Three identical beads of mass lie on a frictionless line

PHYSICS · Multi-part · Fermi University Physics · Mathematical Methods for Physics · Calculus of change and accumulation · Linear algebra and vector spaces

Conditions of Validity, Ch Multiple integrals: Fubini, the change-of-variables theorem and integrals that do not exist

Problem

Three identical beads of mass mm lie on a frictionless line, connected in an open chain by two identical springs of stiffness kk (no walls), with displacements x1,x2,x3x_1,x_2,x_3 from equilibrium and potential energy U(x)=12k[(x2−x1)2+(x3−x2)2]=12xTKx,K=k(1−10−12−10−11).\begin{aligned}U(\mathbf x)&=\tfrac12k\big[(x_2-x_1)^2+(x_3-x_2)^2\big]=\tfrac12\mathbf x^{\mathrm T}\mathsf K\mathbf x,\\\mathsf K&=k\begin{pmatrix}1&-1&0\\-1&2&-1\\0&-1&1\end{pmatrix}.\end{aligned} The unnormalised Boltzmann weight for the three displacements at temperature TT is e−U/kBT=e−xTAxe^{-U/k_BT}=e^{-\mathbf x^{\mathrm T}\mathsf A\mathbf x} with A=K/2kBT\mathsf A=\mathsf K/2k_BT. The matrix K\mathsf K has eigenvalues 00 (eigenvector v0=(1,1,1)/3\mathbf v_0=(1,1,1)/\sqrt3), kk (eigenvector v1=(−1,0,1)/2\mathbf v_1=(-1,0,1)/\sqrt2) and 3k3k (eigenvector v2=(1,−2,1)/6\mathbf v_2=(1,-2,1)/\sqrt6), the standard normal modes of this chain. ((a)) Using the zero eigenvalue and its eigenvector, explain why A\mathsf A is only positive semi-definite, and why the improper integral ∫R3e−xTAx d3x\int_{\mathbb R^3}e^{-\mathbf x^{\mathrm T}\mathsf A\mathbf x}\,\mathrm d^3x fails to converge. ((b)) The physically meaningful partition function integrates only over the two coordinates orthogonal to v0\mathbf v_0. Writing x=y0v0+y1v1+y2v2\mathbf x=y_0\mathbf v_0+y_1\mathbf v_1+y_2\mathbf v_2 with Q=(v0 v1 v2)\mathsf Q=(\mathbf v_0\ \mathbf v_1\ \mathbf v_2), confirm that Q\mathsf Q is orthogonal and state the Jacobian of the change of variables x→(y0,y1,y2)\mathbf x\to(y_0,y_1,y_2) that the change-of-variables theorem assigns it. Hence write xTAx\mathbf x^{\mathrm T}\mathsf A\mathbf x in terms of y0,y1,y2y_0,y_1,y_2, and evaluate Zint=∫−∞∞ ⁣∫−∞∞e−(k/2kBT)y12−(3k/2kBT)y22 dy1 dy2Z_{\rm int}=\int_{-\infty}^{\infty}\!\int_{-\infty}^{\infty} e^{-(k/2k_BT)y_1^2-(3k/2k_BT)y_2^2}\,\mathrm dy_1\,\mathrm dy_2 in terms of kk, kBk_B, TT. ((c)) Evaluate ZintZ_{\rm int} numerically for k=15 N m−1k=15\,\mathrm{N\,m^{-1}} and T=300 KT=300\,\mathrm K, taking kB=1.380649×10−23 J K−1k_B=1.380649\times10^{-23}\,\mathrm{J\,K^{-1}}.

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