Download Proper time. Announcements Today`s class Conservation of

Announcements
Proper time.
An object is moving past you with a relative velocity
u=0.8c (as measured by you). When your watch shows
1 second has elapsed how much time does a clock held
by an observer in the objects rest frame show has
elapsed?
a) 0.6 s
b) 1 s
c) 1.7 s
d) this measurement can’t be made
The proper time is the time in the objects frame and is
shorter than your time by a Lorentz factor t’=t/g
Today’s class
• Reading for Friday: 2.4 - 2.6
• HW 4 Due next Wed. noon. Some of the topics
on this will be on the exam!
• First Midterm is on the 16th. Will cover relativity
(Practice questions are on CULearn later today.
Solutions will be added soon.)
• Extra Office Hours. Duane F619
– Monday 1-4pm
– Tuesday 10-12pm (I will also be in the help room 34:30)
Conservation of Momentum
Problem solved in moving frame
Chapter 2:
• Momentum
• Relativistic forces
• Relativistic energy
m
m
m
v
v
m
v
u’f=0
v
u’f≠0
Depending on where you solve this problem you
get different answers. This is no good!
Relativistic momentum
This means that the (classical) momentum
ptot=Σ(m∙u) of a system is not invariant under
Lorentz transformations.
Classical definition:
 Need to change the definition of the
momentum, but we want that:
Say we measure the mass 'm' in its rest-frame ('proper
mass' or 'rest mass'). Since we measure 'm' it's restframe we agree on the same value for 'm' in all frames.
1. At low velocities the new definition of p must
match the classical definition of momentum.
Assume we take the derivative with respect to the proper
time tproper , which has the same meaning in all frames.
2. We want that the total momentum (Σp) of an
isolated system of bodies is conserved in all
inertial frames.
Relativistic definition:
pm
dr
dt
pm
dr
dt proper
This definition fulfills the conservation of momentum in SR!
To prove it you can apply the Lorentz transformations.
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Relativistic momentum
Relativistic momentum
The time dilation formula tells us that dt = γdtproper for a
frame where the particle is moving. We can therefore
rewrite the definition of the relativistic momentum as
follows:
p g m
1
g 
dr
 g mu
dt
1
u2
c2
An important consequence of the Lorentz-factor γ is, that
no object can be accelerated past the speed of light.

Example:
Classical vs. Relativistic momentum
Particle A has half the mass but twice the speed of
particle B. If the particles’ momenta are p A and pB, then
a) pA > pB
b) pA = pB
c) pA < pB
Newton’s second law is:
F = ma
This is equivalent to:
F = dp/dt
only ~1% change!
u
0.1c
p=m·u
classical
0.1
p=γm·u
relativistic
0.101
difference
[%]
1
0.5c
0.9c
0.99c
0.5
0.9
0.99
0.57
2.06
7.02
14
129
609
Example: Relativistic force
A charged particle (charge q) with mass m is at rest at x = 0
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).
Force acting on the particle: F = q·ℇ
Relativistic dynamics: F ≡ d(γ·m·u)/dt
Therefore:
Or:
g is bigger for the faster particle.
Relativistic “Newton’s 2nd Law”
An electron has a mass m ≈ 9·10-31kg. The table below
shows the classical and relativistic momentum of the
electron at various speeds (units are 2.733·10-22kg·m/s):
~67,000,000 mph
q·ℇ = d(γ·m·u)/dt
q·ℇ dt = d(γ·m·u)
Integrating both sides:  q·ℇ dt =  d(γ·m·u)
yields:
q·ℇ · t = γ·m·u
p  gmu
B
A
Using the definition of the relativistic momentum we
obtain a suitable definition for a relativistic force:
F
dp d
 g m u  , with g 
dt dt
1
1
u2
c2
Example: Relativistic force (cont.)
γ·m·u = q·ℇ·t
Now: Solve
Dividing by
g 
1
1
u2
c2
for the velocity u.
yields: m·u = q·ℇ·t (1-u2/c2)0.5
Square both sides: m2·u2 = q2·ℇ2·t2(1-u2/c2)
Bring u to the left: u2(m2c2 + q2·ℇ2·t2)= q2·ℇ2·t2·c2
Divide by term in bracket and take squareroot:
u
Classical
qℇ ct
c
u
( qℇ t ) 2  ( mc ) 2
Short time: u ≈ q·ℇ·t/m (const acceleration)
Long time: u ≈ c (but just a little smaller) 0
t
2
Ultimate speed demonstration
Ultimate speed demonstration
Clear “table top” demonstration
that the speed of light is
unattainable for massive
particles.
This is a result of Lorentz
transformations.
WILLIAM BERTOZZI
HARVARD
Quiz: On the reading
Energy
0
0
E=
u
d)
E
c)
c
0
c
u
1. At low velocities, the value E of the new
definition must match the classical definition.
E
b)
2
γmc = K
Similar to the definition of the relativistic
momentum we want to find a definition for the
energy E of an object that fulfills the following:
mc2
+0
c
u
c
u
2. The total energy (ΣE) of an isolated system of
bodies must be conserved in all inertial frames.
E
E
Which graph best represents the total energy
of a particle (particle's mass m>0) as a
function of its velocity u, in special relativity?
a)
Larger potential → Bigger force →larger kinetic energy →faster speed?
0
Kinetic energy
Relativistic kinetic energy
The relativistic kinetic energy K of a particle with a rest
mass m is:
The work done by a force F to move a particle
from position 1 to 2 along a path s is defined by:
2
K = γmc2 - mc2 = (γ-1)mc2
W12   F  ds  K 2  K1
1
Note: This is very different from the classical K= ½mv2 .
K1,2 being the particle's kinetic energy at positions
1 and 2, respectively (true only for frictionless
system!).
For slow velocities the relativistic energy equation gives the
same value as the classical equation! Remember the
binomial approximation for γ:
γ ≈ 1+ ½v2/c2
Using relativistic force we can now find the
relativistic kinetic energy of the particle.
 K = (γ-1)mc2 ≈ (1 + ½v2/c2 – 1)mc2 = ½ mv2

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Total energy
Rest energy of a particle
We rewrite the equation for the relativistic
kinetic energy and define the total energy of a
particle as:
E = γmc2 = K + mc2
This definition of the relativistic mass-energy E
fulfills our condition of conservation of total energy.
(Not proven here, but we shall see several
examples where this proves to be correct.)
Relation between Mass and Energy
v
m
E2 = γmc2 = K + mc2
E1 = γmc2 = K + mc2
In the particle's rest frame, its energy is its rest
energy, E0. What is the value of E0?
A:
B:
C:
D:
E:
0
c2
mc2
(γ-1)mc2
½ mc2
Note: This suggests a connection
between mass and energy!
Equivalence of Mass and Energy
v
-v
m
E = γmc2
-v
m
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
Total energy:
Etot,final = Mc2 ≡ 2K + 2mc2 = Etot,initial
Etot = E1+E2 = 2K + 2mc2
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.
Potential energy inside an object contributes to its mass!!!
This argument isn’t a “proof”. But let’s see if it’s correct.
Example:
Rest energy of an object with 1kg
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!
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