Exercises — Noether's theorem — symmetry ↔ conservation law

WHAT to look for: a coordinate whose symbol does not appear in $L$ — only its velocity does. Scan $L$: we see $r$, $\dot{r}$, $\dot{\varphi }$ — but no bare $\varphi$. So $\varphi$ is cyclic. WHY that matters: Euler–Lagrange for $\varphi$ reads $\frac{d}{dt}\frac{\partial L}{\partial \dot{\varphi }}=\underset{=0}{\underbrace{\frac{\partial L}{\partial \varphi }}}\Rightarrow \frac{d}{dt}\left(m{r}^2\dot{\varphi }\right)=0.$ The conserved quantity is the conjugate momentum $p_{\varphi }=\frac{\partial L}{\partial \dot{\varphi }}=m{r}^2\dot{\varphi }=\text{const}.$ This is the angular momentum $L_z$ of the bead. The physical symmetry: rotating the whole system by any angle $\varphi \to \varphi +ϵ$ leaves $L$ unchanged (the table has no preferred direction).

(a) Time-translation invariance ($L$ has no explicit $t$) → energy conserved. (b) Space-translation invariance → total linear momentum $\hat{n}\cdot \vec{P}$ conserved. (c) Rotation invariance → total angular momentum $\hat{n}\cdot \vec{L}$ conserved. The pattern: a "doesn't-matter" in a continuous knob freezes one number.

Step 1 — WHAT the transformation does to velocities. Differentiate the shifts in time: $\delta \dot{x}=-\dot{y},\delta \dot{y}=\dot{x}$. Step 2 — check invariance (WHY $\delta L=0$). $\delta L=\frac{\partial L}{\partial \dot{x}}\delta \dot{x}+\frac{\partial L}{\partial \dot{y}}\delta \dot{y}=m\dot{x}(-\dot{y})+m\dot{y}(\dot{x})=0.✓$ Step 3 — build the charge. $Q=p_x\delta x+p_y\delta y=m\dot{x}(-y)+m\dot{y}(x)=m(x\dot{y}-y\dot{x})=L_z.$ So rotation ↔ $L_z$, exactly as the parent §5 promised.

Read the figure below to see WHY $\delta L=0$ geometrically. The black arrow is the vector $(x,y)$; the red arrow is where the rotation sends it. Notice the red push $\delta \vec{r}=(-y,x)$ is perpendicular to the black arrow (their dot product is $x(-y)+y(x)=0$). A push at right angles to a vector rotates it without stretching it — so the length is frozen, which is exactly why the kinetic energy (built from that length, for the velocity vector) does not change. That perpendicular-push picture is the geometric heart of "rotation is a symmetry," and the conserved charge it produces is $L_z$.

Angular momentum (from cyclic $\varphi$, as in Ex 1.1): $p_{\varphi }=m{r}^2\dot{\varphi }=\text{const}$. Energy (no explicit $t$): $E=\frac{1}{2}m({\dot{r}}^2+r^2{\dot{\varphi }}^2)-\frac{GMm}{r}$. On a circular orbit $\dot{r}=0$, and balancing forces gives $mr{\dot{\varphi }}^2=\frac{GMm}{r^2}$, so ${\dot{\varphi }}^2=\frac{GM}{r^3}=\frac{1}{64}$, $\dot{\varphi }=\frac{1}{8}$.

• $p_{\varphi }=1\cdot 16\cdot \frac{1}{8}=2$.
• $E=\frac{1}{2}\cdot 1\cdot 16\cdot \frac{1}{64}-\frac{1}{4}=\frac{1}{8}-\frac{1}{4}=-\frac{1}{8}$. So $p_{\varphi }=2$ and $E=-0.125$ (bound orbit, hence $E<0$).

