Slide 1

Transcript Slide 1

Relativistic mechanics

--

Scalars -- 4-vectors -- 4-D velocity -- 4-momentum, rest mass -- conservation laws -- Collisions -- Photons and Compton scattering

Velocity addition (revisited) and the Doppler shift -- 4-force

1. Scalars

A scalar is a quantity that is the same in all reference frames, or for all observers. It is an invariant number.

E.g.,

(



)

,

 

(

t rest

, proper time ),



(

l rest

, proper length )

But the time interval ∆t, or the distance ∆x between two events, or the length separating two worldlines are not scalars: they do not have frame-independent

values.

2. 4-vectors



 (



, 

) This 4-vector defined above is actually a frame-independent object, although the components of it are not frame-independent, because they transform by the Lorentz transformation.

E.g., in 3-space, the Different

observers

set up different coordinate systems and assign different coordinates to two points C and L, say Canterbury and London. --They may assign different coordinates to the point of the two cities --They agree on the 3-displacement between the two points, etc.

separating C and L., the distance

With each 4-displacement we can associate a scalar the interval (s) 2 along the vector. The interval associated with the above defined

4-vector

is ( 

) 2  (



) 2  ( 

) 2  (



) 2  ( 

) 2 Because of the similarity of this expression to that of the dot product between 3-vectors in three dimensions, we also denote this interval by a dot product and also by 

2  

 

 (



) 2  ( 

) 2 and we will sometimes refer to this as the magnitude or length of the 4-vector.

--We can generalize this dot product to a dot product between any two 4-vectors 

a



a

  

b

(

a



a

a b

x t

a



a

a b

z x

) 

a and

b



a



b



(

b

) : --When frames are changed, 4-displacement transform according to the Lorentz transformation, and obeys associativity over addition and commutativity :

)





(



 

)

 

 

 

 

;



 

 

 

a ii

) A 4-vector multiplied or divided by a scalar is another 4-vector

3. 4-velocity

In 3-dimensional space, 3-velocity is defined by  lim 

 0  rˆ 

t



d

dt rˆ where ∆t is the time it takes the object in question to go the 3-displacement ∆



the 3-displacement r so that we have 

 

d x dt

However, this in itself won't do, because we are dividing a 4-vector by a non-scalar (time intervals are not scalars); the quotient will not transform according to the Lorentz transformation.

The fix is to replace ∆t by the proper time ∆ 

 the 4-displacement; the 4-velocity is then 

d x



  (



u

dt d

 , 

dx d

 lim   , 

dy d

 0 ,  

 

x



) 

where

(



dt dt

,  :

dx dt

corresponding to the interval of ,    

, 

 

t

)

 ( 

, 

v x

, 

v y

, 

v z

) where

(

v x

,

v y

v z

)

are the components of the 3-velocity 

d

rˆ dt

Although it is unpleasant to do so, we often write 4-vectors as two-component objects with the rest component a single number and the second a 3-vector. In this notation 

 ( 

, 

ˆ ) --What is the magnitude of The magnitude 

2 

must be the same in all frames because 

is a 4-vector.

Let us change into the frame in which the object in question is at rest.

In this frame 



(

, 0 , 0 , 0 ) for



( 0 , 0 , 0 ) and



 2 

or u

 



1

by calculating the dot product of 

2  

 



2 You may find this a little strange. Some particles move quickly, some slowly, but for all particles, the magnitude of the 4-velocity is c. But this is not strange, because we need the magnitude to be a scalar, the same in all frames. If you change frames, some of the particles that were moving quickly before now move slowly, and some of them are stopped altogether. Speeds (magnitudes of 3-velocities) are relative; the magnitude of the 4-velocity has to

be invariant

4. 4-momentum, rest mass and conservation laws

In spacetime 4-momentum 

is mass

times 4-velocity 

--Under this definition, the mass must be a scalar if the 4-momentum is going 







(



,



mv x

,



mv y

,



mv z

)



(



,



m v

)

--The mass

of an object as far as we are concerned is its

rest mass

, or the mass we would measure if we were at rest with respect to the object.

