Chapter 7 Noether’s theorem and Galilean relativity

7.1 Noether’s first theorem

A first integral or conserved quantity is a function \(K(q,\dot q,t)\) that is constant on all natural trajectories,

\begin{equation} \label {eq:60} \frac {dK}{dt}|{\bar q}=0,\quad {\rm with}\ \frac {\delta L}{\delta q^\alpha }|_{\bar q}=0. \end{equation}

An infinitesimal symmetry is an infinitesimal virtual variation that leaves the Lagrangian invariant up to a total derivative,

\begin{equation} \label {eq:61} \boxed {\delta _Q q^\alpha =Q^\alpha (q,\dot q,t)\ {\rm such\ that}\ \delta _Q L=\frac {dF(q,\dot q,t)}{dt}}. \end{equation}

Noether’s first theorem states that:

Indeed,

\begin{equation} \label {eq:62} \delta _Q L=\delta _Q q^\alpha \frac {\partial L}{\partial q^\alpha }+\delta _Q \dot q^\alpha \frac {\partial L}{\partial \dot q^\alpha }=Q^\alpha \frac {\partial L}{\partial q^\alpha }+\frac {dQ^\alpha }{dt}\frac {\partial L}{\partial \dot q^\alpha }=\frac {dF}{dt}, \end{equation}

where the second equality follows because infinitesimal virtual variations commute with the total derivative, while the last equality follows by the assumption that the variation is a symmetry. Using

\begin{equation} \label {eq:63} \frac {dQ^\alpha }{dt}\frac {\partial L}{\partial \dot q^\alpha }=\frac {d}{dt}(Q^\alpha \frac {\partial L}{\partial \dot q^\alpha })-Q^\alpha \frac {d}{dt}\frac {\partial L}{\partial \dot q^\alpha }, \end{equation}

the latter combines with the first term to produce the Euler-Lagrange derivatives, while the former combines with the total derivative on the RHS,

\begin{equation} \label {eq:64} Q^\alpha \frac {\delta L}{\delta q^\alpha }=\frac {d}{dt}(F-\frac {\partial L}{\partial \dot q^\alpha }Q^\alpha ). \end{equation}

Since the LHS vanishes when evaluated at any natural trajectory, it follows that associated the conserved quantity is given by

\begin{equation} \label {eq:65} \boxed {K=\frac {\partial L}{\partial \dot q^\alpha }Q^\alpha -F}. \end{equation}

The overall minus sign is conventional and amounts to writing Noether’s first theorem as

\begin{equation} \label {eq:451} Q^\alpha \frac {\delta L}{\delta q^\alpha }+\frac {d}{dt}K=0. \end{equation}

Remark: Neither the theorem nor the expression depends on the choice of representative for the action: if \(L'=L+\frac {df(q,t)}{dt}\), it follows that \(\delta _Q\) is also an infinitesimal symmetry of \(L'\),

\begin{equation} \label {eq:66} \delta _Q L'=\delta _Q L+\frac {d}{dt}(\delta _Q f)=\frac {d}{dt}(F+\delta _Q f), \end{equation}

with associated conserved quantity

\begin{equation} \label {eq:67a} K'=F+\cancel {\frac {\partial f}{\partial q^\alpha }Q^\alpha }-(\frac {\partial L}{\partial \dot q^\alpha }+\cancel {\frac {\partial f}{\partial q^\alpha }})Q^\alpha =K. \end{equation}

7.2 Applications

7.2.1 Explicit time independence and conservation of energy

Consider a Lagrangian that does not depend explicitly on time, \(L=L(q,\dot q)\), \(\frac {\partial L}{\partial t}=0\). It follows that

\begin{equation} \label {eq:68} \frac {dL}{dt}=\frac {\partial L}{\partial q^\alpha }\dot q^\alpha +\frac {\partial L}{\partial \dot q^\alpha }\ddot q^\alpha =\frac {\delta L}{\delta q^\alpha }\dot q^\alpha +\frac {d}{dt}(\frac {\partial L}{\partial \dot q^\alpha }\dot q^\alpha ), \end{equation}

In other words,

\begin{equation} \label {eq:69} K=\frac {\partial L}{\partial \dot q^\alpha }\dot q^\alpha -L, \end{equation}

is the conserved quantity associated to the infinitesimal symmetry

\begin{equation} \label {eq:70} \delta _Q q^\alpha =\dot q^\alpha . \end{equation}

In particular, if

\begin{equation} \label {eq:71} L=\frac 12 g_{\alpha \beta }(q)\dot q^\alpha \dot q^\beta +A_\alpha (q) \dot q^\alpha -V(q), \end{equation}

the associated conserved quantity

\begin{equation} \label {eq:72} K=g_{\alpha \beta } \dot q^\alpha \dot q^\beta +\cancel {A_\alpha \dot q^\alpha }-\frac 12 g_{\alpha \beta }\dot q^\alpha \dot q^\beta -\cancel {A_\alpha \dot q^\alpha } +V=\frac 12 g_{\alpha \beta }\dot q^\alpha \dot q^\beta +V=E \end{equation}

corresponds to the mechanical energy.

7.2.2 Cyclic variables and conservation of canonical momentum

Suppose that a given variable \(q^\alpha \) is “cyclic” which means that \(L\) does not depend on \(q^\alpha \) but may depend on the associated velocity \(\dot q^\alpha \). To fix ideas, suppose that the first variable \(q^1\) is cyclic, \(\frac {\partial L}{\partial q^1}=0\). It follows that \(\delta _Q q^\alpha =\delta ^\alpha _1\) is an infinitesimal symmetry,

\begin{equation} \label {eq:73} \delta _Q L=\delta ^\alpha _1\frac {\partial L}{\partial q^\alpha }+\frac {d(\delta ^\alpha _1)}{dt}\frac {\partial L}{\partial \dot q^\alpha }=\frac {\partial L}{\partial q^1}+\frac {d(1)}{dt}\frac {\partial L}{\partial \dot q^1}=0. \end{equation}

Since \(F=0\), the associated conserved quantity is

\begin{equation} \label {eq:74} K=\frac {\partial L}{\partial \dot q^\alpha }(\delta ^\alpha _1)=\frac {\partial L}{\partial \dot q^1}, \end{equation}

which is the canonical momentum associated to \(q^1\).