First derive the master identity $\frac{dh}{dt}=-\frac{\partial L}{\partial t}$ — don't just quote it. Start from $h=\dot{x}p-L$ with $p=\frac{\partial L}{\partial \dot{x}}$, and differentiate in time using the product rule and chain rule: $\frac{dh}{dt}=\ddot{x}p+\dot{x}\dot{p}-\frac{dL}{dt}.$ Now expand $\frac{dL}{dt}$ by the chain rule (remember $L$ depends on $x$, $\dot{x}$, and explicitly on $t$): $\frac{dL}{dt}=\frac{\partial L}{\partial x}\dot{x}+\frac{\partial L}{\partial \dot{x}}\ddot{x}+\frac{\partial L}{\partial t}=\frac{\partial L}{\partial x}\dot{x}+p\ddot{x}+\frac{\partial L}{\partial t}.$ Substitute this in: $\frac{dh}{dt}=\ddot{x}p+\dot{x}\dot{p}-\frac{\partial L}{\partial x}\dot{x}-p\ddot{x}-\frac{\partial L}{\partial t}=\dot{x}(\dot{p}-\frac{\partial L}{\partial x})-\frac{\partial L}{\partial t}.$ The bracket vanishes by Euler–Lagrange ($\dot{p}=\frac{d}{dt}\frac{\partial L}{\partial \dot{x}}=\frac{\partial L}{\partial x}$). What survives is $\boxed{\frac{dh}{dt}=-\frac{\partial L}{\partial t}}$ So energy changes only through the explicit time dependence of $L$ — that is the precise Noether statement "time-translation symmetry ⇒ energy conserved." Apply it here. Only the potential carries $t$: $\frac{\partial L}{\partial t}=-\frac{1}{2}\dot{k}{x}^2=-\frac{1}{2}{k}_0\alpha x^2$, so $\frac{dh}{dt}=+\frac{1}{2}{k}_0\alpha x^2\ne 0.$ Energy leaks in at rate $\frac{1}{2}{k}_0\alpha x^2$: you are doing work on the system by turning the knob $k$. Symmetry broken → charge not conserved, exactly as Noether's converse warns.

Step 1 — velocity shift. $\delta \dot{x}=\frac{d}{dt}(t)=1$. Step 2 — vary $L$. $\delta L=m\dot{x}\delta \dot{x}=m\dot{x}=\frac{d}{dt}(mx)$. So it is a total time derivative with $F=mx$. The action is still invariant (boundary term), so Noether still applies — but with the correction. Step 3 — corrected charge. $Q=p\delta x-F=(m\dot{x})(t)-mx=m\dot{x}t-mx.$ Check it is constant: $\frac{dQ}{dt}=m\dot{x}+m\ddot{x}t-m\dot{x}=m\ddot{x}t=0$ because a free particle has $\ddot{x}=0$. ✓ Physically $Q=-m{x}_0$, encoding the initial position — the conserved charge of Galilean boosts. This is the case the parent flagged: $\delta L\ne 0$ but $\delta L=dF/dt$ still gives a law.

WHY $V$ can't stop translation: $V$ depends on ${\vec{r}}_1-{\vec{r}}_2$; under ${\vec{r}}_a\to {\vec{r}}_a+ϵ\hat{n}$ the difference is unchanged, so $\delta V=0$, and kinetic terms only see velocities. Hence $\delta L=0$. Noether charge: $Q={\vec{p}}_1\cdot \hat{n}+{\vec{p}}_2\cdot \hat{n}=\hat{n}\cdot ({\vec{p}}_1+{\vec{p}}_2)=\hat{n}\cdot \vec{P}.$ Take $\hat{n}=\hat{x}$: $P_x=m_1{\dot{x}}_1+m_2{\dot{x}}_2=2(1)+3(-1)=-1$ — a fixed number even though $p_{1x}=2$ and $p_{2x}=-3$ can change through the force. The moral: the center-of-mass coordinate is cyclic; the relative coordinate carries all the interaction.

Step 1 — the corrected charge for a time shift: $Q=p\delta x-F=(m\dot{x})(\dot{x})-L=m{\dot{x}}^2-(\frac{1}{2}m{\dot{x}}^2-V)=\frac{1}{2}m{\dot{x}}^2+V.$ That is exactly $h=T+V=E$. So "shift the clock" ↔ energy, cleanly derived, not assumed. Step 2 — number: $E=\frac{1}{2}(2)(3^2)+5=9+5=14$.

