Ben Gurion University of the Negev www.bgu.ac.il/atomchip , www.bgu.ac.il/nanocenter

Physics 3 for Electrical Engineering

Lecturers: Daniel Rohrlich, Ron Folman Teaching Assistants: Daniel Ariad, Barukh Dolgin Week 2. Special relativity – velocities • kinematic paradoxes relativistic mass and momentum • • addition of relativistic energy • energy-momentum 4-vector • decay and scattering in relativity • nuclear energy Sources: Feynman Lectures

I,

Chap. 15, Sects. 5-6 and Chap. 16; Tipler and Llewellyn, Chap. 1, Sect. 6 and Chap. 2, Sects. 1-4

Kinematic paradoxes The “Twin Paradox” You go out into space for one year at 0.9 of the speed of light. When you come back, how old will your twin brother be? Is there a paradox here? How would you solve it?

The “Triplet Paradox” Grumpy lives in Haifa; Dumpy and Jump live in Tel Aviv. They are identical triplets with synchronized watches. One day, Jump takes a high-speed (

v

=

c

/2) train to visit Grumpy.

Grumpy and Dumpy expect Jump’s watch to run slow during the train ride. But Jump expects

their

watches to run slow! He reasons, “When the train left the station, I saw that my watch was still synchronized with Dumpy’s. So it was still synchronized with Grumpy’s watch. Now I am sitting and they are moving fast relative to me! So their watches must run slower than mine!” Who is right???

Event 1: Jump looks at his watch and at Dumpy’s watch. He sees

t

1

′ x

1

′

=

x

1 = 0.

= 0 and

t

1 = 0. Let their location (Tel Aviv) be Summary of Event 1:

t

1

′

=

t

1 = 0,

x

1

′

=

x

1 = 0.

Event 2: Grumpy looks at his watch and sees

t

2 location (Haifa) be

x

2 = 85 km.

= 0. Let his Summary of Event 2 :

t

2 = 0,

x

2 = 85 km.

Event 1: Jump looks at his watch and at Dumpy’s watch. He sees

t

1

′ x

1

′

=

x

1 = 0.

= 0 and

t

1 = 0. Let their location (Tel Aviv) be Summary of Event 1:

t

1

′

=

t

1 = 0,

x

1

′

=

x

1 = 0.

Event 2: Grumpy looks at his watch and sees

t

2 location (Haifa) be

x

2 = 85 km.

= 0. Let his Summary of Event 2 :

t

2 = 0,

x

2 = 85 km.

Hence ∆

t = t

2 ∆

x = x

2 – –

t

1

x

1 = 0, = 85 km.

Event 1: Jump looks at his watch and at Dumpy’s watch. He sees

t

1

′ x

1

′

=

x

1 = 0.

= 0 and

t

1 = 0. Let their location (Tel Aviv) be Summary of Event 1:

t

1

′

=

t

1 = 0,

x

1

′

=

x

1 = 0.

Event 2: Grumpy looks at his watch and sees

t

2 location (Haifa) be

x

2 = 85 km.

= 0. Let his Summary of Event 2 :

t

2 = 0,

x

2 = 85 km.

Hence ∆

t = t

2 ∆

x = x

2 – –

t

1

x

1 = 0, = 85 km. What is ∆

t′

?

“The barn paradox” The barn is 5 m long to the farmer.

The pole is 10 m long to the runner and 5 m long to the farmer, i.e. we have g = 2 and so b =

v/c

= 0.866.

According to the farmer, the pole fits into the barn at a single instant. But the runner cannot agree! The pole is 10 m long and the barn is only 2.5 long!

What are the events? Event 1: the rear end of the pole reaches the entrance to the barn. Event 2: the front tip of the pole reaches the far end of the barn.

