Download Physics 2170

Review 1
Announcements:
• Last homework is due tomorrow at 12:50pm
• Final exam is on Saturday from 1:30pm – 4:00 pm in G125
(this room)
• I should be in my office from 2-5pm Thursday and 9-12
and 2-5 on Friday if you have questions
– You can also send me email
• The formula sheet for the final is basically the first two
formula sheets put together plus a few more formulae. It
will be put up by 4pm today.
• Today is the last day clickers count although there will be
clicker questions Friday.
• Cover relativity today and quantum mechanics Friday.
http://www.colorado.edu/physics/phys2170/
Physics 2170 – Spring 2009
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Final exam
• It is about 1/3 relativity and the rest is quantum
mechanics with an emphasis on the last few weeks
– There are 3 multipart problems and a total of 25
questions.
– Problems are similar to midterms – maybe a little less
involved.
• I recommend studying the following:
– Old finals on CULearn (don’t worry about questions on
semiconductors)
• I think these finals are a little easy; don’t expect your final
to be that easy.
• Try to take the old finals as if you were taking a real test.
– The 14 homework assignments
– The lectures (including the clicker questions)
– The midterms
http://www.colorado.edu/physics/phys2170/
Physics 2170 – Spring 2009
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Schrödinger’s Cat
A radioactive sample has a 50% chance of emitting an alpha particle.
If it decays, a Geiger detector triggers the release of poison killing a
cat in the box. Before opening the box, the cat is in a superposition
of wave functions:   1
 1
2
dead
2
alive
When does the wave function collapse to either dead or alive?
No clear agreement. Interesting physics/philosophical question.
http://www.colorado.edu/physics/phys2170/
Physics 2170 – Spring 2009
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Quantum entanglement/teleportation
Suppose a particle with angular momentum of 0 decays into two
electrons. By conservation of angular momentum, the sum of the
two electrons angular momentum must also be 0.
If we measure the z-spin of one electron to be +½ then
we know that the other electron must have a z-spin of –½.
But before we measure the first electron, it is in a
mixture of +½ and –½ spin states. The act of
measuring causes the electron to have a definite spin.
We can separate the two electrons, measure the 1st
electron and then measure the 2nd electron before any
possible (light speed) signal can reach the 2nd electron.
And yet the 2nd electron always has the correct spin.
Einstein called this “spooky action at a distance”
http://www.colorado.edu/physics/phys2170/
Physics 2170 – Spring 2009
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Quantum entanglement/teleportation
In 1935 Einstein, Podolsky, and Rosen wrote a paper attempting
to show that quantum mechanics results in a paradox (now
called the EPR paradox).
They proposed a way out: Electrons actually always know their
spin in every direction but experiments can only get the limited
knowledge allowed by quantum mechanics. A better theory
would allow one to get access to this information.
This is called a hidden variable theory.
In 1964, J.S Bell proved that local hidden variable theories
would give a different result in some cases than quantum
mechanics.
Experiments in the 70s & 80s confirmed that quantum mechanics
was correct and local hidden variable theories don’t work.
http://www.colorado.edu/physics/phys2170/
Physics 2170 – Spring 2009
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Special Relativity – Chapter 1
Postulates of relativity:
1. All physical laws are the same in all inertial reference frames
2. The speed of light is the same in all inertial reference frames
Relativity of simultaneity: events which are simultaneous in one
frame may not be in another; even the order may be switched.
Proper time Dt0 between two events is the time measured in
the frame in which both events occur at the same location.
Proper length L0 of an object is the length measured in the
rest frame of the object.
Time dilation (moving clocks run slower): Dt = gDt0
Length contraction (moving objects are shorter in
the direction they are traveling): L = L0/g
http://www.colorado.edu/physics/phys2170/
Physics 2170 – Spring 2009
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Clicker question 1
Set frequency to DA
Jimmy sees two lightning flashes and determines that one
happens 10 ns before the other. Which statement is true:
A. Another inertial observer must find the order of events, the
location of events, and the time difference to be the same.
B. Another inertial observer must find the order of events to be the
same but the location and time difference may be different
C. Another inertial observer must find the order of events and the
locations to be the same but the time difference may be different.
D. Any other inertial observer might find the order of events, the
locations, and the time difference to be different.
E. Say what?
In general, length, time, and order can be
different for different inertial observers.
However, due to causality and finite speed of light, if the events
are far enough apart it is impossible for the order to be reversed.
http://www.colorado.edu/physics/phys2170/
Physics 2170 – Spring 2009
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Special Relativity – Chapter 1 (continued)
Lorentz transformations are used to transform coordinates (position
and time) of an event from a rest frame S to a moving frame S′.
x  g (x  vt)
y'  y
z'  z
t'  g (t  vx / c2 )
The inverse Lorentz transformations go the other way (S′ to S)
and can be obtained by swapping primes and signs.
