Physics 3313 - Lecture 4

13313 Andrew Brandt

Monday February 1, 2010Dr. Andrew Brandt

• HW1 Assigned due Weds 2/3/10 (you can turn it in on Feb.8 , but next HW to be assigned 2/3 will be due 2/10)

• Relativistic momentum and energy

2/1/2010

Speed of Person Revisited

• Reference Frames: It’s all relative, 40 mph/backwards? How fast can a person run?

• http://www.youtube.com/watch?v=By1JQFxfLMM

3313 Andrew Brandt 22/1/2010

Relativistic Momentum• So for relativity Galilean transform replaced by Lorentz transform

• What about Newton’s laws?

• F=ma= and (classically)

• Galilean velocity addition ; taking derivative, with v=const

gives so

• But for Lorentz this has extra velocity dependence in

denominator, so accelerations are not equal and a new

expression is required for relativistic momentum

3313 Andrew Brandt 3

dp

dtp mv

'x xa a

'x xv v v

'

'

dv dv

dt dt

'

'x xdv dv

dt dt

'

'

21

xx

x

v vv

vv

c

2/1/2010

' ( )x x ct ' ( )x

t tc

2

1/ 22

(1 )v

c

Relativistic Momentum

• Relativistic momentum should have the following properties

– should be conserved

– as should reduce to classical momentum

– what about ? It has these properties

– note some texts occasionally define a relativistic mass

which increase with velocity, in order to save

standard momentum formula

In this case the mass becomes infinite as the velocity approaches c ;

we will not do this; the mass is always the rest mass in our approach

3313 Andrew Brandt 4

0p m v0

v

c

0p m v

0m m

2/1/2010

Relativistic Momentum Example

• A meteor with mass of 1 kg travels 0.4c

• What is it made of?

• Find its momentum; what if it were going twice as fast, what would its momentum be? compare with classical case

3313 Andrew Brandt 5

) 0.4v

ac

1.09 8 80 1.09 1 0.4 3 10 1.31 10

sec sec

m kg mp m v kg

) 0.8v

bc

1.67 8 81.67 1 0.8 3 10 4.01 10

sec

kg mp

8) 1.2 10sec

kg mc p mv

8) 2.4 10sec

kg md

**work old Ex. 1.5 on board **2/1/2010

Relativistic Mass and Energy

• The work done by a constant force over a distance S:

• For variable force:

• If object starts from rest and no other forces, work become KE

• With v=ds/dt

• Integration by parts with x = v dx = dv and

3313 Andrew Brandt 6

W F S

( )d mvKE ds

dt

0sW Fds

0 ( )vKE m vd v

xdy xy ydx y v

2/1/2010

• with

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Relativistic Mass and Energy2

0vKE mv m vdv

21/ 2

2(1 )

v

c

2 22 1/ 2 2

2 21/ 2

2

(1 )(1 )

mv vKE mc mc

v cc

22

2 22

1 12 22 2

2 2

(1 )

(1 ) (1 )

vmcmv cKE mc

v vc c

20E mc

2 2E mc KE mc

2( 1)KE mc

Use binomial approximation

for

to show

(1 ) 1

1

nx nx

x

21

2KE mv

2 2 2( 1)mc mc mc

2/1/2010

Relativistic Energy Ex. 1.6• A stationary high-tech bomb explodes into two fragments each with 1.0 kg

mass that move apart at speeds of 0.6 c relative to original bomb. Find the original mass M.

•

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2i iE Mc KE 2 0 fMc E 2 2

1 1 1 2m c m c 2

2 1

2

2

2

1

m cMc

vc

2 / 0.8 2.5M kg kg

2/1/2010

Relativistic Mass and Energy• Mass and energy are related by E=mc2

• Can convert from one to the other • One kg of mass = 9x1016 J of energy (enough to send a 1 million ton payload to

the moon!)• http://blog.professorastronomy.com/2006/01/pluto-and-plutonium.html

• http://www.nasa.gov/mission_pages/newhorizons/launch/newhorizons-allvideos.html

• Is Pluto a planet?• Not anymore, it’s a Dwarf planet (part of Kuiper Belt)• What about 2003UB313? AKA Eris

3313 Andrew Brandt 92/1/2010

More about Energy

• Energy is conserved, that is, in a given reference frame for an isolated system E= constant (it may be a different constant in different reference frames)

• Energy is constant but not invariant, that is, the constant can vary from one frame to another (example an object in its rest frame has less energy than an object in a moving frame).

• What about mass mc2?

• Invariant, but not conserved

3313 Andrew Brandt 102/1/2010

Relativistic Energy and Momentum

• Combining and one can obtain

• These expressions relate the E, p, and m for single particles, but are only

valid for a system of particles of mass M if

which need not be true

• Since mc2 is an invariant quantity, this implies

is also an invariant

• If m=0 this implies E = p = 0, but what if v = c?

• Then =1/ 0 and E = p = 0/0, which is undefined!

• For a massless particle with v = c, then we could have E = pc for example

the photon (which unfortunately has the symbol )

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2 2 2 2( )E mc p c

p mv2E mc

2 2 2E p c

2 2 2 2 2( )E mc p c

2 2i imc Mc

and

2/1/2010

Units

• Use W=qV to define electron volt (eV) a useful energy unit

• 1 eV = 1.6x10-19 C x 1V = 1.6x10-19 J

• Binding energy of hydrogen atom is 13.6 eV

• Uranium atom releases about 200 MeV when it splits in two (fission)

This is not a lot of Joules/fission, but there are a lot of atoms around…

• Rest mass of proton (mp) is 0.938 GeV/c2 which is a fine unit for mass

• Units of momentum, p, MeV/c

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v pc

c E

Relativistic Energy Problems2vE mvc2E mc

2vE pcFrom we can multiply by v to obtain simplifying gives which gives us a useful relation:

An electron and a proton are accelerated through 10 MV, find p, v, and of each(work problem on blackboard)

Electron:

Proton:

2 2 2 2 2

2

( ) (10.51) (0.511) 10.50

10.5

1

10.500.999

10.51

e

e

pc E mc MeV

MeVp

cKE

m c

v pc

c E

2 2

2

(948) (938) 137.4

101 1 1.01

938

137.40.145

948.3

k

p

pc MeV

E

m c

v pc

c E

133313 Andrew Brandt2/1/2010