Step 1 — differentiate, using that $\vec{L}$ is constant. $\frac{d\vec{A}}{dt}=\dot{\vec{p}}\times \vec{L}-mk\frac{d\hat{r}}{dt}.$ Step 2 — set up plane-polar unit vectors. Motion is in a plane; use $\hat{r}$ (radial) and $\hat{\varphi }$ (tangential), with $\hat{z}=\hat{r}\times \hat{\varphi }$. Their time derivatives are the standard results $\frac{d\hat{r}}{dt}=\dot{\varphi }\hat{\varphi }$ and $\frac{d\hat{\varphi }}{dt}=-\dot{\varphi }\hat{r}$. Angular momentum is $\vec{L}=m\vec{r}\times \dot{\vec{r}}=m{r}^2\dot{\varphi }\hat{z}$. Step 3 — expand the first term explicitly (no black box). Insert $\dot{\vec{p}}=-\frac{k}{r^2}\hat{r}$ and $\vec{L}=m{r}^2\dot{\varphi }\hat{z}$: $\dot{\vec{p}}\times \vec{L}=(-\frac{k}{r^2}\hat{r})\times \left(m{r}^2\dot{\varphi }\hat{z}\right)=-mk\dot{\varphi }(\hat{r}\times \hat{z}).$ Now use the right-handed relation $\hat{r}\times \hat{z}=-\hat{\varphi }$ (since $\hat{z}\times \hat{r}=+\hat{\varphi }$). Therefore $\dot{\vec{p}}\times \vec{L}=-mk\dot{\varphi }(-\hat{\varphi })=+mk\dot{\varphi }\hat{\varphi }.$ Step 4 — expand the second term. $mk\frac{d\hat{r}}{dt}=mk\dot{\varphi }\hat{\varphi }.$ Step 5 — combine and cancel. $\frac{d\vec{A}}{dt}=mk\dot{\varphi }\hat{\varphi }-mk\dot{\varphi }\hat{\varphi }=\vec{0}.✓$ Every $r$-dependence cancelled because the two terms produced the same $\hat{\varphi }$-vector — that is the special algebra of the $1/r$ potential. Meaning: $\vec{A}$ points along the orbit's major axis with length $mke$ (eccentricity $e$); its constancy is why Kepler ellipses don't precess. The associated symmetry lives in a rotation group in 4 dimensions — see Symmetry Groups & Lie Algebras.

Step 1 — how each ingredient scales. Under $x\to \lambda x$ and $t\to {\lambda }^{2}t$, the velocity is a ratio: $\dot{x}=\frac{dx}{dt}\to \frac{d(\lambda x)}{d({\lambda }^{2}t)}=\frac{\lambda dx}{{\lambda }^{2}dt}={\lambda }^{-1}\dot{x}\Rightarrow {\dot{x}}^2\to {\lambda }^{-2}{\dot{x}}^2.$ The time measure scales as $dt\to {\lambda }^{2}dt$. Step 2 — assemble the action. $S=\int \frac{1}{2}m{\dot{x}}^{2}dt\to \int \frac{1}{2}m\left({\lambda }^{-2}{\dot{x}}^2\right)\left({\lambda }^{2}dt\right)=\int \frac{1}{2}m{\dot{x}}^{2}dt=S.$ The factor ${\lambda }^{-2}$ from the velocity and ${\lambda }^2$ from the measure cancel exactly — the action is strictly invariant. This is a genuine symmetry of the action. Step 3 — concrete numerical check (reconciled step by step). Take $m=1,v_0=1,T=1$ for the original path $x(t)=t$: $S={\int }_0^1\frac{1}{2}(1)(1{)}^{2}dt=\frac{1}{2}.$ Now apply the scaling with $\lambda =2$. The transformed path is $x^{\prime }=\lambda x=2x$ expressed at scaled time $t^{\prime }={\lambda }^{2}t=4t$; eliminating $t$ gives $x^{\prime }=2t=2\cdot \frac{t^{\prime }}{4}=\frac{t^{\prime }}{2}$, i.e. the new velocity is $\frac{d{x}^{\prime }}{d{t}^{\prime }}=\frac{1}{2}={\lambda }^{-1}{v}_0$ (matching Step 1), and the new interval is $t^{\prime }\in [0,{\lambda }^{2}T]=[0,4]$. Compute the scaled action directly: $S^{\prime }={\int }_0^4\frac{1}{2}(1)(\frac{1}{2}{)}^{2}d{t}^{\prime }=\frac{1}{2}\cdot \frac{1}{4}\cdot 4=\frac{1}{2}.$ $S^{\prime }=\frac{1}{2}=S$ — the shrunken velocity-squared ($\times \frac{1}{4}$) is exactly compensated by the four-times-longer time interval ($\times 4$). The invariance is demonstrated on real numbers, not just symbols.

A coordinate missing from $L$ is called ::: cyclic (ignorable), and its conjugate momentum is conserved. Energy is conserved exactly when $L$ has no explicit dependence on ::: time $t$. The Noether charge for a symmetry $q_i\to q_i+ϵ\delta q_i$ with $\delta L=dF/dt$ is ::: $Q={\sum }_i{p}_i\delta q_i-F$. Rotation invariance conserves ::: angular momentum $\hat{n}\cdot \vec{L}$.