In the farmer ’ s reference frame, Event 1 occurs at time

t

1 and at place

x

1 ; Event 2 occurs at time

t

2 and at place

x

2 ; D

t = t

2 –

t

1 = 0 and D

x

=

x

2 –

x

1 = 5 m. In the runner ’ s reference frame: D

t′

D

x′

= g ( D

t

– = g ( D

x

–

v

D

x

/

c

2 ) = (2)(-0.866)(5 m ) /

c v

D

t

) = 10 m.

= 28.9 ns,

What are the events? Event 1: the rear end of the pole reaches the entrance to the barn. Event 2: the front tip of the pole reaches the far end of the barn.

For the runner, the two events are not simultaneous!

What does it mean that D

t′

is negative?

What are the events? Event 1: the rear end of the pole reaches the entrance to the barn. Event 2: the front tip of the pole reaches the far end of the barn.

Further applications of Lorentz transformations D

x′

D

y′

D

z′

D

t′

= g ( D

x

– = D

y

= D

z

= g ( D

t

–

v

D

t

)

v

D

x

/

c

2 ) How do we derive time dilation from the Lorentz transformations?

If Alice (primed system) moves along the

x

-axis with velocity

v

relative to Bob (unprimed system), then D

x′

D

y′

D

z′

D

t′

= g ( D

x

– = D

y

= D

z

= g ( D

t

–

v

D

t

)

v

D

x

/

c

2 ) How do we derive time dilation from the Lorentz transformations?

c

D

t/2 L c

D

t/2 v

D

t

☺

If Alice (primed system) moves along the

x

-axis with velocity

v

relative to Bob (unprimed system), then D

x′

D

y′

D

z′

D

t′

= g ( D

x

– = D

y

= D

z

= g ( D

t

–

v

D

t

)

v

D

x

/

c

2 ) For Alice, the clock stays in one place. Hence D

x′

= 0.

c

D

t/2 L c

D

t/2

☺

v

D

t

If Alice (primed system) moves along the

x

-axis with velocity

v

relative to Bob (unprimed system), then 0 = D

x′

= g ( D

x

–

v

D

t

) Hence D

x

=

v

D

t

D

t′

= g ( D

t

–

v

D

x

/

c

2 ) = g ( D

t

–

v

2 D

t

/

c

2 ) = g(D

t/

g 2 ) = D

t/

g For Alice, the clock stays in one place. Hence D

x′

= 0.

c

D

t/2 L c

D

t/2

☺

v

D

t

Further applications of Lorentz transformations D

x′

D

y′

D

z′

D

t′

= g ( D

x

– = D

y

= D

z

= g ( D

t

–

v

D

t

)

v

D

x

/

c

2 ) How do we derive length contraction from the Lorentz transformations?

If the runner (primed system) moves along the

x

-axis with velocity

v

relative to the farmer (unprimed system), then D

x′

D

y′

D

z′

D

t′

= g ( D

x

– = D

y

= D

z

= g ( D

t

–

v

D

t

)

v

D

x

/

c

2 ) How do we derive length contraction from the Lorentz transformations?

If the runner (primed system) moves along the

x

-axis with velocity

v

relative to the farmer (unprimed system), then D

x′

D

y′

D

z′

D

t′

= g ( D

x

– = D

y

= D

z

= g ( D

t

–

v

D

t

)

v

D

x

/

c

2 ) For the farmer, D

x

=

L

For the runner, D

x′

= and

L′.

D

t

= 0.

If the runner (primed system) moves along the

x

-axis with velocity

v

relative to the farmer (unprimed system), then

L′ =

D

x′

D

y′

D

z′

D

t′

= g ( D

x

– = D

y

= D

z

= g ( D

t

–

v

D

t

) =

v

D

x

/

c

2 ) g

L

For the farmer, D

x

=

L

For the runner, D

x′

= and

L′.

D

t

= 0.

