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CHAPTER 2 Special Theory of Relativity. 2.1 The Need for Ether 2.2 The Michelson-Morley Experiment 2.3 Einstein’s Postulates 2.4 The Lorentz Transformation 2.5 Time Dilation and Length Contraction 2.6 Addition of Velocities 2.7 Experimental Verification 2.8 Twin Paradox 2.9 Spacetime
It was found that there was no displacement of the interference fringes, so that the result of the experiment was negative and would, therefore, show that there is still a difficulty in the theory itself…
- Albert Michelson, 1907
It is assumed that Newton’s laws of motion must be measured with respect to (relative to) some reference frame.
Para um ponto P
Passo 1. Substituav por -v
Passo 2. Substitua as variáveis com “linha” por quantidades “sem linha” e vice-versa.
Velocidade perpendicular (2) depois do espelho
Velocidade perpendicular (1)
O interferômetro de Michelson
1. AC é paralelo ao movimento da terra induzindo um "vento de éter"2. Luz de fonte S é dividida pelo espelho A e viaja para espelhos, C e D, em direções perpendiculares3. Após a reflexão os feixes recombinam em A ligeiramente fora de fase devido o "vento de éter" como visto pelo telescópio E.
Supondo a transformação de Galileu
Tempo t1 de ida e volta de A para C:
Tempo t2de ida e volta de A para D:
A diferença de tempo é:
Ao girar o aparelho, os comprimentos de caminho óticoℓ1 eℓ2são trocados, produzindo uma alteração diferente no tempo: (note a mudança nos denominadores)
Assim, uma diferença de tempo entre rotações é dada por:
usando uma expansão binomial, assumindo
v/c << 1, chega-se a
V = 3 × 104 m/s
ℓ1 ≈ ℓ2 = 1.2 m
Assim a diferença de tempo torna-se
Δt’− Δt ≈ v2(ℓ1 + ℓ2)/c3 = 8 × 10−17 s
…thus making the path lengths equal to account for the zero phase shift.
With the belief that Maxwell’s equations must be
valid in all inertial frames, Einstein proposes the
Frank at rest is equidistant from events A and B:
−1 m +1 m
Frank “sees” both flashbulbs go off simultaneously.
Mary, moving to the right with speed v, observes events A and B in different order:
−1 m 0 +1 m
Mary “sees” event B, then A.
Step 1: Place observers with clocks throughout a given system.
Step 2: In that system bring all the clocks together at one location.
Step 3: Compare the clock readings.
t = 0
t = d/ct = d/c
The special set of linear transformations that:
known as the Lorentz transformation equations
A more symmetric form:
Recall β = v/c < 1 for all observers.
(note v ≠ c)
Spherical wavefronts in K:
Spherical wavefronts in K’:
Note: these are not preserved in the classical transformations with
We want a linear equation (1 solution!!)
Gives the following result:
from which we derive:
Recalling x’= (x – vt) substitute into x = (x’ + vt’) and solving for t’ we obtain:
t’ may be written in terms of β (= v/c):
Consequences of the Lorentz Transformation:
Clocks in K’run slow with respect to stationary clocks in K.
Lengths in K’ are contracted with respect to the same lengths stationary in K.
To understand time dilation the idea of proper time must be understood:
Not Proper Time
Beginning and ending of the event occur at different positions
Frank’s clock is at the same position in system K when the sparkler is lit in (a) and when it goes out in (b). Mary, in the moving system K’, is beside the sparkler at (a). Melinda then moves into the position where and when the sparkler extinguishes at (b). Thus, Melinda, at the new position, measures the time in system K’ when the sparkler goes out in (b).
with T ’= t’2 - t’1
1) T’ > T0 or the time measured between two events at different positions is greater than the time between the same events at one position: time dilation.
2) The events do not occur at the same space and time coordinates in the two system
3) System K requires 1 clock and K’requires 2 clocks.
To understand length contraction the idea of proper length must be understood:
Each observer lays the stick down along his or her respective x axis, putting the left end at xℓ (or x’ℓ) and the right end at xr(or x’r).
L0 = xr - xℓ
L’0 = x’r – x’ℓ
Thus using the Lorentz transformations Frank measures the length of the stick in K’as:
Where both ends of the stick must be measured simultaneously, i.e, tr = tℓ
Here Mary’s proper length is L’0 = x’r – x’ℓ
and Frank’s measured length is L = xr – xℓWhat Frank and Mary measure
So Frank measures the moving length as L given by
but since both Mary and Frank in their respective frames measure L’0 = L0 (at rest)
and L0 > L, i.e. the moving stick shrinks.
Taking differentials of the Lorentz transformation, relative velocities may be calculated (dv=0 because we are in inertial systems):
Suppose a shuttle takes off quickly from a space ship already traveling very fast (both in the x direction). Imagine that the space ship’s speed is v, and the shuttle’s speed relative to the space ship is u’. What will the shuttle’s velocity (u) be in the rest frame?
defining velocities as: ux = dx/dt, uy= dy/dt, u’x= dx’/dt’, etc. it is easily shown that:
With similar relations for uy and uz:
In addition to the previous relations, the Lorentz velocitytransformations for u’x, u’y, and u’zcan be obtained by switching primed and unprimed and changing v to –v:
Capt. Kirk decides to escape from a hostile Romulan ship at 3/4c, but the Romulans follow at 1/2c, firing a matter torpedo, whose speed relative to the Romulan ship is 1/3c.
