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
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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
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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
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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