PHY.6:5 - Archetypal Grounding

PHY.6:5.1 - Derive coupled motion and identify where mechanical energy goes

Two bodies move along one line. Their masses are m1 and m2. Displacements x1 and x2 are measured from a configuration in which their connecting spring is unstretched, so its extension is x2-x1. Velocities are v1 and v2. A spring and a viscous damper act between them; the connector’s inertia is neglected in the chosen regime.

The momentum balances need the connecting force. Use the ideal response

f = k*(x2-x1) + c*(v2-v1),

where k>0 is stiffness and c>=0 is damping. The force on body 1 is f; that on body 2 is -f.

When a response parameter is missing. Suppose this linear response form is already justified, but its parameters are not yet supplied. A static test with extension 1 m, zero relative velocity and force 3 N determines k=3 N/m; it leaves c undetermined. At extension 1 m and relative velocity -2 m/s, the choices c=0 and c=0.5 N*s/m predict 3 N and 2 N. Both reproduce the static test. A prediction of either body’s acceleration therefore remains conditional on c; a supplied value or useful bound can settle a stronger question. A question about total momentum change can already be answered from the external forces, independently of c. The single test does not establish the assumed linear form.

With the response parameters and external forces u1 and u2 supplied, the connected evolution is

x1' = v1; x2' = v2; m1*v1' = f+u1; m2*v2' = -f+u2.

Adding the momentum balances gives p' = u1+u2 for p=m1*v1+m2*v2. This removes the internal force from the total momentum change while retaining it in the relative motion.

The retained mechanical energy is

H = (m1*v1^2 + m2*v2^2 + k*(x2-x1)^2)/2.

Differentiating and substituting the evolution gives

H' = u1*v1 + u2*v2 - c*(v2-v1)^2.

For the ideal damper whose lost mechanical energy becomes internal energy U, add U' = c*(v2-v1)^2. Then (H+U)' equals the external mechanical power. Omitting U from the motion calculation assumes that its change does not appreciably alter the chosen mechanical response.

Take m1=1 kg, m2=2 kg, k=3 N/m, c=0.5 N*s/m, x1=0 m, x2=1 m, v1=1 m/s, v2=-1 m/s, and no external force. At that instant, f=2 N, the accelerations are 2 m/s^2 and -1 m/s^2, total momentum has zero rate of change, and mechanical energy decreases at 2 W. These are useful initial consequences without computing a full trajectory.

Changed preparation. Body 2 is instead held at x2=1 m from before the initial instant, so v2=0. With the other initial values unchanged, f=2.5 N. The support supplies the reaction u2=f; body 1 accelerates at 2.5 m/s^2. The selected pair now exchanges momentum with the support. Keeping the earlier v2=-1 m/s together with a fixed-position constraint would describe an inconsistent preparation. Suddenly clamping the moving body would be another physical problem, requiring an account of that transition.

The reusable move is to obtain motion by joining balances to a response and preparation, and to revise the relevant exchange when the connection changes.

PHY.6:5.2 - Obtain a total without solving its spatial distribution

Let c(x,t) be the concentration of one chemical species in a fixed region. Its prescribed velocity field is u(x,t). Use diffusion coefficient D>0, a constant first-order consumption rate k>=0, and the flux law

J = c*u - D*grad(c).

The species balance supplies

partial_t(c) = -div(J) - k*c.

These equations state the response assumptions: advection, Fickian diffusion and first-order conversion. Their applicability is a physical premise. Impose no flux of this species through the region’s boundary, J dot n = 0, where n is the outward normal.

For the total amount N(t)=integral_region c(x,t) dx, integration of the balance gives

N' = -integral_boundary J dot n dS - k*N = -k*N,

and therefore N(t)=N(0)*exp(-k*t). The requested total follows from its initial total and the stated boundary and reaction laws. Internal transport need not be solved. Consumption of this species can coexist with conservation of the atoms it transfers into products.

Changed question. A local concentration maximum requires the initial spatial distribution and its evolution. The total alone no longer answers. Alternatively, if the reaction rate varies with position, the total rate becomes -integral_region k(x)*c(x,t) dx; replacing it by a constant times N now needs grounds for that reduction. Return to the spatial distribution or a justified bound when the new use needs it.

This case shows how the intended consequence determines which physical relations must be completed. It also separates a global balance result from a local transport prediction.

PHY.6:5.3 - Detect an incompatible ideal connection before attempting a transient calculation

Two ideal linear capacitors have positive capacitances C1 and C2. Their lower terminals share a reference conductor; their upper terminals are connected through a resistance R>0. The effective description neglects leakage, inductance and radiation. Let v1 and v2 be the upper-terminal potentials relative to the common reference, and let current I flow from capacitor 1 to capacitor 2.

Charge balances and the resistive response give

C1*v1' = -I; C2*v2' = I; R*I = v1-v2.

The total upper-plate charge Q=C1*v1+C2*v2 is constant. The difference delta=v1-v2 obeys

delta' = -(1/C1+1/C2)*delta/R.

Thus the difference decays with time constant tau=R*C1*C2/(C1+C2), and both potentials approach

v_final = (C1*v1(0)+C2*v2(0))/(C1+C2).

For C1=1 F, C2=3 F, R=2 ohm, v1(0)=8 V and v2(0)=0 V, the settled potential is 2 V, tau=1.5 s, and I(t)=4*exp(-t/1.5) A, with time measured in seconds.

Stored electrical energy is H=(C1*v1^2+C2*v2^2)/2. Substitution gives H'=-R*I^2. Its initial and final values are 32 J and 8 J, so 24 J is converted into other energy, here heat in the ideal resistance. More generally, the energy difference is

C1*C2*(v1(0)-v2(0))^2/(2*(C1+C2)).

Changed connection. Set R=0 as an ideal connection from the initial instant. Its constraint v1=v2 conflicts with the supplied unequal initial potentials. The smooth evolution above cannot simply start from them under that constraint. Recover the interaction that establishes the common potential when its current, duration or energy conversion matters. Resistance, inductance or electromagnetic emission may matter depending on the actual arrangement and interval; the ideal connection alone does not specify them.

If only the settled potential is needed, conserved charge together with the premise that the connected system settles can already supply it. A transient description is needed for a different question, such as peak current. Taking the positive-resistance time constant to zero does not remove the finite energy conversion. This is why changing an idealization can require revisiting both the preparation and the requested consequence.