Time Travel Research Center © 2005 Cetin BAL - GSM:+90  05366063183 - Turkey / Denizli 

 

Special Relativity
------------------------------------

 

 


Joseph F. Alward, PhD
Department of Physics
University of the Pacific
        Albert Einstein
         (1879-1955)

 

  Important Terms

Proper time Dt0:  The length of time of
some event as observed by a person at rest
with respect to the event.

Proper length L0:   The length of an object
observed by a person at rest with respect to
the object.

Dilated time Dt:   The length of time of
some event as observed by a person
moving with respect to the event.
 

 Contracted length L:  The length of an
object observed by a person moving
with respect to the object.
----------------------------------------------------
Simple rule:

If there's zero movement relative to the
event or object being measured, use a
subscript zero on the quantities.
----------------------------------------------------
contraction: becoming smaller
dilation: spreading out

 

 

 Important Equations and Concepts

Length Contraction

L = L0 (1-v2 /c2)1/2

If the object is moving with respect
to an observer in an inertial reference
system, it's smaller for that observer.
------------------------------------
Time Dilation

Dt = Dt0 / (1 - v2/c2)1/2

If the event is moving with respect
to an observer in an inertial reference
system, it takes longer to happen for
that observer.

 Mass Transformed
into Energy

DE = Dmc2

Mass gained or lost is equal
to a gain or loss in energy.
--------------------------------------
Energy Transformed
into Mass

Dm = DE / c2

Energy E gained or lost is
equivalent to a gain or loss
of mass.

 Total Energy

E = mc2 / (1 – v2/c2)1/2
-----------------------------
Rest Energy

E0 = mc2
-----------------------------
Kinetic Energy

KE = E – E0

The kinetic energy
is not 1/2 mv2.

 

 

 Einstein's Speed of Light Postulate

If a source of light is not accelerating with respect to
an observer, the speed of light is always
300,000 km/s in a vacuum, no matter what the
relative speed between source and observer may be.

 

   (End preview)

 

  Inertial Systems

Law of Inertia:

If no net force is acting on an
object, it either remains at rest,
or continues to move with
constant velocity.

 

Inertial reference system:

Any place in which Newton’s
law of inertia is valid.

If the airplane is not accelerating,
it’s an inertial reference system.

 

 

 Laws of Physics Same in All Inertial Systems

Both observers say that
the ball’s motion is exactly
as expected from the
acceleration caused by
the pull of the earth.

Both observers see the
ball behaving as Newton’s
laws and gravitational
force predict it should.

 

 Speed of Light in Vacuum Same in All Inertial Systems


                      Both obervers measure the speed of light to be the same; the
                      speed of light is seen to be the same in both inertial systems.

 

 

   Light Clock 

 

 Time Dilation

Astronaut and earth
observer each are in
an inertial system, so
each measures the same
value for the speed of
light, 300,000 km/s.


Which person reports
the longer time interval
between the two events
(departure of light from
source, arrival at the
detector) ?

 

 

 

 Time Dilation Explained: Part One

Each observer sees light
propagate at the same
speed: c = 300,000 km/s
----------------------------------
Since each observer sees the light travel a different distance, each observer measures a different time interval.

Astronaut:  Dt0 = 2D / c

Observer:     Dt = 2s / c

 

 

   Time Dilation Explained:  Part Two



 Pythagoras: s2 = L2 + D2      (1)		 Dt = time interval
        observed by
        earth observer

s = c Dt                  (2)
-----------------------------

Dt0 = time interval
         observed
         by astronaut

D = cDt0                  (3)

 

 Distance traveled by
spacecraft:

L = v Dt                        (4)
----------------------------------
Substituting into (1):

c2 (Dt)2  = v2 (Dt)2 +
                 c2 (Dt0)2       (5)

Solve this for (Dt)2:

 

Dt = Dt0 / (1 - v2/c2)1/2    (6)

 

 

    The Lifetime of a Muon

Muons are unstable particles produced
when protons streaming in from the sun
are absorbed in the atmosphere.  About
one muon strikes each one cm2 of the
earth's surface each second.

Muons travel at a little less than the
speed of light, v = 0.99 c

Observers at rest with respect these
muons measure their average lifetime to
be 2.2 ms; this is the proper time
interval between creation in the upper
atmosphere, and disintegration.

 

 

 Dt0= 2.2 ms

Dt = Dt0 / (1 - v2/c2)1/2

                 v / c = 0.99

                [1 – (.99)2]1/2 = 0.14

Dt = Dt0 / 0.14

    = 2.2 ms / 0.14

    = 15.6 ms

Earth observer measures the lifetime
of the moving muon to be 15.6 ms.

