Transcript Document

Physics 320: Astronomy and
Astrophysics – Lecture IV
Carsten Denker
Physics Department
Center for Solar–Terrestrial Research
NJIT
The Theory of Special Relativity
 The
Failure of the Galilean
Transformations
 The Lorentz Transformation
 Time and Space in Special Relativity
 Relativistic Momentum and Energy
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September 24, 2003
Wave Theory and Ether
Luminiferous Ether  transport light waves, no
mechanical resistance 
 Science of early Greek: earth, air, water, and fire
 heavens composed of fifth element = ether
 Maxwell: There can be no doubt that the
interplanetary and interstellar spaces are not empty,
but are occupied by a material substance or body,
which is certainly the largest, and probably the most
uniform body of which we have any knowledge.
 Measuring absolute velocity?
 Inertial reference systems (Newton’s 1st law)

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Galilean Transformation Equations
x  x  ut
y  y
z  z
t  t
vx  vx  u 


v y  v y   v  v  u and u  const.
vz  vz 
 a  a  F  ma  ma
 Michelson–Morley
Newton’s laws are obeyed in
both inertial reference frames!
experiment:
= 3  108 m/s = const.
 velocity of Earth through ether is zero
c
 Crisis
of Newtonian paradigm for v/c << 1
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September 24, 2003
The Lorentz Transformations
Einstein 1905 (Special Relativity):
On the Electrodynamics of Moving Bodies
 Einstein’s postulates:




The Principle of Relativity: The laws of physics are
the same in all inertial reference frames
The Constancy of the Speed of Light: Light travels
through a vacuum at a constant speed of c that is
independent of the motion of the light source.
Linear transformation equations between space
and time coordinates (x, y, z, t) and (x, y, z, t )
of an event measured in two inertial reference
frames S and S.
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Linear Transformation Equations
x  a11 x  a12 y  a13 z  a14t
y  a21 x  a22 y  a23 z  a24t
z  a31 x  a32 y  a33 z  a34t
t   a41 x  a42 y  a43 z  a44t
u  u( x)iˆ
Principle of Relativity
x  a11 x  a12 y  a13 z  a14t 

y  y
a22  a33  1
 

z  z
 a21  a23  a24  a31  a32  a34  0
t   a41 x  a42 y  a43 z  a44t 
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Linear Transformation Equations
(cont.)
Rotational symmetry
x  a11 x  a12 y  a13 z  a14t 

y  y

  a42  a43  0 (y   y and z   z )
z  z


t   a41 x  a44t
Boundary conditions at origin
x  a11 ( x  ut ) 
t  t   0

y  y
 a12  a13  0  

   x  ut
z  z
  a11u  a14   x  0

t   a41 x  a44t 
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Galilean
Transformations
a11  a44  1
a41  0
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Linear Transformation Equations
(cont.)
Spherically symmetric wave front in S and S
x 2  y 2  z 2  (ct )2  
a11  a44  1/ 1  u 2 / c 2

2
2
2
2
2
x  y  z  (ct )  
a


ua
/
c
41
11

Inverse Lorentz Transform
Lorentz Transform
x 
t 
x  ut
   x  ut 
1  u 2 / c2
y  y
z  z
t  ux / c 2
1 u / c
2
2


  t  ux / c
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x
1
1  u 2 / c2
2

t
x  ut
   x  ut  
1  u 2 / c2
y  y
z  z
t   ux / c 2
1 u / c
2
2

  t   ux / c 2

September 24, 2003
Time and Space in Special Relativity
Intertwining roles of temporal and spatial
coordinates in Lorentz transformations
 Hermann Minkowski: Henceforth space by itself,
and time by itself, are doomed to fade away into mere
shadows, and only a kind of union between the two
will preserve an independent reality.
 Clocks in relative motion will not stay
synchronized
 Different observers in relative motion will
measure different time intervals between the same
two events

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Time Dilation
t1  t2 
t2  t1 
( x2  x1 )u / c2
1 u / c
2
(t2  t1)  ( x2  x1)u / c 2
 t  t2  t1 


Flashbulbs at x1 and x2 at same time t
2
1 u / c
2
2
t 
1 u / c
2
2
Strobe light every t at x1 = x2
 tmoving 
trest
1  u 2 / c2
The shortest time interval is measured by a clock at rest
relative to the two events. This clock measures the
proper time between the two events.
Any other clock moving relative to the two events will
measure a longer time interval between them.
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Length Contraction
x2  x1 
L 
( x2  x1 )  u (t2  t1 )
1 u / c
2
L
1 u / c
2
2
2
Rod along x–axis at rest in S
 L  x2  x1, L  x2  x1 and t1  t2
Lmoving  Lrest 1  u 2 / c 2



The longest length, called the rod’s proper length, is
measured in the rod’s rest frame.
Only lengths or distances parallel to the direction of the
relative motion are affected by length contraction.
Distance perpendicular to the direction of the relative
motion are unchanged.
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Group Assignment
Problem 4.4
 A rod
moving relative to an observer is
measured to have its length Lmoving
contracted to one–half of its original
length when measured at rest. Find the
value of u/c for the rod’s rest frame relative
to the observer’s frame of reference.
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Doppler Shift
obs  rest  vr


rest
rest vs
Sound speed vs and radial velocity vr
tobs  tmoving  tlight
 tobs 
trest
1 u / c
2
2
1 utrest cos 


