In a flexible crosslinked polymer network, deformation restricts the configurations of chain segments between cross-links. The resulting decrease of entropy produces a restoring entropic force. For an ideal affine Gaussian network, the shear modulus is $G={\nu }_s{k}_{B}T$, where ${\nu }_s$ is the number density of elastically active strands. A nearly incompressible isotropic rubber has Young's modulus $E\simeq 3G$. At fixed network density within the rubbery regime, both increase approximately linearly with absolute temperature. This statement does not describe crossing the glass transition or chemical degradation.

A rubbery material based mainly on cis-rich polyisoprene. A lightly crosslinked network provides recoverable deformation through rubber elasticity.

By Albert Einstein. Paper 5, pp. 177–197; originally Annalen der Physik 17, 132–148 (1905).

Cited passages: §§7–8, pp. 191–196.

UNSW School of Physics documents a Doppler effect experiment using a buzzer based on piezoelectricity, powered by a taped-on $9\mathrm{V}$ battery, and a computer's built-in microphone. Audacity records the acoustic wave. First record the stationary source to establish its baseline frequency; then attach it to a bicycle passing the stationary microphone. The published spectrogram shows a rapid decrease in frequency during passage.^[ref]

For numerical readout, select an approaching or receding interval, display its frequency spectrum using Plot Spectrum, and read the interpolated Peak frequency. The protocol recommends repeating with slightly different selections to test stability, and working outdoors to reduce reflections and standing waves. Longer windows refine frequency spacing while averaging over more motion. The baseline controls for the source's actual tone; repeated selections expose analysis-window sensitivity.^[ref]

Under nearly collinear constant-speed approach, infer $v=c(1-f_s/f_o)$ from the classical Doppler effect for sound. For recession the positive speed is $v=c(f_s/f_o-1)$. These inversions need a specified speed of sound. An offset path changes the projection, as in Doppler frequency along a trajectory; neither inversion measures the full speed near closest passage. The protocol's controls cannot remove arbitrary acceleration, background measurement noise or unresolved reflected paths. No new measurements are claimed here.

In the same uniform stationary sound model, let $\tau$ be emission time, $t$ reception time, and ${\mathbf{x}}_s(\tau )$ and ${\mathbf{x}}_o(t)$ source and receiver positions. The direct-path propagation condition and its direction areThe unit vector $\mathbf{n}$ points from the source's old emission position to the receiver's reception position. With velocities ${\mathbf{v}}_s=d{\mathbf{x}}_s/d\tau$ and ${\mathbf{v}}_o=d{\mathbf{x}}_o/dt$, taking a derivative of the propagation condition givesHere the dot products project onto the actual propagation direction. The last equality counts source wave phase cycles at reception for constant emitted frequency $f_s$. It assumes both speeds are below $c$, a unique direct arrival and smooth motion; interpreting a measured short-window frequency spectrum additionally requires little frequency change within that window. The source velocity must be evaluated at emission, not at reception. This delayed geometry matters when the direction changes.

For an original pass-by model, a source follows ${\mathbf{x}}_s(\tau )=(v\tau ,0)$ and a stationary receiver is at $(0,b)$ with $b>0$. The source passes its closest position at emission time $\tau =0$. ThenThese parametric equations give received frequency against reception time. For $\tau <0$ the shift is positive, for $\tau >0$ negative, and the unshifted emission at $\tau =0$ reaches the receiver at $t=b/c$. Thus zero shift occurs after the source's geometric closest passage in the material's reference frame. Increasing $b$ spreads the transition over a longer interval even though $v$ is unchanged. Far before and after passage, the frequency approaches $f_{s}c/(c-v)$ and $f_{s}c/(c+v)$ respectively.

