Three short derivations

PHYSICS · Multi-part · Fermi University Physics · Mechanics & Dynamical Systems · Lagrangian mechanics · Lagrangian mechanics

Action and Symmetry, Ch Symmetry, Noether's theorem and what is conserved

Problem

Three short derivations, each in a system not worked in full elsewhere. ((a)) A charged bead of mass m=0.50m = 0.50 kg slides on a frictionless straight wire under a uniform electric field, which exerts a constant force F0=4.0F_0 = 4.0 N along the wire, so V(x)=−F0xV(x) = -F_0x and L=12mx˙2+F0xL = \frac12 m\dot x^2 + F_0x. Show that the shift x→x+ϵtx \to x + \epsilon t changes LL by ϵ df/dt\epsilon\,\mathrm{d}f/\mathrm{d}t with f=mx+12F0t2f = mx + \frac12F_0t^2, and hence write down the conserved quantity Q=pt−fQ = pt - f. The bead is released from rest at x0=2.0x_0 = 2.0 m at t=0t = 0. Evaluate QQ at t=0t=0, then confirm that it has the same value at t=3.0t = 3.0 s using the equation of motion. ((b)) Two particles of mass m1=1.0m_1 = 1.0 kg and m2=3.0m_2 = 3.0 kg move along a frictionless line and interact only through a spring joining them, an isolated pair with no external force. At t=0t=0 their positions are x1=0x_1 = 0 and x2=1.0x_2 = 1.0 m and their velocities are v1=2.0 m s−1v_1 = 2.0\ \mathrm{m\,s^{-1}} and v2=−1.0 m s−1v_2 = -1.0\ \mathrm{m\,s^{-1}}, whatever state the spring is in. Derive the boost constant K=Pt−MXcmK = Pt - MX_{\rm cm} for this pair, evaluate it at t=0t=0, and use KK together with the (separately conserved) total momentum PP to predict the centre-of-mass position XcmX_{\rm cm} at t=2.0t = 2.0 s, without solving for the motion of either particle. ((c)) A mass m=0.20m = 0.20 kg on a spring of stiffness ks=50 N m−1k_{\rm s} = 50\ \mathrm{N\,m^{-1}} oscillates with amplitude A=0.10A = 0.10 m. Starting from the stationarity of the action under the stretch r(t)→(1+ϵ)r(t)\mathbf{r}(t) \to (1+\epsilon)\mathbf{r}(t) applied over one exact period, derive 2⟨T⟩=ν⟨V⟩2\langle T\rangle = \nu\langle V\rangle for ν=2\nu = 2, then confirm it directly for this oscillator by computing ⟨T⟩\langle T\rangle and ⟨V⟩\langle V\rangle from x(t)=Asin⁡(ωt)x(t) = A\sin(\omega t).

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