PHYSICS - 1 (Lecture -1)

PHYSICS - 1 (Lecture -1) Santabrata Das Department of Physics Indian Institute of Technology Guwahati sbdas@iitg.ernet.in, Room #: 006 30 July, 2014 Santabrata Das (IITG) PH 101, July-November 30 July, 2014 1 / 48

Section 2 Introduction to the Course Go to the link http://www.iitg.ernet.in/sbdas/. Scroll down and click "Welcome to PH101: July-November, 2014". You will see all course related information there. Santabrata Das (IITG) PH 101, July-November 30 July, 2014 2 / 48

Topics Classical Mechanics Energy rest mass energy Relativistic Mechanics Energy rest mass energy and de Broglie wavelength size Quantum Mechanics de Broglie wavelength size and Energy rest mass energy Santabrata Das (IITG) PH 101, July-November 30 July, 2014 3 / 48

Syllabus Classical Mechanics (15 Lectures): (1) Review of Newtonian Mechanics in rectilinear coordinate system. (2) Motion in plane polar coordinates. (3) Conservation principles. (4) Collision problem in laboratory and centre of mass frame. (5) Rotation about fixed axis. (6) Non-inertial frames and pseudo forces. (7) Rigid body dynamics. Special Theory of Relativity (5 Lectures): (1) Postulates of STR. (2) Galilean transformation. (3) Lorentz transformation. (4) Simultaneity, Length Contraction, Time dilation. (5) Relativistic addition of velocities. (6) Energy-momentum relationships. Santabrata Das (IITG) PH 101, July-November 30 July, 2014 4 / 48

Syllabus Quantum Mechanics (8 Lectures): (1) Two-slit experiment. (2) De Broglie s hypothesis. Uncertainty Principle. (3) Wave function and wave packets, phase and group velocities. (3) Schrödinger Equation. Probabilities and Normalization. (4) Expectation values. Eigenvalues and eigenfunctions. (5) Applications in one dimension: Particle in a box, Finite Potential well, Harmonic oscillator. Santabrata Das (IITG) PH 101, July-November 30 July, 2014 5 / 48

Text and Reference Books Texts: D. Kleppner and R. J. Kolenkow, An Introduction to Mechanics, Tata McGraw-Hill, 2007. R. Eisberg and R. Resnick, Quantum Physics of Atoms,Molecules,Solids,Nuclei and Particles, 2nd Ed., John-Wiley, 1985. References: R. P. Feynman, R. B. Leighton, and M. Sands, The Feynman Lectures on Physics, Vol.I, Norosa Publishing House, 1998. J.M. Knudsen and P.G. Hjorth, Elements of Newtonian Mechanics, Springer, 1995. R. Resnick, Introduction to Special Relativity, John Wiley, Singapore, 2000. A. Beiser, Concepts of Modern Physics, Tata McGraw-Hill, New Delhi, 1995. S. Gasiorowicz, Quantum Physics, John Wiley (Asia), 2000. Santabrata Das (IITG) PH 101, July-November 30 July, 2014 6 / 48

Classes Two classes: Tuesday (10.00 AM - 10.55AM and 3.00PM to 3.55PM) Wednesday (11.00 AM - 11.55AM and 2.00PM to 2.55PM) Primarily presented with slides. Slides will be available online after the class. Do not hesitate to interrupt if you have doubts during the class. Santabrata Das (IITG) PH 101, July-November 30 July, 2014 7 / 48

Tutorials Tutorial: Friday (8.00AM - 8.55AM) The main purpose of the tutorial is to provide you with an opportunity to interact with a teacher. The teacher will assist you in clearing your doubts and answer your queries regarding the course topics. A problem sheet will be given to you for your practice. These problems also indicate the difficulty level of the examinations. Your are expected to attempt these problems before you come to tutorial class. Ask your doubts regarding these problems to your teacher during the tutorial class. The teacher may or may not solve all the problems in the tutorial class. In case you find a problem very difficult, do ask your teacher to help you. Santabrata Das (IITG) PH 101, July-November 30 July, 2014 8 / 48

Evolutions Two Quizzes each of 10% weightage Mid-semester Exam of 30% weightage End-semester Exam of 50% weightage Pass marks > 30 Santabrata Das (IITG) PH 101, July-November 30 July, 2014 9 / 48

Class Manners Maintain Silence Attendance Rule Mobile Phone Policy Santabrata Das (IITG) PH 101, July-November 30 July, 2014 10 / 48

Section 2 Review of Kinematics Santabrata Das (IITG) PH 101, July-November 30 July, 2014 11 / 48

