Science 1 M2024 Preparation Notes - PDF

Science-1: prep notes International Institute of Information Technology, Hyderabad Author: Dr Prabhakar Bhimalapuram 2024/2025 1/121, General pointers 1 State of system: 3-dimensional Cartesian coordinate system....

Science-1: prep notes International Institute of Information Technology, Hyderabad Author: Dr Prabhakar Bhimalapuram 2024/2025 1/121, General pointers 1 State of system: 3-dimensional Cartesian coordinate system. scalar quantities (mass, temperature) vector quantities (position vector, velocity and momentum, angular momentum and torque) 2 Vector calculus: vector equalities, addition vector dot product, cross products gradient of a scalar function line integrals: circulation and curl of vector fields surface integrals: flux and divergence of vector fields 2/121, Laws of Newton 1 Ist Law: Inertia. Inertial frames. Uniform rectilinear motion of an object in absence of forces 2 IInd Law: Definition of momentum. Rate of change of momentum equals the net external force on the object 3 IIIrd Law: For central forces, the reaction is equal and opposite to action Great generality of these laws of motion to all kinds of phenomena. Spurred by advances in mathematics. 3/121, Work, kinetic energy. Work done on an object by application of external force ⃗ (t) is given by: F Z t2 W = ⃗ (t) · d⃗r (t) F t1 Using Newton IInd Law, we get R2 ⃗ W = 1 ddtp · ⃗v (t)dt = 12 mv22 − 21 mv12 = K2 − K1 Defining K = 21 mv 2 as kinetic energy, we see force acting on system changes the kinetic energy of the system. 4/121, Conservative forces, potential energy ⃗ (⃗r ) = −∇U(⃗r ), then If the force is conservative, i.e. F R2 W = 1 (−∇U) · d⃗r = −(U2 − U1 ). Thus conservative forces have an associated potential energy function U For conservative forces, the above shows that work W is path-independent, depending only on end-point potential energies W = K2 − K1 = −(U2 − U1 ) Thus, K1 + U1 = K2 + U2. Law of conservation of mechanical energy 5/121, Law of conservation of linear momentum For a vector ⃗s, if F ⃗ · ⃗s = 0, then ⃗ · ⃗s = d ⃗p · ⃗s = d ⃗p · ⃗s
F dt dt ⃗
F · s = 0 implies ⃗p · ⃗s is a constant ⃗ Thus momentum in the direction of ⃗s is conserved 6/121, Torque & Angular momentum and its conservation Angular momentum about origin: ⃗L ≡ ⃗r × ⃗p Torque N ⃗ ≡ d ⃗L = d⃗r × ⃗p + ⃗r × d ⃗p dt dt dt N⃗ = ⃗v × m⃗v + ⃗r × F ⃗ = 0 + ⃗r × F ⃗ ⃗ = d ⃗L = ⃗r × F N ⃗ dt   ⃗ · ⃗s = 0 =⇒ d ⃗L.⃗s = 0; N dt Conservation of Angular momentum along the direction of zero torque. 7/121, For conservative forces, show conservation of energy d d d E = T (⃗v (t)) + U(⃗r (t), t) =⇒ dt E = dt T + dt U d T = m⃗v · dtd ⃗v = m⃗v ⃗ · ⃗a = ⃗v · F dt d P ∂U dxα dt U(⃗r , t) = α ∂x α dt + ∂U∂t = ⃗v · ∇U + ∂U ∂t d E = d (T ⃗ + ∇U) + + U) = ⃗v · (F ∂U dt dt ∂t d ∂U dt E = ∂t =⇒ E =constant when U ≡ U(⃗r ) 8/121, Galilean Relativity All inertial frames have same form of mechanical law. Two inertial frames, K and K ′ having relative velocity v. Then, a particle having ⃗u in K -frame, ⃗u ′ in K ′ -frame. Then ⃗u ′ = ⃗u − ⃗v d ⃗p′ d ⃗p Thus, clearly dt = dt. This gives ⃗′ = F F ⃗ Hence, all mechanical laws remain in same form in both inertial frames. 9/121, Problem solving strategy Identify forces, draw free body diagram “balance forces” to satisfy constraints set up dynamical equations (Use IInd/IIIrd Laws and/or use conservation laws) solve math of dynamical equations, with appropriate boundary conditions Examples: next slide 10/121, Single particle, constrained motion Need to introduce ‘balancing’ forces. Examples: block moving on a horizontal plane. “normal force” block moving down inclined plane. 11/121, For Simple Harmonic Oscillator, find EoM 12/121, For Simple Harmonic Oscillator, find EoM Hooke’s law: F (x) = −kx, EoM as mẍ = −kx q Rewrite as: ẍ(t) + ω02 x(t) = 0, with ω0 = mk Differential equation: second order. Linear in x One standard practice: convert to two first order differential equation. With p = mẋ two equations dtd p(t) = −kx and dtd x(t) = p(t)/m d dt (x, p) = (p/m, −kx), with (x(0), p(0)) = (x0 , p0 ) 1 2 Conservation of energy, E = 2m p + 12 mx 2 Note: solution is x(t) = A+ eiωo t + A− e−iωo t Phase diagram: x(t) vs p(t) 12/121, Show that SHO is applicable for other systems near ‘minima’ 13/121, Limitations of Newtonian Mechanics Handling of constraints (will return to this later) Galilean Invariance (Newtonian Relativity) violated by EM: Special Theory of Relativity 14/121, Limitations of Newtonian Mechanics Handling of constraints (will return to this later) Galilean Invariance (Newtonian Relativity) violated by EM: Special Theory of Relativity Stability of atom: unexplained by classical physics: Quantum Mechanics 14/121, Limitations of Newtonian Mechanics Handling of constraints (will return to this later) Galilean Invariance (Newtonian Relativity) violated by EM: Special Theory of Relativity Stability of atom: unexplained by classical physics: Quantum Mechanics 14/121, Reasons & postulates of Special Theory of Relativity Electromagnetic phenomena seemingly in violation EM wave equation is not Galilean invariant same phenomena seen differently in inertial frames Moving coil, magnet at rest: magnetic field exerts forces (qv × B) on charges in coil, generating current moving magnet, coil at rest: time varying magnetic flux creates electric field on charges (motive emf, Faraday’s Law) Speed of light is same in all inertial frames (expts) 15/121, Reasons & postulates of Special Theory of Relativity Electromagnetic phenomena seemingly in violation EM wave equation is not Galilean invariant same phenomena seen differently in inertial frames Moving coil, magnet at rest: magnetic field exerts forces (qv × B) on charges in coil, generating current moving magnet, coil at rest: time varying magnetic flux creates electric field on charges (motive emf, Faraday’s Law) Speed of light is same in all inertial frames (expts) Postulates of Special Theory of Relativity (Einstein 1905) Same physical laws in all inertial frames Speed of light is same value in all inertial frames 15/121, Events simultaneous in all frames? NO! Concept: Relativity of simultaneity. 