Professor Jasper Halekas Van Allen 70 MWF 12:30-1:20 Lecture
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1 Professor Jasper Halekas Van Allen 70 MWF 1:30-1:0 Lecture
2 Back on regular schedule for the next two weeks until Spring Break! There will be labs and homework due this week and next Labs this week and next are computer labs meet in Van 01 rather than your usual lab room
3 E y Electromagnetic waves: E = x 1 c E t c=speed of light x Vibrations on a string: y = x 1 v y t v=speed of wave x Solutions: E(x,t) Magnitude is non-spatial: = Strength of electric field Solutions: y(x,t) Magnitude is spatial: = Vertical displacement of string
4 EM Waves (light/photons) Amplitude E = electric field E tells you the probability of detec;ng a photon. Maxwell s Equa;ons: E 1 E = x c t Solu;ons are sine/cosine waves: Ma)er Waves (electrons/etc) Amplitude Ψ = mafer field Ψ tells you the probability of detec;ng a par;cle. Schrödinger Equa;on:! Ψ Ψ +U(x,t)Ψ = i! m x t Solu;ons are complex sine/ cosine waves:
5 EM Waves Wave Functions Ext (, ) = Asin( kx ωt) Ext (, ) = Acos( kx ωt) ( ) [ cos( ω ) sin( ω )] Ψ (,) xt = Aexpi kx ωt = A kx t + i kx t In either case, wave packets can be formed by superposing waves with different k values.
6 How to solve a differential equation in physics: 1) Guess functional form for solution ) Make sure functional form satisfies Diff EQ (find any constraints on constants) 1 derivative: need 1 soln à f(x,t)=f 1 derivatives: need soln à f(x,t) = f 1 + f 3) Apply all boundary conditions (find any constraints on constants) The hardest part!
7 y x = 1 v y t 1) Guess functional form for solution Which of the following functional forms works as a possible solution to this differential equation? I. y(x, t) = Ax t, II. y(x, t) = Asin(Bx) III. y(x,t) = Acos(Bx)sin(Ct) a. I b. III c. II, III d. I, III e. None or some other combo
8 y x = 1 v y t 1) Guess functional form for solution Which of the following functional forms works as a possible solution to this differential equation? I. y(x, t) = Ax t, II. y(x, t) = Asin(Bx) III. y(x,t) = Acos(Bx)sin(Ct) a. I b. III c. II, III d. I, III e. None or some other combo
9 II. y(x, t) = Asin(Bx) y( x, t) = Asin( Bx) LHS: RHS: y = AB x 1 y v t 1) Guess functional form for solution AB sin( Bx) = 0 sin( Bx) = 0 = III. y(x,t) = Acos(Bx)sin(Ct) y( x, t) = Acos( Bx)sin( Ct) sin( Bx) y LHS: x 0 AB = AB cos( Bx)sin( Ct) 1 y AC RHS: = cos( Bx)sin( Ct) v t v AC cos( Bx)sin( Ct) = v C B = v cos( Bx)sin( Ct)
10 y(x,t)=asin(kx)cos(ωt) + Bcos(kx)sin(ωt) y(x,t)=csin(kx-ωt) + Dsin(kx+ωt) v k ω = = = ) sin( ) sin( ), ( t kx Ck t kx C t x y ω ω Satisfies wave eqn if: ) sin( t kx v C ω ω λ ω ν f k = = 1 t y v x y = Equivalent
11 y(x,t)=csin(kx-ωt) + Dsin(kx+ωt) What is the wavelength of this wave? Ask yourself àhow much does x need to increase to increase kx-ωt by π? sin(k(x+λ) ωt) = sin(kx ωt + π) k(x+λ)=kx+π kλ=π è k=π/ λ t=0 y k=wave number (radians-m -1 ) What is the period of this wave? Ask yourself àhow much does t need to increase to increase kx-ωt by π? sin(kx-ω (t+τ)) = sin(kx ωt + π ) Speed ωτ=π è ω=π/τ λ ω v = = = πf ω= angular frequency T k x
12 y(x,t) = Asin(kx)cos(ωt) + Bcos(kx)sin(ωt) Boundary conditions? l. y(x,t) = 0 at x=0 and x=l At x=0: y(x,t) = Bsin(ωt) = 0 y(x,t) = Asin(kx)cos(ωt) à only works if B=0 From BC at x = L, what are possible values for k? a. k can have any positive or negative value b. π/(l), π/l, 3π/(L), π/l c. π/l d. π/l, π/l, 3π/L, 4π/L e. L, L/, L/3, L/4,. 0 L
13 y(x,t) = Asin(kx)cos(ωt) + Bcos(kx)sin(ωt) Boundary conditions? l. y(x,t) = 0 at x=0 and x=l At x=0: y(x,t) = Bsin(ωt) = 0 y(x,t) = Asin(kx)cos(ωt) à only works if B=0 From BC at x = L, what are possible values for k? a. k can have any positive or negative value b. π/(l), π/l, 3π/(L), π/l c. π/l d. π/l, π/l, 3π/L, 4π/L e. L, L/, L/3, L/4,. 0 L
14 Which boundary conditions need to be satisfied? I. y(x,t) = 0 at x=0 and x=l y(x,t) = Asin(kx)cos(ωt) +Bcos(kx)sin(ωt) At x=0: y = Bsin(ωt) = 0 à B=0 At x=l: y= Asin(kL)cos(ωt)= 0 à sin(kl)=0 à kl = nπ (n=1,,3, ) à k=nπ/l y(x,t) = Asin(nπx/L)cos(ωt) 0 L n=1 n= n=3
15 With Wave on Violin String: Find: Only certain values of k (and thus λ) allowed à because of boundary conditions for solution Same as for electromagnetic wave in microwave oven: Exactly same for electrons in atoms:
16 Three strings: Case I: no fixed ends y x Case II: one fixed end Case III, two fixed ends: x x For which of these cases, do you expect to have only certain frequencies or wavelengths allowed that is for which cases will the allowed frequencies be quantized. a. I only b. II only c. III only d. more than one
17 Three strings: Case I: no fixed ends y x Case II: one fixed end Case III, two fixed ends: x x For which of these cases, do you expect to have only certain frequencies or wavelengths allowed that is for which cases will the allowed frequencies be quantized. a. I only b. II only c. III only d. more than one
18 Electron bound in atom (by potential energy) PE Free electron Only certain energies allowed Quantized energies Boundary Conditions è standing waves Any energy allowed No Boundary Conditions è traveling waves
19 A confined electron m in a box has wave numbers k = nπ/l. What are the allowed momenta and energy of the particle? A. p = nh/(l), E = nhc/(l) B. p = nh/(l), E = n h /(8m e L ) C. p = hl/(nπ), E = hcl/(nπ) D. p = hl/(nπ), E = h L /(n π )
20 A confined electron m in a box has wave numbers k = nπ/l. What are the allowed momenta and energy of the particle? A. p = nh/(l), E = nhc/(l) B. p = nh/(l), E = n h /(8m e L ) C. p = hl/(nπ), E = hcl/(nπ) D. p = hl/(nπ), E = h L /(n π )
21 At a finite boundary, a wave has to be continuous in both value and slope At an infinite boundary, a wave can be discontinuous in slope but still must be continuous in value
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24 Incident, Reflected, Total Transmitted
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