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Lesson 13 作用量原理

约 1811 字大约 6 分钟

2026-04-08

作用量原理：我们的目的是测地线方程，

d2xμdτ2+Γμνλdxνdτdxλdτ=0\frac{\text{d}^2x^\mu}{\text{d}\tau^2}+\Gamma^\mu{}_{\nu\lambda}\frac{\text{d}x^\nu}{\text{d}\tau}\frac{\text{d}x^\lambda}{\text{d}\tau} = 0

粒子的作用量表达为

IM=−∑nmn∫−∞∞dp[−gμν(xn)dxnμdpdxnνdp]1/2I_M = -\sum_nm_n\int_{-\infty}^\infty\text{d}p\left[-g_{\mu\nu}(x_n)\frac{\text{d}x_n^\mu}{\text{d}p}\frac{\text{d}x_n^\nu}{\text{d}p}\right]^{1/2}

我们想通过其变分得到测地线方程，有

δIM=−∑nmn∫−∞∞dp⋅12[−gμν(xn)dxnμdpdxnνdp]−1/2{−∂gμν∂xnλδxnλdxnμdpdxnνdp−gμν(xn)(ddpδxnμddpxnν+ddpxnμddpδxnν)}=−∑nmn∫−∞∞dτ⋅12{−∂gμν∂xnλδxnλdxnμdpdxnνdp−gμν(xn)(ddpδxnμddpxnν+ddpxnμddpδxnν)}=⋯⋯\begin{aligned} \delta I_M &=-\sum_nm_n\int_{-\infty}^\infty\text{d}p\cdot\frac{1}{2}\left[-g_{\mu\nu}(x_n)\frac{\text{d}x_n^\mu}{\text{d}p}\frac{\text{d}x_n^\nu}{\text{d}p}\right]^{-1/2}\Big\{-\frac{\partial g_{\mu\nu}}{\partial x_n^\lambda}\delta x_n^\lambda\frac{\text{d}x_n^\mu}{\text{d}p}\frac{\text{d}x_n^\nu}{\text{d}p}\\\\ &\quad -g_{\mu\nu}(x_n)\left(\frac{\text{d}}{\text{d}p}\delta x_n^\mu\frac{\text{d}}{\text{d}p}x_n^\nu+\frac{\text{d}}{\text{d}p}x_n^\mu\frac{\text{d}}{\text{d}p}\delta x_n^\nu\right) \Big\}\\\\ &= -\sum_nm_n\int_{-\infty}^\infty\text{d}\tau\cdot\frac{1}{2}\Big\{-\frac{\partial g_{\mu\nu}}{\partial x_n^\lambda}\delta x_n^\lambda\frac{\text{d}x_n^\mu}{\text{d}p}\frac{\text{d}x_n^\nu}{\text{d}p}-g_{\mu\nu}(x_n)\left(\frac{\text{d}}{\text{d}p}\delta x_n^\mu\frac{\text{d}}{\text{d}p}x_n^\nu+\frac{\text{d}}{\text{d}p}x_n^\mu\frac{\text{d}}{\text{d}p}\delta x_n^\nu\right) \Big\}\\\\ &= \cdots\cdots \end{aligned}

后面过程略，但是最后会得到 Christoffel 符号的定义式.

对于一个带电的体系，上面的作用量加一项场的部分和一项电荷之间相互作用的部分. 其中场的部分是

−14∫d4x⋅g1/2(x)Fμν(x)Fμν(x)-\frac{1}{4}\int\text{d}^4x\cdot g^{1/2}(x)F_{\mu\nu}(x)F^{\mu\nu}(x)

这里的 g1/2g^{1/2} 来源于张量密度，d4x\text{d}^4x 不是普通标量.