7.2.3 One-parameter groups of transformations

Suppose that the Lagrangian is invariant under a set of invertible transformations \(q^{\prime \alpha }=q^{\prime \alpha }(q^\beta ,t;s)\) that depend continously on a parameters \(s\), and reduce to the identity at \(s=0\), \(q^{\prime \alpha }(q^\beta ,t;0)=q^\alpha \). By invariance we mean that the transformed Lagragian in terms of the new variables is equal to the old Lagrangian in terms of the same variables,

\begin{equation} \label {eq:75} L'(q',\dot q',t)=L(q(q',t;s),\frac {d}{dt} q(q',t;s),t)=L(q',\dot q',t), \end{equation}

where the first equality merely expresses how the new Lagrangian in terms of the new coordinates is obtained in terms of the old one, while the second equality expresses the invariance of the Lagrangian. When using the inverse transformation, the latter is equivalent to

\begin{equation} \label {eq:76} L(q,\dot q,t)= L(q'(q,t;s),\frac {d}{dt}q'(q,t;s),t). \end{equation}

In this case, the transformation

\begin{equation} \label {eq:77} \boxed {\delta _R q^\alpha =R^\alpha (q,t):=\frac {\partial q'^{\alpha }}{\partial s} |_{s=0}}, \end{equation}

defines an infinitesimal symmetry.

Indeed, \(q^{\prime \alpha }=q^\alpha +sR^\alpha (q,t)+O(s^2)\), while \(\frac {dq^{\prime \alpha }}{dt} =\dot q^\alpha +s\frac {dR^\alpha }{dt}+O(s^2)\). If \(\delta _R \dot q^\alpha =\frac {\partial }{\partial s} \frac {dq^{\prime \alpha }}{dt}|_{s=0}\), it follows that \(\delta _R \dot q^\alpha =\frac {dR^\alpha }{dt}\), as it should for an infinitesimal virtual variation. In these terms, the invariance condition (7.18) becomes

\begin{equation} \label {eq:78} L(q,\dot q,t)=L(q+sR+O(s^2),\dot q+s\frac {dR}{dt}+O(s^2),t),\quad \forall s. \end{equation}

In particular, for \(s=0\), this is an identity, while the terms linear in s on the RHS have to vanish. The latter can be obtained by taking \(\frac {\partial }{\partial s}\) at \(s=0\), and yield

\begin{equation} \label {eq:79} R^\alpha \frac {\partial L}{\partial q^\alpha }+\frac {dR^\alpha }{dt}\frac {\partial L}{\partial \dot q^\alpha }=0, \end{equation}

which can also be written as \(\delta _R L=0\). According to Noether’s theorem, the associated conserved quantity is given by

\begin{equation} \label {eq:80} \boxed {K=\frac {\partial L}{\partial \dot q^\alpha }\delta _Rq^\alpha }. \end{equation}

Note that a cyclic variabe can be interpreted as a particular case. Indeed, the transformation that leaves the Lagrangian invariant is given by \(q^{\prime \alpha }=q^\alpha +s\delta ^\alpha _1\).

Another particular case is a system of particles in three-dimensional Euclidean space, with a Lagrangian of the type

\begin{equation} \label {eq:81} L=T-V,\quad T=\frac 12 \sum _{i=1}^{\mathcal N}m_{(i)}\dot x^\alpha _{(i)}\dot x_{\alpha (i)},\quad V=V(r_{ij}). \end{equation}

In this expression, the summation in the kinetic term over the different particles has been written out explicitly, while the summation over the \(3\) components of the position of a single particle is implicit through the summation convention. Note also that we have defined \(x_{\alpha (i)}=\delta _{\alpha \beta }x^\beta _{(i)}\), and that the potential \(V(r_{ij})\) with \(i<j\) is assumed to depend only on the relative distances between the different particles,

\begin{equation} r_{ij}=\sqrt {(x^\alpha _{(i)}-x^{\alpha }_{(j)}) (x_{\alpha (i)}-x_{\alpha (j)})}\label {eq:82}. \end{equation}

If one now considers translations in Euclidean space,

\begin{equation} \label {eq:83} \boxed {x^{\prime \alpha }_i=x^\alpha _i+a^\alpha }, \end{equation}

for constant \(a^\alpha \) (which correspond to three different parameters \(s\)), invariance of the Lagrangrian in the form (7.18) is manifest since \(\frac {dx^{\prime \alpha }_i}{dt}=\frac {dx^\alpha _i}{dt}\) and \(r'_{ij}=r_{ij}\). The associated infinitesimal transformations are

\begin{equation} \label {eq:84} R^\alpha _{\gamma i}=\frac {\partial x^{\prime \alpha }_i}{\partial a^\gamma }|_{a=0}=\delta ^\alpha _\gamma , \end{equation}

with associated conserved quantities given by the components of total linear momenta,

\begin{equation} \label {eq:85} \boxed {P_\gamma =\sum _{i}\frac {\partial L}{\partial \dot x^\gamma _{(i)}}=\sum _{i}m_{(i)}\dot x_{\gamma (i)}}. \end{equation}

For rotations, we have

\begin{equation} \label {eq:86} \boxed {x^{\prime \alpha }_i={A^\alpha }_\beta x^\beta _i},\quad {A^\alpha }_\beta \delta _{\alpha \gamma } {A^\gamma }_\delta =\delta _{\beta \delta } \iff A^T A=\mathbb {1}, \end{equation}

with \(A\) a constant \(3\times 3\) orthogonal matrix. Again, the invariance of the Lagrangian is manifest since orthogonality implies that \(\frac {dx^{\prime \alpha }_{(i)}}{dt}\delta _{\alpha \gamma } \frac {dx^{\prime \gamma }_{(i)}}{dt}=\frac {dx^{\alpha }_{(i)}}{dt}\delta _{\alpha \gamma } \frac {dx^{\gamma }_{(i)}}{dt}\) while distances are also left invariant. In order to find the parameters, we note that if \({A^\alpha }_\beta =\delta ^\alpha _\beta +{\omega ^\alpha }_\beta +O(\omega ^2)\), with \({\omega ^\alpha }_\beta \) a “small” matrix, the orthogonality condition

\begin{equation} \label {eq:87} (\delta ^\alpha _\beta +{\omega ^\alpha }_\beta +O(\omega ^2))\delta _{\alpha \gamma } (\delta ^\gamma _\delta +{\omega ^\gamma }_\delta +O(\omega ^2))=\delta _{\beta \delta }, \end{equation}

implies to first order in \(\omega \) that

\begin{equation} \label {eq:88} \omega _{\delta \beta }+\omega _{\beta \delta }=0\iff \omega +\omega ^T=0. \end{equation}