Question: does the Enterprise survive?
vtR = 1/3c
vEg = 3/4c
vRg = 1/2c
vRg = velocity of Romulans relative to galaxy
vtR = velocity of torpedo relative to Romulans
vEg = velocity of Enterprise relative to galaxy
We need to compute the torpedo's velocity relative to the galaxy and compare that with the Enterprise's velocity relative to the galaxy.
Using the Galilean transformation, we simply add the torpedo’s velocity to that of the Romulan ship:
Due to the high speeds involved, we really must relativistically add the Romulan ship’s and torpedo’s velocities:
The Enterprise survives to seek out new worlds and go where no one has gone before…
Time Dilation and Muon Decay
The number of muons detected with speeds near 0.98c is much different (a) on top of a mountain than (b) at sea level, because of the muon’s decay. The experimental result agrees with our time dilation equation.
We need to calculate the time needed by the muon to reach the sea (2000 m)
The life time (t) of the muon is 1.5 *10(-6) s
Thus, in order to know how many muons decay, we need to measure the time on the muon frame (the proper time is the time measured on the frame on which the 2 events happen in the same location, i.e. the muon itself).
From earth: T=(2000 m)/ 0.98 c= 6.8 *10(-6) s
From Muon: Tproper = T/ g = 1.36 *10(-6) s
Twins Mary and Frank at age 30 decide on two career paths: Mary decides to become an astronaut and to leave on a trip 8 lightyears (ly) from the Earth at a great speed and to return; Frank decides to reside on the Earth.
Upon Mary’s return, Frank reasons that her clocks measuring her age must run slow. As such, she will return younger. However, Mary claims that it is Frank who is moving and consequently his clocks must run slow.
Who is younger upon Mary’s return?
Since all observers “see” the same speed of
light, then all observers, regardless of their
velocities, must see spherical wave fronts.
s2 = x2 – c2t2 = (x’)2 – c2 (t’)2 = (s’)2
There are three possibilities for the invariant quantity Δs2:
Consider a source of light (for example, a star) and a receiver
(an astronomer) approaching one another with a relative velocity v.
Length of wave train = cT− vT
Viewers at rest everywhere see the waves with their appropriate frequency and wavelength.
A receding source yields a red-shifted wave, and an approaching source yields a blue-shifted wave.
A source passing by emits blue- then red-shifted waves.
cTThe Relativistic Doppler Effect
And the resulting frequency is:
Source frame is proper time.
Use a + sign for v/c when the source and receiver are receding from each other and a – sign when they’re approaching.
Because physicists believe that the conservation of momentum is fundamental, we begin by considering collisions where there do not exist external forces and thus:
dP/dt = Fext= 0
Frank (fixed or stationary system) is at rest in system K holding a ball of mass m. Mary (moving system) holds a similar ball in system K that is moving in the x direction with velocity v with respect to system K.
pFy = mu0
The change of momentum as observed by Frank is
ΔpF = ΔpFy = −2mu0
In order to determine the velocity of Mary’s ball as measured by Frank we use the velocity transformation equations:
Before the collision, the momentum of Mary’s ball as measured by Frank becomes
For a perfectly elastic collision, the momentum after the collision is
The change in momentum of Mary’s ball according to Frank is
Relativistic momentum (2.48)
The mass is then imagined to increase at high speeds.
p = mrv = mgv
The work W12 done by a force to move a particle from position 1 to position 2 along a path is defined to be
where K1 is defined to be the kinetic energy of the particle at position 1.
For simplicity, let the particle start from rest under the influence of the force and calculate the kinetic energy K after the work is done.
The limits of integration are from an initial value of 0 to a final value of .
The integral in Equation (2.57) is straightforward if done by the method of integration by parts. The result, called the relativistic kinetic energy, is
Equation (2.58) does not seem to resemble the classical result for kinetic energy, K = ½mu2. However, if it is correct, we expect it to reduce to the classical result for low speeds. Let’s see if it does. For speeds u << c, we expand in a binomial series as follows:
where we have neglected all terms of power (u/c)4 and greater, because u << c. This gives the following equation for the relativistic kinetic energy at low speeds:
which is the expected classical result. We show both the relativistic and classical kinetic energies in Figure 2.31. They diverge considerably above a velocity of 0.6c.
We rewrite Equation (2.58) in the form
The term mc2 is called the rest energy and is denoted by E0.
This leaves the sum of the kinetic energy and rest energy to be interpreted as the total energy of the particle. The total energy is denoted by E and is given by
Total Energy =
We square this result, multiply by c2, and rearrange the result.
We use Equation (2.62) for β2 and find
The first term on the right-hand side is just E2, and the second term is E02. The last equation becomes
We rearrange this last equation to find the result we are seeking, a relation between energy and momentum.
Equation (2.70) is a useful result to relate the total energy of a particle with its momentum. The quantities (E2 – p2c2) and m are invariant quantities. Note that when a particle’s velocity is zero and it has no momentum, Equation (2.70) correctly gives E0 as the particle’s total energy.
W = (1.602 × 10−19)(1 V) = 1.602 × 10−19 J
W = (1 e)(1 V) = 1 eV
1 eV = 1.602 × 10−19 J
Mass (12C atom)
Mass (12C atom)
The binding energy is the difference between the rest energy of the individual particles and the rest energy of the combined bound system.