Addition of velocities:

u x

=

dx/dt

;

u y = dy/dt ; u z = dz/dt u x ′= dx′/dt′ ; u y ′= dy′/dt′ ; u z ′= dz′/dt′

For inertial reference frames moving with relative velocity

v

along the

x

–axis,

u x ′=

(

u x –v

)

/

(1

–vu x /c 2

)

u y ′= u y u z ′= u z / /

g g (1 (1

–vu –vu x x /c /c 2 2 )

) Thus you can never surpass the speed of light e.g. by shooting a 0.9

c

bullet from a 0.9

c

jet: (0.9

c

+0.9

c

)/[1+(0.9

c) ·

(0.9

c)

/

c

2 ] = (1.80/1.81)

c

= 0.9945

c

Derivation: apply the Lorentz transformations to the differentials,

dx

'

dy dz

' '

dt

' = = = = g

dx dy

g

dz dt

-

v v

g g

dt dx

/

c

2 and substitute them into

u x ′, u y ′

and

u z ′

to obtain

u x

' =

dx

' /

dt

' = g

dx

-

v

g

dt

g

dt

-

v

g

dx

/

c

2 =

dx

-

v dt dt

-

v dx

/

c

2 =

u x

1 -

v u x v

/

c

2

u y

' =

dy

' /

dt

' =

dy

g

dt

-

v

g

dx

/

c

2 =

u y

g ( 1 -

v u x

/

c

2 )

u z

' =

dz

' /

dt

' =

dz

g

dt

-

v

g

dx

/

c

2 = g

u z

( 1 -

v u x

/

c

2 )

Relativistic mass and momentum We now move from

kinematics

(pure description of motion) to

dynamics

(description of forces).

We will discover that if momentum ( mv ) is conserved in one inertial reference frame, it is not conserved in another!

But conservation of momentum is a law of physics. And Einstein postulated that the laws of physics are the same in all reference frames … .

Two identical particles collide in their center-of-mass frame:

y u y

The blue particle has mass

m

and speed

v

and so does the red particle.

x

In the center-of-mass frame, total

u y

momentum is 0 before and after the collision.

u x

Two identical particles collide in their center-of-mass frame:

y u y

The blue particle has mass

m

and speed

v

and so does the red particle.

x

In the center-of-mass frame, total

u y

momentum is 0 before and after the collision.

u x

But now we go to a reference frame moving to the right with speed

u x

:

Two identical particles collide in their center-of-mass frame:

y u y

The blue particle has mass

m

and speed

v

and so does the red particle.

x

In the center-of-mass frame, total

u y

momentum is 0 before and after the collision.

u x y u y ′

But now we go to a reference frame moving to the right with speed

u x

: (

v

' )

u y y ′

=

v u

red

y y

g [ 1 -

u

2

x c

2 ]

u y ′ x

(

v

' ) blue

u y y ′

=

v u

blue

y y

g [ 1 

u

2

x c

2 ]

u x ′ =

0

Since

u y ′ +

conserved…

u y ′ ≠

0, the

y-

component of momentum is not (

v

' )

u y y ′

(

v

' ) blue

u y y ′

=

v u

red

y y

g [ 1 -

u

2

x c

2 ] =

v u

blue

y y

g [ 1 

u

2

x c

2 ]

y u y ′ x u y ′ u x ′ =

0

Since

u y ′ + u y ′ ≠

0, the

y-

component of momentum is not conserved … unless we redefine momentum!

(

v

' )

u y y ′

(

v

' ) blue

u y y ′

=

v u

red

y y

g [ 1 -

u

2

x c

2 ] =

v u

blue

y y

g [ 1 

u

2

x c

2 ]

y u y ′ x u y ′ u x ′ =

0

Since

u y ′ + u y ′ ≠

0, the

y-

component of momentum is not conserved … unless we redefine momentum! Let ’ s try

p

=

m

f(

v

)

v

with f(0)=1.

y u y ′ x

(

v

' )

u y y ′

=

v u

red

y y

g [ 1 -

u

2

x

(

v

' ) blue

u y y ′ c

2 ] =

v u

blue

y y

g [ 1 

u

2

x c

2 ]

u y ′ u x ′ =

0

y u

Boost left

u x y "