16/121, Events simultaneous in all frames? NO! Concept: Relativity of simultaneity. Observer located at the middle of a train, receives flash from front and back of train; flashes set off at same time. Figure: Space-time diagrams in (left) train-fixed frame and (right) platform-fixed frame, with c being same value in both. Red lines are for front, back and mid-points of train. Green lines are for light and have same slope in both plots. 16/121, How do space-time coordinates transform? Frame K ′ with relative velocity v w.r.t. frame K x ′ = γ (x − vt) x = γ (x ′ + vt ′ ) y′ = y y = y′ z′ = z ⇐⇒ z = z′ xv x ′v t ′ = γ (t − 2 ) ′ t = γ (t + 2 ) c c 1 γ=q v2 1− c2 K −frame the event e = (x, t) has transformed K ′ −frame coordinates e′ = (x ′ , t ′ ) Space-time invariant has same value in both frames: (∆s)2 = (∆x)2 + (∆y)2 + (∆z)2 − (c∆t)2 17/121, Notes on Lorentz transform K ′ -frame is moving with velocity v w.r.t. K -frame Given an event e = (x, y, z, t), use forward transform to get e′ = (x ′ , y ′ , z ′ , t ′ ) Strategy is to use difference in events as experimental measurable Note: interchaning prime with unprimed variables and changing v to −v gives Inverse Lorentz Transform. This is an important symmetry! 18/121, What is time difference between ticks of a clock as measured from a moving frame? Take events, e0 = (0, 0) and e1 = (0, τ0 ), e2 = (0, 2τ0 ), · · · (i.e. ticking of the clock stationary in K -frame) 19/121, What is time difference between ticks of a clock as measured from a moving frame? Take events, e0 = (0, 0) and e1 = (0, τ0 ), e2 = (0, 2τ0 ), · · · (i.e. ticking of the clock stationary in K -frame) In K ′ -frame, using Lorentz transform, we get en′ = (−γvnτ0 , γnt0 ). Hence time difference between two consecutive ticks of the stationary clock will be measured in K ′ -frame as γ(n − (n − 1))t0 = γt0. This effect is known as Time dilation 19/121, What is the length of a rod as measured from moving frame? Back end of rod: e1 = (0, t) and front-end of rod e2 = (L0 , t) for any t To measure rod length in K ′ −frame, we need to find e1,2 ′ ′ ′ with t1 = t2 20/121, What is the length of a rod as measured from moving frame? Back end of rod: e1 = (0, t) and front-end of rod e2 = (L0 , t) for any t To measure rod length in K ′ −frame, we need to find e1,2 ′ ′ ′ with t1 = t2 ∆x ′ = γ(∆x − v ∆t) The forward transform involves knowing unprimed quantities , but we do not know which t to use to get x ′ ), we will use inverse Lorentz transform: x = γ(x ′ + vt ′ ), ′ ′ giving us x1,2 = γ(x1,2 + vt1,2 ) and thus ′ ∆x = x2 − x1 = γ(∆x + v · 0) This gives us q v2 ∆x = γ∆x ′ =⇒ L0 = γL′ =⇒ L′ = L0 1− c2 This is known as Length Contraction 20/121, What is the frequency measured by a moving observer? A0 = (0, 0) =⇒ A′0 = (0, 0) ct0 A1 : ct = x+ct0 & x = vt =⇒ ct1 = c−v 2ct0 A2 : ct = x+2ct0 & x = vt =⇒ ct2 = c−v      vct0 ct0 ct0 v vct0 A1 = , =⇒ A′1 = 0, γ − 2 c−v c−v c−v c c−v s 1 1−β ν′ = =ν ∆t ′ 1+β Frequency shift is velocity dependent; velocity of stars 21/121, What about moving rod and a moving clock? Symmetry: both moving and stationary observer will say exactly same thing about other! stationary clock is measured to have “longer time difference between consequtive ticks” by moving observer a stationary observer will measure a moving clock to have ”longer time difference between consecutive ticks” a stationary rod is measured to be shorter than its rest length by moving observer a moving rod is measured to be shorter than its rest length by stationary observer 22/121, Derive relative velocity formulae Particle with x(t) = ut, what is its velocity in K ′ frame? dx ′ dx x ′ = γ(x − vt) =⇒ = γ( − v) dt dt  v dt ′ dx v t ′ = γ t − x 2 =⇒ = γ(1 − ) c dt dt c 2 dx ′ ux − v = dt ′ 1 − ucx2v dy ′ uy ′ = ux v
dt γ 1− c2 dz ′ uz = γ 1 − ucx2v ′
dt 23/121, What is the relativistic momentum? Figure: In K −frame (K ′ -frame), particle A (B) is has u0 velocity. Momentum change of A is −2mu0 , but NOT for B! Having ⃗p = γ(u) m⃗u , solves this issue. 24/121, From momentum, find energy of particle. Show K = (γu − 1)mc 2 , and hence suggest E0 = mc 2. Also E 2 = (pc)2 + (mc 2 )2 25/121, From momentum, find energy of particle. Show K = (γu − 1)mc 2 , and hence suggest E0 = mc 2. Also E 2 = (pc)2 + (mc 2 )2 R2 R2 ⃗ R2 ⃗u ) W = 1 F · dr = 1 ddtp · ⃗u dt = 1 d(γm dt · ⃗u dt R2 After a bit of math, W = 1 dtd (γmc 2 )dt = (γ(u) − 1)mc 2 , to get from rest to velocity ⃗u. Since W = K2 − K1 , we get K = (γ(u) − 1)mc 2 Hence, total energy E = γmc 2 , rest mass energy E0 = mc 2 , gives K = E − E0 Since E = γmc 2 , and p2 = γ 2 m2 u 2 , we get E 2 − p2 c 2 = γ 2 m2 c 4 (1 − u 2 /c 2 ) = m2 c 4 Energy formula: E 2 = (pc)2 + (mc 2 )2 for free particle 25/121, Space-time invariant Events: e1 = (x1 , y1 , z1 , t1 ) and e2 = (x2 , y2 , z2 , t2 ) Define (∆s)2 = (∆x)2 + (∆y)2 + (∆z)2 − c 2 (∆t)2 Lorentz Tranform: x ′ = γ(x − βct), ct ′ = γ(ct − xβ), y ′ = y, z ′ = z with β = v /c Define (∆s′ )2 = (∆x ′ )2 + (∆y ′ )2 + (∆z ′ )2 − c 2 (∆t ′ )2 ′ 2 (∆s  ) =  2 2 2 γ (∆x − βc∆t)  − (c∆t − β∆x)  + (∆y)2 + (∆z)2 =γ (1 − β ) (∆x) + c (∆t) + (∆y)2 + (∆z)2 2 2 2 2 2 = (∆s)2 Space-time 4-vector ⃗s = (x, y, z, ict) If (∆s)2 < 0, an inertial frame exists where both events happen at same position but at different times. Time-like interval. Allows for casual relationships between two events. If (∆s)2 > 0, space-like interval. Both can happen at same time, but at different spatial locations. 26/121, Show electric and magnetic effects are equivalent by STR. Consider a infinite line of charges with density λ on x-axis, and a charge q at a distance d from this ”wire” of charges. All charges at rest. The charge q will experience a force ⃗ = qE F ⃗ directly away from the wire (‘repelling’ force). Consider now a frame K ′ where all the charges are moving to right (+x−axis) with velocity v. Now the charge q will experience two forces (a) ‘repelling’ electric force (b) ’attractive’ magnetic force Due to length contraction in K ′ -frame, density of charges is larger (λ′ > λ) and hence the repelling electric force is larger. But the ’new’ magnetic force exactly cancels the part that is in excess to case of stationary charges! 