Chapter 4 Lagrangian mechanics

4.1 Lagrange coordinates

In the case of \(r\) independent holonomic constraints,

\begin{equation} \label {eq:149} G_m(x^a,t)=0,\quad m=1,\dots , r, \end{equation}

that satisfy the regularity assumptions, there exists locally a change of coordinates

\begin{equation} \label {eq:150} x^a=x^a(q^b,t),\quad q^a=q^a(x^b,t),\quad a,b=1,\dots , 3\mathcal {N}=n, \end{equation}

such that

\begin{equation} \label {eq:151}\left \{ \begin{array}{l} q^a=q^{n-r+m}=G_m,\quad a=n-r+1,\dots ,n,\\ q^a=q^\alpha ,\quad a,\alpha =1,\dots ,n-r. \end {array}\right . \end{equation}

The \(q^\alpha \) are the independent coordinates, also called Lagrange coordinates. Satisfying the constraints in this coordinate system means to put the last \(r\) coordinates to \(0\), \(q^{n-r+m}=0\). Furthermore, virtual displacements that are comptabible with the constraints satisfy \(\delta q^{n-r+m}=0\) while the \(\delta q^\alpha \) are arbitrary and unconstrained. In terms of the old variables, we thus have that, when the constraints hold, variations compatible with the constraints are given by

\begin{equation} \label {eq:152} \delta x^a\approx \frac {\partial x^a}{\partial q^\alpha }\Big |_{q^{n-r+m}=0}\delta q^\alpha . \end{equation}

4.2 Lagrange equations

In terms of Lagrange coordinates, we have

\begin{equation} \label {eq:153} \cancel {\delta } W\approx F_a \delta x^a \approx Q_\alpha \delta q^\alpha ,\quad Q_\alpha = F_a \frac {\partial x^a}{\partial q^\alpha }|_{q^{n-r+m}=0}, \end{equation}

where \(Q_\alpha \) is called the generalized force vector of Lagrange. For notational simplicity, we omit the bar with the indication that one should impose \(q^{n-r+m}=0\). In order to work out the kinetic energy \(T=\frac 12 m_{(a)}\dot x_a\dot x^a\), in terms of the Lagrange coordinates, one uses

\begin{equation} \label {eq:166} \dot x^a=\frac {\partial x^a}{\partial q^\beta }\dot q^\beta +\frac {\partial x^a}{\partial t}, \end{equation}

so that the kinetic energy becomes explicitly

\begin{equation} \label {eq:154} T(q^\alpha ,\dot q^\alpha ,t)=\frac 12 m_{(a)}(\frac {\partial x^a}{\partial q^\alpha }\dot q^\alpha +\frac {\partial x^a}{\partial t})^2 =m_{(a)}[\frac 12 \frac {\partial x_a}{\partial q^\alpha }\frac {\partial x^a}{\partial q^\beta }\dot q^\alpha \dot q^\beta + \frac {\partial x_a}{\partial q^\alpha }\frac {\partial x^a}{\partial t}\dot q^\alpha +\frac 12 \frac {\partial x_a}{\partial t}\frac {\partial x^a}{\partial t}], \end{equation}

while its variation is

\begin{equation} \label {eq:155} \delta T=\frac {\partial T}{\partial q^\alpha }\delta q^\alpha +\frac {\partial T}{\partial \dot q^\alpha }\delta \dot q^\alpha =(\frac {\partial T}{\partial q^\alpha } -\frac {d}{dt}\frac {\partial T}{\partial \dot q^\alpha })\delta q^\alpha +\frac {d}{dt}(\frac {\partial T}{\partial \dot q^\alpha }\delta q^\alpha ). \end{equation}

For any function \(F(q^\alpha ,\dot q^\alpha ,t)\), the Euler-Lagrange derivative is defined by

\begin{equation} \label {eq:165} \frac {\delta F}{\delta q^\alpha }=\frac {\partial F}{\partial q^\alpha } -\frac {d}{dt}\frac {\partial F}{\partial \dot q^\alpha }. \end{equation}

Again, in the context of Hamilton’s or the least action principle, the last term of (4.8) does not contribute when restricting oneself to variations vanishing at the end points. Hamilton’s principle then becomes

\begin{equation} \label {eq:156} \int ^{t_f}_{t_i} dt\ \big [\frac {\delta T}{\delta q^\alpha }+Q_\alpha \big ] \delta q^\alpha (t)=0, \end{equation}

for all unconstrained \(\delta q^\alpha (t)\) (that vanish at the end points). The du Bois-Reymond lemma then implies the Lagrange equations

\begin{equation} \label {eq:157} \boxed {\frac {\delta T}{\delta q^\alpha }=-Q_\alpha }. \end{equation}

