Newton's law of cooling Created 2026-09-29 Updated 2026-10-14
Newton's law of cooling models an object's temperature using a positive constant rate and ambient temperature . The displayed ordinary differential equation assumes one temperature represents the object, the environment remains at , and the proportional rate stays fixed over the modeled range. It also describes warming when . These are modeling assumptions: [ref]
https://ocw.mit.edu/courses/18-03sc-differential-equations-fall-2011/1ce2cf783d5f1d15ee411ec2e5f928ec_MIT18_03SCF11_s1_6text.pdf. MIT's notes on modeling by first-order linear equations
explain the proportional law through a tangent line approximation and caution about large temperature differences.Dimensional analysis requires to have units of inverse time. The product in the exponential function solution is therefore dimensionless. The equilibrium point attracts both warmer and colder starting states: the excess temperature undergoes exponential decay. See initial value problem for computed cooling times, direction field for the family of trajectories, and separable differential equation for a derivation.
Opaque barrier approximation Created 2026-10-14 Updated 2026-10-20
For fixed and , the exact rectangular barrier transmission probability givesThis is the leading term of an asymptotic expansion: the wavefunction amplitude scale is , and the transmission probability scale is , with an interface prefactor. Increasing width by multiplies this approximation by . At fixed incident energy, increasing barrier height increases the tunnelling decay constant as ; heavier nonrelativistic particles also have stronger suppression. For the numerical example in rectangular barrier transmission probability, the approximation gives , compared with the exact .
The regime condition matters. Near at fixed width, tends to zero and the opaque barrier approximation fails; use the exact threshold value. For , expansion of the hyperbolic sine givesThis tends to one as at fixed , but the condition alone does not imply nearly perfect transmission if the remaining factor is large. At , use the oscillatory formula and its scattering resonances, rather than an exponential decay law.
Optical fiber attenuation 2026-10-14
Optical fiber attenuation is the reduction of transmitted optical power along an optical fiber. Define a positive loss in decibels by comparing optical power already launched into the guide with surviving optical power :The factor ten applies to a power ratio; the factor twenty applies to a wave amplitude ratio only when power is proportional to its square under the same conditions.[ref]
Introduction to Optical Fibers, dB, Attenuation and Measurements by Cisco. Introduction to Optical Fibers, dB, Attenuation and Measurements, What is a Decibel?, Base 10 Logarithm Rules and Power/Voltage Gains: attenuation convention
For a uniform illustrative model with constant loss coefficient in and distance in kilometres,For over , the loss is andThus about of the launched optical power survives and is lost. This is an illustrative coefficient, not a property of all optical fibers, and is independent of the earlier illustrative refractive indices.
The exponential decay follows from assuming each equal distance interval transmits the same fraction. For two such intervals, power ratios multiply; their logarithms add, so the losses in decibels add. Two sections each transmit ; together they transmit . The loss curve is straight in decibels and curved on a linear power scale.
Wave absorption and scattering of light can remove guided optical power even in a straight fibre, while bends can let it escape. Entrance losses, joints and connectors contribute additional losses; they belong separately from a uniform distance coefficient. This is why a measured end-to-end power ratio need not equal the straight-fibre model.[ref]
Introduction to Optical Fibers, dB, Attenuation and Measurements by Cisco. Introduction to Optical Fibers, dB, Attenuation and Measurements, Understand Insertion Loss and Calculate a Power Budget: joint losses, launch dependence and distance contribution
Past exam of the advanced fluid dynamics course of the University of Oxford 2017 1 d Solution 2026-10-25
Use the original scanned boundary condition. The scan fixes the normal-field amplitude at the corner . The supplied TeX drops the condition and would instead prescribe a constant normal field on the entire bottom. Those are different boundary-value problems; the solution below uses the legible original scan.
Choose a positive integer , put , and seek a separated decaying mode. A suitable flux function isThis is a decaying linear force-free magnetic arcade. The first and third curl components then satisfy the force-free equation automatically; its second component requires . ThusDirect differentiation gives and . Both horizontal components vanish at , and . The magnetic field on the bottom is , not a uniform .
