Simple Nature - Light and Matter

Solving for w,w = 1 − A1 + A(1 + u)(1 + v) − (1 − u)(1 − v)=(1 + u)(1 + v) + (1 − u)(1 − v)2(u + v)=2(1 + uv)= u + v1 + uv(c) This is all in units where c = 1. The correspondence principle says that we should getw ≈ u + v when both u and v are small compared to 1. Under those circumstances, uv is theproduct of two very small numbers, which makes it very, very small. Neglecting this term inthe denominator, we recover the nonrelativistic result.Page 442, problem 22: At the center of each of the large triangle’s sides, the angles add upto 180 ◦ because they form a straight line. Therefore 4s = S +3×180 ◦ , so S −180 ◦ = 4(s−180 ◦ ).Page 442, problem 28: By the equivalence principle, we can adopt a frame tied to the tossedclock, B, and in this frame there is no gravitational field. We see a desk and clock A go by.The desk applies a force to clock A, decelerating it and then reaccelerating it so that it comesback. We’ve already established that the effect of motion is to slow down time, so clock A readsa smaller time interval.Solutions for Chapter 9Page 544, problem 1: ∆t = ∆q/I = e/I = 0.16 µsPage 545, problem 12: In series, they give 11 kΩ. In parallel, they give (1/1 kΩ +1/10 kΩ) −1 = 0.9 kΩ.Page 548, problem 25: The actual shape is irrelevant; all we care about is what’s connected towhat. Therefore, we can draw the circuit flattened into a plane. Every vertex of the tetrahedronis adjacent to every other vertex, so any two vertices to which we connect will give the sameresistance. Picking two arbitrarily, we have this:This is unfortunately a circuit that cannot be converted into parallel and series parts, andthat’s what makes this a hard problem! However, we can recognize that by symmetry, thereis zero current in the resistor marked with an asterisk. Eliminating this one, we recognize thewhole arrangement as a triple parallel circuit consisting of resistances R, 2R, and 2R. Theresulting resistance is R/2.937