Boost right

u x x u x y u y " u x " =

0

x

Boost right

v

=

u x " u y ′ y u y ′ x u x ′ =

0

Addition of velocities with

v

=

u x "

g = 1 / 1 -

v

2 /

c

2

u y ′ y

' =

u y "

" g [ 1 -

v v u x "

"

/ c c

2 2 ] =

u y y "

" g [ 1 -

v

2 /

c

2 ] = g

u u y y " y u y " = – u y ′ y u y " = – u y ′ u x " =

0

x

Boost right

v

=

u x " u y ′ u x ′ =

0

u y ′ x

So now if we assume

p

=

m

f(

v

)

v

, then

m

f (

u

' )

u y

' =

m

f (

u

" )

u y

" =

m

f   f (

u y

' )

u y

' /

u y

" = f   g f ( g

u y

" ) = f   (

u x

" ) 2  (

u y

" ) 2 (

u x

" ) 2  (

u y

" ) 2   (

u x

" ) 2  (

u y

" ) 2    

u y

" and in the limit

u y "

→ 0 we replace

u x "

with

v

to obtain g = g f ( 0 ) = f (

v

) and therefore

p

=

m

γ

v .

Relativistic momentum:

p

= g

mv

(or “relativistic mass”?)

v/c .

v/c

Relativistic energy We accelerate a body using a force

F

=

dp

/

dt

:

W

= = = 

F vp



dx

 = 

dp



vdt dt pdv

=

v

2

m

g = 

vdp



vm

g

dv mv

2 1 -

v

2 /

c

2 -

m

2 

d

(

v

2 ) 1 -

v

2 /

c

2 =

mv

2 1 -

v

2 /

c

2 

mc

2 1 -

v

2 /

c

2

If we accelerate the body from rest to a final speed

v

, we find that the change in kinetic energy of the body is

W

=

mv

2 1 -

v

2 /

c

2 

mc

2   1 -

v

2 /

c

2 1   =

mc

2 [ g

v

2 /

c

2  g ( 1 -

v

2 /

c

2 ) 1 ] = g

mc

2 -

mc

2 For small

v

/

c

the kinetic energy is

W =

(

γ

–1)

mc

2 ≈

mv

2 /2. But from

W

we do not learn what the total energy is. Could there really be a huge “rest energy”

mc

2 ?

Here’s the answer: consider a head-on collision between two bodies of mass

m

and relativistic momentum ± mγv forming a body of mass

M

at rest:

m m

Momentum is conserved. Now let’s look at the collision in a frame that moves perpendicular to

v

with a small, nonrelativistic velocity

u

:

m m

From conservation of momentum in the

u

-direction, we have

Mu

= 2

m

g '

u

, where the factor

γ

' differs slightly from

γ =

1 / 1 -

v

2 /

c

2 because there is a component of velocity

u

added to

v

.

Now divide both sides by

u

; in the limit

u

→ 0 we get

M

= 2

m

g ,

m m

where

γ

depends only on

v

.

We see that the body at rest having mass

M contains

the kinetic energy of the colliding particles! A body at rest contains energy!

Since

Mc

2 = 2

γmc

2 the energy

E

of a body of mass

m

and speed

v

must equal

E

=

γmc

2 .

 

v c

  2

v

Energy-momentum 4-vector

E

=

mγc

2 , so

E

and pc form a 4-vector

u

= (

E

, pc ) and

u

2 = (

p

c

) 2 (

E

) 2 = (

m

g

v

c

) 2 (

m

g

c

2 ) 2 = -

m

2

c

4 (Note

E

and pc have the same units.) The energy-momentum 4-vector transforms like any other 4-vector under Lorentz transformations. Hence both the energy

E

and the momentum

p

are conserved in each inertial reference frame.

Decay and scattering in relativity A fateful example A heavy particle of mass

M

decays into two lighter particles of mass

m

. What is the velocity of each of the lighter particles? How much kinetic energy is released?