A strictly decaying real normal mode requires , soFor the fundamental, single-arcade choice , this becomes . Equality gives no vertical exponential decay, while a larger magnitude gives an oscillatory rather than decaying vertical dependence for that mode. The data do not explicitly fix the horizontal mode number. If arbitrarily fine higher-mode arcades are permitted, every finite real admits some integer with . The fundamental restriction must therefore not be advertised as a necessary bound for every possible field satisfying only the stated pointwise and wall data.
For a fundamental field line, is constant along its projection in the plane. If denotes its normalized flux-function value, its arch isThe maximum height is at . In the horizontal projection,Thus the vertical projections are arches and the horizontal projections are straight sheared segments joining their paired footpoints. Positive gives positive horizontal slope, negative reverses the shear, and gives an unsheared potential arcade. Higher repeats such cells across the width.
The transcription's alternative uniform-bottom condition would be inconsistent with exponential decay and zero side flux for : solenoidality gives , whereas a uniform nonzero bottom would give a conserved flux that cannot decay to zero. This integral check is an independent reason not to silently solve that mistranscribed problem.
Write . For , exponential decay dominates every fixed power of , so the improper integral converges absolutely and . For , integration by parts givesThe preceding part gives the base case . Repeated application, or mathematical induction, yields
Separate three causal steps: recognition of the modified RNA, occupancy of the target messenger RNA, and reduction of its degradation rate. Call the newly identified RNA-binding protein R. Its recovery in a screen need not mean it contacts the modified ribonucleotide directly; it could recognize a modification-dependent structure or arrive through another protein. Likewise, more target messenger RNA could reflect increased transcription rather than greater stability.
First reproduce the interaction using endogenous R and the endogenous target. Determine R's protein domains, expression and subcellular distribution, then use reciprocal RNA immunoprecipitation and crosslinking and immunoprecipitation to locate the occupied region. Normalize enrichment to input target abundance and R recovery, and include unrelated RNAs, a non-specific antibody and cells lacking R. Crosslink-based occupancy supports close contact, whereas recovery without crosslinking can preserve indirect complexes. Determine whether the relevant modification actually occurs near the binding site: mass spectrometry of purified target fragments can establish chemical identity, supplemented by a modification-specific mapping assay with known modified and unmodified standards. Bulk mass spectrometry of all cellular RNA cannot assign a modification to this particular transcript or position. Do not assume the modification is N6-methyladenosine merely because it is familiar.
Prepare short synthetic RNAs identical in nucleotide sequence, length, ends and labelling except for the single modified ribonucleotide. Measure binding of purified R over a concentration range by electrophoretic mobility shift assay or fluorescence anisotropy, with competition by unlabelled cognate and unrelated RNA. The dissociation constant distinguishes a quantitative preference from a binary pull-down impression. Verify that R is folded and sufficiently pure; include an RNA-binding-defective R mutant, several surrounding sequences and different modification positions. Modification-specific recovery from a cellular extract can disappear with purified R, revealing a missing partner rather than disproving the original screen.
