§A11.1 梯度§A11.2 散度与高斯定理§A11.3 连续性方程§A11.4 旋度与斯托克斯定理§A11.5 拉普拉斯算子§A11.6 常用恒等式与三维分部积分自测题 正文用到本节的地方：刘维尔定理（§2.6 ，相空间中的连续性方程）、电磁波方程（§P3.4 ）、扩散方程与输运（§P5.2 、第25章 ）、福克–普朗克方程中的概率流（§23.4 ）、金兹堡–朗道理论中的梯度项与关联函数（§19.4 ）、卡恩–希利亚德方程（§31.6 ）。
下面用 ∇ = ( ∂ x , ∂ y , ∂ z ) \nabla = (\partial_x,\partial_y,\partial_z) ∇ = ( ∂ x , ∂ y , ∂ z ) 2 f 2f 2 f n n n (A11.4) 、(A11.11) 中的 4 π 4\pi 4 π
§A11.1 梯度 标量场 f ( r ) f(\mathbf r) f ( r )
∇ f = ( ∂ f ∂ x , ∂ f ∂ y , ∂ f ∂ z ) , d f = ∇ f ⋅ d r (A11.1) \nabla f = \left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z}\right),\qquad df = \nabla f\cdot d\mathbf r \tag{A11.1} ∇ f = ( ∂ x ∂ f , ∂ y ∂ f , ∂ z ∂ f ) , df = ∇ f ⋅ d r ( A11.1 ) 第二式就是全微分 (A1.2) 。由它可知：沿单位向量 n \mathbf n n n ⋅ ∇ f \mathbf n\cdot\nabla f n ⋅ ∇ f n \mathbf n n ∇ f \nabla f ∇ f 梯度指向 f f f ，大小是这个最大变化率；沿等值面移动时 d f = 0 df = 0 df = 0 梯度垂直于等值面 。
物理中的梯度：保守力 F = − ∇ V \mathbf F = -\nabla V F = − ∇ V J Q = − κ ∇ T \mathbf J_Q = -\kappa\nabla T J Q = − κ ∇ T J n = − D ∇ n \mathbf J_n = -D\nabla n J n = − D ∇ n §P5.2 ）。"流沿着梯度的反方向"是线性输运的共同形式。
常用梯度 （r = ∣ r ∣ r = \lvert\mathbf r\rvert r = ∣ r ∣ r ^ = r / r \hat{\mathbf r} = \mathbf r/r r ^ = r / r ∇ r = r ^ \nabla r = \hat{\mathbf r} ∇ r = r ^ ∇ f ( r ) = f ′ ( r ) r ^ \nabla f(r) = f'(r)\,\hat{\mathbf r} ∇ f ( r ) = f ′ ( r ) r ^ ∇ ( 1 / r ) = − r ^ / r 2 \nabla(1/r) = -\hat{\mathbf r}/r^2 ∇ ( 1/ r ) = − r ^ / r 2 ∇ ( k ⋅ r ) = k \nabla(\mathbf k\cdot\mathbf r) = \mathbf k ∇ ( k ⋅ r ) = k ∇ e i k ⋅ r = i k e i k ⋅ r \nabla e^{i\mathbf k\cdot\mathbf r} = i\mathbf k\,e^{i\mathbf k\cdot\mathbf r} ∇ e i k ⋅ r = i k e i k ⋅ r ∇ \nabla ∇ i k i\mathbf k i k 附录 A14 ）。
§A11.2 散度与高斯定理 矢量场 v \mathbf v v
∇ ⋅ v = ∂ v x ∂ x + ∂ v y ∂ y + ∂ v z ∂ z (A11.2) \nabla\cdot\mathbf v = \frac{\partial v_x}{\partial x} + \frac{\partial v_y}{\partial y} + \frac{\partial v_z}{\partial z} \tag{A11.2} ∇ ⋅ v = ∂ x ∂ v x + ∂ y ∂ v y + ∂ z ∂ v z ( A11.2 ) 含义：单位体积的净流出量。 取一个边长为 d x , d y , d z dx,dy,dz d x , d y , d z x x x [ v x ( x + d x ) − v x ( x ) ] d y d z ≈ ∂ x v x d x d y d z [v_x(x+dx) - v_x(x)]\,dy\,dz\approx\partial_xv_x\,dx\,dy\,dz [ v x ( x + d x ) − v x ( x )] d y d z ≈ ∂ x v x d x d y d z ( ∇ ⋅ v ) d V (\nabla\cdot\mathbf v)\,dV ( ∇ ⋅ v ) d V V V V 高斯定理 ：