--Again, by switching into the rest frame of the particle, or by calculationg the magnitude we find that 4-momentum, we can show: 



As with 4-velocity, it is strange but true that the magnitude of the 4-momentum does not depend on speed.

Why introduce all these 4-vectors, and in particular the 4-momentum?

--all the laws of physics must be same in all uniformly moving reference frames --only scalars and 4-vectors are truly frame-independent, relativistically invariant conservation of momentum must take a slightly different form. --In all interactions, collisions and decays of objects, the total 4-momentum is conserved (of course we don ’ t consider any external force here).



 

m u



(



,



m v

)

--Furthermore,

E c

We are actually re-defining E and ˆ

to be:

 ˆ



and

 

ˆ You better forget any other expressions you learned for E or p in non-relativistic mechanics.

A very useful equation suggested by the new, correct expressions for E and

Taking the magnitude-squared of 

p v

ˆ  ˆ

c E

2 We get a relation between m, E and



ˆ



2 

2  

 



E c

2 

2 which, after multiplication by c 2 and rearrangement becomes

2 

4 

2 This is the famous equation of Einstein's, which becomes

2 

4 when the particle is at rest  0

In the low-speed limit  

v c

 1



m v

( 1   2 )  1 2 

2 ( 1   2 )  1 2 

m v

 1 2

m v

2 

2  1 2

2 

m v

i.e., the momentum has the classical form, and the energy is just Einstein's famous mc 2 plus the classical kinetic energy mv 2 /2. But remember, these formulae only apply when v << c.

5. Conservation laws

For a single particle: 4-momenum before an action = that after 

 

For a multi-particle system: Summed over All the 4-momenta of all the components of the whole system before interaction 



p i

 



q i

Summed over all the 4-momenta of all the components of the whole system after interaction

5. Collisions

In non-relativistic mechanics collisions divide into two classes: elastic inelastic energy and 3-momentum are conserved.

only 3-momentum is conserved In relativistic mechanics 4-momentum, and in particular the time component or energy, is conserved in all collisions; No distinction is made between elastic and inelastic collisions.



p m



  (  

p m

Before the collision m

,  

p s v mv

, 0 , 0 );  [( m

’ Non-relativistic theory gives: M ’ =2m, v ’   

p s

 1 )

, (

, 0 , 0 , 0 ) 

, 0 , 0 )] After the collision 

 (  '

=v/2 '

, 

' '

' , 0 , 0 ) By conservation of 4-momentum before and after collision, which means that the two 4-vectors are equal, component by component, g '

= [ g + 1]

and g '

' = g

The ratio of these two components should provide v ’ /c; The magnitude of

' 

  

 1 should be M ’ 

2 c; we use 



'

;



2  

' 2  [   1 ] 2

2   2

' 

2  [ 1  2    2 ( 1 

2 )]

2  2 (  2 (   1 )

 2

 1 )

2 --the mass M ’ of the final product is greater than the sum of the masses of its progenitors, 2m.

--So the non-relativistic answers are incorrect, Q: Where does the extra rest mass come from?

A: The answer is energy.

In this classically inelastic collision, some of the kinetic energy is lost.

But total energy is conserved. Even in classical mechanics the energy is not actually lost, it is just converted into other forms, like heat in the ball, or rotational energy of the final product, or in vibrational waves or sound travelling through the material of the ball.

Strange as it may sound, this internal energy actually increases the mass of the product of the collision in relativistic mechanics.

The consequences of this are strange. For example, a brick becomes more massive when one heats it up. Or, a tourist becomes less massive as he or she burns calories climbing the steps of the Effiel Tower. All these statements are true, but it is important to remember that the effect is very very small unless the internal energy of the object in question is on the same order as mc 2 .

For a brick of 1 kg, mc 2 is 10 20 Joules, or 3 *10 13 kWh, a household energy consumption over about ten billion years (roughly the age of the Universe!) For this reason, macroscopic objects (like bricks or balls of putty) cannot possibly be put into states of relativistic motion in Earth-bound experiments.

Only subatomic and atomic particles can be accelerated to relativistic speeds, and even these require huge machines (accelerators) with huge power supplies.

6. Photons and Compton scattering 6.1 properties of photon

i) Can something have zero rest mass?

