Politics.be - Verzoekje

Ja 't is niet mijn fout dat hier geen wiskundige symbolen kunnen getoond worden.

Citaat:

Derivations
[edit] One independent variable

Consider the simplest case, a system with one independent variable, time. Suppose the dependent variables \mathbf{q} are such that the action integral

I = \int_{t_1}^{t_2} L [\mathbf{q} [t], \dot{\mathbf{q}} [t], t] \, dt

is invariant under brief infinitesimal variations in the dependent variables. In other words, they satisfy the Euler–Lagrange equations

\frac{d}{dt} \frac{\partial L}{\partial \dot{\mathbf{q}}} [t] = \frac{\partial L}{\partial \mathbf{q}} [t] .

And suppose that the integral is invariant under a continuous symmetry. Mathematically such a symmetry is represented as a flow, \mathbf{\phi}, which acts on the variables as follows

t \rightarrow t' = t + \epsilon T \!
\mathbf{q} [t] \rightarrow \mathbf{q}' [t'] = \phi [\mathbf{q} [t], \epsilon] = \phi [\mathbf{q} [t' - \epsilon T], \epsilon]

where \epsilon \! is a real variable indicating the amount of flow and T is a real constant (which could be zero) indicating how much the flow shifts time.

\dot\mathbf{q} [t] \rightarrow \dot\mathbf{q}' [t'] = \frac{d}{dt} \phi [\mathbf{q} [t], \epsilon] = \frac{\partial \phi}{\partial \mathbf{q}} [\mathbf{q} [t' - \epsilon T], \epsilon] \dot\mathbf{q} [t' - \epsilon T] .

The action integral flows to

I' [\epsilon] = \int_{t_1 + \epsilon T}^{t_2 + \epsilon T} L [\mathbf{q}'[t'], \dot\mathbf{q}' [t'], t'] \, dt'
= \int_{t_1 + \epsilon T}^{t_2 + \epsilon T} L [\phi [\mathbf{q} [t' - \epsilon T], \epsilon], \frac{\partial \phi}{\partial \mathbf{q}} [\mathbf{q} [t' - \epsilon T], \epsilon] \dot\mathbf{q} [t' - \epsilon T], t'] \, dt'

which may be regarded as a function of ε. Calculating the derivative at ε = 0 and using the symmetry, we get

0 = \frac{d I'}{d \epsilon} [0] = L [\mathbf{q} [t_2], \dot{\mathbf{q}} [t_2], t_2] T - L [\mathbf{q} [t_1], \dot{\mathbf{q}} [t_1], t_1] T +

\int_{t_1}^{t_2} \frac{\partial L}{\partial \mathbf{q}} \left( - \frac{\partial \phi}{\partial \mathbf{q}} \dot{\mathbf{q}} T + \frac{\partial \phi}{\partial \epsilon} \right) + \frac{\partial L}{\partial \dot{\mathbf{q}}} \left( - \frac{\partial^2 \phi}{(\partial \mathbf{q})^2} {\dot\mathbf{q}}^2 T + \frac{\partial^2 \phi}{\partial \epsilon \partial \mathbf{q}} \dot\mathbf{q} - \frac{\partial \phi}{\partial \mathbf{q}} \ddot\mathbf{q} T \right) \, dt .

Notice that the Euler–Lagrange equations imply

\frac{d}{dt} \left( \frac{\partial L}{\partial \dot\mathbf{q}} \frac{\partial \phi}{\partial \mathbf{q}} \dot{\mathbf{q}} T \right) = \left( \frac{d}{dt} \frac{\partial L}{\partial \dot\mathbf{q}} \right) \frac{\partial \phi}{\partial \mathbf{q}} \dot{\mathbf{q}} T + \frac{\partial L}{\partial \dot\mathbf{q}} \left( \frac{d}{dt} \frac{\partial \phi}{\partial \mathbf{q}} \right) \dot{\mathbf{q}} T + \frac{\partial L}{\partial \dot\mathbf{q}} \frac{\partial \phi}{\partial \mathbf{q}} \ddot{\mathbf{q}} \, T
= \frac{\partial L}{\partial \mathbf{q}} \frac{\partial \phi}{\partial \mathbf{q}} \dot{\mathbf{q}} T + \frac{\partial L}{\partial \dot\mathbf{q}} \left( \frac{\partial^2 \phi}{(\partial \mathbf{q})^2} \dot{\mathbf{q}} \right) \dot{\mathbf{q}} T + \frac{\partial L}{\partial \dot\mathbf{q}} \frac{\partial \phi}{\partial \mathbf{q}} \ddot{\mathbf{q}} \, T .

