CP03_2 Lagrange's equation

Lagrange's Equation (2)∂L d ⎛ ∂L − ⎜ ∂q j dt ⎜ ∂q j ⎝ & ⎞ ⎟=0 ⎟ ⎠pj =∂L & ∂q j& H = ∑ qii∂L −L & ∂qi分析力学的发展历程 光学 天文学牛顿力学变分法虚功原理哈密顿原理拉格朗日方程哈密顿方程History of Classical Mechanics1686 Newton: the Principia 1717 Jean Bernoulli: the Principle of Virtual Work 1743 Jean Le Rond d’Alembert: the D’Alembert Principle 1788 Joseph‐Louis Lagrange: Analytical Mechanics3Find the conserved quantity& F=P& N=L⎞ ⎟=0 ⎟ ⎠∂L d ⎛ ∂L − ⎜ & ∂q j dt ⎜ ∂q j ⎝Generalized Momentum(广义动量)∂L d ⎛ ∂L − ⎜ ∂q j dt ⎜ ∂q j ⎝ & ⎞ ⎟=0 ⎟ ⎠d ⎛ ∂L ⎜ dt ⎜ ∂q j ⎝ & ⎞ ⎟=0 ⎟ ⎠∂L =0 ∂q j• The generalized momentum∂L pj ≡ & ∂q j– canonical (正则) or conjugate(共轭) momentum∂L & = mi xi For simple x-y-z coordinates pix = & ∂xiGeneralized MomentumIn spherical coordinates （球坐标）L = T − V ( r , θ, φ)T= 1 & & & m(r 2 + r 2 θ2 + r 2 sin 2 θφ2 ) 2∂L ∂T & pr = = = mr & & ∂r ∂r∂L pj ≡ & ∂q j∂L ∂T & pθ = & = & = mr 2 θ ∂θ ∂θ角动量∂L ∂T & pφ = & = & = mr 2 sin 2 θφ ∂φ ∂φGeneralized Momentum• Generalized momentum may not look like linear momentum– Form may vary if V depends on velocity• Example: a particle in EM field1 2 L = mv − qφ + qA ⋅ v 2∂L & px = = mx + qAx & ∂x– Dimension (unit) may vary, if qj is not a space coordinate• pjqj always has the dimension of action (= work × time)Conservation Laws• pj is conserved if L does not depend explicitly on qj – Such qj is called ignorable or cyclic coordinatedp j dt − ∂L =0 ∂q j∂L =0 ∂q jdp j dt=0p j = constantThis equation is called first integral or circulatory integralConservation Laws& & mi ( yi2 + zi2 ) L = T −V = ∑ − V ( yi , zi , t ) 2 i∂L =0 ∂x jL= 1 & & m( r 2 + r 2 θ 2 ) − V ( r ) 2& pxj = mx j∂L =0 ∂θ& pθ = mr 2θ例题：圆锥面上质量为m的质点，受重力作用，约束在半顶角为α的圆锥面上运动，求质点的运动微分方程。

z = r ( ctgα )z1 2 2 &2 2 1 2 2 & & & & T = m(r + r θ + z ) = m(r csc α + r2θ 2 ) 2 2 1 cscα = 余割函数 sinαmyrα xV = mgz = mgr ( ctgα )1 & & L = T − V = m ( r 2 csc 2 α + r 2θ 2 ) − mgr ( ctgα ) θ 2 ∂L & pθ = & = mr 2θ = 常量 circulatory integral ∂θ& && − rθ 2 sin 2 α + g sin α cos α = 0 rMethod 2: first integral1 & & L = T − V = ma 2θ 2 + ma 2ω (θ + ω )(1 + cos θ ) 2 ∂L & p = & = ma 2θ + ma 2ω (1 + cos θ ) ∂θ & & & H = pθ 2 − L = ma 2θ + ma 2ω (1 + cos θ ) θ()⎡1 2 &2 2 & + ω )(1 + cos θ ) ⎤ − ⎢ ma θ + ma ω (θ ⎥ ⎣2 ⎦1 & = ma 2θ 2 − ma 2ω 2 (1 + cos θ ) = const . 2&& + ω 2 sin θ = 0 θAn easier way:1 1 2 2 & & & & T = m ( x + y ) = ma 2θ 2 + ma 2ω (θ + ω )(1 + cos θ ) 2 2 2 2 2 & + 1 ma 2θ 2 & = ma ω (1 + cos θ ) + ma ω (1 + cos θ )θ 2= T0 + T1 + T2H = T2 − T0 + V = const1 2 & ma 2θ 2 − ma 2ω（ + cos θ）= const 1 2&& θ + ω 2 sin θ = 032[例2]水平面上一斜面B1上有一滑 块B2，质量分别为m1与m2。

