# Question 9106d

Dec 19, 2016

a)
Given $L = {t}^{2} \left({\dot{x}}^{2} / 2 - {x}^{6} / 6\right)$ the movement equations are obtained by doing

$\frac{d}{\mathrm{dt}} \left(\frac{\partial L}{\partial \dot{x}}\right) - \frac{\partial L}{\partial x} = 0$ giving

$\ddot{x} = - \left(2 \frac{\dot{x}}{t} + {x}^{5}\right)$

b)

The condition for invariance of $\int L \mathrm{dt}$ is given by

$\frac{\partial L}{\partial t} \tau + \frac{\partial L}{\partial x} \xi + \frac{\partial L}{\partial \dot{x}} \left(\frac{d \xi}{\mathrm{dt}} - \dot{x} \frac{d \tau}{\mathrm{dt}}\right) + L \frac{d \tau}{\mathrm{dt}} = 0$

Now expanding the total derivatives $\frac{d \xi}{\mathrm{dt}} , \frac{d \tau}{\mathrm{dt}}$ and substituting we arrive at

$\frac{\partial L}{\partial t} \tau + \frac{\partial L}{\partial x} \xi + \frac{\partial L}{\partial \dot{x}} \left(\frac{\partial \xi}{\partial t} + \frac{\partial \xi}{\partial x} \dot{x} - \dot{x} \frac{\partial \tau}{\partial t} - {\dot{x}}^{2} \frac{\partial \tau}{\partial x}\right) + L \left(\frac{\partial \tau}{\partial t} + \frac{\partial \tau}{\partial x} \dot{x}\right) = 0$

From $L$ we have

$\frac{\partial L}{\partial t} = t \left({\dot{x}}^{2} - \frac{1}{3} {x}^{6}\right)$
$\frac{\partial L}{\partial x} = - {t}^{2} {x}^{5}$
$\frac{\partial L}{\partial \dot{x}} = {t}^{2} \dot{x}$

Substituting and grouping the coefficients associated to the powers of $\dot{x}$ and equating to zero we have

{ ((x tau)/3+t xi + t x/6 (partial tau)/(partial t)=0), ((partial xi)/(partial t)-x^6/6 (partial tau)/(partial x) = 0), (t tau+t^2(partial xi)/(partial x) - t^2/2 (partial tau)/(partial t) = 0),((partial tau)/(partial x)=0):}

Now from the last equation, $\tau = \tau \left(t\right)$ substituting into the second equation $\xi = \xi \left(x\right)$ remaining the system of differential equations

{ (x/3 tau+t xi + t/6x (d tau)/(dt)=0), (t tau+t^2 (d xi)/(dx)-t^2/2 (d tau)/(dt)=0):}#

From inspection, the transformation is

$\tau = t$ and $\xi = - \frac{x}{2}$ so

$\int L \mathrm{dt}$ is invariant under the transformation group

$t ' = t + \epsilon t$
$x ' = x - \frac{1}{2} \epsilon x$

c) Suppose that $L \left(t , x , \dot{x}\right)$ is variationally invariant on ${t}_{i} , {t}_{f}$ under the former transformation. Then

$\frac{\partial L}{\partial \dot{x}} \xi + \left(L - \dot{x} \frac{\partial L}{\partial \dot{x}}\right) \tau = {C}^{t e}$

Substituting $\frac{\partial L}{\partial \dot{x}} , L , \xi , \tau$ we get

$\frac{{t}^{2} x \dot{x}}{2} + \frac{{t}^{3} {\dot{x}}^{2}}{2} + \frac{{t}^{3} {x}^{6}}{6} = {C}^{t e}$