Scaling Relations Illustrated by Simple Circuit Elements

A parallel plate capacitor of area A and spacing t gives C = e0A/t, which under isotropic scaling varies as I. The electric field in a charged capacitor is E = a/eot where o is the charge density. This quantity is taken as constant under scaling, so E is also constant. The energy stored in the charged capacitor U = Q2/2C = (1/2) e0E2At, where At is the volume of the capacitor. Thus U scales as I3. If a capacitor is discharged through a resistor R, the time constant is r=RC. Since the resistance R=p£/A, where p is the resistivity, € the length, and A the constant cross section of the device, we see that R scales as I"1. Thus the resistive time constant RC is constant (scales as 1°). The resistive electrical power produced in the discharge is dU/dt= U/RC, and thus scales as I3. The corresponding resistive power density is therefore constant under scale changes.

Consider a long wire of cross section A carrying a current I Ampere's Law gives B =iu0I/2kR as the (encircling) magnetic field B at a radius R from the wire. Consider scaling this system isotropically. If we express I = AEjp, where E is the electric field in the wire, assumed constant in the scaling, and p is the resistivity, then B scales as L. The assumption of a scale-independent current density driven by a scale-independent electric field implies that current I scales as L2. The energy density represented by the magnetic field is fi0B2/2. Therefore the magnetic energy U scales as L5. The time constant for discharge of a current from an inductor V through a resistor R is L'/R. The inductance V of a long solenoid is L'=//0n2A€, where n is the number of turns per unit length, A is the cross section and € the length. Thus inductance V scales as length L, and the inductive time constant L'/R thus scales as L2.

For an LC circuit the charge on the capacitor Q = 0)cos[(C/L)1 /2i]. The radian resonant frequency coLC = (C/L)1/2 thus scales as L°.

2 Systematics of Making Things Smaller, Pre-quantum 2.4

0 0

Post a comment