This implies that \(\omega \) can be written in terms of three independent rotation parameters \(\theta ^\gamma \) as

\begin{equation} \label {eq:89} {\omega ^\alpha }_\beta =\theta ^\gamma {\epsilon ^\alpha }_{\gamma \beta }\iff \omega =\begin{pmatrix} 0 &-\theta ^3& \theta ^2 \\ \theta ^3 & 0 & -\theta ^1 \\ -\theta ^2 & \theta ^1 & 0 \end {pmatrix}, \end{equation}

and that the associated infinitesimal field transformations are

\begin{equation} \label {eq:90} R^{\alpha }_{\gamma i}=\frac {\partial x^{\prime \alpha }_i}{\partial \theta ^\gamma }|_{\theta =0} ={\epsilon ^\alpha }_{\gamma \beta }x^{\beta }_i, \end{equation}

with associated conserved quantities given by the components of the total angular momentum,

\begin{equation} \label {eq:91} \boxed {J_\gamma =\epsilon _{\gamma \alpha \beta }\sum _i x^\alpha _{(i)} m_{(i)}\dot x^\beta _{(i)} }. \end{equation}

We will show below that a rotation matrix that is connected to the identity¹ can be written as a rotation of an angle \(0\leq \theta <2\pi \) around a fixed axis. The rotation may be determined a vector

\begin{equation} \label {eq:92} \vec \theta =\theta \vec n=\theta ^\alpha \vec e_\alpha ,\quad \end{equation}

with \(\vec e_1=\vec 1_x\), \(\vec e_2=\vec 1_y\), \(\vec e_3=\vec 1_z\), of norm

\begin{equation} \label {eq:norm} \theta =\sqrt {(\theta ^1)^2+(\theta ^2)^2+(\theta ^3)^3}, \end{equation}

and unit direction

\begin{equation} \label {eq:434} \hat \theta ^\alpha =\frac {\theta ^\alpha }{\theta }, \end{equation}

as

\begin{equation} \label {eq:93} {A^\alpha }_\beta =\cos \theta (\delta ^\alpha _\beta -\hat \theta ^\alpha \hat \theta _\beta ) +\hat \theta ^\alpha \hat \theta _\beta +\sin \theta \hat \theta ^\gamma {\epsilon ^\alpha }_{\gamma \beta }. \end{equation}

As before, one then finds that

\begin{equation} \label {eq:94} \frac {\partial {A^\alpha }_\beta }{\partial \theta ^\gamma }|_{\theta =0}={\epsilon ^\alpha }_{\gamma \beta }. \end{equation}

7.2.4 Parametrization of a rotation

Consider a rotation matrix \(A\) in 3 dimensions with entries \({A^\alpha }_\beta \), \(\alpha ,\beta =1,2,3\),

\begin{equation} \label {eq:255} A^TA=\mathbb 1\iff A_{\alpha \beta }{A^\alpha }_\gamma =\delta _{\beta \gamma }, \end{equation}

where the Kronecker symbols \(\delta _{\alpha \beta },\delta ^{\alpha \beta }\) are used to lower and raise indices. By taking the determinant, it follows that \({\rm det}A=\pm 1\). In the following, we will concentrate on rotation matrices with unit determminant,

\begin{equation} \label {eq:256} {\rm det}A=1. \end{equation}

Invariant line

For such a matrix, the rotation, i.e., the transformation

\begin{equation} \label {eq:257} x^{\prime \alpha }={A^\alpha }_\beta x^\beta , \end{equation}

leaves a line invariant.

Indeed, such a line is defined by a non trivial solution to the equation

\begin{equation} \label {eq:258} x^\alpha ={A^\alpha }_\beta x^\beta \iff (A-\mathbb 1)\vec x=0. \end{equation}

The existence of a non-trivial solution \(\vec x\neq \vec 0\) requires that

\begin{equation} \label {eq:259} {\rm det}(A-\mathbb 1)=0. \end{equation}

Let us suppose that \({\rm det}(A-\mathbb 1)=k\). From the defining equation of a rotation matrix, we have

\begin{equation} \label {eq:260} A^T(A-\mathbb 1)=\mathbb 1-A^T=(\mathbb 1-A)^T=-(A-\mathbb 1)^T. \end{equation}

By taking the determinant of this equation, it follows that

\begin{equation} \label {eq:261} 1k=(-1)^3k\Longrightarrow k=-k\Longrightarrow k=0. \end{equation}

Let us denote by \(\vec \theta \) a non trivial solution to \(A\vec \theta =\vec \theta \). In other words, the line which is left invariant by the rotation is \(\lambda \vec \theta \), \(\lambda \in \mathbb R\). Furthermore \(\vec \theta =\theta \vec n\), \(\theta =\sqrt {(\theta ^1)^2+(\theta ^2)^2+(\theta ^3)^2}\) is the norm of \(\vec \theta \), while \(\vec n=\frac {1}{\theta }\theta ^\alpha \vec e_\alpha \), with components \(\hat \theta ^\alpha =\frac {\theta ^\alpha }{\theta }\) is the unit vector along the axis of rotation, while \(\vec e_\alpha \) are the unit vectors along the coordinate axis of \(\mathbb R^3\).