∫ V ∇ ⋅ v d V = ∮ ∂ V v ⋅ d A (A11.3) \int_V\nabla\cdot\mathbf v\,dV = \oint_{\partial V}\mathbf v\cdot d\mathbf A \tag{A11.3} ∫ V ∇ ⋅ v d V = ∮ ∂ V v ⋅ d A ( A11.3 ) （d A d\mathbf A d A
例 ：∇ ⋅ r = 3 \nabla\cdot\mathbf r = 3 ∇ ⋅ r = 3 d d d d d d v = g ( r ) r \mathbf v = g(r)\,\mathbf r v = g ( r ) r ∇ ⋅ v = 3 g + r ⋅ ∇ g = 3 g + r g ′ ( r ) \nabla\cdot\mathbf v = 3g + \mathbf r\cdot\nabla g = 3g + rg'(r) ∇ ⋅ v = 3 g + r ⋅ ∇ g = 3 g + r g ′ ( r ) g = r − 3 g = r^{-3} g = r − 3 v = r ^ / r 2 \mathbf v = \hat{\mathbf r}/r^2 v = r ^ / r 2 ∇ ⋅ v = 3 r − 3 − 3 r − 3 = 0 \nabla\cdot\mathbf v = 3r^{-3} - 3r^{-3} = 0 ∇ ⋅ v = 3 r − 3 − 3 r − 3 = 0 r ≠ 0 r\ne0 r = 0 1 r 2 ⋅ 4 π r 2 = 4 π \frac{1}{r^2}\cdot4\pi r^2 = 4\pi r 2 1 ⋅ 4 π r 2 = 4 π
∇ ⋅ r ^ r 2 = 4 π δ 3 ( r ) (A11.4) \nabla\cdot\frac{\hat{\mathbf r}}{r^2} = 4\pi\,\delta^3(\mathbf r) \tag{A11.4} ∇ ⋅ r 2 r ^ = 4 π δ 3 ( r ) ( A11.4 ) （δ 函数见附录 A14 。）这就是点电荷的库仑场满足高斯定律的数学表述。
§A11.3 连续性方程 设某种守恒的量（粒子数、能量、电荷、概率……）的密度为 ρ ( r , t ) \rho(\mathbf r,t) ρ ( r , t ) J \mathbf J J v \mathbf v v J = ρ v \mathbf J = \rho\mathbf v J = ρ v V V V
d d t ∫ V ρ d V = − ∮ ∂ V J ⋅ d A = − ∫ V ∇ ⋅ J d V \frac{d}{dt}\int_V\rho\,dV = -\oint_{\partial V}\mathbf J\cdot d\mathbf A = -\int_V\nabla\cdot\mathbf J\,dV d t d ∫ V ρ d V = − ∮ ∂ V J ⋅ d A = − ∫ V ∇ ⋅ J d V （第二步用了 (A11.3) 。）由于 V V V
∂ ρ ∂ t + ∇ ⋅ J = 0 (A11.5) \frac{\partial\rho}{\partial t} + \nabla\cdot\mathbf J = 0 \tag{A11.5} ∂ t ∂ ρ + ∇ ⋅ J = 0 ( A11.5 ) 物质导数 。当 J = ρ v \mathbf J = \rho\mathbf v J = ρ v ∇ ⋅ ( ρ v ) = v ⋅ ∇ ρ + ρ ∇ ⋅ v \nabla\cdot(\rho\mathbf v) = \mathbf v\cdot\nabla\rho + \rho\nabla\cdot\mathbf v ∇ ⋅ ( ρ v ) = v ⋅ ∇ ρ + ρ ∇ ⋅ v (A11.5) 可以写成
