physics - 1 (lecture -1) - indian institute of technology ... - 1 (lecture -1) santabrata das...

PHYSICS - 1 (Lecture -1)

Santabrata Das

Department of PhysicsIndian Institute of Technology Guwahati

[email protected], Room #: 006

30 July, 2014

Santabrata Das (IITG) PH 101, July-November 30 July, 2014 1 / 48

Section 2

Introduction to the CourseGo 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.


Topics

Classical MechanicsEnergy �rest mass energy

Relativistic MechanicsEnergy ≥ rest mass energy and de Broglie wavelength � size

Quantum Mechanicsde Broglie wavelength & size and Energy � rest mass energy


SyllabusClassical Mechanics (15 Lectures):(1) Review of Newtonian Mechanics in rectilinear coordinatesystem.(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.


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, FinitePotential well, Harmonic oscillator.


Text and Reference BooksTexts:

D. Kleppner and R. J. Kolenkow, An Introduction to Mechanics,Tata McGraw-Hill, 2007.R. Eisberg and R. Resnick, Quantum Physics ofAtoms,Molecules,Solids,Nuclei and Particles, 2nd Ed., John-Wiley,1985.

References:R. P. Feynman, R. B. Leighton, and M. Sands, The FeynmanLectures 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, NewDelhi, 1995.S. Gasiorowicz, Quantum Physics, John Wiley (Asia), 2000.


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 afterthe class.Do not hesitate to interrupt if you have doubts during the class.


Tutorials

Tutorial: Friday (8.00AM - 8.55AM)The main purpose of the tutorial is to provide you with anopportunity to interact with a teacher.The teacher will assist you in clearing your doubts and answer yourqueries regarding the course topics.A problem sheet will be given to you for your practice. Theseproblems also indicate the difficulty level of the examinations.Your are expected to attempt these problems before you come totutorial class. Ask your doubts regarding these problems to yourteacher during the tutorial class.The teacher may or may not solve all the problems in the tutorialclass. In case you find a problem very difficult, do ask your teacherto help you.


Evolutions

Two Quizzes each of 10% weightageMid-semester Exam of 30% weightageEnd-semester Exam of 50% weightage

Pass marks >30


Class Manners

Maintain SilenceAttendance RuleMobile Phone Policy


Section 2

Review of Kinematics


Kinematics in One Dimensions

The motion of the particle is described specifying the positionas a function of time, say, x (t).

The instantaneous velocity is defined as

v (t) =dx(t)

dt= lim

∆t→0

x(t+ ∆t)− x(t)

∆t(1)

and instantaneous acceleration, as

a (t) =dv(t)

dt= lim

∆t→0

v(t+ ∆t)− v(t)

∆t(2)

=d2x(t)

dt2


Simple Example

If x (t) = sin(t), then v (t) = cos(t) and a (t) = −sin(t).

Position

Velocity

Acceleration


Kinematics in One Dimension

Usually the x (t) is not known in advance!

However, the acceleration a (t) is known at all times t > t0(=giventime), and at t0, position x (t0) and velocity v (t0) are known.

The formal solution to this problem is

v (t) = v (t0) +

ˆ t

t0

a(t′)dt′

x (t) = x (t0) +

ˆ t

t0

v(t′)dt′


Constant Acceleration

Let the acceleration of a particle be a0, a constant at all times. If, att = 0 velocity of the particle is v0, then

v (t) = v0 +

ˆ t

0a0dt

= v0 + a0t

And if the position at t = 0 is x0,

x (t) = x0 + v0t+1

2a0t

2 (3)

These are familiar formulae.


Kinematics in One Dimension

More complex situations may arise, where an acceleration isspecified as a function of position, velocity and time, a (x, x, t).In this case, we need to solve a differential equation

d2x

dt2= a (x, x, t)

which may or may not be simple.


Solution methodology

Simple integrationAccleration is function of t alone → dv

dt = a(t)

Accleration is function of x alone → dvdt = dv

dxdxdt = v dv

dx = a(x)

Accleration is function of v alone → dvdt = a(t)→ dt = dv

a(v)

First ordered differential equation

Acceleration is function of v and t → dv

dt= a(v, t)

Acceleration is function of v and x → vdv

dx= a(v, x)

Second ordered differential equation: other cases


Kinematics in Two DimensionsThe position now is specified by a vector in a plane.

r (t) = x (t) i + y (t) j.

xx + dx

y

y + dy

rHtL

rHt+

dtL

Position at t

Position at t+dt

-1 1 2 3 4 5

2

4

6


Kinematics in Two Dimensions

The instantaneous velocity vector is defined as

v (t) =d

dtr(t)