2
2
c 1  u 2 / c2
1 u / c
trest
1  (u / c) cos 
Relativistic Doppler shift
  obs
 rest 1  u 2 / c 2  rest 1  u 2 / c 2


 vr  u cos 
1  (u / c) cos 
1  (vr / c)
  obs   rest
   0 and vr  u
1  vr / c

1  vr / c
  180 and vr  u
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Redshift
Source of light is moving away from the observer:
vr  0  obs  rest Redshift
Source of light is moving toward the observer:
vr  0  obs  rest Blueshift
Redshift parameter:
c    obs  rest
tobs
 z 1 
trest
obs  rest 
z

rest
rest
1  vr / c
1  vr / c
and z 
1
1  vr / c
1  vr / c
Radial motion!
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September 24, 2003
Group Assignment
Problem 4.9
 Quasar
3C 446 is violently variable. Its
luminosity at optical wavelength has been
observed to change by a factor of 40 in as
little as 10 days. Using the redshift
parameter z = 1.404 measured for 3C 446
determine the time for the luminosity
variation as measured in the quasar’s rest
frame.
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September 24, 2003
Relativistic Velocity Transformations
dx 
(vx  u)dt
1  u 2 / c2
, dy  vy dt , dz  vz dt , and dt  
vy 
vy 1  u / c
vz 
2
1  uvx / c 2
vz 1  u / c
1  uvx / c 2
2
1  u 2 / c2
vx  u
vx 
1  uvx / c 2
vx  u
vx 
1  uvx / c 2
2
(1  uvx / c2 )dt
2
vy 1  u 2 / c 2
vy 
1  uvx / c 2
vz 1  u 2 / c 2
vz 
1  uvx / c 2
v  c  v  c
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Relativistic Momentum and Energy
p
mv
1 v / c
2
  mv
2
Relativistic momentum vector
The mass m of a particle has the same value in all reference
frames. It is invariant under a Lorentz tranformation.
xf
xf
xi
xi
K   Fdx  
p f dx
pf
dp
dp
dx  
dp   vdp  F 
pi dt
pi
dt
dt
vf
K  p f v f   pdv 
0

mv 2f
1 v / c
2
f
2
 mc
2

mv 2f
1 v / c
2
f
2

mv
vf
0
1 v / c
2
2
dv
Relativistic
kinetic
energy


1
1  v / c  1  mc 
 1  mc 2 (  1)
 1  v 2f / c 2



2
f
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2

2
September 24, 2003
Relativistic Energy
E
mc 2
1 v / c
2
Erest  mc2
2
  mc 2
Total relativistic energy
Rest energy
E 2  p2c2  m2c2
n
Esys   Ei
Total energy of a system of n particles
i 1
n
psys   pi
i 1
Total momentum of a system of n particles
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September 24, 2003
Group Assignment
Problem 4.16
 Find
the value of v/c when a particle’s
kinetic energy equals its rest energy.
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September 24, 2003
Class Project
Exhibition
Science
Audience
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September 24, 2003
Homework Class Project
 Read
the Storyline hand–out
 Prepare a one–page document with
suggestions on how to improve the
storyline
 Choose one of the five topics that you
would like to prepare in more detail
during the course of the class
 Homework is due Wednesday October 1st,
2003 at the beginning of the lecture!
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September 24, 2003
Homework Solutions
Problem 2.3
dr
a(1  e2 )
d
d 2 dA
L
vr 

e sin 
and
 2
 2
2
dt (1  e cos  )
dt
dt r dt  r
 L  2 a 2 1  e2 / P  A   ab and b  a 1  e 2
d 2 (1  e cos  ) 2


dt
P(1  e2 )3/ 2
2 ae sin 
d 2 a(1  e cos  )
 vr 
and v  r

2
dt
P 1 e
P 1  e2
a(1  e2 )
2 1
2
2
2
2
r
and v  v  vr  v  G(m1  m2 )   
1  e cos 
r a
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September 24, 2003
Homework Solutions
Problem 2.9
2
4

P2 
a3 and a  R  h  6.99 106 m  P  96.6min
G(m1  m2 )
R  3.58 107 m  5.6R
A geosynchronous satellite must be parked over the
equator and orbiting in the direction of Earth’s rotation.
This is because the center of the satellite’s orbit is the
center of mass of the Earth–satellite system (essentially
Earth’s center).
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September 24, 2003
Homework Solutions
Problem 2.11
P2  a3  a  17.9 AU
mcomet
M M
4 2 a3
30

1.98

10
kg
2
GP
rp  a(1 e)  0.585 AU and ra  a(1 e)  35.2 AU
va  0.91 km/s
v p  55 km/s
GM
r av
 7.0 km/s
a
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Kp
Ka

v2p
2
a
v
 3650
September 24, 2003
Homework
is due Wednesday October 1st,
2003 at the beginning of the lecture!
 Homework assignment: Problems 4.5,
4.13, and 4.18
 Late homework receives only half the
credit!
 The homework is group homework!
 Homework should be handed in as a text
document!
 Homework
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September 24, 2003