27/121, Equivalence of electric and magnetic effects (continued) Additional reference: https://tinyurl.com/4x2629nv, where force on two parallel line of charges are analysed from two inertial frames Maxwell laws of electromagnetism showed that magnetic fields result from time-varying electric fields (due to currents, Biot-Savart Law) and electric fields result from time-varying magnetic fields (Faraday’s Law). Maxwell laws demonstrate the connection between electricity and magnetism. Special theory of relativity goes one step further: electric forces and magnetic forces are the same phenomena viewed from different inertial frames. 28/121, Review: One particle Newtonian Mechanics Identify forces, draw free body diagram “balance forces” to satisfy constraints set up dynamical equations (Use IInd/IIIrd Laws and/or use conservation laws) solve math of dynamical equations, with appropriate boundary conditions Examples: get Equations of Motion ⃗ = m(0, 0, −g) = m d 22 (x, y, z) 1 Freely falling body, F dt 2 Charged particle in magnetic field: F ⃗ = q⃗v × B ⃗ 3 Block sliding down an incline (no friction) 4 Simple pendulum Note: In Ex: 4 (5) constraint force is (is not) constant 29/121, From EoM to path properties Newton’s laws: forces change state of system d dt (x, v ) = (v , F /m) for each time point t, Ex: Harmonic Oscillator (1-dim): F (x) = −kx = mẍ Trajectory x(t) = A cos(wt), p(t) = −Amw sin(wt) Alternate description (x/A)2 + (p/B)2 = 1 Describes whole path, not just at single time Ways to get path description: Conservation laws E = T + U along path Straight line path for free particle Also, for free particle R t2 minimum average kinetic energy 1 path Tavg = t2 −t1 t1 dt T (t). Fermat minimum time principle for path of light: Snell’s Law of refraction at interface between medium 30/121, Hand-wavy derivation of a path principle F = ṗ =⇒ [F (t) − ṗ(t)] · η(t) = 0 R2 J ≡ 1 [F − ṗ] · η(t) dt. For true path (x ∗ (t), v ∗ (t)), J = 0 R2 J[η] = 1 [− ∂U ∂x − mv̇ ] · η dt Use dtd (v · η) = v̇ · η + v · η̇ to get R2 J = 1 [−∇U · η − m dtd (v · η) + mv · η̇]dt R2 Set η(t1,2 ) = 0. Giving J = 1 [−∇U · η + mv · η̇]dt Set η(t) = x(t) − x ∗ (t), i.e. small variation of path from true path. δx(t) = η(t), δv = η̇(t) R2 L(x, v ) = −U(x) + mv 2 /2 =⇒ J = 1 δL(x, v )dt R2 Action S ≡ 1 L(x, v )dt, thus J = δS = 0 for true path Principle of stationary action: δS = 0 31/121, Issues with derivation Issue-1: When constraints present, F = −∇U + FC. But FC is not included in above derivation Note that Fc (t) · dx ∗ (t) = 0 on true path Issue-2: Non-simple constraints Ex: Pendulum Mathematically rigorous derivation of ”Euler-Lagrange equations” MT-C7.6 d 2 2d Fk = m dt 2 xk = mk dt 2 xk xk ≡ xk ({q}, t) =⇒ P ∂xk ∂ ẋk ẋk = m ∂q m q̇m + ∂x ∂t k & ∂ q̇n = ∂xk ∂qn  a bit of algebra, using L = T − U After d ∂L ∂L dt ∂ q̇ = ∂q k , ∀k k R t2 Action S = dt L({qk }, {q̇k }, t). Then t1   d ∂L ∂L δS = 0 ⇐⇒ = , ∀k dt ∂ q̇k ∂qk 32/121, Calculus of variations: Find stationary soln. Rx   J = xAB dx f y(x), dy dx (x); x , find y(x) s.t. J is extremum, with fixed y (xA,B ) = yA,B Path space parametrization: y(α, x) = y(0, x) + αη(x) Rx J(α) = x12 f (y(α, x), y ′ (α, x); x) dx =⇒ ∂J for true path y(0, x), we have ∂α |(α=0) = 0 ∂y ′ With y = y0 + αη, we have ∂α = η, ∂y ∂α = dη dx ∂J ∂ R x2 ′ ′ ∂α = ∂α x1 dx f (y, y ; x), with y(α), y (α) x2 x2 ∂f ∂y ′ Z   Z   ∂J ∂f ∂y ∂f ∂f dη = dx + ′ = η+ ′ ∂α x1 ∂y ∂α ∂y ∂α x1 ∂α ∂y dx 33/121, Calculus of variations (cntd.)   d ∂f d ∂f ∂f dη dx ∂y ′ η = η dx ∂y ′ + ∂y ′ dx Thus for η(t1,2 ) = 0, we get Z x2   ∂J ∂f d ∂f = dx − η(x) ∂α x1 ∂y dx ∂y ′ ∂J Setting ∂α = 0, for every η then gives: ∂f d ∂f = ∂y dx ∂y ′ This is the famous Euler-Lagrange equation, with f ≡ f (y(x), y ′ (x); x) 34/121, Euler equation of ’second’ kind f ≡ f (y(x), y ′ (x)), hence ∂f dx = 0, then f − y ′ ∂y ∂f ′ =constant ∂f ∂f ∂f df ∂f ∂f ∂f df = dy + ′ dy ′ + dx =⇒ = y ′ +y ′′ ′ + ∂y ∂y dx dx dy ∂y ∂x d ∂f ∂f Use E-L equation dx ∂y ′ = ∂y to get   df d ′ ∂f ∂f = y ′ + dx dx ∂y ∂x 35/121, Brachistochrone problem MT-C6-Ex:6.2 What is the shortest time curve for the particle travel while starting from rest at point A to reach point B with zA > zB ? p p mv 2 /2 = mgx =⇒ v = 2gx, ds = vdt = dx 2 + dy 2 R 2 √dx 2 +dy 2 R x2 √1+y ′2 T = 1 √ = x dx √ 2gx 1 2gx 36/121, Path Principle: General Pointers Generalize to N-particle PN 1 systems. L = KE − PE = k=1 2 mk vk2 − U({rk }Nk=1 ) R2 S = 1 dt L({rk }, {vk }), δS = 0 over paths. {rk } 7→ qj (transformation of coordinates) still has R2 δS = 0 with S = 1 dt L′ ({qj }, {q̇j }) This is the power of stationary action Now, Z t2     ∂J d ∂L ∂L = dt − · ηk (t) ∂α t1 dt ∂ q̇k ∂qk we can derive equation of motion in qj as d ∂L ∂L L({qk }, {q̇k }) has = , ∀k dt ∂ q̇k ∂qk NOTE: [MT] has direct derivation from F = ma 37/121, Generalised coordinates Simple pendulum MT-E7.2 Wall of death MT-E7.4 block sliding down raising incline MT-P7-11 38/121, Simple Pendulum MT-E7.2 For the simple pendulum, derive E-L EoM single generalised coordinates θ L = 12 m(l θ̇)2 − (−mgl cos θ) = 21 ml 2 θ̇2 + mgl cos θ ∂L ∂ θ̇ = ml 2 θ̇ and ∂L ∂θ = −mgl sin θ 2 E-L: ml θ̈ = −mgl sin θ For small osciallatons, we get θ̈ = −w 2 θ, a simple harmonic oscillator 39/121, Wall of death MT-E7.4 See: https://www.youtube.com/shorts/HKS31wOunY8 (MT-C9-Example 7.4) Particle travelling on a cone of half-angle α subject to gravitational force. Find EoM. z = r cot α, hence (x, y, z) = (r cos θ, r sin θ, r cot α) Clearly, only two generalized coord. required! r and θ 40/121, Wall of death (continued) v 2 = ẋ 2 + ẏ 2 + ż 2 = ṙ 2 +r 2 θ̇2 + ṙ 2 cot2 α = ṙ 2 csc2 α+r 2 θ̇2 U = mgz = mgr cot α L = 12 m(ṙ 2 csc2 α + r 2 θ̇2 − mgr cot α) ∂L ∂θ = 0 and ∂L∂ θ̇ = mr 2 θ̇ =constant [Angular momentum about z-axis is conserved ] ∂L ∂r = mr θ̇2 − mg cot α and ∂L ∂ ṙ = mṙ csc2 α E-L for r is thus, mr̈ csc2 α = mr θ̇2 − mg cot α which is r̈ − r θ̇2 sin2 α + 12 sin(2α) = 0 41/121, Block sliding down raising incline MT-P7.12 A block is located on an incline whose angle is increasing linearly with time. Find EoM Angle of incline θ = αt Distance of block from point of rotation along incline q   1 2 1 2 L = 2 mq̇ + 2 m(q θ̇) − mgq sin θ L = 12 mq̇ 2 + 21 mq 2 α2 − mgq sin(αt) = L(q, q̇, t) ∂L d ∂L ∂ q̇ = mq̇ =⇒ dt ∂ q̇ = mq̈ ∂L ∂q = mqα2 − mg sin(αt) Thus the equation of motion is: q̈ = α2 q − g sin(αt) 42/121, ML: training as variational problem Given data set {xk , yk } find function f s.t. yR = f (x) Standard methodology is to minimize J = dx [y − fθ (x)]2 Alain & Bengio, Journal of Machine Learning Research 15 (2014) 3743 43/121, E-L with undetermined multipliers Constraints fk ({x}, t) = 0 Example of disk rolling down the incline in next slide.   