Important Terms
Proper
time Dt' : The
length of time of some event as observed by a person at rest with respect to the event. Proper
length LP : The length of an
object |
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 |
Important Equations and Concepts
Length Contraction
L = LP (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 = Dt' / (1 - v2/c2)1/2 If the event is moving with respect |
Mass Transformed into Energy DE = Dmc2 Mass gained or lost is equal Energy E gained or lost is |
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. |
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 or 300m/ms in a vacuum, no matter what the relative speed between source and observer may be. |
Inertial Systems
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Law of
Inertia:
If no net force is acting on an
Inertial reference system: Any place in which Newton’s |
Laws of Physics are the Same in All Inertial Systems
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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 |
Speed of Light in Vacuum is the 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. |
Laser Light Clock
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Time Dilation
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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
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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: Dt' = 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) Dt' = time interval D = cDt' (3) |
Distance traveled
by spacecraft: L = v Dt (4) ---------------------------------- Substituting into (1): c2 (Dt)2 = v2 (Dt)2 + c2 (Dt')2 (5) Solve this for (Dt)2:
Dt = Dt' / (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 Observers at rest with respect to these
|
Dt'= 2.2 ms
Dt = Dt' / (1 - v2/c2)1/2 v / c = 0.99
[1 – (.99)2]1/2 = 0.14 = 2.2 ms / 0.14 = 15.6 ms Earth observer measures the lifetime |
Time Dilation
Question:
The star Alpha Centauri is 4.3 light years away. If
a Answer: The traveler is stationary with respect to the
events |
The earth observer says the trip takes Dt = 4.3 / 0.95 = 4.5 years ----------------------------- Dt = Dt' / (1 - v2/c2)1/2 Rearranging: Dt' = Dt (1 - v2/c2)1/2 = 1.4 years |
The Twin Problem: An Apparent Paradox
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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. |
Length Contraction
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Earth observer measures
LP: The earth observer is at rest with respect to the earth-star line, so he measures the proper length LP. Astronaut measures Dt': 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 Dt'. |
Each observer measures the same relative speed: v = v LP / Dt = L / Dt' L = LP (Dt'/Dt) Dt =Dt'(1- v2/c2)1/2 Substituting: L = LP (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 |
LP = 20 m
L = LP (1 -
v2/c2)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 |
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 |
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 |
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 |