斜面倾角为θB2hθvB1设初始时B2在B1的顶点，两物体均无速度，求当滑块B2下 滑离开斜面时 (落差为h)， 求B1的速度以及B2相对斜面的速 度[解]牵连坐标 广义坐标 相对坐标牵连速度(平动) 相对速度2 2 2 1& v1 = x1 ′ &′ v 2 = x2x1 ′ x2r y′ x2r ′ y2O1hr ′ v2B2θx1r v2r v1B1′ 2 − 2v1v2 cos θ O ′ v = v + v2&′ &2 & &′ = x22 + x1 − 2 x1 x2 cos θ1 1 2 & &′ &2 & &′ T = m1 x1 + m2 ( x22 + x1 − 2 x1 x2 cos θ ) 2 2′ V = V1 + V2 = − m2 gx2 sin θ1 1 2 & &′ & & &′ T = m1 x1 + m2 ( x22 + x12 − 2 x1 x2 cos θ ) ′ V = − m 2 gx2 sin θ 2 2 1 ′ &2 &′ & &′ L = T − V = ⎡ ( m1 + m2 ) x1 + m2 x22 − 2m2 x1 x2 cos θ ⎤ + m2 gx2 sin θ ⎣ ⎦ 2∂L ∂x1 = 0∂L ∂t = 0x1为循环坐标& p x1 = ∂L ∂x1 = ( m1 + m 2 ) x1 − m 2 x 2 cos θ = c1 & &′H能量守恒H = T2 − T0 + V = c21 1 2 & 1 − m2 x1 x2 cos θ + m2 x22 & &′ &′ T = ( m1 + m2 ) x 2 2均为2次项′ & &′ & &′ ( m1 + m 2 ) x12 + m 2 x22 − 2m 2 x1 x2 cos θ − 2m 2 gx2 sin θ = c2初积分方程& &′ ( m1 + m2 ) x1 − m2 x2 cos θ = c1′ &2 &′ & &′ ( m1 + m2 ) x1 + m2 x22 − 2m2 x1 x2 cos θ − 2m2 gx2 sin θ = c2定常数t=0r yB2′ x2r1 y& x1 = 0 x1 = 0&′ ′ x2 = 0 x2 = 0r1 xO1Δy 2C2c1 = 0动力学方程c2 = 0θOx1B1r x& &′ ( m1 + m2 ) x1 − m2 x2 cos θ = 0 ′ &2 &′ & &′ ( m1 + m2 ) x1 + m2 x22 − 2m2 x1 x2 cos θ − 2m2 gx2 sin θ = 0m2 & &′ x1 = x2 cos θ & &′ ( m1 + m2 ) x1 − m2 x2 cos θ = 0 ( m1 + m2 ) ′ & &′ & &′ ( m1 + m2 ) x12 + m2 x22 − 2m2 x1 x2 cos θ − 2m2 gx2 sin θ = 0m1 + m2 sin 2 θ 2 ′ &′ x2 = 2 gx2 sin θ m1 + m2滑块离开斜面r yB2′ x2r1 yO1h′ x2 = h / sin θr1 xC2滑块相对斜面的速度m1 + m2 sin 2 θ 2 &′ x2 = 2 gh m1 + m2θ′ &′ v 2 = x2 =斜面的速度2 gh( m1 + m2 ) m1 + m2 sin 2 θOx1B1r x2 gh & v1 = x1 = m2 cos θ ( m1 + m2 )( m1 + m2 sin 2 θ )Summary• Discussed conservation laws– Generalized (conjugate) momentum∂L pj ≡ & ∂q j– Energy function∂L & & H ( q, q, t ) ≡ ∑ q j −L & ∂q j j38。