X ∂L d ∂L ∂fk − + λk (t) =0 ∂qj dt ∂ q̇j ∂qj k Figure: Disk rolling without slipping 44/121, Example: Disk rolling down incline MT-E7.9 Generalized coordinates y and θ ∂g Constraint g(y, θ) = y − Rθ = 0; ∂y = 1, ∂g ∂θ = −R   L = 12 M ẏ 2 + 12 I θ̇2 − Mg(−y sin α) = L(y , ẏ, θ̇) ∂L d ∂L ∂f − +λ =0 ∂y dt ∂ ẏ ∂y ∂L d ∂L ∂f − +λ =0 ∂θ dt ∂ θ̇ ∂θ Mg sin α − M ÿ + λ1 = 0 and 0 − I θ̈ − λR = 0 y − Rθ = 0 Eliminate λ: λ = −I θ̈/R = −I ÿ/R 2 , which gives Mg sin α − M ÿ − Iÿ/R 2 = 0  M ÿ = g sin α M + I/R 2 45/121, Comparison: Newtonian and Lagrangian formulations Newtonian Lagrangian Equation of 2 d ∂L ∂L mk dtd 2 xk = Fk dt ∂ q̇k = Motion ∂qk Order of diff. equa- Second order Second order tion generalised coordi- Coordinates Cartesian only nates Interactions using forces using energies via naturally or by equa- Constraints ‘balancing’ forces tion of constraints 46/121, Time symmetry leads to Conservation of Energy Time symmetry: A path is ’translated’ time, and has no change in value of Lagrangian,i.e. when ∂L ∂t =0 dL ∂L = ∇q L · dq + ∇q̇ L · d q̇ + dt ∂t which gives d
∂L L − q̇ · ∇q̇ L = =0 dt ∂t This shows that H = k qk pk − L = constant when ∂L P ∂t =0 2 Show that H = E when L = mv /2 + U(x) proving energy is conserved 47/121, Translation symmetry leads to Conservation of Momentum Translation symmetry: A path ’translated’ in space, has no change in value of Lagrangian i.e. δL = 0 for δ⃗r in path δL = ∇x L · δx + ∇ẋ L · δ ẋ = 0 Now noting that translation means thatδ ẋ = 0, we get ∂L δL = ∇x L · δx = 0 =⇒ =0 ∂xi d ∂L which from EL equations, give dt ∂vi = 0 =⇒ mvi = constant! 48/121, Rotational symmetry Rotational Symmetry: A path ”rotated” in space, has no change in value of Lagrangian δ⃗r = δ θ⃗ × ⃗r and hence δ⃗v = δ θ⃗ × ⃗v HOMEWORK Show δL = 0 =⇒ ⃗r × ⃗p = constant See MT-7.9 49/121, Hamiltonian Dynamics ∂L pk = ∂ q̇k , whose solution gives q̇k = q̇k (qk , pk , t) P H = k pk q̇k − L(qk , q̇k , t) can be seen as Legrende transform of L to replace variable q̇k H ≡ H({qk }, {q̇k }, t) i.e. natural variables: qk and pk P dH with H(qk , pk , t) and H = k pk q̇k − L(qk , q̇k , t) Hamilton equation of motion ∂H ∂H q̇k = , ṗk = − , ∀k ∂pk ∂qk pk and qk are treated equivalently. for each k, two first-order differential equations 50/121, Two body problem: m1, m2 & gravity [MT:C8] L = 12 m1 v12 + 12 m2 v22 − U(|r1 − r2 |) Two particle system: 6 coordinates (and 6 velocities) Choose appropriate generalised coordinates Center of Mass: R ⃗ = m1 ⃗r1 + m2 ⃗ M M r2 Diff vector: ⃗r = ⃗r1 − ⃗r2 With R ⃗˙ 2 + 1 µ⃗r˙ 2 − U(r ) ⃗ and ⃗r , we find: L = 1 M R 2 2 ˙ ⃗ Set R = 0 (Why?). And ⃗r as 2d vector (Why?): L = 12 µ(ṙ 2 + r 2 θ̇2 ) − U(r ) = L(r , ṙ , θ, θ̇) generalised coordinates r and θ: 1 particle in 2D ∂L ∂θ = 0 =⇒ ∂L ∂ θ̇ = ℓ, a constant. Thus 2 µr θ̇ =⇒ ℓ = r · r θ̇ = ℓ, Kepler’s second law ∂L ∂ ṙ = µṙ and ∂L ∂r = µr θ̇2 − U ′ (r ). Thus, µr̈ = µr θ̇2 − U ′ (r ) =⇒ µr̈ = ℓ2 /(µr 3 ) − U ′ (r ) (a) Circular orbit (b) non-circulur orbit 51/121, Two body problem: continued µr 2 θ̇ = ℓ a constant; AND µr̈ = ℓ2 /(µr 3 ) − U ′ (r ) 2 H = K + U = 12 µ(ṙ 2 + r 2 θ̇2 ) + U(r ) = 12 µṙ 2 + µrℓ 2 − Gmr1 m2 ℓ2 Gm1 m2 Ueff = µr 2 − r Figure: Ueff has a minima 52/121, Two body problem: continued Figure: Scenarios for total energy H ∗ Scenario-1: ṙ = 0 =⇒ H = Ueff. This has ℓ r =constant, θ̇ = µr 2 =constant. This means the trajectory is a circle with constant angular velocity ∗ Scenario-2: H ∈ (Ueff , 0], the particle will be oscillate between two values of r. Closed orbits for n = −1, 2 Scenario-3: H > 0. Particle is NOT confined, Particle Swings by! 53/121, Two-body problem continuedMT:C8-S7 Closed orbits for PE∼ r n , only n=-1 (gravitational) and n=2 (SHO) in general case Gravitational case: q 2 H = 2 µ(ṙ + r θ̇ ) − r = E =⇒ ṙ = µ2 (E + kr ) − µl2 r 2 1 2 2 2 k θ̇ Using dθ = ṙ dr and θ̇ = ℓ/(µr 2 ) we get ℓ/r 2 Z θ(r ) = q dr l2 2µ(E − kr − 2µr 2 ) suggeting u = 1/r , which upon integration gives : s 2 α l 2El 2 = 1 + ϵ cos θ, with α = ,ϵ = 1 + r µk µk 2 Solutions: (a) E < 0 an ellipse (or special case: circle), (b) parabola E = 0 and (c) hyperbola for E > 0 54/121, Two body problem: implications on solar system Kepler ‘derived’ this three laws using incorrect forces (several errors magically cancel to give a law that matches Tycho Brache’s and his observations) Newton proved that his Law of Universal Gravitation will result in Kepler’s law for any two body system and hence solar systems This problem can be considered as one of the origin problems of several fields of calculus and analysis Lagrange and Poincare made significant contributions to analysis of trajectories Perturbations to two body trajectories of planets have been explained by presence of nearby ‘large’ planets. 55/121, Solar system Ancients knew and recognised planets: mercury, venus, earth mars, jupiter and saturn. These are visible to eye and move reasonaly fast. Uranus (which is visible to naked eye) is so slow that it was thought to be a faint star till Hershel’s careful measurements showed it to orbit sun. Unaccounted perturbations to Uranus orbit was postulated to be due to a yet undiscovered planet, which was quickly discovered at the approximate location given by mathematical calculations. Pluto was postulated based on perturbations to Neptune’s orbit starting from 1840’s. Discovery in 1930. Advancements in telescopes (1990’s) showed that several bodies of similar size of Pluto forming the Kupier belt. In 2006, Pluto is ‘downgraded’ to a planetoid. 56/121, Three body problem: no closed solutions A good overview: https://en.wikipedia.org/wiki/Three-body_problem No general solutions possible. However, special situations have known solutions with closed orbits. 