Substituting this into the previous equation, one gets

0 = \frac{d I'}{d \epsilon} [0] = L [\mathbf{q} [t_2], \dot{\mathbf{q}} [t_2], t_2] T - L [\mathbf{q} [t_1], \dot{\mathbf{q}} [t_1], t_1] T - \frac{\partial L}{\partial \dot\mathbf{q}} \frac{\partial \phi}{\partial \mathbf{q}} \dot{\mathbf{q}} [t_2] T + \frac{\partial L}{\partial \dot\mathbf{q}} \frac{\partial \phi}{\partial \mathbf{q}} \dot{\mathbf{q}} [t_1] T +

\int_{t_1}^{t_2} \frac{\partial L}{\partial \mathbf{q}} \frac{\partial \phi}{\partial \epsilon} + \frac{\partial L}{\partial \dot{\mathbf{q}}} \frac{\partial^2 \phi}{\partial \epsilon \partial \mathbf{q}} \dot\mathbf{q} \, dt .

Again using the Euler–Lagrange equations we get

\frac{d}{d t} \left( \frac{\partial L}{\partial \dot\mathbf{q}} \frac{\partial \phi}{\partial \epsilon} \right) = \left( \frac{d}{d t} \frac{\partial L}{\partial \dot\mathbf{q}} \right) \frac{\partial \phi}{\partial \epsilon} + \frac{\partial L}{\partial \dot{\mathbf{q}}} \frac{\partial^2 \phi}{\partial \epsilon \partial \mathbf{q}} \dot\mathbf{q} = \frac{\partial L}{\partial \mathbf{q}} \frac{\partial \phi}{\partial \epsilon} + \frac{\partial L}{\partial \dot{\mathbf{q}}} \frac{\partial^2 \phi}{\partial \epsilon \partial \mathbf{q}} \dot\mathbf{q} .

Substituting this into the previous equation, one gets

0 = L [\mathbf{q} [t_2], \dot{\mathbf{q}} [t_2], t_2] T - L [\mathbf{q} [t_1], \dot{\mathbf{q}} [t_1], t_1] T - \frac{\partial L}{\partial \dot\mathbf{q}} \frac{\partial \phi}{\partial \mathbf{q}} \dot{\mathbf{q}} [t_2] T + \frac{\partial L}{\partial \dot\mathbf{q}} \frac{\partial \phi}{\partial \mathbf{q}} \dot{\mathbf{q}} [t_1] T +

\frac{\partial L}{\partial \dot\mathbf{q}} \frac{\partial \phi}{\partial \epsilon} [t_2] - \frac{\partial L}{\partial \dot\mathbf{q}} \frac{\partial \phi}{\partial \epsilon} [t_1] .

From which one can see that

\left( \frac{\partial L}{\partial \dot\mathbf{q}} \frac{\partial \phi}{\partial \mathbf{q}} \dot{\mathbf{q}} - L \right) T - \frac{\partial L}{\partial \dot\mathbf{q}} \frac{\partial \phi}{\partial \epsilon}

is a constant of the motion, i.e. a conserved quantity. Since \phi [\mathbf{q}, 0] = \mathbf{q}, we get \frac{\partial \phi}{\partial \mathbf{q}} = 1 and so the conserved quantity simplifies to

\left( \frac{\partial L}{\partial \dot\mathbf{q}} \dot{\mathbf{q}} - L \right) T - \frac{\partial L}{\partial \dot\mathbf{q}} \frac{\partial \phi}{\partial \epsilon} .