= limdt→0

r(t+ dt)− r(t)

dt

= limdt→0

x(t+ dt)− x(t)

dti + lim

dt→0

y(t+ dt)− y(t)

dtj

=dx(t)

dti +

dy(t)

dtj

Simillarly, we can show that

a (t) =d2x(t)

dt2i +

d2y(t)

dt2j


Kinematics in Two Dimensions

Velocity and Acceleration are vector quantities.Formal solutions can be written in exactly the same way as in caseof one dimensional motion.Typical problem, however, may specify acceleration as a functionof coordinates, velocities etc.In this case, we have to solve two coupled differential equations

d2x

dt2= ax(x, y, x, y, t)

d2y

dt2= ay(x, y, x, y, t)


Projectile Motion

A ball is projected at an angle θ with a speed u. The net acceleration isin downward direction. Then ax = 0 and ay = −g. The equations are

d2x

dt2= 0

d2y

dt2= −g

We know the solutions.


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

d2x

dt2=

qB

mvy

d2y

dt2= −qB

mvx

Solution is rather simple, that is circular motion in xy plane.


Section 3

Polar Coordinate System


Cartesian CoordinateIn Cartesian coordinate system each point is uniquely specified in aplane by a pair of numerical coordinates, measured in the same unit oflength.

x

y

P

X

Y

O

Each point P is identified with its unique x and y coord: P = P (x, y).Ranges: x, y ∈ (−∞,∞).


Cartesian CoordinateHorizontal lines

X

Y

-1.6

-1.2

- 0.8

- 0.4

0.

0.4

0.8

1.2

1.6

Vertcial lines

X

Y

-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

Clearly, each point has separate coordinates.


Cartesian Coordinate

Geometric definition of magnitude anddirection of vectors allow us to defineoperations: addition, subtraction, andmultiplication by scalars.Often a coordinate system is helpful as itis easier to manipulate the coordinates of avector, than manipulating its magnitudeand direction.To determine vector coordinates, the firststep is to translate the vector so that itstail is at the origin and head will be atsome point P (x, y).

P

xO

y



Position vector of P is r = ix+ jy.Change x only keeping y fixed suchthat P moved to Q.The “direction” of increasing thevector is defined as

ex =dr

dx

This vector is not necessarily normalizedto have unit length, but from it we canalways construct the unit vector as

i =ex|ex|

=1

|dr/dx|dr

dx

P Q

x dx + xO

y



Therefore, unit vectors are defined as

i =1

|dr/dx|dr

dx

and

j =1

|dr/dy|dr

dy

P

j

i

O

y

x


Polar CoordinatesEach point P = (x, y) of a plane can also be specified by its distancefrom the origin, O and the angle that the line OP makes with x−axis.

Transformations:

r = +√x2 + y2

θ = tan−1(yx

)(r, θ) are called Plane Polar Coordinates.

Inverse Transformations:

x = r cos θ

y = r sin θ

Ranges: r ∈ (0, ∞) and θ ∈ (0, 2π).tan θ = tan (θ + π) suggests toreplace θ by θ + π, for x < 0, y < 0.

P

r

Θ

r cosH Θ L

rsinHΘL

Θ

O

y

x

Discontinuity in θ atorigin.


Polar CoordinatesConstant r curves

0.4

0.8

1.2

1.6

Constant r and θ curves

0.4

0.8

1.2

1.6

-

Π2

-

3Π8

- Π

4

-Π

8

0

Π

8

Π

4

3Π

8

Π 2

Unit vectors are perpendicular to constant coordinate curves


Unit Vectors

In Cartesian coordinate, PV of P is r = xi + yj. We find a set of twounit vectors r and θ.

Θ`

Θ`

Θ`

r`

r`

Θ`

Θ

Θ

Θ

i

j

By Coordinate transformation,r = r cos θi + r sin θj

Unit vectors are defined as,

r =1

|drdr |dr

dr= i cos θ + j sin θ

θ =1

|drdθ |dr

dθ= −i sin θ + j cos θ

And

i = cos θr− sin θθ

j = sin θr + cos θθ


Unit Vectors

r`

r`

Θ`

Θ`

30

45

Unit vectors at a point P depend onθ coordnate of PExample: If P ≡ (1, π/6) , then

r =1

2

(√3i + j

)θ =

1

2

(−i +

√3j)

Example: Let P ≡ (1, π/4) , then

r =1√2

(i + j

)θ =

1√2

(−i + j

)Note that everywhere, r ⊥ θ .