 

 

  Time Dilation

Question:

The star Alpha Centauri 4.3 is light years away. If a
spacecraft could travel at a speed v = 0.95 c, how
long would an earth observer say a trip to that star
would it take? What would the traveler say?

Answer:

The traveler is stationary with respect to the events
which define the time interval: the departure from
earth, and arrival at the star. Thus, the traveler
measures the proper time, Dt0, while the earth
observer measures the dilated time, Dt.

 The earth observer
says the trip takes

Dt  = 4.3 / 0.95

      = 4.5 years

-----------------------------

Dt = Dt0 / (1 - v2/c2)1/2

Rearranging:

Dt0 = Dt (1 - v2/c2)1/2

       = 1.4 years

 

 

 The Twin Problem: An Apparent Paradox

Person on earth
remained in an inertial
reference system.
--------------------------------
The space- traveler
accelerated, and thus
wasn't in an inertial
reference system.
-------------------------------
The special theory of
relativity is valid for the
person on earth, but not
for the space traveler
because he was in
a non-inertial reference
system.(Twin paradox)

 

 Length Contraction

Earth observer measures L0:
The earth observer is at rest
with respect to the earth-star
line, so he measures the
proper length L0.


Astronaut measures  Dt0:

Since the astronaut's clock is
at the beginning and end of the event--at his departure from
earth, and at his arrival at the
star-- he measures the proper
time Dt0.		 Each observer measures
the same relative speed:

v = v

L0 / Dt = L / Dt0

L = L0 (Dt0 /Dt)    

Dt =Dt0(1- v2/c2)1/2            

Substituting:

L = L0 (1- v2 /c2)1/2  

 

 

 

 

  Length Contraction

A rod is 20 meters long when
observed while stationary with
respect to an observer.

How long is it to an observer
moving at a speed v = 0.98 c
with respect to the rod?

 

 L0 = 20 m

L = L0 (1 - v2/c2)1/2
   = 20 [ 1 - (0.98)2 ]1/2

   = 3.98 m

 

 

 

  Time Dilation and Length Contraction


            Both observers measure the same relative speed, v.		 Event begins with birth of muon
and ends with its disintegration.
--------------------------------------------
An observer on earth is moving
with respect to the event, so he
measures the dilated (longer)
time, Dt = 5 x 10-5 s.

Same observer measures the
proper (longer) distance, L0.
--------------------------------------------
Observer riding with the muon
measures the proper (shorter)
time Dt0 = 2 x 10-6 s, and
the contracted (shorter)
distance L.
 

 

 

   Energy-Mass Equivalence 


Mass is energy:  E = mc2   
Energy is mass:  m = E /c2		 When a flashlight radiates a
quantity of light energy, it
loses mass.
---------------------------------------
When a flower absorbs
sunlight, its mass increases.
---------------------------------------
When a uranium nucleus
splits, the mass of the
remnants is less than the
original mass. The difference
appears as light, heat, and
kinetic energy.

 

 

  Energy to Mass Transformation 


  From the sun:  0.10 watts /cm2 		By how much would the
mass of a butterfly increase
after one hour facing the
sun, assuming all of the
energy is absorbed?
--------------------------------------
Assume area of butterfly
is 75 cm2.

Time = 3600 seconds
 		 DE = (0.10)(3600)(75)

    = 27,000 Joules

Dm = DE / c2

c = 3 x 108 m/s

Dm = 3 x 10-13 kg

 

  Mass to Energy Transformation

What is the energy-equivalent
of one gram* of matter?

 E = mc2

    = (0.001 kg)(3.0 x 108 m/s)2  

    = 9 x 1013 J


---------------------------------------------
A penny has a mass of about 3 g.

 One joule is the work done in lifting a one
newton weight one meter.

10,000 newtons = 2500 pounds
  1,000 meters = 3280 feet (0.6 mile)
(10,000 N) (1,000 m) = 107 J
----------------------------------------------------------
About how many one-ton cars could be
blasted about 1/2 mile upward if one
gram of matter were completely converted
to energy?

 

  Total Energy              Rest Energy           Kinetic Energy

What is the total energy of
a particle of mass m = 2 kg
moving with speed v = 0.9 c?
----------------------------------------
E = mc2 / (1-v2/c2)1/2

    = 4.13 x 1017 J

 What is the particle’s rest
energy?

------------------------------------
E0 = mc2

    = 1.80 x 1017 J

 What is the particle’s kinetic
energy?

--------------------------------------
Kinetic Energy = E - E0

KE = 2.33 x 1017 J

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