Distinguish chemical recognition from structural recognition. Compare RNA secondary structure with and without the modification by RNA structure probing. Then disrupt the stem without changing the binding motif, or restore pairing through compensatory mutations. If an unmodified but constitutively exposed motif binds R, and restoring occlusion suppresses binding even when the modification remains, an RNA modification switch is plausible. Direct recognition is supported when modification discrimination persists after accessibility is matched and is lost upon a specific R binding-pocket mutation. Structural work can identify the contacts, but binding and structure controls remain essential. Modification-dependent exposure of a binding motif has an experimental precedent in the N6-methyladenosine-dependent RNA modification switch for HNRNPC.[ref]
N6-methyladenosine-dependent RNA structural switches regulate RNA–protein interactions by Nian Liu and others. Abstract; Figures 1–3: structural accessibility, binding and methylation dependence
Establish stability by acute depletion of R, followed by a metabolic RNA labelling pulse and chase. Measure loss of labelled target molecules using appropriate spike-ins and a time course spanning the expected half-life, while independently estimating newly synthesized target RNA. SLAM-seq is one possible implementation: chemical treatment makes incorporated 4-thiouridine detectable through characteristic sequencing changes.[ref]
Thiol-linked alkylation of RNA to assess expression dynamics by Veronika A. Herzog and others. Abstract: metabolic RNA labelling and sequencing-based measurement of expression dynamics
Include unlabelled controls, check label toxicity and correct for label reutilization or incomplete chase. When reporting molecules per biological cell, separate dilution through increasing cell number from chemical degradation. A transcription inhibitor time course provides an orthogonal check, but broad inhibition itself can change decay. Under a constant first-order degradation model, labelled molecules obeyA shorter half-life after R depletion, reversed by normal R but not its RNA-binding-defective mutant, establishes the proposed effect. Allow a more complex fit if the exponential decay model is inconsistent with the measured time course. At steady state, abundance is synthesis rate divided by degradation rate, so abundance alone cannot identify which rate changed.Use a controlled target reporter gene carrying the occupied region, then remove or mutate that region and measure reporter messenger RNA decay. Preserve polyadenylation, RNA splicing and coding potential where possible, and distinguish site loss from altered RNA secondary structure. Manipulate the relevant modification-producing enzyme acutely, rescuing with normal or catalytically inactive enzyme; verify target-site modification after the manipulation. Cross the two perturbations: if R stabilizes the target specifically through the modification, removing the modification should reduce R occupancy and eliminate or substantially weaken the extra effect of removing R. This epistasis supports a shared pathway, but floor effects or redundant sites can obscure it. A site-specific chemically modified reporter helps separate local recognition from the many indirect consequences of removing a global modification enzyme.
Identify the affected decay step using poly(A) tail measurements, cap-state assays and mapping of decay intermediates. Test whether R excludes a microRNA-loaded Argonaute complex or a decay-promoting RNA-binding protein, blocks an endonuclease, inhibits deadenylation or decapping, or recruits a protective partner such as poly(A)-binding protein. Use protein interaction analysis with and without ribonuclease to distinguish RNA-bridged association from a more direct interaction. R mutants that retain target binding but lose a partner interaction separate recruitment from recognition. Depleting the implicated decay factor should suppress the destabilization caused by R loss; purified-system decay assays can test whether R alone physically protects the substrate.
Measure biological translation and subcellular location in parallel, using ribosome profiling or a controlled biological translation reporter and single-molecule fluorescence in situ hybridization. R might stabilize the target indirectly by altering ribosome loading or compartment access. Inhibiting biological translation can itself alter decay, and colocalization in a granule does not establish protection. The reported stabilizing action of IGF2BP proteins on modified targets provides a candidate mechanism to compare with, rather than an identity to assign to R.[ref]
Recognition of RNA N6-methyladenosine by IGF2BP proteins enhances mRNA stability and translation by Huilin Huang and others. Abstract and Results: modification-associated target recognition, stability and translation
The strongest mechanism links a verified target-site modification to R binding, a defined protective activity and a rescued change in target half-life.The measured decay is approximately consistent with the predicted loss rate. Following rapid singlet-to-triplet population transfer, the unquenched phosphorescence intensity obeysA straight-line fit of to the four times, expressed in seconds, givesThis differs from by about . For example, using the first point and the predicted rate gives subsequent intensities about , close to the supplied . The data support the expected millisecond exponential decay to this precision. With no measurement uncertainties supplied, a formal statistical acceptance or rejection is not justified.
Original solutions to every part of the elasticity and plasticity course of the University of Oxford Part C examination, held on 30 May 2023. Source: released examination PDF, bundle pages 317–320; paper pages 1–4. Numbering follows the original paper. In Question 1(c), the printed count needs a strict-cutoff qualification: at the newly counted state has no exterior exponential decay and must be excluded. The solution includes that correction and the exceptional small-parameter balances. In Question 3(d), purely elastic unloading is valid only until reverse yielding begins.
