Extend Noether's theorem to non-autonomous lagrangian systems. Hint. Let \(M_{1}=M \times \mathbb{R}\) be the extended configuration space (the direct product of the configuration manifold \(M\) with the time axis \(\mathbb{R}\) ). Define a function \(L_{1}: T M_{1} \rightarrow \mathbb{R}\) by $$ L \frac{d t}{d z} \text { : } $$ i.e., in lucal coordinates \(q, t\) on \(M_{1}\) we define it by the formula $$ L_{1}\left(\mathbf{q}, t, \frac{d \mathbf{q}}{d \tau}, \frac{d t}{d \tau}\right)=L\left(\mathbf{q}, \frac{d \mathbf{q} / d \tau}{d t, d \tau}, t\right) \frac{d t}{d \tau} $$ We apply Noether's theorem to the lagrangian system \(\left(M_{2}, L_{1}\right)\). If \(L_{1}\) admits the transformations \(h^{s}: M_{1} \rightarrow M_{1}\). we obtain a first integral \(l_{1}: T M_{1} \rightarrow \mathbb{R}\). Since \(\int L d t=\int L_{1} d \tau_{\text {. this reduces }}\) to a first integral \(I: T M \times\) of \(\rightarrow \mathbb{R}\) of the viginal system. If. in local coordinates \((\boldsymbol{q}, t)\) on \(M_{1}\), we have \(I_{1}=I_{1}(\mathbf{q}, t, d \mathbf{q} / d \tau, d l / d \tau)\), then \(I(\mathbf{q}, \mathbf{q}, t)=l_{1}\left(\mathbf{q}, I_{1}, \dot{\mathbf{q}}, 1\right)\). In particular, if \(L\) does not depend on time, \(L_{1}\) admits translations along time, \(h^{s}(q, t)=\) (q. \(t+5\) ). The corresponding first integral \(l\) is the energy integral.

Step by step solution

01

Define the extended configuration space#

First, let's define the extended configuration space \(M_1\) which is the direct product of the configuration manifold \(M\) with the time axis \(\mathbb{R}\). This will help us extend Noether's theorem to non-autonomous lagrangian systems.

02

Define the function \(L_1\)#

Define a new function \(L_1: T M_1 \rightarrow \mathbb{R}\) as: \[ L_1\left(\mathbf{q}, t, \frac{d \mathbf{q}}{d \tau}, \frac{d t}{d \tau}\right)=L\left(\mathbf{q}, \frac{d \mathbf{q} / d \tau}{d t, d \tau}, t\right) \frac{d t}{d \tau}. \] This new function will help us apply Noether's theorem to the extended lagrangian system \((M_2, L_1)\).

03

Apply Noether's theorem to the extended lagrangian system#

Suppose \(L_1\) admits transformations \(h^s: M_1 \rightarrow M_1\). By applying Noether's theorem to the extended lagrangian system, we can obtain a first integral \(l_1: T M_1 \rightarrow \mathbb{R}\).

04

Relate the first integral \(l_1\) to a first integral \(I\) of the original system#

Since $\int L d t=\int L_{1} d \tau\(, the first integral \)l_1\( reduces to a first integral \)I: T M \times \rightarrow \mathbb{R}$ of the original system. In local coordinates \((\boldsymbol{q}, t)\) on \(M_{1}\), if we have $I_{1}=I_{1}(\mathbf{q}, t, d \mathbf{q} / d \tau, d l / d \tau)\(, then \)I(\mathbf{q}, \mathbf{q}, t)=l_{1}\left(\mathbf{q}, I_{1}, \dot{\mathbf{q}}, 1\right)$.

05

Find the energy integral for \(L\) not depending on time#

In particular, if the original lagrangian \(L\) does not depend on time, \(L_1\) admits translations along time, \(h^{s}(q, t)=\) (\(q\), \(t+5\)). The corresponding first integral \(l\) is the energy integral, which is a conserved quantity for the original non-autonomous lagrangian system.

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