PeroK said:
I've just been learning about Noether's theorem, so I thought I'd try this one. The notation is a bit different from what I'm used to.
a) The Lagrangian is independent of time, so energy is conserved (Euler-Lagrange equation for time coordinate):
##\dot{H} = - \frac{\partial L}{\partial t} = 0##
Hence, the total energy ##E## is equal to the Hamiltonian:
##E = H = p_i \dot{x^i} - L = \frac{\partial L}{\partial \dot{x^i}}\dot{x^i} - L = m\dot{x^i}\dot{x^i} - L = \frac12 mv^2 + \frac{U_0}{r^2}##
Hence:
##E = \frac12 mv_0^2 + \frac{U_0}{r_0^2}##
b) We have the transformation:
##x^* = (1 + \epsilon)x; \ y^* = (1 + \epsilon)y; \ z^* = (1 + \epsilon)z; \ t^* = (1 + \epsilon)^2t##
Hence: ##\psi^x = x; \ \psi^y = y; \ \psi^z = z; \ \phi = 2t##
The transformed Lagrangian is:
##L^* = \frac12 m(v^*)^2 - \frac{U_0}{(r^*)^2}##
Where:
##(v^*)^2 = (\frac{dx^*}{dt^*})^2 + \dots = (1+\epsilon)^{-2}(\frac{dx}{dt})^2 + \dots = (1+\epsilon)^{-2}v^2##
And:
##(r^*)^2 = (1+\epsilon)^{2}r^2##
Therefore, we have:
##L^* = L(1+\epsilon)^{-2}##
Hence:
##L^* \frac{dt^*}{dt} = L##
Which is the condition for invariance under the transformation.
c) The corresponding conserved Noether Charge is given by:
##Q = p_i\psi^i - H\phi = m(\dot{x}x + \dot{y}y + \dot{z}z) - 2Et = m\vec{v} \cdot \vec{r} - 2Et##
At ##t=0##, we have ##Q = m\vec{v_0} \cdot \vec{r_0}##, which leads to:
##m\frac{d}{dt}(r^2) = 2m\vec{v} \cdot \vec{r} = 2Q + 4Et##
And
##mr^2 = 2Et^2 + 2Qt + mr_0^2##
Which implies only unstable (circular) orbits.
Thank you for solving this! I already began to believe it would survive Emmy's centenary!
There is nothing to add to your solutions. Mine (taken from an exam for engineers!) is a bit closer in notation to the one given by the problem statement and the hints, so I will add it here - mostly because it took so long to solve this one, so people might choose what they find easier to grasp, so please don't interpret it as patronizing or so.
a)
(i) Energy is
homogeneous in time, so we chose ##\psi_i=0 , \varphi=1##
(ii) and check our equation by
\begin{equation*}
\left. \dfrac{d}{d\varepsilon}\right|_{\varepsilon = 0} \left(\mathcal{L}^*\,\cdot\,\dfrac{d}{dt}\,(t+\varepsilon )\right)=\left. \dfrac{d}{d\varepsilon}\right|_{\varepsilon = 0} \left(\mathcal{L}^*\,\cdot\,1\right) = 0
\end{equation*}
since ##\mathcal{L}^*## doesn't depend on ##t^*## and thus not on ##\varepsilon##, and calculate
(iii) the Noether charge as
\begin{align*}
Q(t,x,\dot{x})&=\mathcal{L}- \sum_{i=1}^N\dfrac{\partial \mathcal{L}}{\partial \dot{x}_i} \,\dot{x}_i\\
&=T-U-\dfrac{m}{2}\left( \dfrac{\partial}{\partial \dot{x}_i}\left( \sum_{i=1}^3 \dot{x}^2_i \right)\,\dot{x}_i \right)\\
&=\dfrac{m}{2}\, \dot{\vec{r\,}}^2 - U -m\,\dot{\vec{r\,}}^2\\
&=-T-U\\
&=-E\\
&=-\dfrac{m}{2}\, \dot{\vec{r\,}}^2- \dfrac{U}{\vec{r\,}^2}\\
&=-\dfrac{m}{2}\, \dot{\vec{r\,}}_0^2- \dfrac{U}{\vec{r\,}_0^2}
\end{align*}
by time invariance.
