Iterated collinear boosts with equal rapidity Created 2026-10-07 Updated 2026-10-31
Starting from rest, compose positive collinear Lorentz boosts, each with the same finite rapidity . Each step has relative velocity , measured in the preceding rest inertial frame. The relativistic velocity-addition formula givesThese are equal increments of rapidity, not equal increments of velocity measured in the original inertial frame. They can describe a chain of relatively moving inertial frames or successive idealized changes of a massive particle's motion; no particular acceleration schedule is assumed. [ref]
https://davidtong.org/teaching/general-relativity/grhtml/S1. David Tong, General Relativity, §1.2.4: collinear boost composition and rapidity
.For a relative velocity at each step,The first four velocities are , , , and . At step eight, . The rapidities grow by the same amount at each step, while the successive velocity increments become smaller.
Equal collinear boosts add rapidities linearly while their composed velocities approach the speed of light
. The illustrative sequence uses relative velocity per step and . The horizontal coordinate counts boosts, rather than elapsed time. The line at marks the speed of light; every plotted velocity is strictly below it. Details.A short derivation applies the hyperbolic-function identity for to the recurrence, giving . Substituting and gives the displayed closed expression. In particular,for every finite nonnegative integer . The limit is as , but no finite sequence reaches the speed of light. This illustrates the beauty of rapidity: unbounded additive parameters coexist with bounded physical velocities. Replacing every step by a negative rapidity gives the corresponding approach to .
Put for and , using . Since the zero mode has zero kinetic energy, the exact kinetic energy in this trial state isTogether with part (d), the fixed-mean-number Bogoliubov variational energy isFor repulsive contact interaction, , each . Using the double-angle hyperbolic-function identities,soThe stationary point therefore satisfiesThis is a minimum, not just a vanishing derivative. Define the positive Bogoliubov quasiparticle dispersionAt the displayed solution , and . HenceThe hyperbolic cosine has its unique minimum at zero, proving global minimization of each retained term. Treating equal-magnitude wavevectors together simply multiplies the same function by their multiplicity. At a fixed ultraviolet cutoff, the resulting energy change from isIndeed . More directly, for small positive ,The pair vacuum expectation value lowers the interaction energy at first order in , whereas the nonzero occupation number and its kinetic energy cost start at second order. The two-mode squeezed vacuum admixes opposite-momentum pairs into the condensate, introducing spatial correlations that reduce the chance of two bosons coinciding. These correlations lower repulsive contact interaction enough to outweigh the added kinetic energy. A small quantum depletion of a Bose condensate can therefore lower the energy below the uncorrelated condensate value.
Two qualifications delimit this result. First, the paper does not state the sign of . More generally a finite stable minimum exists only when for every retained nonzero mode; repulsion ensures this. For example, with in some mode, tends to as , so an unconditional minimum claim for arbitrary is false. Also the formula minimizes the truncated Bogoliubov theory energy, not the exact quartic Hamiltonian including its discarded terms, and it must remain in the regime of small quantum depletion of a Bose condensate.
Second, a three-dimensional unregulated contact interaction cannot give a finite absolute beyond-leading-order ground-state energy by this bare sum. For large , , whose three-dimensional sum has an ultraviolet divergence. A finite ultraviolet cutoff, or renormalization of a three-dimensional dilute-gas contact coupling matched to a physical scattering length, is needed. This does not alter the modewise minimization or the finite-cutoff lowering mechanism just derived.
Past exam of the mathematics course of the University of Cambridge 2016 iii Paper 318 2 Solution Created 2026-10-03 Updated 2026-10-06
During the positive half-period, write and , with . Then . Its two eigenvalues are , and corresponding eigenvectors are and . Since , they are independent. Thus the general half-period solution isUsing the even and odd terms in the matrix exponential, . Because , the fundamental matrix of a linear differential equation givesThe lower-left denominator is , as printed in the PDF. In the negative half-period and . The analogous matrix exponential has the stated , replacing by its complex conjugate . Both matrices satisfywith in the second determinant. Equivalently, their generators have zero trace, so the determinant of each matrix exponential is one.
At each complete period the state is multiplied by the monodromy matrix of a periodic linear system , so . The determinant is therefore . Let and . Multiplying the two matrices and taking the trace givesThe ratio sum vanishes because the two ratios are and . With , the hyperbolic-function identity for consequently yieldsFor , : differentiating gives , since this derivative has derivative and vanishes initially. The characteristic polynomial is . Hence the two Floquet multipliers are positive, distinct and reciprocal, with dominant multiplier and growth rateThe finite-time propagation within each half-period is bounded independently of the number of cycles, so it does not change this asymptotic Floquet growth rate. The expression is the maximal rate, attained for generic nonzero initial data. The exceptional initial state in the reciprocal multiplier's eigenspace decays with rate ; the zero state stays zero. Thus a vanishing mean alpha effect does not preclude dynamo action in this periodically switched Parker dynamo wave model.