From

2 

4 

2  E = p c (p is the magnitude of the 3-momentum) Substitute E=pc into v = p c 2 /E = c So massless particles would always have to travel at v = c, the speed of light. Strange??

Photons, or particles of light, have zero rest mass, and this is why they always travel at the speed of light.

ii) The magnitude of a photon's 4-momentum 

p E

2 

2 

2  

2  0 

2  0 ; E

but this does not mean that the components are all zero.  pc --The time component squared, E 2 / c 2 , is exactly cancelled out by the sum of the space components squared,



p z

2  ˆ 2 --Thus the photon may be massless, but it carries momentum and energy, and it should obey the law of conservation of 4-momentum.

6.2 Compton scattering.

The idea of the experiment is to beam photons of known momentum Q at a target of stationary electrons,and measure the momenta Q ’ of the scattered photons as a function of scattering angle.

We therefore want to derive an expression for Q ’ as a function of  .

Before the collision the 4-momenta of the photon and electron are: 

  (

, 0 , 0 ); 

p e

 (

, 0 , 0 , 0 ) after they are: The conservation law is 

  

  

 

  

(





'



,

' cos

2 

 

 

,

' sin



,

  

 

p e

  

p e



 

0 );

 

p e



e q e

  

q e

(

  

,

( 

q e



cos



   2 

  )

p e

2   

,

 (

q e



q e

   For all photons 

 

 0 ; and for all electrons  



sin



p e

( ) 2 

) (

) Also, in this case And: 

p e

 

q e

 

 

Equation (a) becomes: 2

 2 ; 

  

' 

' 2

' ( 1  cos  cos  )  

2 ( 1   )

, 0 )

But by conservation of energy,

(



−1)mc is just Q−Q

’

and (a − b)/ab is just 1/b−1/a, so we have what we are looking for: 1

'  1

 1

( 1  cos  ) This prediction of special relativity was confirmed in a beautiful experiment by Compton (1923) and has been reconfirmed many times since by undergraduates in physics lab courses.

In addition to providing quantitative confirmation of relativistic mechanics, this experimental result is a demonstration of the fact that photons, though massless, carry momentum and energy.

Quantum mechanics tells The energy E of a photon is related to its wave frequency by E = h  Then 1



c E



c hv

 

; 1

'   '

so we can rewrite the Compton scattering equation in its traditional form:  '   

h mc

( 1  cos  )

7. Particle decay and pair production 7.1 Particle decay:

An elementary particle of rest mass M decays from rest into a photon and a new particle of rest mass M/2. Find its velocity. 

p M

M 

 (

, 0 , M/ 2 0 , 0 ) 

p ph

 (

hv c

,  For 3-momentum conservation, the particle moves in x direction, and the photon moves in –x direction.

hv c

, 0 , 0 ); 

p M

2  ( 

,  2

, 0 , 0 ) 2 By momentum conservation: 

p M

 

p M

 

p ph

2      

0 0 0          

    0

c hv c

 0                   0  0

2         ; (2) Into (1): Solve for u:

Mc u

  

0 .

6

 



hv c

 

 

c Mu

2  

(



) 2 

(



) 2 1 

2 

2 ( 1 ) ( 2 ) 1 

1 

/ /

c c

7.2 Pair production - gamma photon can not be converted to e and e+ Show that the following pair production cannot occur without involvement of other particles. e Let m be the rest mass of electrons and u, v the 3-velocities of electron 

p ph c c



hv hv

       

hv hv

0 0

c c

                    0 1 1 1

m c mu x mu y

       e+          0 2 2 

m mv

c m x

and positron.

v y

     



1 (

u x

2  Sub. (3) into (2):  1 ( 1 

v x v y

Sub. (3) into (1):

u v y y u

)

)  

hv hv

 1 

1 

mc hv

(

u c

(

u c



2 (  1



u y



  (  1

u x

v y

) 2 ) 2   (

u x

  

( 1    2 ) 2

v x

)

v x u y v y

)

u y v y

( 1 ) ) ( 2 ) ( 3 ) ( 5 ) ( 4 ) Compare (4) and (5): (

u x



v x u y v y

)

( 1 

u y v y

) ( 6 ) For u x and v x < c (6) can not be satisfied Pair production needs an additional particle to carry off some momentum.