Work in the material's rest reference frame, with constant speed of sound $c>0$. For one-dimensional propagation, choose the positive $x$ direction along the travelling wave from its emission point toward the receiver. Define $u_s=d{x}_s/dt$ and $u_o=d{x}_o/dt$ as the signed source and receiver velocities along that same axis. Both use the same sign convention: positive $u_s$ moves the source toward the receiver ahead, whereas positive $u_o$ moves that receiver away from the incoming wave. Assume $|u_s|<c$ and $|u_o|<c$, constant velocities, and no source-receiver crossing during the interval considered. Then

To derive this, let $T_s=1/f_s$ be the emission wave period. During $T_s$, one wave crest travels $c{T}_s$ while the source advances $u_s{T}_s$. Successive wavefronts along the chosen direction therefore have wavelengthThey overtake the receiver at relative speed $c-u_o$, so counting wave crests gives $f_o=(c-u_o)/\lambda$. This separates the source's effect on wavelength from the receiver's effect on crossing rate. Even with a moving source, a stationary receiver obeys $f_o\lambda =c$ because the wave speed through the material remains $c$.

For a stationary receiver and a source approaching at speed $v$, set $u_o=0$ and $u_s=v$: $f_o=f_{s}c/(c-v)>f_s$. For recession choose the positive axis along the departing wave toward the receiver again, so the source has $u_s=-v$: $f_o=f_{s}c/(c+v)<f_s$. For the deliberately chosen values $c=343\mathrm{m}{\mathrm{s}}^{-1}$, $v=20\mathrm{m}{\mathrm{s}}^{-1}$ and $f_s=500\mathrm{Hz}$, the two predictions areThe corresponding wavelengths are $0.646\mathrm{m}$ and $0.726\mathrm{m}$, compared with $c/f_s=0.686\mathrm{m}$ for a stationary source. These are calculations for chosen parameters, not reported measurements.

A receiver approaching a stationary source at $20\mathrm{m}{\mathrm{s}}^{-1}$ instead has $u_s=0$, $u_o=-20\mathrm{m}{\mathrm{s}}^{-1}$, giving $f_o=500(363/343)\mathrm{Hz}\simeq 529.15\mathrm{Hz}$. Moving away gives $f_o=500(323/343)\mathrm{Hz}\simeq 470.85\mathrm{Hz}$. The equal closing speeds do not make these equal to the moving-source values. If both source and receiver move together with $u_s=u_o$, however, the numerator and denominator cancel and $f_o=f_s$. For speeds small compared with $c$, the first-order shift is $(f_o-f_s)/f_s\simeq (u_s-u_o)/c$; the difference between moving-source and moving-receiver experiments appears beyond this first order.

A shallow donor impurity in an isotropic semiconductor can be approximated by an electron of effective mass $m^{\ast }$ in a Coulomb potential screened by relative permittivity ${ϵ}_r$. Its ground-state binding energy and radius areThe donor level lies $E_B$ below the conduction band. These are magnitudes: the bound electron's energy relative to the band edge is negative. Band anisotropy, short-distance impurity structure and screening by mobile carriers can invalidate the simple Bohr model rescaling.

For shallow hydrogenic donor states of radius $a_B^{\ast }$ and random donor number density $N_D$, the mean spacing is of order $N_D^{-1/3}$. Taking this spacing comparable to $a_B^{\ast }$ gives $N_D\sim (a_B^{\ast }{)}^{-3}$; a geometric diameter criterion instead gives $N_D\sim (2a_B^{\ast }{)}^{-3}$. The orbitals have exponentially decaying tails rather than hard boundaries, so there is no exact geometric onset. Appreciable overlap also limits the independent-donor ionisation model; a genuine transport transition needs additional electronic physics.

Splitting two integers in positional notation gives $x=aq+b$ and $y=cq+d$, where $q=B^m$ shifts a block by $m$ numeral digits in fixed radix $B\ge 2$. The distributive property givesComputing $ac,ad,bc,bd$ separately needs four block multiplications. Merely halving the operands therefore gives an ideal recurrence relation $F(n)=4F(n/2)+O(n)$, still with quadratic time complexity. The Karatsuba three-product identity instead computes $ac$, $bd$ and $(a+b)(c+d)$, recovering the middle coefficient by two subtractions. The MIT notes present precisely this four-product versus three-product distinction.^[ref]

For an original small example, multiply $47$ by $63$. Set $B=q=10$, $a=4$, $b=7$, $c=6$, $d=3$. Four separate block products would give $ac=24$, $ad=12$, $bc=42$, $bd=21$. The three-product calculation isThe crucial reuse is that $u$ and $v$ supply the outer coefficients and also remove the unwanted terms from $w$. There is no need to compute $ad$ and $bc$ separately. Expanding $w=ac+ad+bc+bd$ proves the identity. Induction extends it to Karatsuba multiplication once each smaller block product is correct and Karatsuba recursion termination supplies decreasing sizes and complete base cases.