Geometry of a rotation in 3 dimensions

If \(\overrightarrow {OP}=x^\alpha \vec e_\alpha =\vec x\) denotes the vector relating the origin \(O\) with coordinates \(x^\alpha =0\) to an arbitrary point \(P\) with coordinates \(x^\alpha \), the image of this vector under the rotation (7.41) is the vector \(\overrightarrow {OP'}=x^{\prime \alpha } \vec e_\alpha =\vec x'\), where the coordinates of \(P'\) are \(x^{\prime \alpha }\). Furthermore, by suitably choosing the orientation of \(\vec n\), one may arrange that \((\vec n,\overrightarrow {OP},\overrightarrow {OP'})\) is oriented positively.

A rotation preserves the scalar products and thus the norms of vectors,

\begin{equation} \label {eq:262} \vec u'\cdot \vec v'=u'_\alpha v'^\alpha =A_{\alpha \beta }u^\beta {A^\alpha }_\gamma v^\gamma =u^\beta \delta _{\beta \gamma }v^\gamma =u_\beta v^\beta . \end{equation}

Let us define the point \(Q\) to be the orthogonal projection of \(P\) on the line defined by \(\vec n\), \(\overrightarrow {OQ}=(\overrightarrow {OP}\cdot \vec n)\vec n\), and show that \(Q,P,P'\) lie in the plane perpendicular to \(\vec n\). Writing \(\overrightarrow {OP}=\overrightarrow {OQ}+\overrightarrow {QP}\), it follows that

\begin{equation} \overrightarrow {QP}\cdot \vec n=(\overrightarrow {OP}-\overrightarrow {OQ})\cdot \vec n=\overrightarrow {OP}\cdot \vec n-(\overrightarrow {OP}\cdot \vec n)\vec n\cdot \vec n=0\label {eq:263} \end{equation}

Furthermore,

\begin{equation} \overrightarrow {QP'}\cdot \vec n=\overrightarrow {OP'}\cdot \vec n-\overrightarrow {OQ}\cdot \vec n=A(\overrightarrow {OP})\cdot A(\vec n)-\overrightarrow {OQ}\cdot \vec n=\overrightarrow {OP}\cdot \vec n-\overrightarrow {OQ}\cdot \vec n=0\label {eq:264}. \end{equation}

Furthermore

\begin{equation} ||\overrightarrow {QP}||=||\overrightarrow {QP'}||.\label {eq:268} \end{equation}

Indeed, starting from \(||\overrightarrow {OP}||^2=||\overrightarrow {OP'}||^2\), by using that

\begin{equation} ||\vec a+\vec b||^2=a^2+b^2+2 ab\cos \phi ,\label {eq:277} \end{equation}

with \(a,b\) the norms of \(\vec a,\vec b\) and \(\phi \) the angle between them, one develops the left hand side as

\begin{equation} \label {eq:267} ||\overrightarrow {OP}||^2=||\overrightarrow {OQ}+\overrightarrow {QP}||^2= ||\overrightarrow {OQ}||^2+||\overrightarrow {QP}||^2+\cancel { 2\overrightarrow {OQ}\cdot \overrightarrow {QP}}. \end{equation}

The last term vanishes since both vectors are orthogonal. The result follows by applying the same reasoning to the right hand side, that is to say decomposing \(\overrightarrow {OP'}=\overrightarrow {OQ}+\overrightarrow {QP'}\), and after having computed the norm, cancelling the common term \(||\overrightarrow {OQ}||^2\) on both sides of the equation.

The triangle \(QPP'\) thus has two sides of equal length, \(QP\) and \(QP'\) so that, if \(M\) is the point in the middle of \(PP'\), \(QM\) bisects the angle at \(Q\) and \(\overrightarrow {QM}\) is orthogonal to \(\overrightarrow {PP'}\). In the plane orthogonal to the axis of rotation, the angle at \(Q\), which will be called \(\theta \), is the angle of rotation that brings you from \(P\) to \(P'\).

By using that

\begin{equation} \vec a\times \vec b=ab\sin \phi \ \vec n,\label {eq:278} \end{equation}

where \(\vec n\) is the normal to the plane formed by \(\vec a,\vec b\), it follows that

\begin{equation} \label {eq:269} \overrightarrow {QM}\times \overrightarrow {PP'}=QM\, PP'\ \vec n, \end{equation}

since the angle bewteen the two vectors is \(\frac {\pi }{2}\) radians. Furthermore,

\begin{equation} \label {eq:270} QM=QP'\cos \frac {\theta }{2},\quad \frac {PP'}{2}=MP'=QP'\sin \frac {\theta }{2}. \end{equation}

Taking the quotient of these two equations yields

\begin{equation} \label {eq:271} 2\tan {\frac {\theta }{2}}=\frac {PP'}{QM}. \end{equation}

Since \((\vec n,\overrightarrow {OP},\overrightarrow {OP'})\) is positively oriented, it follows for the unit vectors that \(\vec n\times \overrightarrow {1_{QM}}= \overrightarrow {1_{PP'}}\). Multiplying by the previous equation yields

\begin{equation} \label {eq:272} 2\tan {\frac {\theta }{2}}\ \vec n\times \overrightarrow {1_{QM}}=\frac {PP'}{QM}\ \overrightarrow {1_{PP'}}. \end{equation}

Defining

\begin{equation} \label {eq:273} \overrightarrow {\Omega }=\tan {\frac {\theta }{2}}\vec n,\quad \Omega =\tan \frac {\theta }{2}, \end{equation}

one may rewrite the former equation as

\begin{equation} \label {eq:274} 2 \overrightarrow {\Omega }\times \overrightarrow {QM}=\overrightarrow {PP'}. \end{equation}

Analytic expression of the rotation matrix

We want an expression of \(\overrightarrow {OP'}\) in terms of \(\overrightarrow {OP}\).