Unit Vectors

These unit vectors are functions of the polar coordinates, only of θ infact.

r = i cos θ + j sin θ

θ = −i sin θ + j cos θ

∂r

∂θ= θ

∂θ

∂θ= −r

Time derivates

˙r = θθ˙θ = −θr


Section 4

Motion in Plane Polar Coordinates


Motion in Polar Coordinates

Suppose a particle is travellingalong a trajectory given by r(t).Now the position vector

r (t) = x (t) i + y (t) j

= r (t) (i cos (θ (t)) + j sin (θ (t)))

= r (t) r (θ (t))

r (t) depends on θ implicitlythrough r vector.

r(t+dt)

r(t)

θ

θ + δθ

We need to know θ to see how r is oriented. We need to know r to seehow far we are from the origin.



Then the velocity vector

v =dr

dt

=d

dt[r (t) r (θ (t))]

=dr

dtr + r

dr

dθ

dθ

dt

= rr+rθθ

Radial velocity = rr

Tangential velocity = rθθ

θ is called the angular speed.



The Acceleration vector

a =dv

dt

= rr + rdr

dt+ rθθ + rθθ + rθ

dθ

dt

= rr + rθθ + rθθ + rθθ − rθ2r

=(r − rθ2

)r +

(rθ + 2rθ

)θ

Radial component: r is the linear acceleration in radial direction.The term −rθ2r is usual centripetal acceleration.Tangential component: rθ is the linear acceleration in thetangential direction. The term 2rθθ is called Coriolis acceleration.


Uniform Linear Motion

-6 -4 -2 0 2 4 6

1

2

3

4

Position Vector: r(t) = 2ti + j.Polar coordinates

r(t) =√x(t)2 + y(t)2

=√

4t2 + 1

andtan θ(t) = 1/2t -2 -1 0 1 2

1

2

3

4

5

Tangential

Radial


Uniform Linear Motion

-6 -4 -2 0 2 4 6

1

2

3

4

-6 -4 -2 0 2 4 6

1

2

3

4

-6 -4 -2 0 2 4 6

1

2

3

4

-6 -4 -2 0 2 4 6

1

2

3

4

-6 -4 -2 0 2 4 6

1

2

3

4

-6 -4 -2 0 2 4 6

1

2

3

4

Velocity Vector: v(t) = 2i.

r =2t√

4t2 + 1

θ =2

4t2 + 1

v(t) =2t√

4t2 + 1r− 2√

4t2 + 1θ

-2 -1 1 2

-2

-1

1

2

Tangential

Radial


Circular Motion

In a circular motion, r = R = Constant . Then, r = r = 0 . Thus

v = Rθθ

a = −Rθ2r +Rθθ

Consider a special case of uniform circular motion: θ = 0 .Speed is constant, and velocity is tangentialAcceleration is radial (central)


Circular MotionConsider a case of a non-uniform circular motion in whichθ = α = Constant

movie

Here,

v = Rθθ = Rαtθ

a = −Rθ2r +Rθθ

= −Rα2t2r +Rαθ


Circular MotionConsider a case of a non-uniform circular motion in whichθ = α = Constant

Here,

v = Rθθ = Rαtθ

a = −Rθ2r +Rθθ

= −Rα2t2r +Rαθ


Circular Motion: Hammer Throw


Spiral MotionConsider a particle moving on a spiral given by r = aθ with a uniformangular speed ω. Then r = aθ = aω.

v = aωr + aω2tθand a = −aω3tr + 2aω2θ


Central Acceleration

Central Accelerations

When the acceleration of a particle points to the origin at all times andis only function of distance of the particle from the origin, theacceleration is called central acceleration.

a = f(r)r

Examples are 1/r2 (Gravitational and Electrostatic), 1/r6

(Van-dar-Waals), kr (Spring) etc.Angular momentum of the particle remains constant.

rθ + 2rθ =d

dt(r2θ) = 0

r2θ = constant during motion.


ExampleAcceleration of a particle is given by a (x) = −ω2x. And at time t = 0,position is A and velocity is zero.

dv

dt= −ω2x

dv

dx

dx

dt= −ω2x

vdv

dx= −ω2x

After integrating this, we get

v = ω√A2 − x2 ≡ dx

dt(4)

Finally the solution for x is

x(t) = A sin(ωt) (5)

We have considered this motion earlier.Santabrata Das (IITG) PH 101, July-November 30 July, 2014 46 / 48

Air ResistanceSuppose a ball is falling under gravity in air, resistance of which isproportional to the velocity of the ball.

a (y) = −g − ky

If the ball was just dropped, velocity of the ball after time t then

v(t) = −gk

(1− e−kt

)

time

g/k

velocity


Polar Coordinates


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