D ρ D t ≡ ∂ ρ ∂ t + v ⋅ ∇ ρ = − ρ ∇ ⋅ v (A11.6) \frac{D\rho}{Dt}\equiv\frac{\partial\rho}{\partial t} + \mathbf v\cdot\nabla\rho = -\rho\,\nabla\cdot\mathbf v \tag{A11.6} D t D ρ ≡ ∂ t ∂ ρ + v ⋅ ∇ ρ = − ρ ∇ ⋅ v ( A11.6 ) D ρ / D t D\rho/Dt D ρ / D t ρ \rho ρ (A1.3) ：d d t ρ ( r ( t ) , t ) = ∂ t ρ + r ˙ ⋅ ∇ ρ \frac{d}{dt}\rho(\mathbf r(t),t) = \partial_t\rho + \dot{\mathbf r}\cdot\nabla\rho d t d ρ ( r ( t ) , t ) = ∂ t ρ + r ˙ ⋅ ∇ ρ ∇ ⋅ v = 0 \nabla\cdot\mathbf v = 0 ∇ ⋅ v = 0 不可压缩 流动），跟随运动时密度不变。
正文中的连续性方程 ：
刘维尔定理 （§2.6 ）。相空间的"速度"是 ( q ˙ i , p ˙ i ) = ( ∂ H / ∂ p i , − ∂ H / ∂ q i ) (\dot q_i,\dot p_i) = (\partial\mathcal H/\partial p_i,\,-\partial\mathcal H/\partial q_i) ( q ˙ i , p ˙ i ) = ( ∂ H / ∂ p i , − ∂ H / ∂ q i ) ∑ i [ ∂ 2 H ∂ q i ∂ p i − ∂ 2 H ∂ p i ∂ q i ] = 0 \sum_i\left[\frac{\partial^2\mathcal H}{\partial q_i\partial p_i} - \frac{\partial^2\mathcal H}{\partial p_i\partial q_i}\right] = 0 ∑ i [ ∂ q i ∂ p i ∂ 2 H − ∂ p i ∂ q i ∂ 2 H ] = 0 (A2.4) ）。哈密顿流是不可压缩的，由 (A11.6) 即得 (2.22) 。若有摩擦（p ˙ = − ∂ H / ∂ q − γ p \dot p = -\partial\mathcal H/\partial q - \gamma p p ˙ = − ∂ H / ∂ q − γ p − f γ < 0 -f\gamma<0 − f γ < 0 第23章 ）。扩散 （§P5.2 ）：J = − D ∇ n \mathbf J = -D\nabla n J = − D ∇ n (A11.5) ，得 ∂ n / ∂ t = D ∇ 2 n \partial n/\partial t = D\nabla^2n ∂ n / ∂ t = D ∇ 2 n 附录 A15 ）。福克–普朗克方程 （§23.4 ）：(23.11) 就是一维的 ∂ P / ∂ t + ∂ J / ∂ x = 0 \partial P/\partial t + \partial J/\partial x = 0 ∂ P / ∂ t + ∂ J / ∂ x = 0 J J J (23.12) 给出，包含漂移与扩散两部分。卡恩–希利亚德方程 （§31.6 ）：守恒的组分 ϕ \phi ϕ ∂ ϕ / ∂ t = − ∇ ⋅ J \partial\phi/\partial t = -\nabla\cdot\mathbf J ∂ ϕ / ∂ t = − ∇ ⋅ J J = − M c ∇ μ \mathbf J = -M_{\mathrm c}\nabla\mu J = − M c ∇ μ
§A11.4 旋度与斯托克斯定理 ∇ × v = ( ∂ y v z − ∂ z v y , ∂ z v x − ∂ x v z , ∂ x v y − ∂ y v x ) (A11.7) \nabla\times\mathbf v = \left(\partial_yv_z - \partial_zv_y,\ \partial_zv_x - \partial_xv_z,\ \partial_xv_y - \partial_yv_x\right) \tag{A11.7} ∇ × v = ( ∂ y v z − ∂ z v y , ∂ z v x − ∂ x v z , ∂ x v y − ∂ y v x ) ( A11.7 ) 含义：单位面积的环量。 绕法线为 n \mathbf n n ∮ v ⋅ d l ≈ ( ∇ × v ) ⋅ n d A \oint\mathbf v\cdot d\mathbf l\approx(\nabla\times\mathbf v)\cdot\mathbf n\,dA ∮ v ⋅ d l ≈ ( ∇ × v ) ⋅ n d A 斯托克斯定理 ：
