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Tutorial Reminder

Please download the tutorial from the course web page and try them outTutorial class will be conducted on 12 DEC 03 (Friday)Submit solutions for PROBLEMS (1),(2), (7) and (9) to your tutors [NOT Conceptual questions]Tutors will discuss the solutions with youPassword to solution 1 will be provided after the tutorial class

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Recap

The length of a body measured at rest = proper length L = improper length = length of an object when it’s in relative motion wrp to the observer

Length contraction, L = L’/, L’ = proper length

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Lorentz transformationConsider an O frame and an O’ frame; O’ is moving with velocity +u wrp to O:

O’O

+u

Object M

I see O’ moving with a velocity +u; I use {x,t} to denote the coordinates of object M; I see M is moving with velocity ux

I see O moving with a velocity -u; I use {x’,t’} to denote the coordinates of object M; I see M is moving with velocity u’x

-u

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Lorentz transformation relates {x’,t’} {x,t}; u’x ux

)''( utxx

)(' utxx xcutt )/(' 2

')/(' 2 xcutt

21'

cuuuu

ux

xx

2

'1

'

c

uuuu

ux

xx

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ExampleTwo events are seen to be occurring at the same point in a rest frame O.

O

t1

The two events are denoted as t1 ,t2 such that t = t2- t1 = 1 s, and x =0.

t2

In the moving frame O’, what is the time interval between the two events?

O’

u

6

ss

cc

t

xcut

xxcutt

ttt

09.7199.0

1

1

)/(

))(/()(

'

2

2

122

12

'1

'2

The time t’ as seen by O’ in terms of t is simply related by

xcutt )/(' 2Hence the time interval in O’, t’ in terms of t is simply

This is nothing but just the time dilation effect

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Example (relativistic velocity addition)

Rocket 1 is approaching rocket 2 on ahead-on collision course. Each is moving at velocity 4c/5 relative to an independent observer midway between the two. With what velocity does rocket 2 approaches rocket 1?

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C.f. In GT, their relative speed would just be 4c/5 + 4c/5 = 1.6 c – which violates constancy of speed of light postulate. See how LT handle this situation:

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Choose the observer in the middle as in the stationary frame, O

Choose rocket 1 as the moving frame O‘Call the velocity of rocket 2 as seen from rocket 1 u’x. This is the quantity we are interested in

Frame O' is moving in the +ve direction as seen in O, so u = +4c/5The velocity of rocket 2 as seen from O is in the

-ve direction, so ux = - 4c/5

Now, what is the velocity of rocket 2 as seen from frame O', u ’x = ? (intuitively, u ’x must be in the negative direction)

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Use the LT

c

c

cc

cc

cuuuu

ux

xx 41

40

54

54

1

54

54

1'

2

2

i.e. the velocity of rocket 2 as seen from rocket 1 (the moving frame, O’) is –40c/41, which means that O’ sees rocket 2 moving in the –ve direction (to the left in the picture), as expected.

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Relativistic DynamicsStill remember conservation of linear momentum? It’s one of the very fundamental law in physics – due to the translational symmetry of spaceThe total linear momentum before and after collision must be the same

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m1u1

m2u2

m1v1 m2v2

Conservation of momentum means

m1u1 +m2u2 = m1v1 +m2v2

Conservation of momentum must also hold true in other inertial reference frames, according to Einstein postulate:

m’1u’1 +m’2u’2 = m’1v’1 +m’2v’2 (m’ = m in Newton’s mechanics)

where the velocities transform according to LT, i.e. u1 u’1 = LT(u1), v1 v’1 = LT(v1), etc.

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Classically, p = mu. In the other frame, p’ = m’u’; the mass m’ (as seen in frame O’) is the same as m (as seen in O frame) – this is according to Newton’s mechanicsHowever, simple consideration will reveal that in order to preserver the consistency between conservation of momentum and the LT, the definition of momentum (as according to Newton’s view) has to be re-defined, such that m’ is not equal to m. That is, the mass of an moving object, m, is different from its value when it’s at rest, m0

Read lecture notes (or Krane) for a more rigorous illustration why the definition of classical momentum is inconsistent with LT

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Relativistic definition of linear momentum

Consider a mass is moving with speed v in a rest frameClassically, p = mv , m = mass is constant and not changing with its state of motionRelativistically, p = m0v

O

O’

vM

I see M is at rest. Its mass is m0, momentum, p’ = 0

I see the momentum of M as p = mv=m0v

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m0 = rest mass = the mass measured in a frame where the object is at rest. It’s value is a constant

m = m0 :the mass of a moving object changes as its speed changes

(v/c)2]-1/2; where v = | v |

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Example

The rest mass of an electron is m0 = 9.11 x 10-31kg.

p = m0u

Compare it with that calculated with classical definition.

If it moves with u = 0.75 c, what is its relativistic momentum?

m0

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Solution

The Lorentz factor is = [1-(u/c)2] -1/2 = [1-(0.75c/c)2] -1/2=1.51Hence the relativistic momentum is simply

p = x m0 x 0.75c = 1.51 x 9.11 x 10-31kg x 0.75 x 3 x 108 m/s = 3.1 x 10-22 kg m/s = Ns

In comparison, classical momentum gives pclassical = m0 x 0.75c = 2.5 x 10-22

Ns – about 34% lesser than the relativistic value

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For 2-D case (object moving in the x-y plane)

The combined velocity of m is v = vx + vy (vector), with magnitude given by v = [(vx)2

+ (vy)2 ]-1/2

p = px + py = px i + py jpy = m0 vy

px = m0vx

px = m0vx

For object moving only in 1-D (say in the x-direction) with speed vx, its momentum is simply

For a mass moving in x-y plane, it has two components of momentum

The mass changes according to m = m0where [1 – (v / c)2] -1/2, depend on the magnitude of the speed the mass is moving (the combined speed)

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Relativistic Energy

Recall the law of conservation of mechanical energy:

Work done by external force on an object (W) = the change in kinetic

energy of the object, ()

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K1

F

K2

F

s

= K2 - K1

W = F s Conservation of mechanical energy: W = The total energy of the object, E = K + U. Ignoring potential energy, E of the object is solely in the form of kinetic energy. If K1 = 0, then E = K2. But in general, U also needs to be taken into account for E.

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in SR we have redefined the classical mass and momentum to that of relativistic version

mclass(cosnt) –> mSR = m0

pclass = mclass u –> pSR = (m0)u

We will also see how E = mc2 comes about in the following

E.g, in classical mechanics, K = mv2/2. However, this relationship has to be supplanted by the relativistic version K = mv2/2 –> K = E – m0c2 = mc2 - m0c2

Hence we must also modify the relation btw work and energy so that the law conservation of energy is consistent with SR