Past exam of the elasticity and plasticity course of the University of Oxford 2023 1 b Solution 2026-10-25
Substitution of the proposed normal mode into the wave equation gives the ordinary differential equationsA nonzero exterior displacement field has exponential decay at infinity only when . Withits wave amplitude is . At the exterior wave amplitude is affine and cannot decay; at it is oscillatory and cannot decay. Setting the exterior wave amplitude identically zero would impose both and , forcing the entire normal mode to vanish.
We also need . Indeed, if , the rigid boundary condition gives with . At the interface,which contradicts continuity of traction because both values of the shear modulus are positive. At , the layer wave amplitude is and the same contradiction follows from .
Therefore . Define the positive transverse wavenumberThe rigid boundary condition now gives . The interface boundary conditions giveNeither nor can vanish for a nontrivial normal mode with . Dividing therefore gives the rigid-backed shear-wave dispersion relation:Conversely, positive satisfying these relations generate a nonzero normal mode through the wave amplitudes above. Its physical displacement field is the real part of the complex expression.
Past exam of the elasticity and plasticity course of the University of Oxford 2023 1 c Solution 2026-10-25
Eliminating the phase velocity from the first two relations yieldsWith , putThen . The dispersion relation becomesHere follows from . The parameter has dimensions of length; and are dimensionless.
For the strict mode count for a rigid-backed elastic layer, considerThe right side is negative, so intersections occur only on the negative branches of the tangent,On each such branch is strictly increasing. If , its limit at is negative infinity. At the other end of the available interval, it is positive: either with , or with . The intermediate value theorem and strict monotonicity give exactly one intersection. No such interval exists when .
Intersections of the negative tangent branches with the interface loading curve. A branch beginning exactly at the cutoff contributes no trapped mode.
Details. Thus the exact answer, including all cutoff values, isAway from , this equals the printed floor function expression . At , the printed expression is one too large: the th interval is empty and the limiting state has , contradicting exponential decay. This exception is present in the native paper, rather than introduced by transcription.
Now hold fixed and let . Neglecting the right side gives and hence the candidate phase velocitiesThe zero is excluded: it would require , already ruled out. For every , the candidate is a valid leading asymptotic expansion. A first correction, obtained by expanding the tangent about , isThe big O notation here assumes a fixed positive gap between and . In particular, when , all leading approximations are valid. The phase velocity increases strictly with , so only the last branch can cause an endpoint difficulty.
If , the formal last equation root lies outside the allowed interval. The actual last equation root approaches , where the small factor is balanced by the vanishing denominator. This is a singular perturbation. For fixed strictly inside that interval, write . Since ,The valid endpoint approximation for a rigid-backed shear mode consequently givesHere we used and . Thus this normal mode approaches the upper wave speed from below; it does not approach the inadmissible value obtained from .
There is a further exceptional fixed value . Put again. Now , so the balance changes toIn the last formula the bracket specifies the first nonzero correction. The zeroth-order candidate is correct as a limit but misses this correction and the positive exterior exponential decay rate. More generally, close to , let and . For , a transition approximation valid on both sides isIt recovers the balance at and the ordinary expansion above the transition. The fixed-endpoint formula involving should be used when is below, and bounded away from, .
Using the probability density function of the Erlang distribution, the expected value isHere , and two applications of integration by parts give . The boundary terms vanish because exponential decay dominates these powers.
Past exam of the mathematical methods I course of the University of Oxford 2016 resit 11 b Solution 2026-10-25
Set . The real solution of the damped harmonic oscillator equation is . The initial displacement fixes , and release from rest gives . ThereforeThe exponential decay rate of the amplitude is , while the weak-damping energy decay rate is .
Past exam of the mathematical methods I course of the University of Oxford 2018 3 Solution 2026-10-25
The characteristic equation of the damped harmonic oscillator is . In the underdamped case its roots are , where . Thus the exact real solution is . For , , giving the requested leading approximationFor mass and spring constant , the mechanical energy is . Differentiating and using the differential equation gives . In particular, energy decay of a viscously damped harmonic oscillator is not an exact pointwise exponential decay: at a turning point. To leading order in , the period average isThe quality factor is . Using the period and the small fractional loss gives . Equivalently, the energy-decay definition yields the same leading result. Replacing by in the oscillatory phase is not uniform over arbitrarily long times; its phase error is .
