

  【4】正Lorentz变换,不包括空间反演变换( ∣ Λ ∣ = − 1 |\Lambda|=-1 ∣Λ∣=−1, Λ 0 0 ≥ 1 \Lambda^0_0\ge 1 Λ00​≥1),也不包括时间反演变换( ∣ Λ ∣ = − 1 |\Lambda|=-1 ∣Λ∣=−1, Λ 0 0 = − 1 \Lambda^0_0=-1 Λ00​=−1)

洛伦兹变换的简单分类

纯转动(以绕z轴转动为例)

Λ β α = ( 1 0 0 0 0 cos ⁡ θ sin ⁡ θ 0 0 − sin ⁡ θ cos ⁡ θ 0 0 0 0 0 ) \Lambda^\alpha_\beta=\left(\begin{matrix}1 & 0 & 0 & 0\\ 0 & \cos\theta & \sin\theta & 0\\ 0 & -\sin\theta & \cos\theta & 0\\ 0 & 0 & 0 & 0\end{matrix}\right) Λβα​=⎝⎜⎜⎛​1000​0cosθ−sinθ0​0sinθcosθ0​0000​⎠⎟⎟⎞​

推动(boosts)

  两个坐标系 ∑ \sum ∑、 ∑ ′ \sum' ∑′之间有相对运动,其中 ∑ ′ \sum' ∑′系相对于 ∑ \sum ∑系的速度为 v → \overrightarrow v v 。考虑速度 v → \overrightarrow v v 沿x轴方向,则:

Λ β α ( b o o s t ) = ( cosh ⁡ ϕ − sinh ⁡ ϕ 0 0 − sinh ⁡ ϕ cosh ⁡ ϕ 0 0 0 0 1 0 0 0 0 1 ) \Lambda^\alpha_\beta(boost)=\left(\begin{matrix}\cosh\phi & -\sinh\phi & 0 & 0\\ -\sinh\phi & \cosh\phi& 0 & 0\\ 0 & 0 & 1& 0\\ 0 & 0 & 0 & 1\end{matrix}\right) Λβα​(boost)=⎝⎜⎜⎛​coshϕ−sinhϕ00​−sinhϕcoshϕ00​0010​0001​⎠⎟⎟⎞​

  使用此变换后:

{ t ′ = t cosh ⁡ ϕ − x sinh ⁡ ϕ x ′ = − t sinh ⁡ ϕ + x cosh ⁡ ϕ \left\{\begin{aligned}t'&=t\cosh\phi-x\sinh\phi\\ x'&=-t\sinh\phi+x\cosh\phi\end{aligned}\right. {t′x′​=tcoshϕ−xsinhϕ=−tsinhϕ+xcoshϕ​

  令 tanh ⁡ ϕ = v \tanh\phi=v tanhϕ=v,则 cosh ⁡ ϕ = 1 1 − v 2 \cosh\phi=\dfrac{1}{\sqrt{1-v^2}} coshϕ=1−v2 ​1​, sinh ⁡ ϕ = v 1 − v 2 \sinh\phi=\dfrac{v}{\sqrt{1-v^2}} sinhϕ=1−v2 ​v​,则可以得到:

{ t ′ = t − v x 1 − v 2 x ′ = x − v t 1 − v 2 \left\{\begin{aligned}&t'=\dfrac{t-vx}{\sqrt{1-v^2}}\\ &x'=\dfrac{x-vt}{\sqrt{1-v^2}}\end{aligned}\right. ⎩⎪⎪⎨⎪⎪⎧​​t′=1−v2 ​t−vx​x′=1−v2 ​x−vt​​

  通常定义 γ \gamma γ因子为 γ ≡ ( 1 − v 2 ) − 1 / 2 \gamma\equiv(1-v^2)^{-1/2} γ≡(1−v2)−1/2,则boost可以写为如下形式:

Λ β α ( b o o s t ) = ( γ − v γ 0 0 − v γ γ 0 0 0 0 1 0 0 0 0 1 ) \Lambda^\alpha_\beta(boost)=\left(\begin{matrix}\gamma & -v\gamma & 0 & 0\\ -v\gamma & \gamma& 0 & 0\\ 0 & 0 & 1& 0\\ 0 & 0 & 0 & 1\end{matrix}\right) Λβα​(boost)=⎝⎜⎜⎛​γ−vγ00​−vγγ00​0010​0001​⎠⎟⎟⎞​

  还可以写为:

Λ 0 0 = γ ,   Λ 0 i = − γ v i ,   Λ j 0 = − γ v j ,   Λ j i = δ i j + v i v j γ − 1 v 2 \Lambda^0_0=\gamma,\ \Lambda^i_0=-\gamma v_i,\ \Lambda^0_j=-\gamma v_j,\ \Lambda^i_j=\delta_{ij}+v_iv_j\dfrac{\gamma-1}{v^2} Λ00​=γ, Λ0i​=−γvi​, Λj0​=−γvj​, Λji​=δij​+vi​vj​v2γ−1​

  任何正齐次Lorentz变换都可以表示为推动和转动的乘积

以上内容记录于2021-9-18

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