Kinematics in One Dimensions The motion of the particle is described specifying the position as a function of time, say, x (t). The instantaneous velocity is defined as v (t) = dx(t) dt x(t + t) x(t) = lim t 0 t (1) and instantaneous acceleration, as a (t) = dv(t) dt = d2 x(t) dt 2 v(t + t) v(t) = lim t 0 t (2) Santabrata Das (IITG) PH 101, July-November 30 July, 2014 12 / 48

Simple Example If x (t) = sin(t), then v (t) = cos(t) and a (t) = sin(t). Position Velocity Acceleration Santabrata Das (IITG) PH 101, July-November 30 July, 2014 13 / 48

Kinematics in One Dimension Usually the x (t) is not known in advance! However, the acceleration a (t) is known at all times t > t 0 (=given time), and at t 0, position x (t 0 ) and velocity v (t 0 ) are known. The formal solution to this problem is ˆ t v (t) = v (t 0 ) + a ( t ) dt t 0 x (t) = x (t 0 ) + ˆ t t 0 v ( t ) dt Santabrata Das (IITG) PH 101, July-November 30 July, 2014 14 / 48

Constant Acceleration Let the acceleration of a particle be a 0, a constant at all times. If, at t = 0 velocity of the particle is v 0, then v (t) = v 0 + And if the position at t = 0 is x 0, These are familiar formulae. ˆ t 0 = v 0 + a 0 t a 0 dt x (t) = x 0 + v 0 t + 1 2 a 0t 2 (3) Santabrata Das (IITG) PH 101, July-November 30 July, 2014 15 / 48

Kinematics in One Dimension More complex situations may arise, where an acceleration is specified as a function of position, velocity and time, a (x, ẋ, t). In this case, we need to solve a differential equation which may or may not be simple. d 2 x = a (x, ẋ, t) dt2 Santabrata Das (IITG) PH 101, July-November 30 July, 2014 16 / 48

Solution methodology Simple integration Accleration is function of t alone dv dt = a(t) Accleration is function of x alone dv dt = dv dx Accleration is function of v alone dv dt First ordered differential equation dx dt = v dv dx = a(x) = a(t) dt = dv a(v) Acceleration is function of v and t dv = a(v, t) dt Acceleration is function of v and x v dv = a(v, x) dx Second ordered differential equation: other cases Santabrata Das (IITG) PH 101, July-November 30 July, 2014 17 / 48

Kinematics in Two Dimensions The position now is specified by a vector in a plane. r (t) = x (t) î + y (t) ĵ. 6 4 y dy y 2 r t dt r t Position at t+dt Position at t x dx x 1 1 2 3 4 5 Santabrata Das (IITG) PH 101, July-November 30 July, 2014 18 / 48

Kinematics in Two Dimensions The instantaneous velocity vector is defined as v (t) = d dt r(t) r(t + dt) r(t) = lim dt 0 dt x(t + dt) x(t) y(t + dt) y(t) = lim î + lim ĵ dt 0 dt dt 0 dt = dx(t) dt î + dy(t) dt ĵ Simillarly, we can show that a (t) = d2 x(t) dt 2 î + d2 y(t) dt 2 ĵ Santabrata Das (IITG) PH 101, July-November 30 July, 2014 19 / 48

Kinematics in Two Dimensions Velocity and Acceleration are vector quantities. Formal solutions can be written in exactly the same way as in case of one dimensional motion. Typical problem, however, may specify acceleration as a function of coordinates, velocities etc. In this case, we have to solve two coupled differential equations d 2 x dt 2 = a x (x, y, ẋ, ẏ, t) d 2 y dt 2 = a y (x, y, ẋ, ẏ, t) Santabrata Das (IITG) PH 101, July-November 30 July, 2014 20 / 48

Projectile Motion A ball is projected at an angle θ with a speed u. The net acceleration is in downward direction. Then a x = 0 and a y = g. The equations are We know the solutions. d 2 x dt 2 = 0 d 2 y dt 2 = g Santabrata Das (IITG) PH 101, July-November 30 July, 2014 21 / 48

Charged Particle in Magnetic Field A particle has a velocity v in xy plane. Magnetic field is in z direction. The acceleration is given by q mv B. The equations are d 2 x dt 2 = qb m v y d 2 y dt 2 = qb m v x Solution is rather simple, that is circular motion in xy plane. Santabrata Das (IITG) PH 101, July-November 30 July, 2014 22 / 48

Section 3 Polar Coordinate System Santabrata Das (IITG) PH 101, July-November 30 July, 2014 23 / 48

Cartesian Coordinate In Cartesian coordinate system each point is uniquely specified in a plane by a pair of numerical coordinates, measured in the same unit of length. Y x P y O X Each point P is identified with its unique x and y coord: P = P (x, y). Ranges: x, y (, ). Santabrata Das (IITG) PH 101, July-November 30 July, 2014 24 / 48