Velocity addition formula is also different
u  u  v 2
1  uv / c
than the Galilean velocity addition formula:
Spacetime diagrams can be used to
understand where & when events
occur and how they are related in
the space-time continuum.
Future
Relativistic Doppler shift for observer
and source approaching at .
1 
fobs  f source
1 
http://www.colorado.edu/physics/phys2170/
(forward
light cone)
ct
Physics 2170 – Spring 2009
x
Elsewhere
Past
(backward
light cone)
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Clicker question 2
Set frequency to DA
Two events take place 90 m apart with an intervening time interval of
0.60 ms in reference frame S. What is the speed of the reference
frame S′ which measures the proper time between the two events?
A. 0
B. 0.25c
C. 0.50c
D. 0.90c
E. c
x  g ( x  vt)
y  y
z  z
t  g (t  v2 x)
c
Proper time Dt0 between two
events is the time measured in
the frame in which both events
occur at the same location.
So in the S′ frame, both events must occur at point 0.
That is, x′ = 0 when x = 90 m and t = 0.60 ms.
requires x – vt = 0, i.e. x = vt.
x  g ( x  vt)  0
x  90 m  150 m/ ms  0.5c
v

Solving for v:
t .6 ms
To get
http://www.colorado.edu/physics/phys2170/
Physics 2170 – Spring 2009
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Clicker question 3
Set frequency to DA
A person on Earth sees two spaceships heading toward each other.
The earthling sees rocket A moving at 0.8c. In the Earth frame, how
fast are the two spaceships moving toward each other?
A. 0
B. 0.6c
C. 0.8c
D. 0.98c
E. 1.6c
u 
u v
1  uv / c2
u
u  v
1  uv / c2
The person on Earth sees that they are closing at
1.6c. This does not mean that they see anything
moving faster than the speed of light. It is just that
they are approaching each other at that speed.
http://www.colorado.edu/physics/phys2170/
Physics 2170 – Spring 2009
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Clicker question 4
Set frequency to DA
A person on Earth sees two spaceships heading toward each other.
The earthling measures ship A moving at 0.6c and ship B at 0.8c. In
the reference frame of ship A, how fast is ship B moving?
A. 0
B. 0.6c
C. 0.8c
D. 0.95c
E. 1.6c
u 
u v
1  uv / c2
u
u  v
1  uv / c2
u is the speed of something in the S frame.
u′ is the speed of the same thing in the S′ frame.
v is the relative speed between the S and S′ frames.
We know u: the speed of rocket B in the Earth frame (S frame)
We know v: the relative speed between the S frame (Earth)
and the S′ frame (rocket A)
So we need u′: the speed of rocket B in the S′ frame
u
u  v  1.4c  0.95c
1  uv / c2 1  0.48
http://www.colorado.edu/physics/phys2170/
Physics 2170 – Spring 2009
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Special Relativity – Chapter 2
Conservation of momentum and energy continues to hold for
isolated systems if we redefine momentum and energy as:
p  g u mu
E  g u mc2
for mass m with velocity u and g u  1 / 1  u2 / c2
This gives a rest energy of
Erest  mc2
Define kinetic energy: KE  E  Erest  g u mc2  mc2  (g u  1)mc2
Energy-momentum relationship:
E2  ( pc)2  (mc2)2
For a massless particle like a photon, this reduces to E  pc
One other equation is   pc / E
This is it for equations in Chapter 2. Usually use conservation
of energy and conservation of momentum to solve problems.
http://www.colorado.edu/physics/phys2170/
Physics 2170 – Spring 2009
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Clicker question 5
Set frequency to DA
A L particle with mass 1116 MeV/c2 decays at rest to a proton with
mass 938 MeV/c2 and a pion with mass 140 MeV/c2. What can we
say about the proton and pion momentum and energy?
p  g u mu
A. The proton and pion have the same
magnitude momentum and energy.
E  g u mc2
B. The proton and pion have the same energy
Erest  mc2
but the proton has more momentum.
C. The proton and pion have the same
E2  ( pc)2  (mc2)2
momentum but the proton has more energy.
D. The proton and pion have equal and opposite
  pc / E
momenta and the proton has more energy.
E. Need more information
KE  E  E  (g  1)mc2
rest
u
Conservation of momentum requires the proton and pion
momentum add to 0 so must be equal and opposite. Since E2 =
(pc)2 + (mc2)2, the proton has more energy due to larger mass.
http://www.colorado.edu/physics/phys2170/
Physics 2170 – Spring 2009
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