57/121, Multi-particle systems: Center of Mass Consider a system of N particles, with masses {mk }, position vectors {⃗rk }, velocities {⃗vk } etc.. Total mass of system M = Nk=1 mk P Center of Mass (CoM) is defined as: R⃗ = 1 PN mk r⃗k M k=1 Momentum of particle k is ⃗pk = mk ⃗vk Total linear momentum of system, P ⃗ ≡ P p⃗k k Total angular momentum of system, ⃗L ≡ k ⃗rk × ⃗pk P Net Torque about origin of system, N ⃗ ≡ P ⃗rk × F ⃗k k Check that N ⃗ ≡ d ⃗L gives above expression for N ⃗ dt 58/121, Collection of particles Forces: Fk = Fke + Fki (external and internal forces, resp.) Fki = j Fk←j (internal interactions) P For central forces Fk←j = −Fj←k P i P k Fk = ⟨j,k ⟩ Fk ←j = 0! i P P k rk × Fk = ⟨j,k⟩ rk × Fk←j = 0! Effects of these forces: Net force: F = k Fk = k Fke P P ⃗ Effective mass M at Center of Mass position R 2 ⃗ =F Overall motion: M dtd 2 R ⃗ Angular momentum: ⃗L = ⃗LCoM + ⃗Li CoM Internal forces are central =⇒ the total internal torque is zero! 59/121,  ⃗ conserved, when no external forces Show P 60/121,  ⃗ conserved, when no external forces Show P d ⃗ d 2 mk dtd 2 ⃗r P P dt P = [ dt k mk ⃗vk ] = k ⃗ = 1 R P 2 ⃗ = 1 d 22 P mk⃗rk = 1 d 22 P mk⃗rk =⇒ dtd 2 R ⃗ M k M dt k M dt d ⃗ d ⃗ P  P dt P= dt ⃗ k pk = k Fk = net force on system If no external forces then ⃗k = P F F ⃗ j̸=k k←j , inter-particle interactions  only  P ⃗ P P ⃗ P ⃗ ⃗ k Fk = k j̸=k Fk←j = ⟨j,k⟩ Fk←j + Fj←k = 0! Hence d P ⃗ = 0, P⃗ is a conserved quantity dt When external forces exist, following math similar to above argument, it can be shown that: ⃗ ext = P F Define net external force F ⃗ k k,ext d ⃗ ⃗ dt P = Fext d2 ⃗ ⃗ And hence M dt 2 R = Fext 60/121, Show ⃗L conserved when no external forces 61/121, Show ⃗L conserved when no external forces rk = R + rk′ =⇒ vk = V + vk′ + rk′ ) × (V + vk′ ) P P L = k rk × pk = k mk (R P which gives L = MR × V + k rk′ × pk′ Angular momentum= L of CoM + Lsystem about CoM d d P d d⃗  ⃗ ⃗ ⃗ ⃗ ⃗ P dt L = dt k rk × p k = k dt rk × p k + rk × dt pk d⃗ P P dt L = k [0 + rk × Fk ] =⇒ N = k rk × Fk P for every particle k, force Fk = Fk,ext + j̸=k Fk←j d L = k rk × Fke = k Nke P P dt Rate of change of L = sum external torque If no enxternal torque, angular momentum of system is conserved 61/121, Show energy is conserved 62/121, Show energy is conserved Rt P P Rt W = t12 k Fk (t)·drk (t) = k t12 mk dvdtk ·vk dt = T2 −T1 where T ≡ k 12 mk vk2 P Center of Mass: rk = R + rk′ , vk = V + vk′ vk2 = V 2 + 2vk′ · V + vk′ 2 T = k 12 mk vk2 = 21 MV 2 + 0 + k 12 mk vk′ 2 P P KE of system = CoM KE + System KE in CoM frame Fk = Fke + j̸=k Fk←j P P R2 W = k 1 (Fke + Fki ) · drk i If Uj,k = U(rk , rj ) then Fj←k − = −F k ← k unfinished 62/121, Summary: Multi-particle systems MT:C9.1-9.5 Center of Mass of collection of particles Total momentum of system: P ⃗ ≡ P pk = P ⃗ CoM k Net force on system F ⃗ =MdP ⃗ dt If there are no external forces, linear momentum of system is conserved and equals the CoM momentum Total angular momentum of system about origin: ⃗L ≡ P Lk = P rk × pk = RCoM × PCoM + P r ′ × p′ k k k k k = CoM L about origin + sum of particle L′k about CoM Net torque on system N = dtd L For central forces, net interal torque must vanish KE of system equals sum of CoM KE and sum of KE of each partcle in CoM frame For a conservative system, the total energy is constant in absence of external forces. 63/121, Light 1 Huygens proposed waver theory of light (1690AD) 2 Newton’s treatise (Opticks 1705AD): proposed corpuscular theory of light 3 Refraction, diffraction, internal reflection etc were of considerable importance 4 Spectacles were ‘normal’ technology of the time 5 Young (1801AD): Double slit experiment support for wave theory of light. Also find wavelength of light source. 6 Fresnel’s mathematical Huygens-Fresnel Theory (1818AD) conclusively sets up light as wave 7 Maxwell Equations (1860’s AD): In vacuum, give wave equations for E ⃗ and B⃗ 64/121, Light in vacuum d2 ⃗ ⃗ and same for B ⃗ 1 dt 2 = µ01ϵ0 ∇2 E E 2 PLANE WAVE SOLUTIONS 3 ⃗ =E E ⃗ 0 ei(kz−wt) , propagating along z-axis 4 ⃗ = k (ẑ × E B ⃗ 0 ), that is B ⃗0 = E ⃗0 w 2 1 5 Transverse waves, c = µ0 ϵ0 and νλ = c 6 Superposition of plane waves is also solution: Polarization. 65/121, Light: color depends on wavelength 1 Wavelength from using Diffraction gratings (like Youngs Double slit) 2 Visible range (Human): Red 800 nm, Blue 400 nm 3 Sunlight, after refraction from prism, is found to contain non-visible infra-red radiation and some ultra-violet radiation 4 When metals are heated to high temperature, they emit light and cool off (even beyond convective cooling, if present) 5 Spectrum at a temperature is independent of metal used 6 Kirchhoff: formulated Black body problem: What is the spectrum of black-body? Major technological importance, scientific enquiry 66/121, Black body radiation 1 Stefan-Boltzamann Radioactive power law: P = σT 4 2 Wein Displacement Law: λmax T =constant 3 Wein Distribution law (1896AD): I(ν) ∝ ν 3 e−bν/T , 4 Rayleigh-Jeans law(1899AD): I(ν) ∝ ν 2 T Figure: Wein’s formula is good fit for large ν and Ralyeigh’s for small ν 67/121, Planck’s Black body radiation formula 1 Planck (1894-1900AD): Attempted to get an expression / model that could fit all frequencies: fresh data from his experimental colleagues for small ν 2 Planck’s model required statistical mechanics, and that energy levels of a particular frequency be discrete for good expt fit: 2hν 3 1 I(ν) = 2 hν c e kT − 1 3 Planck: a better model will remove quantization of energy 4 First instance of quantization of energy E(ν) = nhν 68/121, Photoelectric Effect (Hertz 1887AD) In vacuum, charged metal objects discharge their charge when (UV) light falls on it Counter-intuitive results unexplained by contemporary physics Einstein (1905AD): proposes a model that uses Planck energy quantization. Light packet (photon) falling on metal knocks-off an electron with remainder energy into kinetic energy of electron hν = KEmax + W 69/121, X-ray production: inverse photoelectric effect 1 High KE electrons impinge on metal, light is emitted 1 X-Ray’s are light of very small wavelength (0.1nm) 2 X-Ray Diffraction probes chemical structure 70/121, Compton Expt: Photon as particle: confirmation 71/121, Summary: Light 1 Light as wave: Young’s Double slit,, X-Ray Diffraction 2 Light as particle: Photoelectric, Compton Effects 3 LIGHT WAVE-PARTICLE DUALITY: Light is both a wave and a particle 4 wavelength , frequency ν, speed c = λν 5 Mass m = 0, E = pc. Planck’s E = hν =⇒ p = λh 72/121, de Broglie hypothesis: matter waves (1924AD) 1 Inspired by wave-particle duality of light, de Broglie proposed wave-particle duality of particles. Specifically, that a wave of wavelength λ = ph is associated with matter having momentum p 2 Einstein strongly supported this idea, and experiments conclusively showed that particles (electrons) show diffraction pattern when scattered off crystals (just like X-Rays). Also, show interference patterns in double slit experiment. 