Past exam of the mathematics course of the University of Cambridge 2001 iii Paper 38 1 Solution Created 2026-10-03 Updated 2026-10-14
Work with the complex representation and take its real part to obtain the physical magnetic field. For constant magnetic diffusivity and the prescribed circular velocity , the resistive induction equation and giveTo see the last equation directly, . Taking its curl reproduces the induction equation if , up to a time-dependent additive constant removable by the gauge transformation. This is the scalar form of advection and magnetic diffusion in cylindrical coordinates.
Put . The material derivative of is , because . A radial derivative instead acts on both amplitude and phase:Also . Substitution gives the shearing-coordinate magnetic flux equation, here for azimuthal mode number one:Writing it with on the left is important when considering zero magnetic diffusivity: one must not literally divide by . In the ideal limit, , so . This is magnetic flux freezing: the scalar flux pattern is carried with each rotating fluid ring. It does not make the magnetic field stationary. Indeed,so differential rotation winds the field and can make its azimuthal component grow linearly in time. Rigid rotation, for which , only rotates the pattern.
For the specified interior angular velocity, and away from the circulation edge. Let , and . The amplitude equation becomesFor a smooth initial amplitude on the cell scale, at bounded away from zero and the edge, the term dominates when . Its accumulated damping is order one at , whereas the integrated terms proportional to are only and the terms independent of are . The resulting cubic-time resistive damping in a differentially rotating cell isThe magnetic field has this leading exponential decay envelope. Its components also contain the algebraic winding factor visible in , so the claim concerns the cubic exponent, not an exact amplitude with no prefactor. Equivalently the radial wavenumber grows as and the integrated resistive damping is .
This is phase mixing in magnetic flux expulsion. The circulation creates ever finer gradients, allowing weak magnetic diffusion to act much sooner than its unsheared time :Rapid winding followed by resistive smoothing expels the imposed field from the cell interior. The maintained exterior magnetic field is redistributed into a circulation-edge boundary layer. The result is an interior high-magnetic Reynolds number transient, not a uniform assertion at the axis and the nonsmooth edge, nor the exact solution with a continuously imposed far field. The diffusion length at is , identifying where a local bulk approximation can fail. The linear profile also needs smoothing at a regular rotation axis and at if used as a fully smooth global flow.
In photospheric magnetoconvection, circulating and diverging convection transports weak magnetic flux towards converging downflow regions. Solar granulation concentrates field in intergranular lanes, while larger-scale supergranulation helps form the solar magnetic network. Flux expulsion explains why a highly conducting photosphere can have relatively weak field in cell interiors and intermittent concentrations at their edges: high electrical conductivity delays diffusion on the original scale, but does not prevent diffusion on the fine scales created by advection. A long-lived circulation must last long enough for the expulsion time; rapidly changing cells need not reach this idealized state. Once the field becomes strong, the Lorentz force modifies the convection, so this prescribed-flow calculation alone neither determines the final field strength nor describes strong-field sunspots.
Past exam of the mathematics course of the University of Cambridge 2004 ia Paper 4 10A Solution Created 2026-09-24 Updated 2026-10-14
Write and . The gravitational central force has zero torque, so the specific angular momentumis constant. For a nonradial orbit , motion lies in the plane normal to , and plane polar coordinates give . With and primes denoting derivatives,The radial equation is therefore the Binet equationIts general solution, after choosing the polar origin of angle, isThe origin is a focus of this conic section. To identify which conic section occurs, use the conserved specific energyA bound, nonradial orbit has , hence and is an ellipse. For example, putting , in givesan explicit ellipse equation. This is the intended low, bound Kepler orbit. The inverse-square equation alone does not imply an ellipse: an initially tangential speed larger than gives , and a hyperbolic Kepler orbit. Zero angular momentum instead gives radial motion. Those qualifications are necessary for the elliptic conclusion.
In a circular orbit, centripetal acceleration balances the gravitational acceleration:Here is the positive angular speed; either sense of circulation has this period.