为求电荷相互作用的项，首先写出狭义相对论下的流：

Jμ(x)=∑nendxnμdtδ3(x−xn(t))=∑nendxnμdτdτdtδ3(x−xn(t))J^\mu(x) = \sum_ne_n\frac{\text{d}x_n^\mu}{\text{d}t}\delta^3(x-x_n(t)) = \sum_ne_n\frac{\text{d}x_n^\mu}{\text{d}\tau}\frac{\text{d}\tau}{\text{d}t}\delta^3(x-x_n(t))

简单地把 33 改成 44 就是四维形式，但是并不广义相对论协变，因为 δ4\delta^4 并不是标量，还需要乘一个张量密度的系数，

Jμ(x)=∑nen∫dτ⋅g1/2(xn(τ))dxnμdτδ4(x−xn(τ))J^\mu(x) = \sum_ne_n\int\text{d}\tau\cdot g^{1/2}(x_n(\tau))\frac{\text{d}x_n^\mu}{\text{d}\tau}\delta^4(x-x_n(\tau))

由流，得到粒子之间的电磁相互作用，

∫d4x⋅g1/2(x)Aμ(x)Jμ(x)=∫dτ∫d4x∑nδ4(x−xn(τ))Aμ(x)endxnμdτ=∑nen∫−∞∞dpdxnμ(p)dpAμ(xn(p))\begin{aligned} &\int\text{d}^4x\cdot g^{1/2}(x)A_\mu(x)J^\mu(x)\\\\ =&\int\text{d}\tau\int\text{d}^4x\sum_n\delta^4(x-x_n(\tau))A_\mu(x)e_n\frac{\text{d}x_n^\mu}{\text{d}\tau}\\\\ =&\sum_ne_n\int_{-\infty}^\infty\text{d}p\frac{\text{d}x_n^\mu(p)}{\text{d}p}A_\mu(x_n(p)) \end{aligned}

相对于之前无电荷的作用量，这里多了两项：

IM=−∑nmn∫−∞∞dp[−gμν(xn)dxnμdpdxnνdp]1/2−14∫d4x⋅g1/2(x)Fμν(x)Fμν(x)−∑nen∫−∞∞dpdxnμ(p)dpAμ(xn(p))\begin{aligned} I_M &=-\sum_nm_n\int_{-\infty}^\infty\text{d}p\left[-g_{\mu\nu}(x_n)\frac{\text{d}x_n^\mu}{\text{d}p}\frac{\text{d}x_n^\nu}{\text{d}p}\right]^{1/2}-\frac{1}{4}\int\text{d}^4x\cdot g^{1/2}(x)F_{\mu\nu}(x)F^{\mu\nu}(x)\\\\ &\quad -\sum_ne_n\int_{-\infty}^\infty\text{d}p\frac{\text{d}x_n^\mu(p)}{\text{d}p}A_\mu(x_n(p)) \end{aligned}

变分后，第一项和第三项都有 xnμx_n^\mu，对于第一项变分之后得到测地线方程，下面考虑对第三项变分，

δI3=∑n∫−∞∞dp⋅[dδxnμdpAμ(xn(p))+dxnμdp∂Aμ∂xν∣x=xn(p)δxnμ(p)]=∑nen∫−∞∞dp(−Fμνdxnμdp)δxν+boundary terms\begin{aligned} \delta I_3 &= \sum_n\int_{-\infty}^\infty\text{d}p\cdot\left[\frac{\text{d}\delta x_n^\mu}{\text{d}p} A_\mu(x_n(p))+\frac{\text{d}x_n^\mu}{\text{d}p}\left.\frac{\partial A_\mu}{\partial x^\nu}\right|_{x=x_n(p)}\delta x_n^\mu(p)\right]\\\\ &= \sum_ne_n\int_{-\infty}^\infty\text{d}p\left(-F_{\mu\nu}\frac{\text{d}x_n^\mu}{\text{d}p}\right)\delta x^\nu+\text{boundary terms} \end{aligned}

这里的第一项可以通过 integral by part，

dδxnμdpAμ(xn(p))⟶−δxnμ∂Aμ∂xν∣x=xn(p)dxnν(p)dp\frac{\text{d}\delta x_n^\mu}{\text{d}p}A_\mu(x_n(p))\longrightarrow -\delta x_n^\mu\left.\frac{\partial A_\mu}{\partial x^\nu}\right|_{x=x_n(p)}\frac{\text{d}x_n^\nu(p)}{\text{d}p}

