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5 Dec 2008: x µy) (x µy) = 0 , x = |a|2b |b|2a , (2x y) (a b) = λ. ... 2. (a) Show that. a (b c) = b (c a) = c (a b).

26 Feb 2008: constant k:. mẍ = kx. so x = A sin(. k/m t) B cos(. ...  = E B|B|2. The advantage of writing the results in vector notation, rather than in coordinates, is that they are thentrue in any coordinate system (and not just the

26 Feb 2008: Hence E = 1. 2mV 2 and h = V b (because h = |r ṙ|);. ... Φ, Φ δΦ). Let. k =qQ. 8π0E. so that. b = k cot 12Φ,.

25 Jul 2008: Spheroid: axes a, b, b, aspect ratio r = ab. rod r > 1 disk r < 1. ... A B C. r r 22(ln 2r3. 2). 6 ln 2r11r 2.

30 Oct 2008: We add and subtractf(b),. I[f,g,θ] = f(b). b. a. g(sin[ωθ(x)]) dx. ... b. a. [f(x) f(b)]g(sin[ωθ(x)]) dx. = f(b). b. a. g(sin[ωθ(x)]) dx.

8 Apr 2008: From the fundamental theorem of calculus we. have(. Ib. ). c= f (b, c). ... Also, from direct differentiation. (. Ic. ). b=. b. 0. (. fc. ). xdx. Therefore, if. b = b(x), c = c(x), then.

3 Jul 2008: φν (φν B)) (B φν Bφν) (60)where. J νi = φν iφν κνφνi j j uν (61)uν = ( jφν σ). ... η (φ φ (φ φ)) η((φ (φ B)) (B φ Bφ)) = 0.

3 Jul 2008: ψ Cδ(x)‖Mβ := ‖ψ‖L1β |C|. (7.2). and let B = B(Mβ,Lβ) be a Banach space of continuous linear maps Mβ Lβ. ... and B = B. From (7.7), (7.10) and (7.11) the statement of theorem follows.

8 Apr 2008: F |B〉 = |B〉 , ()F |F〉 = |F〉. The new operator ()F anticommutes with Qα since. ... FQα|F〉 = ()F |B〉 = |B〉 = Qα|F〉 = Qα()F |F〉 ={. ()F , Qα}. = 0. Next, consider the trace. Tr. {. ()F{. Qα , Q̄β̇. }}. = Tr{. ()FQα︸ ︷︷ ︸.

23 Sep 2008: a) 0. (z) (time integrated) Final I. (b) D. Fig. 2.