By using that \(\overrightarrow {QM}=\frac 12 ( \overrightarrow {QP}+ \overrightarrow {QP'})\), it follows from the previous equation that

\begin{equation} \label {eq:276} \overrightarrow {PP'}=\overrightarrow {\Omega }\times (\overrightarrow {QP}+ \overrightarrow {QP'})=\overrightarrow {\Omega }\times (\overrightarrow {OP}+ \overrightarrow {OP'}-\cancel {2\overrightarrow {OQ}}), \end{equation}

where the last term does not contibute since it is parallel to \(\overrightarrow {\Omega }\). When using that

\begin{equation} \label {eq:279} \vec a\times (\vec b\times \vec c)=(\vec a\cdot \vec c) \vec b-(\vec a\cdot \vec b) \vec c, \end{equation}

it follows by applying \(\vec \Omega \times \) to (7.59) that

\begin{equation} \label {eq:280} \overrightarrow {\Omega }\times \overrightarrow {PP'}=(\overrightarrow {\Omega }\cdot (\overrightarrow {OP}+ \overrightarrow {OP'}))\overrightarrow {\Omega }-(\overrightarrow {\Omega }\cdot \overrightarrow {\Omega })(\overrightarrow {OP}+ \overrightarrow {OP'}). \end{equation}

When using on the left hand side that

\begin{equation} \overrightarrow {PP'}=\overrightarrow {QP'}-\overrightarrow {QP}= \overrightarrow {OP'}-\overrightarrow {OP},\label {eq:275} \end{equation}

and on the right hand side that

\begin{equation} \overrightarrow {\Omega }\cdot \overrightarrow {OP}= \overrightarrow {\Omega }\cdot \overrightarrow {OQ}=\overrightarrow {\Omega }\cdot \overrightarrow {OP'}\label {eq:281} \end{equation}

equation (7.61) becomes

\begin{equation} \label {eq:282} \overrightarrow {\Omega }\times \overrightarrow {OP'}-\overrightarrow {\Omega }\times \overrightarrow {OP}=2(\overrightarrow {\Omega }\cdot \overrightarrow {OP})\overrightarrow {\Omega }-\Omega ^2 (\overrightarrow {OP}+ \overrightarrow {OP'}). \end{equation}

Using (7.62) in the left hand side of (7.59) and (7.64) in the right hand side yields

\begin{equation} \label {eq:283} \overrightarrow {OP'}-\overrightarrow {OP}=2 \overrightarrow {\Omega }\times \overrightarrow {OP}+2(\overrightarrow {\Omega }\cdot \overrightarrow {OP})\overrightarrow {\Omega }-\Omega ^2 (\overrightarrow {OP}+ \overrightarrow {OP'}), \end{equation}

or equivalently

\begin{equation} \label {eq:284} \overrightarrow {OP'}=\frac {1}{1+\Omega ^2}\Big [(1-\Omega ^2)\overrightarrow {OP}+2 \overrightarrow {\Omega }\times \overrightarrow {OP}+2(\overrightarrow {\Omega }\cdot \overrightarrow {OP})\overrightarrow {\Omega } \Big ]. \end{equation}

The trigonometric relations

\begin{equation} \label {eq:285} \frac {1-\Omega ^2}{1+\Omega ^2}=\frac {1-\tan ^2\frac {\theta }{2}}{1+\tan ^2\frac {\theta }{2}} =\frac {\frac {\cos ^2\frac {\theta }{2}-\sin ^2\frac {\theta }{2}}{\cos ^2\frac {\theta }{2}}}{ \frac {\cos ^2\frac {\theta }{2}+\sin ^2\frac {\theta }{2}}{\cos ^2\frac {\theta }{2}}}= \cos ^2\frac {\theta }{2}-\sin ^2\frac {\theta }{2}=\cos \theta , \end{equation}

\begin{equation} \label {eq:286} \frac {2\Omega }{1+\Omega ^2}=2\cos ^2\frac {\theta }{2}\tan \frac {\theta }{2}=2\sin \frac {\theta }{2} \cos \frac {\theta }{2}=\sin \theta , \end{equation}

\begin{equation} \label {eq:287} \frac {2\Omega ^2}{1+\Omega ^2}=\frac {2\tan ^2\frac {\theta }{2}}{1+\tan ^2\frac {\theta }{2}} =2\sin ^2\frac {\theta }{2}=1-\cos \theta , \end{equation}

then allow one to rewrite

\begin{equation} \label {eq:288} \vec x'=A\vec x=\cos \theta \vec x+\sin \theta \vec n\times \vec x+(1-\cos \theta )(\vec n\cdot \vec x)\vec n=\cos \theta \vec x+ \frac {\sin \theta }{\theta } \vec \theta \times \vec x+\frac {{1-\cos \theta }}{\theta ^2}(\vec \theta \cdot \vec x)\vec \theta . \end{equation}

When expressed in components, this gives

\begin{equation} \label {eq:93a} x^{\prime \alpha }={A^\alpha }_\beta x^\beta =\cos \theta x^\alpha +\frac {\sin \theta }{\theta }{\epsilon ^\alpha }_{\beta \gamma }\theta ^\beta x^\gamma +(\frac {1-\cos \theta }{\theta ^2})\theta ^\alpha \theta _\beta x^\beta . \end{equation}

When identifying the coefficient of \(x^\beta \) on the right hand side, this gives the rotation matrix,

\begin{equation} \label {eq:93b} {A^\alpha }_\beta =\cos \theta \delta ^\alpha _\beta +(\frac {1-\cos \theta }{\theta ^2})\theta ^\alpha \theta _\beta -\frac {\sin \theta }{\theta }{\epsilon ^\alpha }_{\beta \gamma }\theta ^\gamma , \end{equation}

or, equivalently, its expression as in (7.37),

\begin{equation} \label {eq:435} \boxed {{A^\alpha }_\beta =\cos \theta (\delta ^\alpha _\beta -\hat \theta ^\alpha \hat \theta _\beta ) +\hat \theta ^\alpha \hat \theta _\beta +\sin \theta \hat \theta ^\gamma {\epsilon ^\alpha }_{\gamma \beta }}, \end{equation}