The operation model matters. Let $n$ count numeral digits in fixed radix $B$, and count single-digit multiplication, addition, subtraction and copying as constant-cost operations. The nonrecursive work on size-$n$ operands is $O(n)$: it includes forming sums, positional shifts and Karatsuba carry handling. Treating an arbitrarily large integer multiplication as one operation would hide the very saving being studied. The coefficients $24,54,21$ above must be shifted and added; concatenating them does not give $2961$.

In the ideal equal-half model, Karatsuba multiplication has $K(n)=3K(n/2)+O(n)$. At depth $j$ its additional work is at most a constant times $n(3/2{)}^j$, and its leaf count is $3^k=n^{{\log }_{2}3}$ for $n=2^k$. A geometric series gives $K(n)=O(n^{{\log }_{2}3})$, where ${\log }_{2}3\approx 1.585$. Four recursive products would instead give $4^k=n^2$ leaves. Saving one block product repeatedly changes the exponent, even though the combination requires extra addition and subtraction.

For the literal sum-based version, each operand sum can occupy one extra numeral digit. Its worst-case upper recurrence relation is $K(n)\le 3K(\lceil n/2\rceil +1)+Cn$ for a nondecreasing cost envelope, above a fixed base case threshold of at least three digits. Karatsuba digit-operation cost proves that this extra digit preserves the exponent; Karatsuba operand splitting handles odd lengths and leading zero padding. For signed integers, apply the nonnegative procedure to their absolute values and restore the sign. The small ledger establishes an exact identity, while Karatsuba practical crossover explains why it alone cannot predict a speedup for two-digit operands.

The combination cost is the work needed to reconstruct an answer from the returned smaller answers. In a time complexity calculation, also count partitioning the input, preparing arguments, and other work outside recursive calls. Write their total as $g(n)$ for a size-$n$ call. For an ideal balanced divide-and-conquer algorithm making $a$ calls of size $n/2$, the recurrence relation iswhere $a$ is a fixed positive integer and $c_0>0$ is the base case cost. Take $n=2^k$, a power of two, so the size becomes one after exactly $k$ divisions. This is a model of total sequential work: all $a$ smaller calls contribute, even if a machine could run some simultaneously.

At level $j$, there are $a^j$ calls, each of size $n/2^j$. Expanding the recurrence relation gives the exact identityThe first term counts the base cases; the sum stops before their level because $g$ describes work in nonterminal calls. This separates depth, number of calls, and local work. Merge sort has two smaller calls and a linear merge algorithm, giving $T(n)=2T(n/2)+O(n)=O(n\log n)$ in Big O notation. The MIT notes derive this by summing equal work across levels, and contrast it with the same two-way split followed by quadratic local work.^[ref]

For the first constructed model, set $a=2$, $c_0=1$ and $g(s)=s$. Every internal level costs $2^j(n/2^j)=n$, while the leaves cost $n$. Thus $T(n)=n{\log }_{2}n+n$, giving $T(16)=80$. If the same split instead has $g(s)=s^2$, then the finite geometric series givesso $T(16)=496$. The root alone already costs $n^2$ in this second model. Smaller arguments do not eliminate expensive reconstruction. These are stipulated operation counts, rather than measured running times or exact merge sort instruction counts.

Binary search has a different saving: after its comparison, only one half remains a candidate. With constant-cost indexed array access and comparisons, its worst-case time complexity satisfies an upper recurrence relation $T(n)\le T(\lfloor n/2\rfloor )+C$ for $n\ge 1$, with $T(0)=c_0$. There is one continuing branch, rather than two. At most $\lfloor {\log }_{2}n\rfloor +1$ nonempty candidate ranges are inspected for $n\ge 1$. Copying the retained half into a fresh array on every step would add linear work at that step and lose this logarithmic bound. The shrinking range and its representation both matter.