Transcript Week 9

Announcements
• Homework: Supplemental Problems
• 2nd Project is due at the final exam which is
Monday May 2 at 4:00 – 6:00pm. A list of
potential projects is posted.
• Final Exam will be another hour exam
covering the material we covered since the
last exam. It will be the second thing during
the final exam period (project presentation
first)
The Need for Relativity
c
1
 0 0
First published in a four part paper by James
Clerk Maxwell between 1861 and 1862. Maxwell
showed that light was an electromagnetic wave
whose propagation speed was equal to the
inverse square root of the product of the two
fundamental constants in his equations.
The Need for Relativity II
Given that light was
an electromagnetic
wave, a medium for
the wave was
postulated: the
luminiferous aether.
Performed in 1887 by Albert
Michelson and Edward Morley at
Case Western Reserve University
to detect the motion of Earth
through the luminiferous aether
In 1905 Albert Einstein
figured out the problem
He published a paper
titled “On the
Electrodynamics of
Moving Bodies” which
laid out the Special
Theory of Relativity. In it
he showed how things
behave differently if their
relative speeds are close
to the speed of light.
Newton assumed motions
transformed according to
equations Galileo developed
Galileo assumed t = t’. Turns out that
isn’t correct. Since x’ also depends
on t, it is incorrect, too.
The Lorentz
Transformation
x  x  0 at t   t  0
x 
x  vt
 c
1 v
x
2
z  z
z  z
vx
t  2
c
t
2
v
1
c2
2
x  vt 
 c
2
v
1
y  y
vx
2
c
t 
2
v
1
c2
2
y  y
t
 
 
In relativity, we are describing
an “event”
An event is described by eight quantities
 x, y, z, t 
and
 x, y, z, t
Example
A rocket is moving in the +x direction at 0.7c to an
observer on the ground. At t = t’ = 0 an observer in
the rocket has the same position as an observer
on the ground (x = x’ = 0). At t = 1.0x10-7 s the
observer in the rocket sees a flash goes off 3.0m
from that him. Describe the event in each
reference frame.
Solution 1
Easiest to choose the rocket as the unprimed
reference frame. Then x = 3.0m, y = 0m and z = 0m.
The light had to travel 3.0m to reach the observer at
7.0x10-7s so the event occurred earlier by
3.0m
8

1.0

10
s
8 m
3.0 10 s
Thus
t  1.0 107 s  1.0 108 s  9.0 108 s
So
 x, y, z, t    3.0m, 0m, 0m,9.0 10
8
s
Solution 2
Since we used the rocket as the unprimed
reference frame, the velocity of the primed
reference frame is -0.7c. Since the relative
motion is in the x and x’ direction only, y’
and z’ are both zero.
x 
x  vt
1 v

2
c
3.0m   0.7  3.0 108 m s  9.0 108 s 
1
2
9.0 10
vx
t 2
c
t 

2
1 v 2
c
8
 0.7c 
2
 30.7m
c2
0.7  3.0 10   3.0m 

s
 3.0 10 
 1.36  10
8 m
8 m
1
 0.7c 
2
c2
s
2
s
7
s
Final Solution
 x, y, z, t    3.0m, 0m, 0m,9.0 10 s 
7




 x , y , z , t    30.7m, 0m, 0m,1.36 10 s 
8
Consequences of Special
Relativity
Length contraction: two observers do not see the
length of a moving object as the same. If L0 is the
length of the object an observer that is stationary with
respect to it measures then the moving observer will
see its length as
2
v
L  L0 1 
c2
This length contraction is only in the direction of the
relative motion
Consequences of Special
Relativity 2
Time dilation: to observers moving with respect to
each other do not see time moving at the same rate.
If Dt0 is the time interval as measured by an
observer stationary with respect to the clock, a
moving observer will measure the time interval as
Dt  
Dt0
2
v
1
c2
Watch videos on Simultaneous Events in
Relativity and Time Dilation in Relativity
Example
John sits on a asteroid in space and observes his
friend Jane fly by him in a rocket moving at 0.8c. He
sees Jane holding a meter stick and a clock but they
don’t seem to match his meter stick and clock. How
long does he see Jane’s meter stick as and at what
rate does he see her clock moving?
Solution
Since we have the velocity of the rocket with respect
to John, use him as the unprimed reference frame.
For the calculation use L0 = 1.0m and Dt0 = 1.0s
v  0.8c
L  L0 1  v
Dt  
2
c2
Dt0
1 v
 1.0m
1.0 s

2
c
2
 0.8c 
1
1
 0.8c 
2
c2
 1.0m 1  .64  0.6m
 1.67 s
2
c2
Velocity Transformations
Given the velocity u of a particle
with respect to observer O in
the unprimed system.
u v
u x  x
vu
1  2x
c
u y 
 c
2
uy 1 v
2
u z 
vu
1  2x
c
 c
2
uz 1  v
1
With the reverse transforms from primed to unprimed
u  v
ux  x
vu 
1  2x
c
uy 
 c
2
uy 1  v
vu
1  2x
c
2
uz 
 c
2
uz 1  v
1
vux
c2
2
vu x
c2
2
Example
John sits on a asteroid in space and observes his
enemy Jane fly by him in a rocket moving at 0.8c.
After Jane flies by he fires his super cannon that
hurls a projectile at Jane at 0.9c (muzzle speed he
measures). How fast does Jane see the projectile
moving toward her?
Solution
Let John be the unprimed reference frame. The v = 0.8c
and ux = 0.9c. Since there is no uy or uz, we only need to
transform the x velocity.
ux  v
0.9c  0.8c
u x 

 0.357c
vu x
0.9c  0.8c 

1 2 1
c
c2
So, Jane sees the projectile approaching her at a
speed on 0.357c, not 0.1c as you would expect from
classical physics