Worked exercise: Rotation around the \(\vec 1_ z\) axis

In order to find the analytic expression for a rotation matrix around the \(z\) axis by an angle \(\theta \) in the \((x,y)\) plane from the expression (7.71), we note that \(\theta ^1=\theta ^2=0\), while \(\theta ^3=\theta \). It follows that \(-\frac {\sin \theta }{\theta }{\epsilon ^\alpha }_{\beta \gamma }\theta ^\gamma = -\frac {\sin \theta }{\cancel {\theta }}{\epsilon ^\alpha }_{\beta 3}\cancel {\theta }\), so that

\begin{equation} \begin{split} \label {eq:289} A_{\vec 1_ z}&=\begin{pmatrix} \cos \theta & 0 & 0\\ 0 &\cos \theta & 0\\ 0 & 0 & \cos \theta \end {pmatrix}+ \begin{pmatrix} 0 & 0 & 0\\ 0 &0 & 0\\ 0 & 0 & \frac {\cancel {\theta ^2}(1-\cos \theta )}{\cancel {\theta ^2}} \end {pmatrix}+ \begin{pmatrix} 0 & -\sin \theta & 0\\ \sin \theta & 0 & 0\\ 0 & 0 & 0 \end {pmatrix}\\&= \begin{pmatrix} \cos \theta & -\sin \theta & 0\\ \sin \theta &\cos \theta & 0\\ 0 & 0 & 1 \end {pmatrix}, \end {split} \end{equation}

which is the expected result from the viewpoint of a rotation matrix in the \((x,y)\) plane.

7.2.5 One-parameter groups of transformations involving time

Consider now a one-parameters group of invertible transformations that affects time as well,

\begin{equation} \label {eq:95} \left \{\begin{array}{l}q^{\prime \alpha }=q^{\prime \alpha }(q^\beta ,t;s),\\ t'=t'(t;s)\end {array}\right . \Longrightarrow q^{\prime \alpha }(t') =q^{\prime \alpha }(q^\beta (t(t';s)),t(t';s);s), \end{equation}

with \(q^{\prime \alpha }=q^{\prime \alpha }(q^\beta ,t;0)=q^\alpha \), \(t'(t;0)=t\). According to the equation (6.26) of the previous section, the transformed Lagrangian is

\begin{equation} \label {eq:96} L'(q'(t'),\frac {dq'(t')}{dt'},t')=\frac {dt}{dt'}L \big (q(q'(t'),t'),\frac {dt'}{dt}\frac {d}{dt'}q(q'(t'),t'),t(t')\big ), \end{equation}

where we have temporarily suppressed the \(s\) dependence. Let us assume that the transformed Lagrangian is equal to the starting point Lagrangian (in terms of the new variables), up to a total derivative,

\begin{equation} \label {eq:97} L'(q'(t'),\frac {dq'(t')}{dt'},t') =L(q'(t'),\frac {d}{dt'}q'(t'),t')-\frac {d}{dt'}G'(q'(t'),t'), \end{equation}

with \(G'|_{s=0}=0\). When using the inverse transformation and multiplying by \(\frac {dt'}{dt}\), the generalized invariance condition (7.77), where the left hand side has been expressed in terms of \(L\) by using (7.76), can also be written as

\begin{equation} \label {eq:98} L(q(t),\frac {d}{dt}q(t),t)=\frac {dt'}{dt}L\big (q'(q,t),\frac {dt}{dt'} \frac {d}{dt}q'(q(t),t),t'(t)\big )- \frac {d}{dt}G(q(t),t), \end{equation}

with \(G(q(t),t)=G'(q'(q(t),t),t'(t))\).

In terms of the action, this generalized invariance condition means that as a functional of the new trajectories and the transformed end-points, the

\begin{equation} \label {eq:99} S[q']=\int ^{t'=t'(t_f)}_{t'=t'(t_i)}dt L\big (q'(t),\frac {d}{dt}q'(t),t), \end{equation}

equals the action as a functional of the old trajectories, up to a boundary term. Indeed, by first changing the name of the dummy integration variable, and then doing the change of variables from primed to unprimed ones, we get

\begin{equation} \label {eq:100} S[q']=\int ^{t'=t'(t_f)}_{t'=t'(t_i)}dt' L(q'(t'),\frac {d}{dt'} q'(t'),t')=\int ^{t=t_f}_{t=t_i}dt\frac {dt'}{dt} L\big (q'(q(t),t),\frac {dt}{dt'}\frac {d}{dt} q'(q(t),t),t'(t)\big ). \end{equation}

When using the generalized invariance condition (7.78), this gives

\begin{equation} \label {eq:101} S[q']=\int ^{t_f}_{t_i}dt L(q(t),\frac {d}{dt} q(t),t)+\Big [G(q(t),t)\Big ]^{t_f}_{t_i}. \end{equation}

Note that the boundary term does not contribute when computing the variation of trajectories with fixed end-points.

Expanding the transformed trajectories to first order in \(s\) gives

\begin{equation} \label {eq:95a} \left \{\begin{array}{l}q^{\prime \alpha }(t')=q^{\alpha }(t) +s R^\alpha (q^\beta ,t)+O(s^2),\\ t'=t+s\xi (t)+O(s^2),\end {array}\right .,\quad G=sG_1(q,t)+O(s^2), \end{equation}

the associated equal-time infinitesimal virtual variation is

\begin{multline} \label {eq:102} \boxed {\delta _S q^\alpha (t)}=\lim _{s\to 0}\frac {1}{s}\big (q^{\prime \alpha }(t)-q^\alpha (t)\big )= \lim _{s\to 0}\frac {1}{s}\big (q^{\prime \alpha }(t'-s\xi (t)+O(s^2))-q^\alpha (t)\big )\\= \lim _{s\to 0}\frac {1}{s}\big (\cancel {q^\alpha (t)}+s R^\alpha (q^\beta (t),t)-s\frac {dq^{\prime \alpha }}{dt'}(t')\xi (t) +O(s^2)-\cancel {q^\alpha (t)}\big )=\boxed {R^\alpha -\dot q^\alpha \xi }. \end{multline} Such a variation leaves the Lagrangian invariant up to a total derivative. Indeed, the generalized invaraince condition (7.78) can be written as

\begin{equation} \label {eq:103} L(q,\dot q,t)=(1+s\dot \xi )\big (L(q+sR,(1-s\dot \xi )\frac {d}{dt}(q+sR),t+s\xi \big )-s\frac {d}{dt} G_1+O(s^2). \end{equation}

