PHY.5:5 - Archetypal Grounding

PHY.5:5.1 - Keep a motor’s torque while eliminating fast current dynamics

Consider an ideal linear motor over a range in which resistance R, inductance L, inertia J, damping b and conversion constant k are positive and constant. In consistent SI units, use the same k for torque per current and back voltage per angular speed:

L*i' = V - R*i - k*omega

J*omega' = k*i - b*omega

The electrical and mechanical balances describe the assumed device, including its load in J and b. Saturation, variable load or a different drive would require the corresponding physical relations.

Suppose the work needs the slow speed response. The electrical relaxation time is tau=L/R. The current toward which the electrical part relaxes is q(t)=(V(t)-k*omega(t))/R. Replacing i by q gives

J*omega' = (k/R)*V - (b+k^2/R)*omega.

The eliminated current still supplies driving torque and additional damping. The slow response time of this candidate is J/(b+k^2/R). Compare tau with that time and the drive’s variation time.

To examine the neglected response, set e=i-q. The full electrical equation gives tau*e'=-e-tau*q'. If |q'|<=K on the interval, integration yields

|e(t)| <= |e(0)|*exp(-t/tau) + tau*K*(1-exp(-t/tau)).

The bound on q’ can come from |q'|<=(|V'|+k*|omega'|)/R and the allowed drive and acceleration, using the full physical account where needed. A bound inferred only by assuming the proposed approximation would leave that assumption unresolved.

For example, let R=2 ohms, L=0.02 henry, k=0.1 in the stated SI convention, J=0.02 kg m² and b=0.01 N m s per radian. Then tau=0.01 s and the candidate slow time is about 1.33 s. With |e(0)|<=0.5 ampere and K=1 ampere per second, the current departure at 0.05 s is at most 0.01331 ampere, giving a torque departure at that instant of at most 0.001331 N m relative to kq.

The slow speed error is a different output. With the same initial speed and drive, let delta be full speed minus reduced speed. It satisfies

J*delta' + (b+k^2/R)*delta = k*e.

Let B(s) denote the current-departure bound above. Since the initial speed difference is zero and the response kernel is positive, integration gives

|delta(t)| <= (k/J)*integral_0^t exp(-(b+k^2/R)*(t-s)/J)*B(s) ds.

For the stated values, this speed-departure bound at 0.05 s is about 0.026064 rad/s, hence less than 0.02607 rad/s. An allowed error of 0.03 rad/s therefore permits the reduced calculation for that speed reading. The integral carries the earlier transient into the answer; the small current departure at the final instant alone would not give this bound.

Changed work. If the next question concerns torque immediately after switching, the bound includes the initial current departure. Use the electrical transient. If the drive varies on the electrical relaxation time, recompute its departure instead of extending the slow-drive approximation. These returns change which physical response is retained.

PHY.5:5.2 - Decide when a chain can be treated as a continuous medium

Consider an infinite ideal one-dimensional chain with identical masses m, spacing a and linear springs of stiffness kappa. Each mass moves a small distance u_j from its reference position. The balance is

m*u_j'' = kappa*(u_(j+1)-2*u_j+u_(j-1)).

For waves with wavenumber q in 0<q*a<pi, substitution of a sinusoidal wave gives

omega^2 = (4*kappa/m)*sin^2(q*a/2).

For wavelengths long compared with a, expanding the neighboring displacements gives the continuum equation u_tt=c^2*u_xx with c=a*sqrt(kappa/m). It predicts both phase and group speed c. The chain’s phase speed divided by c is sin(q*a/2)/(q*a/2); its group speed divided by c is cos(q*a/2). Expanding those ratios gives v_phase/c = 1-(q*a)^2/24+O((q*a)^4) and v_group/c = 1-(q*a)^2/8+O((q*a)^4), exposing their different first corrections.

At q*a=0.2 these ratios are about 0.99833 and 0.99500. A half-percent allowance for these speeds accommodates this ideal comparison, subject to the question’s waveform and other physical premises. The group-speed difference is already larger than the phase-speed difference.

Changed work. A disturbance containing wavelengths near the shortest traveling waves of the chain probes q*a near pi. The chain’s group speed tends to zero, while the continuum account keeps c. A question about that disturbance’s propagation requires retaining the discrete dispersion or an adequate extension. Making the computation of the uncorrected continuum equation more accurate cannot recover the omitted physical dependence.

This case concerns an ideal linear chain. It demonstrates choosing spatial resolution from the wave that matters. It supplies no claim that every material, boundary or large deformation obeys the same chain law.

PHY.5:5.3 - Retain fluctuations after fast velocity has relaxed

For a one-dimensional Brownian particle in a uniform equilibrium bath, take mass m, drag coefficient gamma and temperature T, with no applied force. The underdamped account has position x, velocity v and thermal forcing. Let tau=m/gamma and D=k_B*T/gamma. With an initially equilibrated velocity, its velocity covariance is (k_B*T/m)*exp(-|t-s|/tau).

Integrating that covariance over the two times gives

E[(x(t)-x(0))^2] = 2*D*(t-tau*(1-exp(-t/tau))).

At times large compared with tau, the overdamped diffusion description gives 2*D*t. Its omitted contribution to this mean-square displacement is bounded by 2*D*tau. This calculation states which long-time consequence the reduction preserves.

Setting the mean velocity to zero and deleting the forcing instead gives no displacement spread. The unresolved bath continues to transfer random impulses after the velocity’s preparation has relaxed. For the displacement distribution, retain their diffusion effect.

Changed work. Change the bath to a spatially varying temperature and ask about entropy production. The uniform-bath calculation no longer answers the question. Celani and coauthors show a further distinction: under their smooth-temperature and small-inertia conditions, the overdamped position process has the appropriate limit, while the mean rate of entropy production retains an additional positive contribution absent from the naive overdamped expression. Return to the thermodynamic observable and its limiting calculation. Position accuracy alone cannot decide that use. Their 2012 paper states the preparation and the contribution.