3 Extending this idea, Schrodinger found the ”wave equation” for such a wave, that is now called ”Schrodinger Equation” (1926AD). This equation is for ALL quantum mechanics currently in use, it is QM equivalent to Newtons ”F=ma” 73/121, Davisson-Gremer Experiment: Particle Diffraction Figure: Davisson-Gremer expt. Diffraction peaks of electrons from the Nickel crystal surface For Nickel crystal d = 0.091 nm, θ = 65deg. Using nλ = 2d sin θ gives λ = 0.165 nm. 54 eV electrons have λ = ph = √2mh KE = 0.166 nm 74/121,  h Matter waves, λ = p 1 For 1µg mass moving at 1µm/s, λ ∼ 10−34 /(10−9 ∗ 10−6 ) = 10−20 m! (electron, me = 9.1 × 10−31 kg, proton=1.6 × 10−27 kg). 2 Regime of small: in energy/ momentum / mass 3 NOTE: MACROSCOPIC SYSTEMS also 4 Localization of wave: particle behavior: group wave 5 1D wave: A cos(2πx/λ + 2πνt) = A cos(kx + wt) Amplitude A, k = 2π/λ, w = 2πν 75/121, Matter waves: group wave Superposition is also a solution: A cos(wt − kx) + A cos((w + ∆w)t − (k + ∆k)x) ∆w ∆k
∼ 2A cos(wt − kx) cos 2 t − 2 x w Figure: Individual waves of velocity vp = k. Resulting gruop wave vg = ∆w∆k 76/121, Matter waves: group wave (2) Many individual waves, more localization in group wave E 2πγmc 2 2π 2π 2πγmv w = 2πν = 2π =. And k = = p= h h λ h h p with γ = 1/ 1 − v 2 /c 2 w c2 Phase velocity vp = k = v dw dw/dv Group velocity vg = dk = dk/dv =v 77/121, Uncertainity Principle 1 Waves have a fundamental property: ∆x∆k > 2 Using de Broglie: k = 2π λ = 2πp h which gives 1 ∆x · ∆p ≥ ℏ 2 This is the famous Heisenberg Uncertainty principle (1927AD) “It is impossible to determine precisely both the position and momentum of a particle simultaneously” Such uncertainty exists for every pair of complementary variables: E and t Example: Particle in the box problem: 78/121, Old Quantum Theory Plank (1900AD) hypothesized: energy of oscillator is quantized E = nhν Bohr (1913AD) hypothesized: angular momentum of electron quantized mvr = nh/2π, n = 1, 2, 3, · · · for electron in Hydrogen atom; this is the famous Bohr model Somerfield (and his group) attempted to explain multi-electron atoms by postulating additional quantization conditions; this is extension of Bohr’s model for Hydrogen Quantization conditions were proposed ad hoc (pulled out of a hat like a magician), and was quite unsatisfactory 79/121, Bohr Model of Hydrogen atom (1913AD) Electron orbiting the fixed nucleus: 1 Postulate: mvr = nℏ: Angular momentum is quantized 2 Orbit of electron around hydrogen: 2πr = nλ with de h Broglie wave λ = mv , which gives mvr = nℏ 2 3 E = 12 mv 2 − K er and balancing electrostatic force with centrifugal force gives: mv 2 2 2 r = K er 2 =⇒ mv 2 = Ke2 /r. Thus E = − 21 K er 2 (nℏ) 2 4 mvr = nℏ and mv 2 = Ke2 /r , gives: v = Ke nℏ , r = mKe 2, 1 K 2 me4 thus E = −R n2 , with R = 2ℏ2 En − Em = −R n12 − m12. Gives good fit to Lyman,
5 Balmer, Paschen and Pfund series for Hydrogen atom 80/121, Frank-Hertz experiment(1913AD) Electrons, in a vacuum tube, sent through vapor (of mercury). Light of 253.6nm is emitted; this corresponds exactly to KE of 4.9eV electron. Confirmation of the basic ideas behind Bohr model for hydrogen atom 81/121, Quantum Theory: math formulation 1 Insists that there a “wave function” Ψ(x, t) that is state of the system; it can be complex function 2 It satisfies Schrodinger equation (evolution of Ψ) d iℏ Ψ(x, t) = ĤΨ(x, t) dt 3 Born Postulate: |Ψ( x, t)|2 dx is probability of finding system at x at time t 4 Every experimental observable has an operator, and expectation value can be calculated by Z ⟨A⟩ = dxΨ∗ (x, t)Â Ψ(x, t) 5 Measurement and collapse of wave function 82/121, Motivate proof of Schrodinger Equation Consider a matter wave for a ‘free’ particle 2 2 1 Waves satisfy wave equation: ∇ ψ(x) + k ψ(x) = 0 where k = 2π/λ h 2 p 2 2 de Broglie: λ = , gives us: ∇ ψ(x) + ( ) ψ(x) = 0 p ℏ 3 Suggestion: ψ(x) = exp (ipx/ℏ) p2 4 For free particle E = which using (2) can be 2m written as: 1 ℏ2 2 ℏ2 2 E =− ∇ ψ(x) =⇒ − ∇ ψ(x) = Eψ(x) ψ 2m 2m 2p and for a particle in a potential as: E = 2m + V (x) giving us Time Independent Schrodinger Equation: ℏ2 2   − ∇ + V (x) ψ(x) = Eψ(x) 2m 83/121, Another motivating derivation 1 Wave can be represented as Ψ(x, t) = exp i(kx − wt) 2 de Broglie: k = 2π λ = pℏ ν E 3 Planck-Einstein: E = hν =⇒ w = 2π = ℏ 4 Ψ(x, t) = exp ℏi (px − Et), matter wave of particle with momentum p and energy E 2 5 ∂Ψ ∂t = − ℏi E ψ, and ∂∂xΨ2 = − ℏ12 p2 Ψ p2 2 2 6 Now using E = 2m , we get iℏ ∂Ψ ∂t = (− 2mℏ ∂ Ψ ∂x 2 ) 7 Extending for particle in a potential V (x), with p2 E = 2m + V (x): ℏ2 ∂ 2   d iℏ Ψ(x, t) = − + V (x) Ψ(x, t) dt 2m ∂x 2 84/121, Features of Schrodinger equation 1 Linearity and Superposition: a1 Ψ1 (x, t) + a2 Ψ2 (x, t) also satisfies Schrodinger equation, if iℏ ∂t∂ Ψi (x, t) = ĤΨi (x, t) for i = 1, 2 2 Normalization: a1 Ψ1 + a2 Ψ2 is normalized if |a1 |2 + |a2 |2 = 1 3 Stationary Solutions: Ψ(x, t) = ψ(x)χ(t) satisfy Ĥψ(x) = Eψ(x) and χ(t) = χ(0) exp (−iEt/ℏ) ∂ 4 Operators: 2  x̂ = x and p̂x =  −iℏ ∂x (next slide), ∂2 ∂2 ∂2 K̂ = − 2m ℏ ∂x 2 + ∂y 2 + ∂z 2 and finally Ê = Ĥ = K̂ + V (x) 5 Expectation values Z ⟨Â⟩ = ψ ∗ Âψ 6 Correspondence principle: CM limit for QM problems 85/121, Momentum operator Z ∂ ∂ ⟨px ⟩ = m ⟨x⟩ = m dx Ψ∗ (x, t) x Ψ(x, t) ∂t ∂t 1 SE gives iℏΨt = −CΨxx + V Ψ and −iℏΨ∗t R= −CΨ∗xx + V Ψ∗ ∗ 2 ⟨vx ⟩ = R dx x(Ψ Ψt + Ψ∗t Ψ) = 1 −C iℏ dx x(Ψ Ψxx − ΨΨ∗xx ) ∗ C dx x∂x (Ψ∗ Ψx − ΨΨ∗x ) R =− iℏ 3 Using uv-rule for integration, we get C C dx (Ψ∗ Ψx − ΨΨ∗x ) =2 iℏ dx Ψ∗ ∂x ∂ R R ⟨vx ⟩ = + iℏ Ψ 2C ∂ 2 Hence vˆx = p̂x /m = iℏ ∂x , with C = ℏ /(2m), we get p̂x = hi ∂x ∂ All operators can be expressed in position and momentum operators! K̂ (r̂ , p̂) 86/121, Particle in 1-D box 1 Particle confined to x ∈ [0, L]. So potential V (x) = 0∀x ∈ [0, L] and V (x) = ∞∀x ∈ / [0, L] 2 Region-1 R1 : x ∈ (−∞, 0), R2 : x ∈ [0, L] and R3 : x ∈ [L, ∞) 3 SE is −CDxx ψ(x) + V (x)ψ(x) = Eψ(x), C = ℏ2 /2m 4 For R1 and R3 , we have LHS of SE having V (x) = ∞ while RHS is finite. Only possible solution: ψ(x) = 0! 