∮ C v ⋅ d l = ∫ S ( ∇ × v ) ⋅ d A (A11.8) \oint_C\mathbf v\cdot d\mathbf l = \int_S(\nabla\times\mathbf v)\cdot d\mathbf A \tag{A11.8} ∮ C v ⋅ d l = ∫ S ( ∇ × v ) ⋅ d A ( A11.8 ) 二维版本就是附录 A2 中的格林公式 (A2.6) 。
无旋场与势 。∇ × ( ∇ f ) = 0 \nabla\times(\nabla f) = 0 ∇ × ( ∇ f ) = 0 x x x ∂ y ∂ z f − ∂ z ∂ y f = 0 \partial_y\partial_zf - \partial_z\partial_yf = 0 ∂ y ∂ z f − ∂ z ∂ y f = 0 ∇ × F = 0 \nabla\times\mathbf F = 0 ∇ × F = 0 F = − ∇ V \mathbf F = -\nabla V F = − ∇ V 附录 A2 中"恰当微分"判据 (A2.7) 的矢量说法：∇ × F = 0 \nabla\times\mathbf F = 0 ∇ × F = 0 ∂ F i / ∂ x j = ∂ F j / ∂ x i \partial F_i/\partial x_j = \partial F_j/\partial x_i ∂ F i / ∂ x j = ∂ F j / ∂ x i ∇ ⋅ ( ∇ × v ) = 0 \nabla\cdot(\nabla\times\mathbf v) = 0 ∇ ⋅ ( ∇ × v ) = 0
§A11.5 拉普拉斯算子 ∇ 2 f = ∇ ⋅ ∇ f = ∂ 2 f ∂ x 2 + ∂ 2 f ∂ y 2 + ∂ 2 f ∂ z 2 (A11.9) \nabla^2f = \nabla\cdot\nabla f = \frac{\partial^2f}{\partial x^2} + \frac{\partial^2f}{\partial y^2} + \frac{\partial^2f}{\partial z^2} \tag{A11.9} ∇ 2 f = ∇ ⋅ ∇ f = ∂ x 2 ∂ 2 f + ∂ y 2 ∂ 2 f + ∂ z 2 ∂ 2 f ( A11.9 ) 它衡量 f f f (A3.6) ，f ( x + h ) + f ( x − h ) − 2 f ( x ) h 2 ≈ ∂ x 2 f \frac{f(x+h) + f(x-h) - 2f(x)}{h^2}\approx\partial_x^2f h 2 f ( x + h ) + f ( x − h ) − 2 f ( x ) ≈ ∂ x 2 f ∇ 2 f > 0 \nabla^2f>0 ∇ 2 f > 0 ∂ t n = D ∇ 2 n \partial_tn = D\nabla^2n ∂ t n = D ∇ 2 n ∇ 2 e i k ⋅ r = − k 2 e i k ⋅ r \nabla^2e^{i\mathbf k\cdot\mathbf r} = -k^2e^{i\mathbf k\cdot\mathbf r} ∇ 2 e i k ⋅ r = − k 2 e i k ⋅ r
球对称函数的拉普拉斯 。对 f ( r ) f(r) f ( r ) ∇ f = f ′ r r \nabla f = \frac{f'}{r}\mathbf r ∇ f = r f ′ r §A11.2 的公式（g = f ′ / r g = f'/r g = f ′ / r
∇ 2 f = 3 f ′ r + r ( f ′ r ) ′ = f ′ ′ + 2 r f ′ = 1 r d 2 ( r f ) d r 2 (A11.10) \nabla^2f = 3\frac{f'}{r} + r\left(\frac{f'}{r}\right)' = f'' + \frac2rf' = \frac1r\frac{d^2(rf)}{dr^2} \tag{A11.10} ∇ 2 f = 3 r f ′ + r ( r f ′ ) ′ = f ′′ + r 2 f ′ = r 1 d r 2 d 2 ( r f ) ( A11.10 ) （d d d f ′ ′ + d − 1 r f ′ f'' + \frac{d-1}{r}f' f ′′ + r d − 1 f ′