That these differential equations are equivalent to Newton’s equations for ideal holonomic constraints can also be seen directly as follows. One starts from

\begin{equation} \label {eq:158}\left \{ \begin{array}{l} m_{(a)}\ddot x_a=F_a+\mu ^m\frac {\partial G_m}{\partial x^a}\\ G_m=0 \end {array}\right .. \end{equation}

The first set of equations are contracted with \(\frac {\partial x^a}{\partial q^\alpha }\), and then the constrained are imposed \(q^{n-r+m}=0\). The constraints are then trivially fulfilled. When using that

\begin{equation} \label {eq:160} \frac {\partial G_m}{\partial x^a}\frac {\partial x^a}{\partial q^\alpha }=0, \end{equation}

and the definition of the \(Q_\alpha \), the first set of equations reduces to

\begin{equation} \label {eq:159} m_{(a)}\ddot x_a \frac {\partial x^a}{\partial q^\alpha }=Q_\alpha , \end{equation}

with the understanding that \(q^{n-r+m}=0\). It remains to be shown that

\begin{equation} \label {eq:161} \frac {d}{dt}\frac {\partial T}{\partial \dot q^\alpha }-\frac {\partial T}{\partial q^\alpha }=m_{(a)}\ddot x_a \frac {\partial x^a}{\partial q^\alpha }\iff \frac {d}{dt}(\dot x_a\frac {\partial \dot x^a}{\partial \dot q^\alpha })-\dot x_a\frac {\partial \dot x^a}{\partial q^\alpha } =\frac {d}{dt}(\dot x_a \frac {\partial x^a}{\partial q^\alpha })-\dot x_a\frac {d}{dt}\frac {\partial x^a}{\partial q^\alpha }. \end{equation}

This is satisfied since it follows directly from (4.6) that

\begin{equation} \label {eq:162} \frac {\partial \dot x^a}{\partial \dot q^\alpha }=\frac {\partial x^a}{\partial q^\alpha },\quad \frac {\partial \dot x^a}{\partial q^\alpha }= \frac {\partial ^2 x^a}{\partial q^\alpha \partial q^\beta }\dot q^\beta +\frac {\partial ^2 x^a}{\partial q^\alpha \partial t}=\frac {d}{dt}\frac {\partial x^a}{\partial q^\alpha }. \end{equation}

4.3 Dissipative systems

In some dissipative systems there exists, besides the applied and the reaction forces, also friction forces proportional to the velocities. In this case,

\begin{equation} \label {eq:158b}\left \{ \begin{array}{l} m_{(a)}\ddot x_a=F_a+\mu ^m\frac {\partial G_m}{\partial x^a}-\zeta _{ab}\dot x^b\\ G_m=0 \end {array}\right ., \end{equation}

When contracting with \(\frac {\partial x^a}{\partial q^\alpha }\), the additional term on the right hand side is

\begin{equation} \label {eq:168} - \zeta _{ab}\dot x^b \frac {\partial x^a}{\partial q^\alpha }. \end{equation}

If the constant friction coefficients are symmetric, \(\zeta _{ab}=\zeta _{ba}\), one may write the associated Lagrange equations in terms of the Rayleigh’s dissipation function

\begin{equation} \label {eq:163} D=\frac 12 \zeta _{ab}\dot x^a\dot x^b, \end{equation}

as

\begin{equation} \label {eq:164} \boxed {\frac {\delta T}{\delta q^\alpha }=-Q_\alpha -\frac {\partial D}{\partial \dot q^\alpha }}, \end{equation}

because

\begin{equation} \label {eq:167} \frac {\partial D}{\partial \dot q^\alpha }=\zeta _{ab} \frac {\partial \dot x^a}{\partial \dot q^\alpha }\dot x^b= \zeta _{ab}\dot x^b \frac {\partial x^a}{\partial q^\alpha }. \end{equation}

4.4 Euler-Lagrange equations

In the absence of such friction forces and if all applied forces derive from a potential \(V(x^a)\), natural trajectories have to satisfy the Euler-Lagrange equations,

\begin{equation} \label {eq:169} \boxed {\frac {\delta L}{\delta q^\alpha }=0}. \end{equation}