b) ##\dot{\vec{r}}\,^*=\dfrac{d\vec{r}\,^*}{dt^*}=\dfrac{(1+\varepsilon)\,d\vec{r}}{(1+\varepsilon)^2\, dt }=\dfrac{1}{1+\varepsilon}\,\dot{\vec{r}}\,## and thus ##\,\mathcal{L}^*=\dfrac{1}{(1+\varepsilon)^2}\,\mathcal{L}\, ##, i.e.
\begin{align*}
\left. \dfrac{d}{d\varepsilon}\right|_{\varepsilon =0}\left(\mathcal{L}\left(t^*,x^*,\dot{x}^*\right)\cdot \dfrac{dt^*}{dt} \right) &= \left. \dfrac{d}{d\varepsilon}\right|_{\varepsilon =0} \mathcal{L}^*\,\dfrac{dt^*}{dt}\\ &=\left. \dfrac{d}{d\varepsilon}\right|_{\varepsilon =0} \dfrac{\mathcal{L}}{(1+\varepsilon)^2}\cdot (1+\varepsilon)^2\\ &=\left. \dfrac{d}{d\varepsilon}\right|_{\varepsilon =0}\mathcal{L} \\&= 0
\end{align*}
and the condition of Noether's theorem holds.
c) For the given transformations we have
\begin{align*}
x &\longmapsto x^* = (1+\varepsilon)x & \Longrightarrow \quad& \psi_x=x\\
y &\longmapsto y^* = (1+\varepsilon)y & \Longrightarrow \quad& \psi_y=y\\
z &\longmapsto z^* = (1+\varepsilon)z & \Longrightarrow \quad& \psi_z=z\\
t &\longmapsto t^* = (1+2\varepsilon)t & \Longrightarrow \quad& \varphi=2t\\
\end{align*}
and so the Noether charge is given by
\begin{align*}
Q(t,x,\dot{x})&= \sum_{i=1}^N \dfrac{\partial \mathcal{L}}{\partial \dot{x}_i}\,\psi_i + \left(\mathcal{L}-\sum_{i=1}^N \dfrac{\partial \mathcal{L}}{\partial \dot{x}_i}\,\dot{x}_i\right)\varphi\\
&=\sum_{i=1}^3 \dfrac{\partial}{\partial \dot{x}_i}\left(\dfrac{m}{2}\,\dot{\vec{r}\,}^2-\dfrac{U_0}{\vec{r\,}^{2}}\right)\,\psi_i \,+\\
&+ \left(\dfrac{m}{2}\,\dot{\vec{r}}\,^2-\dfrac{U_0}{\vec{r\,}^{2}}-\sum_{i=1}^3 \dfrac{\partial }{\partial \dot{x}_i}\,\left(\dfrac{m}{2}\,\dot{\vec{r}}\,^2-\dfrac{U_0}{\vec{r\,}^{2}}\right)\dot{x}_i\right)\varphi\\
&=m(\dot{x}x+\dot{y}y+\dot{z}z) \,+ \\
&+\left( \dfrac{m}{2}\dot{\vec{r\,}}^2-\dfrac{U_0}{\vec{r\,}^{2}}-m(\dot{x}^2+\dot{y}^2+\dot{z}^2)\right)2t\\
&=m\, \dot{\vec{r}}\,\vec{r}\,+\left( -\dfrac{m}{2}\dot{\vec{r\,}}^2-\dfrac{U_0}{\vec{r\,}^{2}} \right)2t\\
&=m\, \dot{\vec{r}}\,\vec{r}\, -(T+U)2t\\
&=m\, \dot{\vec{r}}\,\vec{r}\, -2Et\\
&\stackrel{t=0}{=} m\, \dot{\vec{r}}_0\,\vec{r}_0
\end{align*}
which shows that invariance under different transformations result in different conversation quantities.