f = 1 / r f = 1/r f = 1/ r r f = 1 rf = 1 r f = 1 ∇ 2 ( 1 / r ) = 0 \nabla^2(1/r) = 0 ∇ 2 ( 1/ r ) = 0 r ≠ 0 r\ne0 r = 0 (A11.4) ，∇ 2 1 r = ∇ ⋅ ( − r ^ r 2 ) = − 4 π δ 3 ( r ) \nabla^2\frac1r = \nabla\cdot\left(-\frac{\hat{\mathbf r}}{r^2}\right) = -4\pi\delta^3(\mathbf r) ∇ 2 r 1 = ∇ ⋅ ( − r 2 r ^ ) = − 4 π δ 3 ( r )
∇ 2 1 4 π r = − δ 3 ( r ) (A11.11) \nabla^2\frac{1}{4\pi r} = -\delta^3(\mathbf r) \tag{A11.11} ∇ 2 4 π r 1 = − δ 3 ( r ) ( A11.11 )
f = e − r / ξ / r f = e^{-r/\xi}/r f = e − r / ξ / r r f = e − r / ξ rf = e^{-r/\xi} r f = e − r / ξ ∇ 2 f = f / ξ 2 \nabla^2f = f/\xi^2 ∇ 2 f = f / ξ 2 r ≠ 0 r\ne0 r = 0 r → 0 r\to0 r → 0 f ≈ 1 / r f\approx1/r f ≈ 1/ r 1 / r 1/r 1/ r ( − ∇ 2 + ξ − 2 ) e − r / ξ 4 π r = δ 3 ( r ) (-\nabla^2 + \xi^{-2})\frac{e^{-r/\xi}}{4\pi r} = \delta^3(\mathbf r) ( − ∇ 2 + ξ − 2 ) 4 π r e − r / ξ = δ 3 ( r ) §19.4 中验证奥恩斯坦–泽尼克解 (19.13) 的步骤。d d d ∇ 2 r 2 − d = 0 \nabla^2r^{2-d} = 0 ∇ 2 r 2 − d = 0 r ≠ 0 r\ne0 r = 0 ( 2 − d ) ( 1 − d ) r − d + ( d − 1 ) ( 2 − d ) r − d = 0 (2-d)(1-d)r^{-d} + (d-1)(2-d)r^{-d} = 0 ( 2 − d ) ( 1 − d ) r − d + ( d − 1 ) ( 2 − d ) r − d = 0 d d d r 2 − d r^{2-d} r 2 − d d = 2 d = 2 d = 2 ln r \ln r ln r G ∝ r − ( d − 2 + η ) G\propto r^{-(d-2+\eta)} G ∝ r − ( d − 2 + η ) d − 2 d - 2 d − 2
§A11.6 常用恒等式与三维分部积分 ∇ ⋅ ( f v ) = f ∇ ⋅ v + v ⋅ ∇ f , ∇ × ( ∇ × v ) = ∇ ( ∇ ⋅ v ) − ∇ 2 v (A11.12) \nabla\cdot(f\mathbf v) = f\,\nabla\cdot\mathbf v + \mathbf v\cdot\nabla f,\qquad \nabla\times(\nabla\times\mathbf v) = \nabla(\nabla\cdot\mathbf v) - \nabla^2\mathbf v \tag{A11.12} ∇ ⋅ ( f v ) = f ∇ ⋅ v + v ⋅ ∇ f , ∇ × ( ∇ × v ) = ∇ ( ∇ ⋅ v ) − ∇ 2 v ( A11.12 ) 第二式的验证（x x x [ ∇ × ( ∇ × v ) ] x = ∂ y ( ∂ x v y − ∂ y v x ) − ∂ z ( ∂ z v x − ∂ x v z ) = ∂ x ( ∂ y v y + ∂ z v z ) − ( ∂ y 2 + ∂ z 2 ) v x [\nabla\times(\nabla\times\mathbf