As , , , and . ThereforeIn particular, at the rate is asymptotic to , as required. The growth calculation uses the product , rather than averaging the two generators: the two generators do not commute.
Past exam of the mathematics course of the University of Cambridge 2024 iii Paper 303 1 b ii Solution Created 2026-09-24 Updated 2026-09-25
Differentiating the mean-field free energy and imposing stationarity givesThe hyperbolic-function identity turns this into the self-consistency equation
Set in Laplace's equation. Separation of variables givesThe Sturm-Liouville problem in has the constant mode with , and the eigenfunctions with , . Indeed, integration by parts makes nonnegative; for the first Neumann boundary condition removes the sine term and the second requires to be an integer. For , both derivative conditions make constant.
Expand the top data using a Fourier cosine series:The constant coefficient isFor , the constant part has zero integral against , and two integrations by parts give
Since the bottom value is zero, a convenient factor depending on for the nonconstant modes is , which vanishes at . Its value at is . The zero mode is linear in , so matching both constant boundary values givesTo obtain exactly the paper's form, use the hyperbolic-function identityThus
Each product of hyperbolic sine or hyperbolic cosine with has . The zero mode is also a harmonic function. On , the factor lies between zero and one. Since converges, the Weierstrass M-test gives uniform convergence on the closed rectangle. The series therefore gives zero at the bottom and the continuous Fourier cosine series of at the top, including its endpoints. On compact subsets with , the factors decay exponentially with , justifying termwise partial derivatives. In particular on both open vertical edges because every differentiated cosine factor vanishes there. The solution has finite Dirichlet energy: Fourier orthogonality gives the th nonconstant mode's energy as , whose sum is finite. Here is the hyperbolic cotangent. Thus a proves uniqueness in that class.
There is a corner compatibility condition for mixed boundary data in the printed wording. The top data have , but the paper writes even at . The series solves the interior equation and all the open-edge boundary conditions; the literal derivative condition at is incompatible with the top data. A function differentiable in along the top edge would necessarily have . Hence it cannot simultaneously satisfy the printed zero derivative at that corner, and there is no solution with all those literal corner requirements. The standard interpretation of this mixed boundary condition imposes the Neumann boundary condition on the open sides and the Dirichlet boundary condition continuously on the horizontal edges. Differentiating the series term by term on the top edge would be invalid at this corner; the uniform convergence of the undifferentiated series does not license that operation.
Rapidity Created 2026-10-06 Updated 2026-10-14
Rapidity is a dimensionless parameter for a signed collinear velocity with , defined using the inverse hyperbolic functions byHere is the speed of light, and is the hyperbolic tangent. Zero velocity has zero rapidity; negative velocity has negative rapidity. Every finite real rapidity gives , while requires . Thus a light ray has no finite rapidity in this parametrization. The header's formula uses natural units : its is the numerical velocity .
For the inertial frames in the velocity-addition formula, moves at signed velocity along the -axis of . Set . Re-expressing the same particle in the moving inertial frame givesThe subtraction is a passive change of reference frame; the inverse adds the particle's rapidity in to the frame's rapidity relative to . For example, and give , so and , exactly as in the velocity-addition formula.
For a massive particle of rest mass moving in one spatial dimension, its energy and signed momentum areThe hyperbolic cosine is the particle's Lorentz factor, and the hyperbolic sine is . The hyperbolic Pythagorean identity immediately gives the relativistic energy-momentum relation . In the older notation, writing the rest mass as recovers , , and . These expressions use natural units in which the factors of are numerically one; the preceding formulas restore those factors explicitly. [ref]
https://davidtong.org/pdfs/teaching/dynamics-and-relativity/dynrel7.pdf. David Tong, Dynamics and Relativity, §7.3.2: rapidity; §7.4.3: energy and momentum
.The useful simplification follows from the hyperbolic-function identity : the nonlinear velocity-addition formula becomes ordinary addition of rapidities. See iterated collinear boosts with equal rapidity for a plotted example. A common collinear Lorentz boost shifts every rapidity by the same constant, so a rapidity difference is invariant. This makes rapidity useful for relativistic factorized scattering in one spatial dimension, where two-body S-matrices depend on the difference of the incoming rapidities, with particle species fixed.