8 Velocity addition (revisited) and the Doppler shift 8 .1 Velocity addition revisited

In S, a particle of mass

is In S ’ moving at speed v , the 4-momentum of the particle:

= 

 (  1 g g g g ' ' ' '

mc mv mv mv

' ' '

x y z mc

, ù ú ú ú û =  1 moves in the

-direction at speed

v x

, so its 4-momentum

g ê ê ê ê 0 2 g 2

, b 0 , 0 ) g g 2 0 b 2 where 0  1  1 1  0 0 0 0 1 0 0 1 ù ú ú ú û é ê ê ê ë g 1

g 1

mv x

0 0

v c

ù ú ú ú û

2 2   ' = ' g ê ê ê ë 2 g g 1

2 g 1

b 0 0 g 2 g 1

mv x

b + g 1 g 2

mv x

mv mc

   1 1   2 2

mv mc

   1 1   2 2

mcv mv

v

/ /

c

ù ú ú ú û

c

with



v

c

   1    1  2

c

v c

v

 

and

 1   1 2

v

This is a much simpler derivation than that found before.

  2

v

/ 2

c

 

v

' 1  1 

v

1 

2 

v v v

c

8.2 Photon makes an angle from x axis

S ’ y ’  v Q ’ (f rest ) S

Q(f obs )  ’ 

x ’ 

'



(

' ,

      0

'

Q Q Q

cos

 cos sin       

' ,

' sin

        0 0 

' , 0 )

  0 0 Equate each component on both side: z

 

'  

' cos  ' 

 0 0 

' ( 1  

1 0   0 0

(

cos  ' )

,



cos

f obs

 

,

0 1       

 

Q Q

' ' '  0 cos sin   ' '      ;

f em

sin

1     1   

, 0

v c

);

cos 

em Q

cos 

sin    

'  

' cos  '

' sin  '   1 cos       cos cos

  ' '  (    cos cos  

obs

 ' )    ( ( 1     cos cos   ' ' ) ) 1     cos cos  

em em

Doppler effect from:

f obs



i) If



the light



0 ; is

f obs



blue

f em

1

 

1

  2 

shited , when



f em

the light

f em

1   cos 

1   2

1



1

  

source is moving toward you.

ii) If the When 

light v  c :  is  ; red

f obs

1   v c

f em

 1 1 1   v c  1   2 ;  shited , when the 

f ob f em

light 

f em

/( 1 

v c

) 1  1    source is

classic

moving



away from you.

When v  c : ii) If 

  1  v c / 2 ;  1 ;

f obs

1  v c 

f em f ob



f em

/( 1 

v c

) 1 1   2

classic D



This predicts a very small transvers e Doppler effect.

if if Aberration of light from: if 

 0 ; cos 

obs

cos 

obs

  1   1 ;   cos cos  

em em



obs



  ;   / 2 ; cos 

obs

  1 ; cos 

obs

  ; 

obs

 0   0  

obs

  / 2 1 3 1 3 2 2 Light rays emitted by source in S ’ Light rays observed in S When v is very large so that  =0.9, and cos  obs =0.9,  obs =26 

http://www.anu.edu.au/Physics/Savage/TEE/site/tee/learning/aberration/aberration.html

9. 4-force

We recall the 4-velocity and 4-momentum are defined in terms of derivatives 

with respect to proper time rather than coordinate time t . The definitions are 

d x d

  we will need to use Because

and



 ( 

p E c

, 

) 



 

 

d u d

( 

dE cd



, 

and d d p

 ) 

 



d p d d



ˆ   

is rest mass 

Also, if the

rest mass m of the object in question is a constant

(not true if the object in question is doing work, because then it must be using up some of its rest energy!), 

p d



d p d





 (   



 

  

p m

) 2 

2 0 

  0 



d d



  orthogonality 

are orthogonal. In 3+1-dimensional spacetime, 

K

0 is something quite different from orthogonality in 3-space: it has nothing to do with 90  angles.