With the stated drag force, use the full angular momentum . Gravitational torque is still zero, while the drag torque isTherefore for constant . Its direction remains fixed, and its magnitude has exponential decay with time constant ; the orbit remains planar although its shape is no longer a fixed ellipse.
Past exam of the mathematics course of the University of Cambridge 2006 iii Paper 51 1 b Solution Created 2026-10-03 Updated 2026-10-14
Use the connected correlation function , so that the constant ordered contribution is removed. Away from a thermodynamic critical point its long-distance decay defines the correlation length: when this limit exists. In a massive rotation-invariant phase the exponential decay is typically multiplied by an algebraic factor.
Expand the Landau-Ginzburg theory free energy about a stable uniform minimum , writing . To quadratic order,integration of the Gaussian field theory gives the Fourier-space connected correlation function . Thus the Ornstein--Zernike correlation function hasFor an ordinary quartic transition, above the transition and below, so and , respectively. The correlation length diverges as on both sides, with different amplitudes. At criticality the mass vanishes and the Gaussian correlation function is proportional to for . Interactions can change this power to .
Past exam of the mathematics course of the University of Cambridge 2006 iii Paper 51 2 Solution Created 2026-10-03 Updated 2026-10-14
Use for inverse temperature, to distinguish it from the order-parameter critical exponent . In a translation-invariant pure thermodynamic phase, the connected two-point correlation function isFor finite correlation length, at it decays exponentially, possibly with an algebraic prefactor; is the asymptotic exponential decay rate. Below an ordered transition the pure phase is selected by taking the thermodynamic limit before or . A zero-field symmetric mixture of the two ordered phases has a nondecaying connected term and does not describe the usual pure-phase magnetic susceptibility.
Let denote extensive physical free energy and . Since the field contributes to the statistical Hamiltonian, differentiation givesExpanding the variance and using translation invariance proves the correlation-function susceptibility sum rule:The sum includes . If the source is instead defined as the dimensionless quantity , the corresponding derivative omits the displayed inverse-temperature factor.
A real-space renormalization group step of length factor uses a local blocking kernel satisfying . For example, a deterministic majority-spin rule assigns one coarse Ising spin to each block; tied blocks can be resolved with equal probabilities. Define the blocked statistical Hamiltonian byThe spacing becomes . All generated interactions, including multispin terms and the field-independent constant, must be retained for an exact transformation. Summing over uses normalization to prove . Iteration gives and .
The coarse probability law is the exact marginal law of these block variables. Indeed, for any coarse observable , its expected value is the microscopic expected value of . A blocking rule that retains the long-wavelength order-parameter mode, with source and observable transformations tracked, therefore retains long-distance response and correlation functions. Their field normalization may change, while their physical length scales and critical singularities do not. Partition-function equality alone would not justify this assertion for an arbitrary kernel that discards all slow variables. Truncating the generated operator family is likewise an approximation.
For the free-energy recursion use dimensionless free energy per site,in the thermodynamic limit. Exact partition-function invariance gives . If one step sends to , the coefficient records the identity-operator contribution from eliminated fluctuations and normalization. ConsequentlyRepeated substitution derives the additive free-energy recursion under blocking:In physical energy units, divide this convention by . The additive term is essential for absolute free energies even when it does not affect the leading critical powers.
A renormalization-group fixed point satisfies . Linearize in scaling coordinates so that . A relevant operator has , an irrelevant operator has , and a marginal operator has with behaviour decided by nonlinear terms. Thus the eigenvalues of the discrete linear map are ; the sign criterion applies to the generator exponents , not to the discrete multipliers. The critical surface is the stable manifold on which every relevant scaling field has been tuned away. Irrelevant coordinates flow to the renormalization-group fixed point along it. A repulsive renormalization-group trajectory leaves the renormalization-group fixed point along an unstable relevant direction, ultimately entering a noncritical phase or another scaling regime.