v)]_x = \partial_y(\partial_xv_y - \partial_yv_x) - \partial_z(\partial_zv_x - \partial_xv_z) = \partial_x(\partial_yv_y + \partial_zv_z) - (\partial_y^2 + \partial_z^2)v_x [ ∇ × ( ∇ × v ) ] x = ∂ y ( ∂ x v y − ∂ y v x ) − ∂ z ( ∂ z v x − ∂ x v z ) = ∂ x ( ∂ y v y + ∂ z v z ) − ( ∂ y 2 + ∂ z 2 ) v x ∂ x 2 v x \partial_x^2v_x ∂ x 2 v x ∂ x ( ∇ ⋅ v ) − ∇ 2 v x \partial_x(\nabla\cdot\mathbf v) - \nabla^2v_x ∂ x ( ∇ ⋅ v ) − ∇ 2 v x §P3.4 由它从麦克斯韦方程推出波动方程 (P3.10) 。
三维分部积分 。在第一式中取 v = ∇ g \mathbf v = \nabla g v = ∇ g
∫ V f ∇ 2 g d V = ∮ ∂ V f ∇ g ⋅ d A − ∫ V ∇ f ⋅ ∇ g d V (A11.13) \int_Vf\,\nabla^2g\,dV = \oint_{\partial V}f\,\nabla g\cdot d\mathbf A - \int_V\nabla f\cdot\nabla g\,dV \tag{A11.13} ∫ V f ∇ 2 g d V = ∮ ∂ V f ∇ g ⋅ d A − ∫ V ∇ f ⋅ ∇ g d V ( A11.13 ) 若边界项为零（函数在无穷远处衰减，或采用周期性边界条件），则 ∫ f ∇ 2 g = − ∫ ∇ f ⋅ ∇ g \int f\nabla^2g = -\int\nabla f\cdot\nabla g ∫ f ∇ 2 g = − ∫ ∇ f ⋅ ∇ g ∫ f g ′ ′ = − ∫ f ′ g ′ \int fg'' = -\int f'g' ∫ f g ′′ = − ∫ f ′ g ′ §19.4 中 δ ∫ g ∣ ∇ m ∣ 2 = − ∫ 2 g ( ∇ 2 m ) δ m \delta\int g\lvert\nabla m\rvert^2 = -\int2g(\nabla^2m)\,\delta m δ ∫ g ∣ ∇ m ∣ 2 = − ∫ 2 g ( ∇ 2 m ) δ m (A11.13) ；附录 A12 中的泛函导数也要用到它。
自测题
计算 ∇ ⋅ ( r ^ / r n ) \nabla\cdot(\hat{\mathbf r}/r^n) ∇ ⋅ ( r ^ / r n ) r ≠ 0 r\ne0 r = 0 ［答：( 2 − n ) / r n + 1 (2-n)/r^{n+1} ( 2 − n ) / r n + 1 n = 2 n = 2 n = 2
验证 ∇ 2 ( e − r / ξ / r ) = e − r / ξ / ( ξ 2 r ) \nabla^2(e^{-r/\xi}/r) = e^{-r/\xi}/(\xi^2r) ∇ 2 ( e − r / ξ / r ) = e − r / ξ / ( ξ 2 r ) r ≠ 0 r\ne0 r = 0
对 H = p 2 / 2 m + V ( q ) \mathcal H = p^2/2m + V(q) H = p 2 /2 m + V ( q ) ( p / m , − V ′ ( q ) ) (p/m,\,-V'(q)) ( p / m , − V ′ ( q )) p ˙ = − V ′ − γ p \dot p = -V' - \gamma p p ˙ = − V ′ − γ p ［答：− γ -\gamma − γ
扩散发生在一个封闭容器中，器壁上 J ⋅ d A = 0 \mathbf J\cdot d\mathbf A = 0 J ⋅ d A = 0 (A11.5) 与高斯定理证明容器内的总粒子数不变。
证明 ∇ ⋅ ( ∇ × v ) = 0 \nabla\cdot(\nabla\times\mathbf v) = 0 ∇ ⋅ ( ∇ × v ) = 0 ［提示：写开后每一项都以"混合偏导数之差"的形式出现。］
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