电磁场部分，

δIA=∫d4xg(−14FμνFμν)\begin{aligned} \delta I_A &= \int\text{d}^4x\sqrt{g}\left(-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}\right) \end{aligned}

其中后面 Fμν=∂μAν−∂νAμF^{\mu\nu}=\partial^\mu A^\nu-\partial^\nu A^\mu，是反对称的，所以可以写成 2∂μAν2\partial^\mu A^\nu. 之后对这个 AνA^\nu 做变分，

δIA=∫d4xg(−1)(∂μAν−∂νAμ)∂μδAν=∫d4xgFμν;ν\delta I_A = \int\text{d}^4x\sqrt{g}(-1)(\partial_\mu A_\nu-\partial_\nu A_\mu)\partial^\mu\delta A^\nu = \int\text{d}^4x\sqrt{g} F_{\mu\nu}{}^{;\nu}

下面用作用量原理推导场方程.

场的相互作用是 local 的，因为没有长程关联，也就没有双重的积分. 取一个标量的作用量，

IG=−116πG∫d4x⋅g1/2RI_G = -\frac{1}{16\pi G}\int\text{d}^4x\cdot g^{1/2} R

变分，

δ(g1/2R)=δ(g1/2gμνRμν)=δgμνg1/2Rμν+δg1/2R+g1/2gμνδRμν\begin{aligned} \delta\left(g^{1/2}R\right) &= \delta\left(g^{1/2}g^{\mu\nu}R_{\mu\nu}\right)\\\\ &= \delta g^{\mu\nu}g^{1/2}R_{\mu\nu}+\delta g^{1/2}R+g^{1/2}g^{\mu\nu}\delta R_{\mu\nu} \end{aligned}

最难处理的是最后这一项，因为根本不知道 RR 是什么. 但是我们知道 RμνR_{\mu\nu} 没缩并之前是由 Γ\Gamma 构成的，而且 δΓ\delta\Gamma 是 tensor (因为使得 Γ\Gamma 不是 tensor 的那一项变分的时候没有了). 因此

δRμν=something’s derivative\delta R_{\mu\nu} = \text{something's derivative}

这一项的贡献是

∫d4xg⋅DμJ~μ\int\text{d}^4x\sqrt{g}\cdot D_\mu \tilde J^\mu

由 Stokes 可以化为面积分，将面积趋于无穷大，此式得零. 因此这一项对运动方程没有贡献.

接下来处理第一项，

gνρgρσ=δμσ⟹δgμρgρσ+gμρδgρσ=0g_{\nu\rho}g^{\rho\sigma} = \delta_\mu{}^\sigma \Longrightarrow \delta g_{\mu\rho}g^{\rho\sigma}+g_{\mu\rho}\delta g^{\rho\sigma} = 0

所以 δgρσ=−gρμδgμνgνσ\delta g^{\rho\sigma}=-g^{\rho\mu}\delta g_{\mu\nu}g^{\nu\sigma}. 第二项是 δg1/2=12g1/2gμνδgμν\delta g^{1/2}=\displaystyle{\frac{1}{2}}g^{1/2}g^{\mu\nu}\delta g_{\mu\nu}. 最终得到

δIG=−116πG∫d4x(−Rρσ+12gρσR)g1/2δgρσ\delta I_G = -\frac{1}{16\pi G}\int\text{d}^4x\left(-R^{\rho\sigma}+\frac{1}{2}g^{\rho\sigma}R\right)g^{1/2}\delta g_{\rho\sigma}

这个变分要和物质场的变分合在一起等于零. 如果场方程要正确，那么物质场对 gρσg_{\rho\sigma} 变分必须得是能动张量. 这听起来很奇怪，不过这件事是正确的. 我们以一群粒子为例，

IM=−∑nmn∫−∞∞dp(−gμνdxnμdpdxnνdp)1/2I_M = -\sum_n m_n\int_{-\infty}^\infty\text{d}p\left(-g_{\mu\nu}\frac{\text{d}x_n^\mu}{\text{d}p}\frac{\text{d}x_n^\nu}{\text{d}p}\right)^{1/2}