Local renormalization-group flow in the zero-field slice: attraction along the critical surface and repulsion in the thermal direction
. Details. To separate the singular free-energy density from its regular background, suppose the eliminated-mode contribution is analytic near the renormalization-group fixed point. Look for an analytic background satisfying . Then satisfies . In linear thermal and field coordinates, an analytic monomial has background coefficientprovided the denominator is nonzero. This explicitly explains when analytic subtraction of an inhomogeneous renormalization recursion works. The printed inhomogeneous equation may describe the free-energy function before this subtraction, or a singular part defined only up to analytic terms. The inhomogeneous sum need not be small compared with the full free energy: an analytic constant can dominate a vanishing singular contribution. It can be omitted in deriving leading singular powers after regular subtraction, provided no renormalization-group free-energy resonance, marginal logarithm or dangerously irrelevant coupling obstructs the homogeneous critical asymptotics. At a renormalization-group free-energy resonance , additive logarithms can remain, so simply dropping the term would miss them.
Suppose there are exactly two relevant scaling fields, thermal and magnetic , with positive exponents , and the remaining fields approach a regular limit. Physical reduced temperature and magnetic field agree with these coordinates to leading order, up to nonuniversal metric factors. The homogeneous scaling hypothesis for critical phenomena isas an asymptotic scaling equation. Choose to make the thermal argument or . Defining givesThe two functions describe respectively and : approaching the transition from disordered and ordered sides need not give the same function. In particular the ordered zero-field function has one-sided derivatives associated with spontaneous magnetization. Metric factors multiply this formula in different microscopic normalizations.
The correlation length in Bravais lattice units obeys . At , the same choice of gives . Two thermal derivatives of give . One and two field derivatives give, respectively, and , assuming the relevant branch amplitudes are nonzero. At , choose ; then , so . HenceSimple algebra now gives the Rushbrooke scaling relation and the hyperscaling relation . Also , giving the Widom scaling relation. These statements refer to leading singular contributions with the homogeneous two-field scaling assumptions. Above the upper critical dimension, the quartic coupling is dangerously irrelevant because it is still needed to stabilize the ordered phase; ordinary mean-field critical exponents then satisfy Rushbrooke and Widom but generally not hyperscaling. Marginal logarithms also require qualifications at the upper critical dimension.
For the connected-correlation scaling form , the function tends to a finite nonzero value as and decays exponentially at large , up to powers of , in a massive short-range phase. The scaling ansatz alone does not fix that prefactor. If the usual massive Gaussian field correlation tail applies, matching the powers gives , with a correlation-length-dependent amplitude in fixed by matching the critical scaling form.
In physical coordinates the lattice sum has site-density factor . Its long-distance singular part therefore giveswhere is the area of the unit sphere in dimensions. For , finite , an integrable exponential tail and a nonzero integral, the last factor tends to a finite nonzero constant as . Microscopic distances contribute a regular background. Thus , proving the Fisher scaling relationEquivalently the field has scaling dimension , so its conjugate field has , which gives the same identity from the exponent formulas above.
Past exam of the mathematics course of the University of Cambridge 2007 ib Paper 4 15F i Solution Created 2026-09-24 Updated 2026-10-14
For an integer and , the definition of the Laplace transform gives . For , integration by parts givesThe endpoint term vanishes because exponential decay dominates every fixed power of . Induction therefore gives . Multiplication by replaces by in the defining integral, soThe second formula holds for and is the Laplace transform shift theorem.
Past exam of the mathematics course of the University of Cambridge 2016 ia Paper 2 5A b ii Solution Created 2026-09-24 Updated 2026-10-06
At the first half-temperature time, the solution of Newton's law of cooling givesSince , this time is positive and finite. The instantaneous temperature increase supplies the new initial condition . Restart the same first-order linear differential equation at :This is Newton cooling with an instantaneous temperature jump. The time translation in the exponential decay is essential: its prefactor is the excess immediately after the jump.
Past exam of the mathematics course of the University of Cambridge 2016 ia Paper 2 5A b i Solution Created 2026-09-24 Updated 2026-10-06
With a positive cooling rate , Newton's law of cooling gives . The temperature excess satisfies the first-order linear differential equation . Applying the initial condition to its exponential decay solution givesThe temperature approaches the ambient value , rather than zero. The intended cooling situation has and .