5 For R2 , we pget −CDxx ψ = Eψ =⇒ Dxx ψ = −k 2 ψ where k = 2mE/ℏ2 is a real number. This has solutions ψ(x) = A cos(kx) + B sin(kx). 6 Boundary conditions-1: boundary R1 : R2 gives 0 = A.1 + B.0 =⇒ A = 0 7 Boundary conditions-2: boundary R2 : R3 gives B sin(kL) = 0 =⇒ kL = πn where n=1,2,3,4... 8 Solution: ψ(x) = 0∀x ∈ / [0, L] and ψ(x) = B sin( πnx L )∀x ∈ [0, L] with 87/121, Particle in 1-D box (2) Above condition for k gives En = n2 E1 , with E1 = h2 /(8mL2 ) and the wave function R∞ ψn (x) = B sin(nπx/L). Requiring −∞ |ψ(x)|2 dx = 1 gives p B = 2/L Figure: Particle in 1-D box. Left ψ(x) and Right ψ 2 (x). Notice the number of nodes in the wave function. 88/121, Particle in 3-D box Box is defined to have x ∈ [0, Lx ], y ∈ [0, Ly ], z ∈ [0, Lz ]. Potential V (x, y, z) = 0 inside the box and V = ∞ outside. −C(Dxx +Dyy +Dzz )ψ(x, y, z)+V (x, y , z)ψ(x, y, z) = Eψ(x, y, z) Outside the box, V = ∞, so ψ = 0! Now for inside the box, we use the ansatz of Separation of variables: ψ(x, y, z) = ψ1 (x)ψ2 (y)ψ3 (z) gives us −C(ψ2 ψ3 Dxx ψ1 + ψ1 ψ3 Dyy ψ2 + ψ1 ψ2 Dzz ψ3 ) = Eψ1 ψ2 ψ3 Dxx ψ1 Dyy ψ2 Dzz ψ3 E + + =− ψ1 ψ2 ψ3 C And thus PiB in x, y and z variables. So s 23 n1 πx n2 πy n3 πz ψ(x, y, z) = sin sin sin Lx Ly Lz Lx Ly Ly  2  n n2 n2 h2 with Energy E = L21 + L22 + L32 8m 89/121, Harmonic Oscillator 1 SE: −Cψxx + 12 kx 2 ψ(x) = Eψ(x), 2 Choose x = by to get −Cψyy /b2 + 12 kb2 y 2 ψ = Eψ, 4 2 3 Rearrange: ψyy − kbC y 2 ψ + EbC ψ = 0. Now choosing b so that kb4 /C = 1 and α = Eb2 /C = α, we get ψyy + (α − y 2 )ψ = 0 4 When y → ±∞, we get ψyy − y 2 ψ = 0, which in turn suggests ψ ∼ exp dy 2 , with 4d 2 = 1. We choose ψ ∼ exp(−y 2 /2) 5 Ansatz: ψ(y) = f (y) exp(−y 2 /2). This gives us ψy = exp(−y 2 /2)(fy − yf ) and ψyy = exp(−y 2 /2)(−y(fy − yf ) + (fyy − yfy − f )) 6 Thus SE ψyy − y 2 ψ + αψ = 0 becomes: 2 exp(−y P /2) (fyy − 2yfy + (α − 1)f ) = 0. Set 2n+1−α f (y) = n an y n , to get an+2 = (n+1)(n+2) an 90/121, Harmonic (2) 1 Thus ψ = exp(−y 2 /2)f (y) converges when 2n + 1 − α = 0 for integer n. For α = 2n + 1, ψn (y) = exp(−y 2 /2)fn (y) , fn is polynomial of order n 2 n = 0, α = 1, we get a2 = 0 and hence f0 = 1 3 n = 1, α = 3, we get f1 = 2y 4 n = 2, α = 4 , we get f2 = 4y 2 − 2 91/121, Harmonic Oscillator (2) 92/121, Tunneling 1 In SHO, non-zero probability beyond classical turning points (with KE¡0!) 2 Starting in Region-1 with energy E < U, classically the particle cannot penetrate Region-2. 3 Conditions at boundaries: x = 0 and x = L: 1 ψ must be continuous 2 Derivative dψ/dx must be continuous 93/121, The tunnel effect 1 2m/ℏ2 Region-1: Dxx ψI + C1 Eψ1 = 0, with C1 =p 1 ψI = A exp (ik1 x) + B exp (−ik1 x) , k1 = 2mE/ℏ2 2 Region-2: Dxx ψII + C1 (E − U)ψII = 0 1 C exp (k2 x) + D exp (−k2 x) , ψII = p k2 = 2m(U − E)/ℏ2 3 Region-3: Dxx ψIII + CEψIII = 0 1 ψIII = F exp (ik1 x) (only the +ve direction wave) 4 Boundary conditions: 1 ψI (x = 0) = ψII (x = 0) =⇒ A + B = C + D 2 Dx ψI (x = 0) = Dx ψII (x = 0) =⇒ ik1 (A − B) = k2 (C − D) 3 ψII (x = L) = ψIII (x = L) =⇒ C exp (k2 L) + D exp (−k2 L) = F exp (ik1 L) 4 Dx ψII (x = L) = Dx ψIII (x = L) =⇒ k2 (C exp (k2 L) − D exp (−k2 L) = k1 (F exp (ik1 L) 94/121, Tunnel effect (3) Transmission coefficient T ∼ e−2k2 L Scanning Electron Microscope (SEM) uses principle of tunneling: surface topography and composition 95/121, Hydrogen Atom ℏ2 SE: − (ψxx + ψyy + ψzz ) + V (x, y, z)ψ = Eψ 2m ∂ 2ψ     1 ∂ 2 ∂ψ 1 ∂ ∂ψ 2m 2 (r )+ 2 sin θ + 2 + 2 (E−V ) ψ = 0 r ∂r ∂r r sin θ ∂θ ∂θ ∂ϕ ℏ Ansatz: Separation of variables: ψ(r , θ, ϕ) = R(r )Θ(θ)Φ(ϕ) sin2 θ d     2 dR sin θ d dΘ r + sin θ + R dr dr Θ dθ dθ 1 d 2 Φ 2mr 2 sin2 θ  2  e + +E =0 Φ dϕ2 ℏ2 4πϵ0 r Note Φ appears in only one term! 1 d 2Φ 2 = −ml2 =⇒ Φ(ϕ) = A eiml ϕ Φ dϕ Note: Φ(ϕ) = Φ(ϕ + 2πn) requires integer squared −ml2 ! 96/121, Hydrogen atom (2) Note that the third term depends on Φ(ϕ) alone, so equating it to −ml2 , we get are small rearrangement: 1 d 2 dR 2mr 2 e2 m2 1 d dΘ (r )+ 2 (K +E) = 2l − (sin θ ) R dR dr ℏ r sin θ Θ sin θ dθ dθ which gives, after equating both sides to l(l + 1): ml2   1 d dΘ − sin θ = l(l + 1) sin2 θ Θ sin θ dθ dθ 1 d 2 dR 2mr 2 e2 (r )+ (K + E) = l(l + 1) R dR dr ℏ2 r 97/121, Hydrogen atom (3) 1 Φ(ϕ) = Aeiml ϕ , where ml = 0, ±1, ±2, · · · 2 Θ(θ) depends on both ml2 and l , it can be shown that ml = 0, ±1, ±2, · · · , ±l 3 R(r ) depends on both l and a new quantum number n, such that l = 0, 1, 2, · · · (n − 1) 4 So ψ(r , θ, ϕ) = Rn,l (r )Θl,ml (θ)Φml (ϕ) 5 n : Principal quantum number (shell): Energy 6 l: Orbital quantum number (subshell): Angular momentum magnitude 7 m: Magnetic quantum number: direction of angular momentum 98/121, Hydrogen (4) 99/121, Hydrogen atom (5) 100/121, Hydrogen atom (6): Transitions Figure: Energy level diagram for hydrogen atom, showing allowed transitions ∆l = ±1, given by dx ψf∗ (ex)ψi ̸= 0 R 101/121, Zeeman effect: magnetic dipole of atom interacts with external magnetic field eℏ τ = µB sin θ, Um = ml 2m B 102/121, Multi-electron atoms: from Hydrogen atom solution 1 SE for multi-electron atoms: Only numerical methods. 