Lecture 2 8.251 Spring 2007 Lecture 2 - Topics • Energy and momentum • Compact dimensions, orbifolds • Quantum mechanics and the square well Reading: Zwiebach, Sections: 2.4 - 2.9 x± = √12(x 0 ± x 1) x+ l.c. time Leave x2 and x3 untouched. −ds2 = −(dx0)2 + (dx1)2 + (dx2)2 + (dx3)2 = ηµvdxµdxv u, v = 0, 1, 2, 3 2dx+dx− = (dx0 + dx1)(dx0 − dx1) = (dx0)2 − (dx1)2 −ds2 = −2dx+dx− + (dx2)2 + (dx3)2 = ηˆµvdxµdxv u, v = +, −, 2, 3 ⎤⎡ ηˆµν = ⎢⎢⎣ 0 −1 0 0 −1 0 0 0 0 ⎥⎥⎦0 1 0 0 0 0 1 1Lecture 2 8.251 Spring 2007 ηˆ++ = ηˆ−− = ηˆ+I = ηˆ = −I I = 2, 3 ηˆ+− = ηˆ−+ = −1 η22 = η33 = 1 Given vector aµ, transform to: a± := √12(a 0 ± a 1) Einstein’s equations in 3 space-time dimensions are great. But 2 dimensional space is not enough for life. Luckily, it works also in 4 dimensions (d5, d6, ...). Why don’t we live with 4 space dimensions? If we lived with 4 space dimesnions, planetary orbits wouldn’t be stable (which would be a problem!) Maybe there’s an extra dimension where we can unify gravity and ... Maybe if so, then the extra dimensions would have to be very small – too small to see. String theory has extra dimensions and makes theory work. Though caution: this is a pretty big leap. Trees in a Box Look at trees in a box Move a little and see another behind it 2� Lecture 2 8.251 Spring 2007 In fact, see ∞ row that are all identical! Leaves fall identically and everything. Dot Product 3a b = −a◦b◦ + a ibi · i=1 = −a +b− − a−b+ + a 2b2 + a 3b3 = ηˆµν aµbν aµ = ηˆµν a ν a+ = ηˆ+ν a ν = ηˆ+−a− = −a− a+ = −a− a− = −a + dx− vlc = dx+ Light rays a bit like in Galilean physics - go from 0 to ∞. 3Lecture 2 8.251 Spring 2007 Energy and Momentum µEvent 1 at x Event 2 at xµ + dxµ (after some positive time change) dxµ is a Lorentz vector The dimension along the room, row is actually a circle with one tree, so not actually infinity. See light rayws that goes around circle multiple times to see multiple trees. Crazy way to define a circle This circle is a topological circle - no “center”, no “radius” Identify two points, P1 and P2. Say the same (P1 ≈ P2) if and only if x(P1) = x(P2) + (2πR)n (n ∈ Z) Write as: x ≈ x + (2πR)n Define: Fundamental Domain = a region sit. 1. No two points in it are identified 2. Every point in the full space is either in the fundamental domain or has a representation in the fundamental domain. So on our x line, we would have: 4� Lecture 2 8.251 Spring 2007 −ds2 = −c 2dt2 + (d�x)2 = −c 2dt2 + v 2(dt)2 = −c 2(1 − β2)(dt)2 ds2 is a positive value so can take square root: ds = 1 − β2dt In to co-moving Lorentz frame, do same computation and find: −ds2 = −c 2(dtp)2 + (d�x)2 = −c 2(dtp)2 dtp: Proper time moving with particle. Also greater than 0. ds = cdtp dxµ = Lorentz Vector ds Define velocity u-vector: cdcxµ uµ = dx Definite momentum u-vector: m dxµ dxµ pµ = muµ = � = mγ 1 − β2 dt dt 1 γ = � 1 − β2 Rule to get the space we’re trying to construct: Take the f d, include its boundary, and apply the identification · 5Lecture 2 8.251 Spring 2007 Note: Easy to get mixed up if rule not followed carefully. Consider �2 with 2 identifications: (x, y) ≈ (x + L1, y) (x, y) ≈ (x, y + L2) Blue: Fundamental domain for first identification Red: Fundamental domain for second identification 6� � � � �� � Lecture 2 8.251 Spring 2007 dx0 d�x pµ = mγ ,dt dt = (mcγ, mγ�v) E = , �p c 2 E: relativistic energy = µc√1−β2 p: relativistic momentum Scalar: µp pµ = (p 0)2 + (p�)2 · E2 = − c2 + �p 2 2 2 2 2m c m v= − 1 − β2 +1 − β2 2 2 1 − β2 c= −m 1 − β2 2 2 c= −m Every observer agrees on this value. Light Lone Energy x0 = time, Ec = p0 +x = time, Ec lc = p+? –¿ Nope! hJustify using QM: Ψ(t, �x) = e −i (Et−p�0�x) Can think of the IDs as transformations - points “move.” Here’s something that “moves” some points but not all. Orbfolds 1. ID: x ≈ −x FD: 7� � Lecture 2 8.251 Spring 2007 Think of ID as transformation x → −x This FD not a normal 1D manifold since origin is fixed. Call this half time �/Zz the quotient. 2. ID: x ≈ x rotated about origin by 2π/n In polar coordinates: z = x + iy 2πi z n z ≈ e Fundamental domain can be chosen to be: 8� � Lecture 2 8.251 Spring 2007 Cone! We focus on these two since quite solvable in string theory. p�ˆ= h�/i SE: ∂Ψ E ih = Ψ ∂x0 c ih ∂ Ψ = EΨ c ∂t So for our x+, want ih ∂Ψ ∂x+ = ElccΨ E Et − �p �x = − − ct + p� �x· ·c = −p · x = −(p+x + p−x− + . . .) Now have isolated dependence on x+, so can take derivative: +i hΨ = e (p+x + + . . .) ∂Ψ ih = −p+Ψ ∂x+ So: Elc − p+ = p− = Suppose have line segment of length a. Particle constrained to this: 9� � � � � � � � � � � � Lecture 2 8.251 Spring 2007 Compare to physics of world with particle constrained to thin cylinder of radius R and length a (2D) Can be defined as: with ID (x, y) ≈ (x, y + 2πR) So: SE = −h2 ∂2 + ∂2 = EΨ2m ∂x2 ∂y2 1. kπx Ψk = ���� sin a � �2h2 kπ Ek = 2m a 2. Ψ�k,l = ���� sin kπx cos ly a R Ψk,l = ���� sin kπx sin ly a R If states with l = 0 then get same states as case 1, but if l = 0 get different E� �2 �value from Rl contribution. Only noticeable at very high temperatures.