Cartesian Coordinate Horizontal lines Vertcial lines 1.6 1.2 0.8 0.4 0. 0.4 0.8 1.2 1.6 Y X 1.6 1.2 0.8 0.4 0. 0.4 0.8 1.2 1.6 1.6 1.2 0.8 0.4 0. 0.4 0.8 1.2 1.6 Y X Clearly, each point has separate coordinates. Santabrata Das (IITG) PH 101, July-November 30 July, 2014 25 / 48

Cartesian Coordinate Geometric definition of magnitude and direction of vectors allow us to define operations: addition, subtraction, and multiplication by scalars. Often a coordinate system is helpful as it is easier to manipulate the coordinates of a vector, than manipulating its magnitude and direction. To determine vector coordinates, the first step is to translate the vector so that its tail is at the origin and head will be at some point P (x, y). y O x P Santabrata Das (IITG) PH 101, July-November 30 July, 2014 26 / 48

Cartesian Coordinate Position vector of P is r = îx + ĵy. Change x only keeping y fixed such that P moved to Q. The direction of increasing the vector is defined as y P Q e x = dr dx This vector is not necessarily normalized to have unit length, but from it we can always construct the unit vector as O x dx x î = e x e x = 1 dr/dx dr dx Santabrata Das (IITG) PH 101, July-November 30 July, 2014 27 / 48

Cartesian Coordinate Therefore, unit vectors are defined as 1 dr î = dr/dx dx j P i and y ĵ = 1 dr dr/dy dy O x Santabrata Das (IITG) PH 101, July-November 30 July, 2014 28 / 48

Polar Coordinates Each point P = (x, y) of a plane can also be specified by its distance from the origin, O and the angle that the line OP makes with x axis. Transformations: r = + x 2 ( + y 2 θ = tan 1 y ) x (r, θ) are called Plane Polar Coordinates. Inverse Transformations: x = r cos θ y = r sin θ r sin Θ O r Θ x r cos Θ P y Ranges: r (0, ) and θ (0, 2π). tan θ = tan (θ + π) suggests to replace θ by θ + π, for x < 0, y < 0. Discontinuity in θ at origin. Santabrata Das (IITG) PH 101, July-November 30 July, 2014 29 / 48

Polar Coordinates Constant r curves Constant r and θ curves Π 2 3Π 8 Π 4 1.6 1.2 0.8 0.4 1.6 1.2 0.8 0.4 Π 8 0 Π 8 Π 4 Π 2 3Π 8 Unit vectors are perpendicular to constant coordinate curves Santabrata Das (IITG) PH 101, July-November 30 July, 2014 30 / 48

Unit Vectors In Cartesian coordinate, PV of P is r = xî + yĵ. We find a set of two unit vectors ˆr and ˆθ. By Coordinate transformation, r = r cos θî + r sin θĵ Unit vectors are defined as, ˆr = 1 dr dr dr = î cos θ + ĵ sin θ dr And ˆθ = 1 dr dr dθ = î sin θ + ĵ cos θ dθ Θ î ĵ = cos θˆr sin θˆθ = sin θˆr + cos θˆθ Santabrata Das (IITG) PH 101, July-November 30 July, 2014 31 / 48

Unit Vectors Θ Θ r r Unit vectors at a point P depend on θ coordnate of P Example: If P (1, π/6), then ˆr = 1 ( ) 3 î + 2 ĵ ˆθ = 1 ( î 2 + ) 3ĵ 45 30 Example: Let P (1, π/4), then ˆr = 1 ) (î + ĵ 2 ˆθ = 1 ) ( î + ĵ 2 Note that everywhere, ˆr ˆθ. Santabrata Das (IITG) PH 101, July-November 30 July, 2014 32 / 48

Unit Vectors These unit vectors are functions of the polar coordinates, only of θ in fact. ˆr = i cos θ + j sin θ ˆθ = i sin θ + j cos θ ˆr θ ˆθ θ = ˆθ = ˆr Time derivates ˆr = θˆθ ˆθ = θˆr Santabrata Das (IITG) PH 101, July-November 30 July, 2014 33 / 48

Section 4 Motion in Plane Polar Coordinates Santabrata Das (IITG) PH 101, July-November 30 July, 2014 34 / 48

Motion in Polar Coordinates Suppose a particle is travelling along a trajectory given by r(t). Now the position vector r (t) = x (t) î + y (t) ĵ = r (t) (î cos (θ (t)) + ĵ sin (θ (t))) = r (t)ˆr (θ (t)) r(t+dt) r(t) r (t) depends on θ implicitly through ˆr vector. θ θ δθ We need to know θ to see how r is oriented. We need to know r to see how far we are from the origin. Santabrata Das (IITG) PH 101, July-November 30 July, 2014 35 / 48