2 Wave function can have even/odd symmetry: ψ(x1 = a, x2 = b) = ± ψ(x1 = b, x2 = a) 3 Electron spin is also quantized: quantun number ms 4 Exclusion Principle: No two electrons can have same set of quantum numbers (n, l, ml , ms ) 103/121, Periodic Table 104/121, Spin-orbit coupling 105/121, Energy level ordering 106/121, Molecular Covalent Bond 107/121, Statistical Mechanics Atoms / molecules approximated as interacting particles (Lennard-Jones, electrostatic, bonds (as springs) etc) Classical mechanics reasonably model dynamics ISSUE: N ∼ 1023 atoms. Too large for tracking numerically Statistics is the way to go Statistical Mechanics: child of Boltzmann, based on strong belief on reality of atoms Statistical Mechanics gives a probabilistic description of system, moving away from deterministic Classical Mechanics. This was major sticking point/contention of its acceptability among Boltzmann’s contemporary scientists (thermodynamics, mechanics) 108/121, Simple Harmonic oscillator p mẍ = −kx =⇒ x = A cos(wt), w = k/m x = A cos(wt) =⇒ p = mẋ = −Amw sin(wt) E = 12 kx 2 + 2m 1 2 p = 12 kA2 phase plot (x vs p): point on ‘circle’ with angle wt. Phase-point x = (x, p) visits all points equally Make a probabilistic statement: ”Equal probability for system to be in any of the possible phase points” A mathematically rigorous proof for Hamiltonian systems: Louivlle theorem for phase space density (in an ensemble) dtd ρ(qk , pk , t) = 0 109/121, Postulates of statistical mechanics
State of system x = {⃗rk }Nk=1 , {⃗pk })Nk=1 , x ∈ R6N. Phase-space point POSTULATE: ”Equal a priori probability of states with equal energy” EQUIVALENTLY: ”probability of state x is proportional to e−E(x)/kT ”. This is commonly known as Boltzmann Law. Ensemble: a collection of states consistent with thermodynamic constraint Micro-canonical: particles, volume and energy Canonical: particles, volume and temperature Constant pressure, temperature and particles Grand canonical: chemical potential, volume and temperature POSTULATE: (Ergodicity) Time average equals Ensemble average 110/121, Canonical ensemble: general setup phase point x ∈ R3N × R3N (generalized coordinate, generalized momentum space) p(x) ∝ e−E(x)/kT = Z1 e−E(x)/kT Partition function Z = dx e−E(x)/kT = Z1 e−βE , R β = 1/kT ∂ R ⟨E⟩ = dx p(x)E(x) = − ∂β ln Z Reminder Gibbs-Helmholtz relation in Thermodynamics:   ∂(A/T ) U= ∂(1/T ) V ,N We identify A/kT = − ln Z + ϕ(V , N), setting ϕ = 0 A = −kT ln Z , all thermodynamics from this relation! 111/121, Degeneracy of energy level, Entropy A = −kT ln Z = −kT dx exp − E(x) R R∞ kT Z = E0 dE W (E) exp(−E/kT ) Integrand is gaussian with peak Ē Z = e−Ē/kT W (Ē)∆E where ∆E is spread   S = (U − A)/T = (Ē + kT ln Z )/T = k ln W (Ē)∆E Entropy S is thus a measure of number of states assailable to system given by S(T ) = k ln W (Ē)∆E where Ē and spread ∆E both are T dependent Microcanonical (const E,V,N): has S(E, V , N) = k ln W (E) uncertainty principle ∆x∆px ≥ h. So gives volume of each state is given by ∆x∆px = h 112/121, System of non-interacting particles Ideal system i.e. identical non-interacting (independent) particles at temperature T P Total energy E = Ek where Ek is KE of particle k Probability density p(x) = ΠNk=1 p1 (xk ), p1 (xk ) = Z11 e−E1 (xk )/kT with Z1 = dxk e−E1 (xk )/kT R Barometric Law: E1 = mgh =⇒ ρ(h) ∝ exp( mgh kT ) Maxwell-Boltzmann distribution: 2 E1 = 12 m⃗v · ⃗v =⇒ p1 (⃗v ) ∝ exp(− mv 2kT ) 113/121, System of non-interacting particles (cntd. Equi-partition theorem: 2 Z1 = dx1 e−p1 /2mkT R R R R R R R dx = d⃗r d ⃗p = dx dy dz dpx dpy dpz  hR 2 i d ⃗p e−p /2mkT = R Z1 = d⃗r 1 hR i3 √ 3 +∞ 2 V −∞ dpx e−px /2mkT = V 2πmkT Z1 ∝ T 3/2 ⟨E1 ⟩ = − ∂lnZ ∂β 1 = 32 kT for N Particle system ⟨E⟩ = 32 Nk T = 23 RT   ZN = (Z1 )N =⇒ p = − ∂A ∂V = kT VN T ,N Every quadratic term in energy contributes 12 kT 114/121, Grand-canonical ensemble statistics: Ideal systems i.e. non-interacting particles Constant chemical potential µ, volume V and Temperature T. Varying number of particles N each state can have occupancy of: ns = 0, 1 =⇒ Fermi-Dirac statistics ns = 0, 1, 2, · · · , N =⇒ Bose-Einstein statistics 1 Occupation probability: prob(ns ) = −µ exp( EskT , with )±1 + sign for Fermi-Dirac statistics − sign for Bose-Einstein statistics When 1 is removed in denominator, we get Boltzmann statistics 115/121, ± sign in occupation statistics 1 Maxwell-Boltzmann works well for system of identical particles that can be distinguished 2 ψ1 = ψa (1)ψb (2) and ψ2 = ψa (2)ψb (1) violates non-distinguishably 3 ψB = √12 (ψa (1)ψb (2) + ψa (2)ψb (1)) has even symmetry ψB (1, 2) = ψB (2, 1) 1 Apparent increment of probability! 4 ψF = √12 (ψa (1)ψb (2) − ψa (2)ψb (1)) has odd symmetry ψF (1, 2) = −ψF (2, 1) 1 Apparent decrement of probability! (infact zero probability at same positions) 116/121, Black-body radiation: Boson photon gas 8πν 2 kT Ralyeigh-Jeans: u(ν)dν = dν c2 2L 2
Standing wave: jx2 + jy2 + jz2 = λ , set q j = jx2 + jy2 + jz2 1 Number of standing waves: g(j)dj = 2 · 8 · 4πj 2 dj j = 2L λ = 2Lν c Density of standing waves: 2 G = V1 g =⇒ G(ν)dν = 8πν c3 dν If energy per wave is kT , the energy density is thus: 2 u(ν)dν = 8πν c3 kT dν R Note that this has a major issue! Utotal = u(ν) dν = ∞ 117/121, Planck Radiation Formula Planck assumed that walls of black body are made of oscillators; to calculate average energy of an oscillator of frequence ν (called Ē(ν)) that could keep energy finite Planck after several attempts was ‘cornered’ to assume that energy of oscillator is not continuous but is given by nhν. This gives: X e−nhν/kT hν Ē = nhν = hν/kT n Z e −1 This average energy per oscillator gives energy of the cavity as: 8πh ν3 u(ν)dν = 3 hν/kT dν c e −1 matching experimental data over the whole range of ν 118/121, LASER Ni→j = Nj→i EQUILIBRIUM Ni→j = Ni Bij u(ν) and Nj→i = Nj (Aij + Bij u(ν)) Aji /Bji u(ν) = Ni Bij Nj · Bji −1 Stimulated emission: same probability as absorption (i.e Bji = Bij ) Ratio of spontaneous and stimulated emission is ν 3 , 3 from Aji = Bji · 8πhν c3 119/121, Specific heat of solids CM model of solid: Particles oscillate about their mean position, i.e.each atom is a 3-d oscillator. Thus each has energy 3 · 2 · 21 kT For E = 3NkT , the system has Cv = 3Nk , a constant hν Einstein (1907AD) suggested: Ē = ehν/kT −1 and hν 2 e−hν/kT
hence for T → 0 , we get Cv =∝ kT 120/121, Electrons in metal 1 1 Electrons are fermions, thus nFD (ϵ) = (ϵ−ϵ )/kT √e F +1 2L 2m √ 2 g(j)dj = πj 2 dj, with j = 2L λ = 2Lp h =h ϵ 3 This gives us √ √ ϵ g(ϵ)dϵ ∝ ϵ dϵ =⇒ n(ϵ) ∝ (ϵ−ϵ )/kT dϵ e F +1 121/121,