These equations can either be obtained from the Lagrange equations (4.11) by using that \(L=T-V\) and

\begin{equation} \label {eq:170} Q_\alpha =F_a\frac {\partial x^a}{\partial q^\alpha }=-\frac {\partial V}{\partial x^a}\frac {\partial x^a}{\partial q^\alpha }=-\frac {\partial V}{\partial q^\alpha }, \quad \frac {\delta V}{\delta q^\alpha }= \frac {\partial V}{\partial q^\alpha }, \end{equation}

where \(V(q^\alpha ,t)=V(x^a(q^\alpha ,t))\). Alternatively, one derive the Euler-Lagrange equations directly from the least action principle:

\begin{equation} \label {eq:171} 0= \delta S=\int ^{t_f}_{t_i}dt\ (\frac {\partial L}{\partial q^\alpha }\delta q^\alpha +\frac {\partial L}{\partial \dot q^\alpha }\delta \dot q^\alpha )= \Big [\frac {\partial L}{\partial \dot q^\alpha }\delta q^\alpha \Big ]^{t_f}_{t_i}+\int ^{t_f}_{t_i}dt\ \frac {\delta L}{\delta q^\alpha }\delta q^{\alpha }. \end{equation}

The passage from the first to the second equality involves an integrations by parts, the boundary term vanishes because of the assumption that the variations vanish at the end points, and the Euler-Lagrange equations then follow by using the fact that the variations \(\delta q^\alpha (t)\) are arbitrary and by the du Bois-Reymond lemma.

Remarks:  

• • In some problems, the potential may depend on velocities, \(V=V(q^\alpha ,\dot q^\alpha ,t)\) or, more generally, the Lagrangian may be of form \(L(q^\alpha ,\dot q^\alpha ,t)\), without a direct interpretation in terms of \(T\) and \(V\). In this case, the least action principle for natural trajectories still gives rise to the Euler-Lagrange equations for these trajectories.

• • When making explicit the second order time derivatives in the left hand side of the Euler-Lagrange derivatives,

  \(\seteqnumber{0}{4.}{24}\)

  \begin{equation} \label {eq:175} -\frac {\delta L}{\delta q^\alpha }= \frac {d}{dt}\frac {\partial L}{\partial \dot q^\alpha }- \frac {\partial L}{\partial q^\alpha }=\frac {\partial ^2L}{\partial \dot q^\alpha \partial \dot q^\beta }\ddot q^\beta -f_\alpha (q^\beta ,\dot q^\beta ,t), \end{equation}

  it follows that the Euler-Lagrange equations are equivalent to differential equations in normal form if and only if

  \(\seteqnumber{0}{4.}{25}\)

  \begin{equation} \label {eq:176} \Big |\frac {\partial ^2 L}{\partial \dot q^\alpha \partial \dot q^\beta }\Big |\neq 0. \end{equation}

  We call this the regular case. In this case, the inverse matrix \(M^{\gamma \alpha }\frac {\partial ^2 L}{\partial \dot q^\alpha \partial \dot q^\beta }=\delta ^\gamma _\beta \) where \(M^{\gamma \alpha }=M^{\gamma \alpha }(q^\beta ,\dot q^\beta ,t)\), exists and the normal form of the Euler-Lagrange equations is

  \(\seteqnumber{0}{4.}{26}\)

  \begin{equation} \label {eq:177} \ddot q^\gamma =F^\gamma ,\quad F^\gamma (q^\alpha ,\dot q^\alpha ,t)=M^{\gamma \alpha }f_\alpha . \end{equation}

  Unless otherwise specified, we will always be in the regular case in the rest of the course.

• • It follows from (4.24) that one may define the functional derivative of the action with respect to a trajectory \(q^\alpha (t)\) by

  \(\seteqnumber{0}{4.}{27}\)

  \begin{equation} \label {eq:172} \frac {\delta S}{\delta q^\alpha (t)}=\frac {\delta L}{\delta q^\alpha } \big |_{q^\alpha =q^\alpha (t),\dot q^{\alpha }=\frac {d}{dt}q^\alpha (t)}. \end{equation}

• • If \(L'=cL\), \(c\in \mathbb R\), the solutions of the Euler-Lagrange equations are unchanged.