To order \(0\) in \(s\), this is an identity, while the terms of order \(1\) yield

\begin{equation} \label {eq:104} \dot \xi L(q,\dot q,t)+\frac {\partial L}{\partial q^\alpha } R^\alpha + \frac {\partial L}{\partial \dot q^\alpha }(\frac {d}{dt}R^\alpha -\dot \xi \dot q^\alpha )+\cancel {\frac {\partial L}{\partial t}(q,\dot q,t)\xi }-\frac {d}{dt} G_1=0. \end{equation}

The first term can be integrated by parts

\begin{equation} \label {eq:105} \dot \xi L=\frac {d}{dt}(\xi L)-\xi (\frac {\partial L}{\partial q^\alpha }\dot q^\alpha +\frac {\partial L}{\partial \dot q^\alpha }\ddot q^\alpha +\cancel {\frac {\partial L}{\partial t}}), \end{equation}

and, when cancelling (as indicated) and combining like terms, we get

\begin{equation} \label {eq:106} \frac {\partial L}{\partial q^\alpha }(R^\alpha -\dot q^\alpha \xi )+\frac {\partial L}{\partial \dot q^\alpha }\frac {d}{dt}(R^\alpha -\dot q^\alpha \xi )=\frac {d}{dt}(G_1-\xi L) \iff \delta _S L=\frac {d}{dt}( G_1-\xi L). \end{equation}

The associated conserved quantity is

\begin{equation} \label {eq:107} \boxed {K=\frac {\partial L}{\partial \dot q^\alpha }\delta _S q^\alpha +\xi L-G_1}. \end{equation}

7.2.6 Time-translation symmetry and conservation of energy

In particular, a Lagrangian does not depend explicit on time, \(\frac {\partial L}{\partial t}=0\) iff it possesses the symmetry \(q^{\prime \alpha }=q^\alpha \), \(t'=t-s\). In this case \(G=0\), \(\xi =-1\), \(R^\alpha =0\), \(\delta _S q^\alpha =\dot q^\alpha \). The associated conserved quantity is

\begin{equation} \label {eq:109} E=\frac {\partial L}{\partial \dot q^\alpha }\dot q^\alpha -L, \end{equation}

discussed previously, which now appears from the viewpoint of the symmetry of the system under constant time translations.

7.2.7 Galilean boosts

Galilean transformations on a system of particles are described by

\begin{equation} \label {eq:108} x^{\prime \alpha }_i={A^\alpha }_\beta x^\beta _i+v^\alpha t+a^\alpha , t'=t+b, \quad A^TA=\mathbb {1}. \end{equation}

They consists of a combination of spatial translations, rotations, time-translations, described by the constant parameter \(b\), and changes of observers with a constant velocity in a given direction, sometimes called “Galilean boosts” and described by the \(3\) parameters \(v^\alpha \).

Sometimes it is useful to provide the following \(5\times 5 \) matrix form of these transformations (suppressing the particle label)

\begin{equation} \label {eq:254} \begin{pmatrix} x^{\prime \alpha }\\ t'\\ 1 \end {pmatrix} =\begin{pmatrix} {A^\alpha }_\beta & v^\alpha & a^\alpha \\ 0 0 0 & 1 & b\\ 0 0 0 & 0 & 1 \end {pmatrix} \begin{pmatrix} x^\beta \\ t\\ 1 \end {pmatrix}. \end{equation}

We have already analyzed invariance of the particle Lagrangian (7.23) under spatial translations (parameters \(a^\alpha \)) and rotations (parameters \(\omega ^\alpha \) of the rotation matrix \(A\)) and the associated conserved quantities \(P_\alpha ,J_\alpha \). For the time-translation, explicit time-indepence implies that the associated conserved conserved quanity is the energy, \(E=\sum _i\frac {\partial L}{\partial \dot x^\alpha _{(i)}}\dot x^\alpha _{(i)}-L\). So only the Galilean boosts,

\begin{equation} \label {eq:266} x^{\prime \alpha }_i= x^\alpha _i+v^\alpha t,\quad t'=t, \end{equation}

remain to be analyzed explicitely.

More generally, for the whole set of Galilean transformations, we have \(\frac {dt'}{dt}=1\), \(\frac {dx^{\prime \alpha }_{i}(dt')}{t'}={A^\alpha }_\beta \frac {dx^{\beta }_{i}(t)}{dt} +v^\alpha \). For the kinetic energy of the particle Lagrangian (7.23), this gives

\begin{multline} \label {eq:110} T(\frac {dx^{\prime \alpha }_i(t')}{dt'})=\frac 12 \sum _im_{(i)}({A^\alpha }_\beta \frac {dx^\beta _{(i)}}{dt} +v^\alpha )(A_{\alpha \gamma }\frac {dx^\gamma _{(i)}}{dt} +v_\alpha )\\= \frac 12 \sum _im_{(i)} \frac {dx^\alpha _{(i)}}{dt}\frac {dx_{\alpha (i)}}{dt} +\sum _i m_{(i)} v_\alpha {A^\alpha }_\beta \frac {dx^\beta _{(i)}}{dt}+\frac 12 \sum _i m_{(i)} v^\alpha v_\alpha \\=T(\frac {dx^\alpha _{i}}{dt})+\frac {d}{dt}(\sum _i m_{(i)} v_\alpha ({A^\alpha }_\beta x^\beta _{(i)}+\frac 12 t v^\alpha ). \end{multline} Since relative distances are invariant even if one translates the individual coordinates linearly in time (by the same amount), the potential is also invariant. It follows that

\begin{equation} \label {eq:111} G=\sum _i m_{(i)} v_\alpha ({A^\alpha }_\beta x^\beta _{(i)}+\frac 12 t v^\alpha ), \end{equation}

and, for Galilean boosts parametrized by \(v^\gamma \), \(R^\alpha _{\gamma i}=\delta ^\alpha _\gamma t\), \(\xi =0\), \(\frac {\partial G}{\partial v^\gamma }|_{a,b,\theta ,v=0}=\sum _im_{(i)} x_{\gamma (i)}\). The associated conserved quantities are thus

\begin{equation} \label {eq:112} K_\gamma =\sum _im_{(i)} (\dot x_{\gamma (i)} t-x_{\gamma (i)}). \end{equation}

7.2.8 Worked Exercise 4: Geodesic motion and Killing vectors

Consider the generalized coordinates \(q^\alpha \) and the Lagrangian

\begin{equation} \label {eq:445} L=\frac 12 g_{\alpha \beta }\dot q^\alpha \dot q^\beta , \end{equation}

where the “metric” \(g_{\alpha \beta }=g_{\alpha \beta }(q)\) is a given function of the coordinates.