对 gμνg_{\mu\nu} 变分，

δIM=−∑nmn∫−∞∞dp⋅12(−gμνdxnμdpdxnνdp)−1/2(−dxnμdpdxnνdpδgμν(xn))=∑nmn∫−∞∞dp⋅12dxnμdτdxnνdτδgμν(xn)=∫d4xg(x)∑nmn∫−∞∞dτ⋅12dxnμdτdxnνdτ⋅g(x)δ4(x−xn(τ))δgμν(x)\begin{aligned} \delta I_M &= -\sum_nm_n\int_{-\infty}^\infty\text{d}p\cdot\frac{1}{2}\left(-g_{\mu\nu}\frac{\text{d}x_n^\mu}{\text{d}p}\frac{\text{d}x_n^\nu}{\text{d}p}\right)^{-1/2}\left(-\frac{\text{d}x_n^\mu}{\text{d}p}\frac{\text{d}x_n^\nu}{\text{d}p}\delta g_{\mu\nu}(x_n)\right)\\\\ &=\sum_nm_n\int_{-\infty}^\infty\text{d}p\cdot\frac{1}{2}\frac{\text{d}x_n^\mu}{\text{d}\tau}\frac{\text{d}x_n^\nu}{\text{d}\tau}\delta g_{\mu\nu}(x_n)\\\\ &= \int\text{d}^4x\sqrt{g(x)}\sum_nm_n\int_{-\infty}^\infty\text{d}\tau\cdot\frac{1}{2}\frac{\text{d}x_n^\mu}{\text{d}\tau}\frac{\text{d}x_n^\nu}{\text{d}\tau}\cdot\sqrt{g(x)}\delta^4(x-x_n(\tau))\delta g_{\mu\nu}(x) \end{aligned}

而这个体系的能动张量是

Tμν(x)=∫dτ∑nmndxnμdτdxnνdτg1/2(x)δ4(x−xn(τ))T^{\mu\nu}(x) = \int\text{d}\tau\sum_nm_n\frac{\text{d}x_n^\mu}{\text{d}\tau}\frac{\text{d}x_n^\nu}{\text{d}\tau} g^{1/2}(x)\delta^4(x-x_n(\tau))

验证了结果.

提示

现在来思考一下为什么这件事情是对的.

S=∫dt⋅L(q,q˙)S = \int\text{d}t\cdot L(q,\dot{q})

的变分是

δS=∫t1t2(δLδqδq+δLδq˙δq˙)dt+L(q(t2),q˙(t2))Δt−L(q(t1),q˙(t1))‾δS2=∫t1t2[(δLδqδq−ddtδLδq˙)‾0−ddt(δLδq˙δq)]+δS2=δLδq˙δq∣t1t2+δS2=[−δLδq˙q˙+L(q(t),q˙(t))]t1t2=0\begin{aligned} \delta S &= \int_{t_1}^{t_2}\left(\frac{\delta L}{\delta q}\delta q+\frac{\delta L}{\delta \dot{q}}\delta\dot{q}\right)\text{d}t +\underset{\delta S_2}{\underline{L(q(t_2),\dot{q}(t_2))\Delta t-L(q(t_1),\dot{q}(t_1))}}\\\\ &= \int_{t_1}^{t_2}\left[\underset{0}{\underline{\left(\frac{\delta L}{\delta q}\delta q - \frac{\text{d}}{\text{d}t}\frac{\delta L}{\delta \dot{q}}\right)}}-\frac{\text{d}}{\text{d}t}\left(\frac{\delta L}{\delta \dot{q}}\delta q\right)\right]+\delta S_2\\\\ &= \left.\frac{\delta L}{\delta\dot{q}}\delta q\right|^{t_2}_{t_1}+\delta S_2\\\\ &= \left[-\frac{\delta L}{\delta \dot{q}}\dot{q}+L(q(t),\dot{q}(t))\right]^{t_2}_{t_1} = 0 \end{aligned}

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2026/4/8 14:51
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