• • Suppose that \(L'=L+\frac {d f}{dt}\) for some \(f=f(q^\alpha ,t)\). It follows that \(\frac {\delta L'}{\delta q^\alpha }=\frac {\delta L}{\delta q^\alpha }\). Indeed,

  \(\seteqnumber{0}{4.}{28}\)

  \begin{equation} \label {eq:173} \frac {\delta }{\delta q^\alpha }(\frac {d f}{dt})=\frac {\delta }{\delta q^\alpha } (\frac {\partial f}{\partial q^\beta }\dot q^\beta +\frac {\partial f}{\partial t})=\frac {\partial ^2 f}{\partial q^\alpha \partial q^\beta }\dot q^\beta +\frac {\partial ^2 f}{\partial q^\alpha \partial t}- \frac {d}{dt}{\frac {\partial f}{\partial q^\alpha }}=0. \end{equation}

  Conversely, suppose that \(\frac {\delta L'}{\delta q^\alpha }=\frac {\delta L}{\delta q^\alpha }\), or if \(M(q^\alpha ,\dot q^\alpha ,t)=L'-L\), that¹ \(\frac {\delta M}{\delta q^\alpha }=0\). It follows that \(M=\frac {d f}{dt}\) with \(f=f(q^\alpha ,t)\). Indeed, in \(\frac {\delta M}{\delta q^\alpha }\) the only terms that involve second order derivatives are given by \(-\frac {\partial ^2M}{\partial \dot q^\alpha \partial \dot q^\beta }\ddot q^\beta \) and have to vanish separately. It thus follows that \(\frac {\partial ^2M}{\partial \dot q^\alpha \partial \dot q^\beta }=0\). When writing

  \(\seteqnumber{0}{4.}{29}\)

  \begin{multline} \label {eq:178} M(q^\alpha ,\dot q^\alpha ,t)=M(0,0,t)+\int ^1_0d\tau \frac {d}{d\tau }M(\tau q^\alpha ,\tau \dot q^\alpha ,t)\\ =\frac {d}{dt}\int ^t dt' M(0,0,t') +\int ^1_0\frac {d\tau }{\tau }(q^\alpha \frac {\partial M}{\partial q^\alpha }+ \dot q^\alpha \frac {\partial M}{\partial \dot q^\alpha })(\tau q^\alpha ,\tau \dot q^\alpha ,t). \end{multline} By the usual manipulation, we may write

  \(\seteqnumber{0}{4.}{30}\)

  \begin{equation} \label {eq:179} q^\alpha \frac {\partial M}{\partial q^\alpha }+\dot q^\alpha \frac {\partial M}{\partial \dot q^\alpha }=q^\alpha \frac {\delta M}{\delta q^\alpha } +\frac {d}{dt}(q^\alpha \frac {\partial M}{\partial \dot q^\alpha }), \end{equation}

  where the first term vanishes by assumption and \(N=q^\alpha \frac {\partial M}{\partial \dot q^\alpha }\) does not depend on \(\dot q^\beta \) since we have already shown that \(\frac {\partial ^2M}{\partial \dot q^\alpha \partial \dot q^\beta }=0\). Hence \(N=N(q^\alpha ,t)\). Furthermore

  \(\seteqnumber{0}{4.}{31}\)

  \begin{multline} (\frac {d}{dt} N)(\tau q^\alpha ,\tau \dot q^\alpha ,t) =(\dot q^\alpha \frac {\partial N}{\partial q^\alpha } +\frac {\partial N}{\partial t})(\tau q^\alpha ,\tau \dot q^\alpha ,t) =\tau \dot q^\alpha \frac {\partial N}{\partial q^\alpha }(\tau q^\alpha ,t) +\frac {\partial N}{\partial t})(\tau q^\alpha ,t)\\ =\frac {d}{dt} N(\tau q^\alpha ,t). \label {eq:180} \end{multline} Now \(\frac {d}{dt}\) can be taken out of the integral and

  \(\seteqnumber{0}{4.}{32}\)

  \begin{equation} \label {eq:181} M=\frac {df}{dt},\quad f(q^\alpha ,t)=\int ^t dt' M(0,0,t')+\int ^1_0\frac {d\tau }{\tau }N (\tau q^\alpha ,t). \end{equation}

4.4.1 Worked Exercice 3: A Lagrangian of sufficient generality

Consider a Lagrangian of the form

\begin{equation} \label {eq:182} L=\frac {1}{2}g_{\beta \gamma }\dot q^\beta \dot q^\gamma +A_\beta \dot q^\beta -V, \end{equation}

where \(g_{\beta \gamma }=g_{\gamma \beta },A_\beta ,V\) are functions of \(q^\delta \) and \(t\) only.