Motion in Polar Coordinates Then the velocity vector Radial velocity = ṙˆr v = dr dt Tangential velocity = r θˆθ θ is called the angular speed. = d [r (t)ˆr (θ (t))] dt = dr dˆr dθ + r dtˆr dθ dt = ṙˆr+r θˆθ Santabrata Das (IITG) PH 101, July-November 30 July, 2014 36 / 48

Motion in Polar Coordinates The Acceleration vector a = dv dt = rˆr + ṙ dˆr dt + ṙ θˆθ dˆθ + r θˆθ + r θ dt = ( rˆr + ṙ θˆθ + ṙ θˆθ + r θˆθ r θ 2ˆr = r r θ 2)ˆr ( θ) + r θ + 2ṙ ˆθ Radial component: r is the linear acceleration in radial direction. The term r θ 2ˆr is usual centripetal acceleration. Tangential component: r θ is the linear acceleration in the tangential direction. The term 2ṙ θˆθ is called Coriolis acceleration. Santabrata Das (IITG) PH 101, July-November 30 July, 2014 37 / 48

Uniform Linear Motion 3 2 1 Position Vector: r(t) = 2tî + ĵ. Polar coordinates r(t) = x(t) 2 + y(t) 2 = 4t 2 + 1 5 4 3 2 Radial Tangential and tan θ(t) = 1/2t 1 Santabrata Das (IITG) PH 101, July-November 30 July, 2014 38 / 48

Uniform Linear Motion 3 2 1 Velocity Vector: v(t) = 2î. Radial ṙ = 2t 4t 2 + 1 Tangential 1 θ = v(t) = 2 4t 2 + 1 2t 4t 2 + 1ˆr 2 ˆθ 4t 2 + 1 2 1 1 2 1 Santabrata Das (IITG) PH 101, July-November 30 July, 2014 39 / 48

Circular Motion In a circular motion, r = R = Constant. Then, ṙ = r = 0. Thus v = R θˆθ a = R θ 2ˆr + R θˆθ Consider a special case of uniform circular motion: θ = 0. Speed is constant, and velocity is tangential Acceleration is radial (central) Santabrata Das (IITG) PH 101, July-November 30 July, 2014 40 / 48

Circular Motion Consider a case of a non-uniform circular motion in which θ = α = Constant Here, v = R θˆθ = Rαtˆθ a = R θ 2ˆr + R θˆθ = Rα 2 t 2ˆr + Rαˆθ movie Santabrata Das (IITG) PH 101, July-November 30 July, 2014 41 / 48

Circular Motion Consider a case of a non-uniform circular motion in which θ = α = Constant Here, v = R θˆθ = Rαtˆθ a = R θ 2ˆr + R θˆθ = Rα 2 t 2ˆr + Rαˆθ Santabrata Das (IITG) PH 101, July-November 30 July, 2014 42 / 48

Circular Motion: Hammer Throw Santabrata Das (IITG) PH 101, July-November 30 July, 2014 43 / 48

Spiral Motion Consider a particle moving on a spiral given by r = aθ with a uniform angular speed ω. Then ṙ = a θ = aω. v = aωˆr + aω 2 tˆθand a = aω 3 tˆr + 2aω 2 ˆθ Santabrata Das (IITG) PH 101, July-November 30 July, 2014 44 / 48

Central Acceleration Central Accelerations When the acceleration of a particle points to the origin at all times and is only function of distance of the particle from the origin, the acceleration is called central acceleration. a = f(r)ˆr Examples are 1/r 2 (Gravitational and Electrostatic), 1/r 6 (Van-dar-Waals), kr (Spring) etc. Angular momentum of the particle remains constant. r θ + 2ṙ θ = d dt (r2 θ) = 0 r 2 θ = constant during motion. Santabrata Das (IITG) PH 101, July-November 30 July, 2014 45 / 48

Example We have considered this motion earlier. Santabrata Das (IITG) PH 101, July-November 30 July, 2014 46 / 48 Acceleration of a particle is given by a (x) = ω 2 x. And at time t = 0, position is A and velocity is zero. After integrating this, we get Finally the solution for x is dv dt dv dx dx dt = ω 2 x = ω 2 x v dv dx = ω2 x v = ω A 2 x 2 dx dt (4) x(t) = A sin(ωt) (5)

Air Resistance Suppose a ball is falling under gravity in air, resistance of which is proportional to the velocity of the ball. a (ẏ) = g kẏ If the ball was just dropped, velocity of the ball after time t then v(t) = g (1 e kt) k g/k velocity time Santabrata Das (IITG) PH 101, July-November 30 July, 2014 47 / 48

Polar Coordinates Santabrata Das (IITG) PH 101, July-November 30 July, 2014 48 / 48