Lecture Spring ηˆ++ = ηˆ = ηˆ+i = ηˆ = I I = 2, 3 ηˆ+ = ηˆ + = 1 η 22 = η 33 = 1. a ± := 1 2 (a 0 ± a 1 )

Transcription

1 Lecture 2 - Topics Energy and momentum Compact dimensions, orbifolds Quantum mechanics and the square well Reading: Zwiebach, Sections: x ± = 1 2 (x 0 ± x 1 ) x + l.c. time Leave x 2 and x 3 untouched. ds 2 = (dx 0 ) 2 + (dx 1 ) 2 + (dx 2 ) 2 + (dx 3 ) 2 = η µv dx µ dx v u, v = 0, 1, 2, 3 2dx + dx = (dx 0 + dx 1 )(dx 0 dx 1 ) = (dx 0 ) 2 (dx 1 ) 2 ds 2 = 2dx + dx + (dx 2 ) 2 + (dx 3 ) 2 = ηˆµv dx µ dx v u, v = +,, 2, 3 ηˆµν =

2 Given vector a µ, transform to: ηˆ++ = ηˆ = ηˆ+i = ηˆ = I I = 2, 3 ηˆ+ = ηˆ + = 1 η 22 = η 33 = 1 a ± := 1 2 (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

3 In fact, see row that are all identical! Leaves fall identically and everything. Dot Product 3 a b = a b + a i b i i=1 = a + b a b + + a 2 b 2 + a 3 b 3 = ηˆµν a µ b ν a µ = ηˆµν a ν a + = ηˆ+ν a ν = ηˆ+ a = a a + = a a = a + dx v lc = dx + Light rays a bit like in Galilean physics - go from 0 to. 3

4 Energy and Momentum µ Event 1 at x Event 2 at x µ + dx µ (after some positive time change) dx µ 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, P 1 and P 2. Say the same (P 1 P 2 ) if and only if x(p 1 ) = x(p 2 ) + (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

5 ds 2 = c 2 dt 2 + (d x) 2 = c 2 dt 2 + v 2 (dt) 2 = c 2 (1 β 2 )(dt) 2 ds 2 is a positive value so can take square root: ds = 1 β 2 dt In to co-moving Lorentz frame, do same computation and find: ds 2 = c 2 (dt p ) 2 + (d x) 2 = c 2 (dt p ) 2 dt p : Proper time moving with particle. Also greater than 0. ds = cdt p Define velocity u-vector: dx µ = Lorentz Vector ds Definite momentum u-vector: cdcx µ u µ = dx m dx µ dx µ p µ = mu µ = = mγ 1 β 2 dt dt 1 γ = 1 β 2 Rule to get the space we re trying to construct: Take the f d, include its boundary, and apply the identification 5

6 Note: Easy to get mixed up if rule not followed carefully. Consider R 2 with 2 identifications: (x, y) (x + L 1, y) (x, y) (x, y + L 2 ) Blue: Fundamental domain for first identification Red: Fundamental domain for second identification 6

7 dx 0 d x p µ = mγ, dt dt = (mcγ, mγ v) E =, p c E: relativistic energy = p: relativistic momentum Scalar: 2 µc 1 β 2 µ p p µ = (p 0 ) 2 + (p ) 2 E 2 = c 2 + p m c m v = 1 β β β 2 = m c 1 β = m c Every observer agrees on this value. Light Lone Energy x 0 = time, E c = p 0 + x = time, E lc c = p +? Nope! Justify using QM: Ψ(t, x) = e i h (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

8 Think of ID as transformation x x This FD not a normal 1D manifold since origin is fixed. Call this half time R/Z z the quotient. 2. ID: x x rotated about origin by 2π/n In polar coordinates: z = x + iy z e Fundamental domain can be chosen to be: 2πi n z 8

9 Cone! We focus on these two since quite solvable in string theory. SE: p ˆ = h /i So for our x +, want ih Ψ x + = E lc cψ E Et p x = ct + p x c = p x = (p + x + p x +...) Ψ E ih = Ψ x 0 c ih Ψ = EΨ c t Now have isolated dependence on x +, so can take derivative: So: +i h Ψ = e (p + x ) Ψ ih = p + Ψ x + E lc p + = p = Suppose have line segment of length a. Particle constrained to this: 9

10 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 = h = EΨ 2m x 2 y 2 kπx Ψ k = }{{} sin a 2 h 2 kπ E k = 2m a kπx ly Ψ k,l = }{{} sin cos a R kπx ly Ψ k,l = }{{} sin sin a R If states with l = 0 then get same states as case 1, but if l = 0 get different E 2 l value from contribution. Only noticeable at very high temperatures. R 10