Show that in the regular case, the “metric” \(g_{\beta \gamma }\) is invertible. Show that the left hand sides of the Euler-Lagrange equations may be written as

\begin{equation} \label {eq:183} -\frac {\delta L}{\delta q^\alpha }=g_{\alpha \beta } \ddot q^\beta +\Gamma _{\alpha \beta \gamma } \dot q^\beta \dot q^\gamma -F_{\alpha \beta }\dot q^\beta +\frac {\partial V}{\partial q^\alpha } +\frac {\partial }{\partial t}(g_{\alpha \beta }\dot q^\beta + A_\alpha ), \end{equation}

where the “Christoffel symbols of the first kind” \(\Gamma _{a\beta \gamma }\) and the “field strengths” \(F_{\alpha \beta }\) are given by

\begin{equation} \label {eq:184} \Gamma _{\alpha \beta \gamma }=\frac 12(\frac {\partial g_{\alpha \beta }}{\partial q^\gamma } +\frac {\partial g_{\alpha \gamma }}{\partial q^\beta } -\frac {\partial g_{\beta \gamma }}{\partial q^\alpha }),\quad F_{\alpha \beta }=\frac {\partial A_\beta }{\partial q^\alpha } -\frac {\partial A_\alpha }{\partial q^\beta }. \end{equation}

Solution: The first part follows from

\begin{equation} \label {eq:185} \frac {\partial ^2 L}{\partial \dot q^\alpha \partial \dot q^\beta }=g_{\alpha \beta }. \end{equation}

For the second, we compute

\begin{equation} \label {eq:186} -\frac {\partial L}{\partial q^\alpha }= -(\frac 12 \frac {\partial g_{\beta \gamma }}{\partial q^\alpha }\dot q^\beta \dot q^\gamma +\frac {\partial A_\beta }{\partial q^\alpha }\dot q^\beta -\frac {\partial V}{\partial q^\alpha }), \end{equation}

\begin{equation} \label {eq:187} \frac {d}{dt}\frac {\partial L}{\partial \dot q^\alpha }=g_{\alpha \beta } \ddot q^\beta +\frac {\partial g_{\alpha \beta }}{\partial q^\gamma }\dot q^\gamma \dot q^\beta +\frac {\partial A_\alpha }{\partial q^\beta }\dot q^\beta +\frac {\partial }{\partial t}(g_{\alpha \beta }\dot q^\beta + A_\alpha ). \end{equation}

The result follows by adding the terms on the right hand sides of these two equations and using

\begin{equation} \label {eq:188} \frac {\partial g_{\alpha \beta }}{\partial q^\gamma }\dot q^\gamma \dot q^\beta =\frac 12 (\frac {\partial g_{\alpha \beta }}{\partial q^\gamma } +\frac {\partial g_{\alpha \gamma }}{\partial q^\beta })\dot q^\gamma \dot q^\beta . \end{equation}

4.5 Points of equilibrium

Consider a Lagrangian of the form of the previous section, but assume now that there is no explicit time dependence in the coefficients \(g_{\beta \gamma },A_\beta ,V\). Equilibrium points are points \(\bar q^\beta \) such that

\begin{equation} \label {eq:174} \frac {\partial V}{\partial q^\alpha }(\bar q^\beta )=0. \end{equation}

In this case, the trajectories \(q^\beta (t)=\bar q^\beta \), with \(\bar q^\beta \) constant in time, are natural trajectories.

Indeed, these trajectories satisfy \(\dot q^\beta =0=\ddot q^\beta \) and, according to the previous section, the left hand side of the equations of motion are given by

\begin{equation} \label {eq:189} -\frac {\delta L}{\delta q^\alpha }=g_{\alpha \beta } \ddot q^\beta +\Gamma _{\alpha \beta \gamma }\dot q^\beta \dot q^\gamma -F_{\alpha \beta }\dot q^\beta +\frac {\partial V}{\partial q^\alpha }. \end{equation}

When evaluating (4.42) at \(q^\beta =\bar q^\beta \) and using in addition (4.41), we thus get zero.