Let us treat this problem in the rest frame of the heavy particle. Energy-momentum conservation: (

Mc

2 ,

0

) = (

mγ

1

c

2 ,

mγ

1

v

1

c

) + (

mγ

2

c

2 ,

mγ

2

v

2

c

) = (

E

1 ,

p

1

c

) + (

E

2 ,

p

2

c

) Hence

p

1 = – p 2

Mc

2 and

γ

1

= γ

2

= γ

from momentum conservation, and = 2

mγc

2 from energy conservation. Therefore

γ

=

M

/2

m

and

M

2

m

= g = 1 -

v

1 2 /

c

2

v

=

M

2 -

M

4

m

2 we get .

So in the rest frame of the heavy particle, the lighter particles come out back to back with opposite velocities and equal speeds

v

=

M

2 4

m

2

c M

There is no kinetic energy initially. The final kinetic energy is 2

mc

2 ( g 1 ) = 2

mc

2  

M

2

m

1   =

Mc

2 2

mc

2 More generally, if the heavy particle decays into several lighter particles of masses

m i

, the kinetic energy released is 

i m i

( g

i

1 )

c

2 = (

M



i m i

)

c

2

Application: fission of U 235 into Ba 141 and Kr 92 and also two neutrons yields how much kinetic energy?

mass (a.m.u.) U 235 Ba 141 Kr 92 n 235.043923

140.914411

91.926156

1.008665

Kinetic energy yield = (235.043923 – 140.914411 – 91.926156 – 2 × 1.008665) a.m.u. × (1.660539 × 10 – 27 kg/a.m.u.) × (2.997925 × 10 8 m/s) 2 = 2.77629 × 10 – 11 J

One gram of U 235 contains 2.56 × 10 21 atoms of U 235 , so fission of one gram of pure U 235 yields (2.56 × 10 21 ) × (2.78 × 10 – 11 J) = 7.12 × 10 10 J of kinetic energy.

By way of comparison, consider two cars, each with 1500 kg mass, at speed 90 km/hour = 25 m/s, in a head-on collision. The total kinetic energy lost in the collision is (3000 kg) × (25 m/s) 2 = 1.875 × 10 6 J. So fission of one gram of U 235 has the destructive power of about 40,000 such collisions.

The bomb dropped on Hiroshima caused about 800 grams of U

235

undergo fission

…

to

...and about 80,000 deaths on that day, two to three times as many within five years.

Relativistic scattering: another fateful example Two deuterium ( 1 H 2 ) nuclei collide and produce a helium-3 ( 2 He 3 ) nucleus and a neutron. What are the velocities and energies of the helium nucleus and the neutron?

The easiest way to solve this problem is to go to the “center-of momentum” frame, i.e. the inertial frame in which the total momentum is zero:

The easiest way to solve this problem is to go to the “center-of momentum” frame, i.e. the inertial frame in which the total momentum is zero:

In this case the problem reduces to the previous problem, with the initial mass

M

of the decaying particle replaced by the initial energy of the two deuterium nuclei divided by

c

2 .

What is the minimum amount of kinetic energy produced by this scattering process (in the center-of-momentum frame)?

What is the minimum amount of kinetic energy produced by this scattering process (in the center-of-momentum frame)?

It is [2 ×

m

(deuterium) –

m

(helium-3) –

m

(neutron)] ×

c

2

Note: In fission of uranium-235, the fraction of mass converted into energy is 235.043923

– 140.914411

– 91.926156

– 2  1.008665

235.043923

= 0 .

000791

Note: In fission of uranium-235, the fraction of mass converted into energy is 235.043923

– 140.914411

– 91.926156

– 2  1.008665

235.043923

= 0 .

000791 In fusion of two deuterium nuclei to helium-4, it is

2



2.01410778

– 4.00260325

2



2.01410778

=

0 .

00636

Nuclear energy: the curve of binding energy Average binding energy ∆

E

/

A

vs. mass number A

Nuclear energy: the curve of average nucleon mass Average mass of nucleon

m

/

A

vs. mass number A