Download Lecture 11

Announcements
• Reading for Review chapters 1 and 2
A proton (mass 1.6e-27 kg) is travelling at 0.99 c. How
much Kinetic energy does it have?
• HW 4 Due Wed. noon. Some of the topics on
this will be on the exam! Don’t be late.
K = (γ-1)mc2
a) 6e-11 Joules
• First Midterm is on the 16th. Will cover relativity
b) 2e-10 Joules
– I can put some other practice exams (not multiple
choice up if it will help).
c) 9e-10 Joules
d) 1e-9 Joules
• Extra Office Hours. Duane F619
e) 4e-8 Joules
– Today 1-4pm
– Tomorrow 10-12pm (I will also be in the help room 34:30)
Today’s class
Review: Relativistic mechanics
The good and bad of nuclear
energy:
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•
•
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Momentum and energy review
Applications
Massless particles
Review (more next time)
Or:
Or:
Relativistic force:
F
dr
  mu
dt proper
dp d
  m u 
dt dt
Etot = γmc2 = K + mc2
These definitions fulfill the momentum and energy
conservation laws. And for u<<c the definitions for p, F,
and K match the classical definitions. But we found that
funny stuff happens to the proper mass ‘m’.
Equivalence of Mass and Energy
We use conservation of energy because it often simplifies
calculations. Remember the charged particle (charge -e)
with mass m in a uniform electric field ℇ ? Plot the velocity u
of the particle as a function of time t (assume the particle is
released at t = 0).
Potential energy: U = -eV
Therefore:
pm
Total energy of a
particle with mass ‘m’:
Example simpler: Energy
Relativistic Kinetic Energy: K ≡
Relativistic momentum:
(γ-1)mc2
m
Conservation of the total energy requires that the final
energy Etot,final is the same
2 as the energy Etot, before 2
thetot
collision. final
Therefore:
initial
E =γ
Mc = 2K
+ 2mc
We find that the total mass M of the final system is larger
than the sum of the masses of the two parts! M>2m.
qV = (γ-1)mc2
qV /mc2 +1= γ
-v
Etot,final = Mc2 ≡ 2K + 2mc2 = Etot,initial
0-qV = K-0
 
v
m
1
1
Then just solve for u. No integration necessary.
u2
c2
Potential energy inside an object contributes to its mass!!!
This argument isn’t a “proof”. But let’s see if it’s correct.
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Example:
Rest energy of an object with 1kg
Turns out this is very useful
E0 = mc2 = (1 kg)·(3·108 m/s )2 = 9·1016 J
9·1016 J = 2.5·1010 kWh = 2.9 GW · 1 year
This is a very large amount of energy! (Equivalent to the
yearly output of ~3 very large nuclear reactors.)
Enough to power all the homes in Colorado for a year!
E=mc2  Convert mass to energy?
Atomic cores are built from neutrons and protons. There
are very strong attractive forces between them. The
potential energy associated with the force keeping them
together in the core is called the binding energy E B.
We now know that the total rest energy of the particle
equals the sum of the rest energy of all constituents minus
the total binding energy EB (the binding energy is what
counters repulsion and lowers internal potential energy):
Mc2 = Σ(mi c2) – EB
Mc2 = Σ(mi c2) – EB
Or in terms of Mass per nucleon
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Example: Deuterium fusion
Example: Deuterium fusion
Isotopes of Hydrogen:
Isotope mass:
Deuterium: 2.01355321270 u
Helium 4: 4.00260325415 u
(1 u ≈ 1.66·10-27 kg)
 1kg of Deuterium yields ~0.994 kg of Helium 4.
Energy equivalent of 6 grams:
E0 = mc2 = (0.006 kg)·(3·108 m/s )2 = 5.4·1014 J
Enough to power ~20,000 American households for 1 year!
Rest energy is real: Why does the sun
shine?
Rest Energy is real: Particle creation
The bubble chamber shows
trajectories of charged
particles.
Hans Bethe
first proposed this
mechanism
4 Hydrogen → 1 Helium 4
Important comments
In this bubble chamber
picture, a photon hits an
electron so hard that two
new massive particles, an
electron and a “positron”,
are created from the extra
energy. The remaining
energy shows up as kinetic
energy of motion.
Chain reactions
Modern experiments probe
higher and higher energies.
Initial mass of a neutron
and a 235U nucleus.
More and more particles!
Intermediate nucleus is
unstable 236U
Particle tracks from LHC.
Final products have less
mass, but much more
kinetic energy. Conversion
of mass to kinetic energy.
As an experimental result: These
quantities (momentum and energy as
above) ARE conserved. Seen over and
over again in high energy experiments.
Oh yes, and more neutrons,
so the reaction can run wild
(chain reaction!).
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Relationship of Energy and momentum
Recall:
Total Energy: E
Momentum:
γmc2
=
p = γmu
Application: Massless particles
From the momentum-energy relation E2 = p2c2 + m2c4
we obtain for mass-less particles (i.e. m=0):
Therefore: p2c2 = γ2m2u2c2 = γ2m2c4 · u2/c2
use:
u2  2  1
 2
c2