Defining the Christoffel symbols of the second kind by

\begin{equation} \label {eq:446} \Gamma ^\delta _{\beta \gamma }=g^{\delta \alpha }\Gamma _{\alpha \beta \gamma }, \end{equation}

where \(g^{\delta \alpha }\) denotes the inverse metri, \(g^{\delta \alpha }g_{\alpha \beta }=\delta ^\delta _\beta \), show that the equations of motion are equivalent to the geodesic equation

\begin{equation} \label {eq:449} \ddot q^\delta +\Gamma ^{\delta }_{\beta \gamma }\dot q^\beta \dot q^\gamma =0. \end{equation}

Solution: This follows directly by multiplying the two non-vanshing first terms in (4.35) by the inverse metric.

Let us introduce the notation \(\partial _\alpha =\frac {\partial }{\partial q^\alpha }\). Suppose that there is a \(\xi ^\alpha (q)\) that satisfies the “Killing equation”

\begin{equation} \label {eq:447} \xi ^\gamma \partial _\gamma g_{\alpha \beta }+\partial _\alpha \xi ^\gamma g_{\gamma \beta }+\partial _\beta \xi ^\gamma g_{\alpha \gamma }=0. \end{equation}

Show that the transformation \(\delta _\xi ^1 q^\alpha =\xi ^\alpha (q)\) defines a symmetry with associated Noether charge

\begin{equation} \label {eq:448} K_\xi ^1=g_{\alpha \beta }\dot q^\alpha \xi ^\beta . \end{equation}

Solution: The variation of the Lagrangian vanishes on account of (7.99),

\begin{equation} \label {eq:450} \delta _\xi ^1 L=\frac 12 \partial _\gamma g_{\alpha \beta } \xi ^\gamma +\dot q^\alpha \dot q^\beta +g_{\alpha \gamma }\dot q^\alpha \partial _\beta \xi ^\gamma \dot q^\beta =0. \end{equation}

The result then follows from the formula for the Noether charge.

7.2.9 Worked Exercise 5: Geodesic motion and Killing tensors

This exercise requires an understanding of the covariant derivative \(D_\mu \) defined in terms of the Levi-Civita connection \(\Gamma ^{\delta }_{\beta \gamma }\). It is metric compatible \(D_\alpha g_{\beta \gamma }=0\). The Killing equation (7.99) can then be shown to be equivalent to

\begin{equation} \label {eq:452} D_{(\alpha _0}\xi _{\alpha _1)}=0, \end{equation}

where indices are lowered and raised with the metric \(g_{\mu \nu }\) and its inverse, and a round parenthesis denotes complete symmetrization of the enclosed indices,

\begin{equation} \label {eq:453} M_{(\alpha _1\dots \alpha _k)}=\frac {1}{k!}\sum _{\sigma \in S_k}M_{\alpha _{\sigma (1)}\dots \alpha _{\sigma (k)}}. \end{equation}

A Killing tensor is a symmetric tensor \(\xi _{\alpha _1\dots \alpha _k}=\xi _{(\alpha _1\dots \alpha _k)}\) that satisfies

\begin{equation} \label {eq:454} D_{(\alpha _0}\xi _{\alpha _1\dots \alpha _k)}=0. \end{equation}

In particular, a Killing vector corresponds to the particular case where \(k=1\).

Show that

\begin{equation} \label {eq:455} K_\xi ^k=\frac {1}{k!}\xi _{\alpha _1\dots \alpha _k}\dot q^{\alpha _1}\dots \dot q^{\alpha _k} \end{equation}

is a conserved quantity and that the associated symmetry is

\begin{equation} \label {eq:456} \delta _\xi ^kq^\alpha =\frac {1}{(k-1)!}\xi ^\alpha _{\alpha _2\dots \alpha _k}\dot q^{\alpha _2}\dots \dot q^{\alpha _k}. \end{equation}

Solution:

\begin{multline} \label {eq:457} \frac {d}{dt} K^k_\xi =\frac {1}{k!}\partial _{\alpha _0}\xi _{\alpha _1\dots \alpha _k}\dot q^{\alpha _0}\dot q^{\alpha _1}\dots \dot q^{\alpha _k}+\frac {1}{(k-1)!} \xi _{\alpha _1\dots \alpha _k}\ddot q^{\alpha _1}\dot q^{\alpha _2}\dots \dot q^{\alpha _k}\\= \frac {1}{k!}\partial _{\alpha _0}\xi _{\alpha _1\dots \alpha _k}\dot q^{\alpha _0}\dot q^{\alpha _1}\dots \dot q^{\alpha _k}+\frac {1}{(k-1)!} \xi _{\alpha _1\dots \alpha _k}(-g^{\alpha _1\beta }\frac {\delta L}{\delta q^\beta })\dot q^{\alpha _2}\dots \dot q^{\alpha _k}\\-\frac {1}{(k-1)!} \xi _{\beta \alpha _2\dots \alpha _k}\Gamma ^\beta _{\alpha _0\alpha _1}\dot q^{\alpha _0}\dot q^{\alpha _1}\dots \dot q^{\alpha _k}. \end{multline} The first and the third term vanish on account of (7.105). Since the second term is proportional to the field equations, \(K^k_\xi \) is a conserved quantity, and one may read off the characteristic of the symmetry (7.106).

7.2.10 Worked Exercise 6: Laplace-Runge-Lenz vector

7.2.11 Worked Exercise 7: Harmonic oscillator conserved quantities