E = pc , (if m=0)
 p2c2 = γ2m2c4 – m2c4
p=γmu and E=γmc2  p/u = E/c2
=E2
This leads us the momentum-energy relation:
E2 =
or:
(pc)2
(mc2)2
+
E2 = (pc)2 + E02
Using E=pc leads to:
u=c
, (if m=0)
Massless particles travel at the speed of light!!
… no matter what their total energy is!!
Example:
Electron-positron annihilation
Do neutrinos have a mass?
Positrons (e+, aka. antielectron) have exactly the same mass
as electrons (e-) but the opposite charge:
me+ = me-= 511 keV/c2 (1 eV ≈ 1.6·10-19J)
Neutrinos are elementary particles. They come in three
flavors: electron, muon, and tau neutrino (e,μ, τ). The
standard model of particle physics predicted such particles.
The prediction said that they were mass-less.
E1, p1
eBAM!
e+
E2, p2
At rest, an electron-positron pair has a total energy
E = 2 · 511 keV. Once they come close enough to each
other, they will annihilate one other and convert into two
photons.
Conservation
momentum:
1 = -ptwo
2.
What can of
you
tell about pthose
photons?
Conservation of energy: E1+E2 = 2mc2 ,  E1 = E2 = 511 keV
Do neutrinos have a mass? (cont.)
Bruno Pontecorvo predicted the ‘neutrino oscillation,’ a
quantum mechanical phenomenon that allows the neutriono
to change from one flavor to another while traveling from the
sun to the earth.
The fusion reaction that takes place in the sun (H + H  He)
produces such e. The standard solar model predicts the
number of e coming from the sun.
All attempts to measure this number on earth revealed only
about one third of the number predicted by the standard
solar model.
What about those faster than
light neutrinos?
The OPERA collaboration (hundreds of scientists) has
recently reported the discovery of faster than light neutrino’s
What did they do?
Why would this imply that the neutrinos have a
mass?
Massless particles travel at the speed of light!
i.e. γ  ∞, and therefore, the time seems to be standing
still for the neutrino:
ΔtEarth = γ ∙ Δtneutrino(“proper”)
They shot
a beam of
neutrinos
from
CERN
Measured
their
arrival
time in
Italy
In the HW: muon or pion experiments. The half-live time of the
muons/pions in the lab-frame is increased by the factor γ.
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What about those faster than
light neutrinos?
What about those faster than
light neutrinos?
This is a simple of a measurement one can do. Measure the
Time of Flight.
velocity = distance/time
Same result with simple
proton distribution
When the protons (and hence
neutrion’s) left CERN
When neutrino’s detected
What about those faster than
light neutrinos?
With neutrino’s getting
there 60nS before light
would
Summary SR
• Classical relativity  Galileo transformation
• Special relativity (consequence of 'c' is the same in all inertial
Is this result right?
We know that neutrino’s can travel
at sub luminal speeds (supernova
1987a)
We know that other particles are
governed by the constancy of
speed of light.
What could have gone wrong?
What are the implications?
frames; remember Michelson-Morley experiment)
– Time dilation & Length contraction, events in
spacetime  Lorentz transformation
– Spacetime interval (invariant under LT)
– Relativistic forces, momentum and energy
– Lot's of applications (and lot's of firecrackers)
…
Everything we have discussed to this point will be part of the first
mid-term exam (including reading assignments and homework.)
If you have questions ask as early as possible!!
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