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Transcript
By: Jennifer Doran
What was Known in 1900
• Newton’s laws of
motion
• Maxwell’s laws of
electromagnetism
Contradiction Between Laws
Newton’s Laws
Predicted that the
speed of light should
depend on the motion
of the observer and
the light source
Maxwell’s Laws
Predicted that light in
a vacuum should
travel at a constant
speed regardless of
the motion of the
observer or source
Two Postulates of
Special Relativity
• The laws of physics are the same for all
non-accelerating observers
• The speed of light in a vacuum is constant
for all observers, regardless of motion
Inconsistencies with Classical
Mechanics
• Newton’s Laws state
that v+u=V
• Einstein’s postulate
says that the speed of
light is independent of
the motion of all
observers and
sources.
Classical Lorentz Transformation
x = x' + u t' ; y = y' ; z = z' ; t = t' OR
x' = x – u t ; y = y' ; z = z' ; t = t'
Then v = v' + u
And a = a'
BUT this is only good for u<<v.
To make the transforms relativistic, assume:
x = Γ (x' + u t')
x' = Γ (x – u t)
Finding Γ
A light pulse starts at S at t = 0; S' at t' = 0
So in S, x = ct, and in S‘, x’ = c t', by
Einstein’s second postulate
Then ct = Γ (c t' + u t') = Γ (c + u) t'
And ct' = Γ (c - u) t
Substitute for t', then ct = [Γ(c + u)Γ(c - u)t]/c
Then Γ2 = c2 / (c2-u2) = 1/(1- u2/c2)
Γ = 1/√(1- u2/c2)
Time Dilation
• Δt = Γ•Δtp
• Δtp is the proper time, the time between
events which happen at the same place.
• Since Γ is always greater than 1, all clocks
run more slowly according to an observer
in relative motion.
Time Dilation Example
• An astronaut in a spaceship traveling
away from the earth at u = 0.6c decides to
take a nap. He tells NASA that he will call
them back in 1 hour. How long does his
nap last as measured on earth?
Time Dilation Example
•
•
•
•
Δtp = 1 hour
1 – (u/v)2 = 1 – (0.6)2 = 0.64
Therefore, Γ = √(1/0.64) = 1.25
Δt = 1.25 hours
Length Contraction
• L = (1/Γ)•Lp
• Lp is the length of the object in the
reference frame in which the object is at
rest.
• All observers in motion relative to the
object measure a shorter length, but only
in the direction of motion.
Length Contraction
Length Contraction Example
In the reference frame of a muon traveling
at u = 0.999978c, what is the apparent
thickness of the atmosphere? (To an
observer on earth, the height of the
atmosphere is 100 km.)
Length Contraction Example
•
•
•
•
L = (1/Γ)•Lp
Lp = 100 km
L = 100 km•√(1-(0.999978)2)
L = 0.66 km = 660 m
Relativistic Velocity Addition
• When objects are moving at relativistic
speeds, classical mechanics cannot be
used.
• v = (v' + u) / (1 + v'u/c2)
• v = velocity addition
• v' = velocity of object moving in the
reference frame of u.
• u = motion of object
Relativistic Velocity Addition
Example
• A spaceship moving away from Earth at a
speed of 0.80c fires a missile parallel to its
direction of motion. The missile moves at
a speed of 0.60c relative to the ship. What
is the speed of the missile as measured by
an observer on Earth?
Relativistic Velocity Addition
Example
•
•
•
•
Cannot use classical mechanics.
v is the velocity we are looking for.
u = 0.80c = the velocity of the spaceship
v' = 0.60c = the velocity of the missile in
the reference frame of the spaceship
• v = (v' + u) / (1 + v'u/c2)
• v = (0.6c + 0.8c) / (1 + (0.6c)(0.8c)/c2)
• v = 1.40c/1.48 = 0.95c
Relativistic Velocity Addition
Example
• A star cruiser is moving away from the
planet Mars with a speed of 0.58c and
fires a rocket back towards the planet at a
speed of 0.69c as seen from the star
cruiser. What is the speed of the rocket as
measured by an observer on the planet?
Relativistic Velocity Addition
Example
• v is the velocity we are looking for.
• u = 0.58c = the velocity of the spaceship
• v' = -0.69c = the velocity of the rocket in
the reference frame of the star cruiser
• v = (v' + u) / (1 + v'u/c2)
• v = (0.58c - 0.69c) / (1 + (0.58c)(0.69c)/c2)
• v = -0.11c/0.5998 = -0.18c
• Compared to –0.11c
Twin Paradox
• A clock in motion runs slower than one at
rest, including biological clocks.
• Apparent paradox with one twin staying on
earth and other going on a relativistic
journey.
• Solution
Twin Paradox Example
• Twin sisters Betty and Ann decide to test
the relativity theory. Ann stays on earth.
On January 1, 2000, Betty goes off to a
nearby dwarf star that is 8 light years from
Earth. Betty travels there and back at
0.8c.
Twin Paradox Example
• What is Betty’s travel time to the star
according to Ann?
• t = d/v = 8 light years / 0.8c = 10 years
• So Betty’s total travel time according to
Ann is 20 years.
Twin Paradox Example
• As soon as Betty reaches the star, she
sends her sister an e-mail message
saying, “Wish you were here!” via radio.
When, from Ann’s perspective, does Ann
receive the e-mail from Betty?
• 2010 + 8 yrs = 2018
Twin Paradox Example
• When Ann receives Betty’s e-mail
message, what is the date on it?
• Time dilation Γ = 1/√(1- (0.8c)2/c2)
• So Γ = 0.6c
• 10 years • 0.6c = 6 years have passed
• So the e-mail will have the date of January
2006.
Twin Paradox Example
• How much younger is Betty than Ann
when Betty returns from the star?
• For Betty, her return date is 2012, but for
Ann, she returns in 2020.
• Ann is now 8 years older than Betty.
Twin Paradox Example
Newtonian Gravity
• Quote from Newton
• Two masses in Newton’s theory — inertial
mass and gravitational mass
• Newton saw no reason why these masses
should be equal
• BUT they are!!!
Einstein on Gravity
“I was sitting in a chair in the patent office at
Bern when all of a sudden a thought
occurred to me: if a person falls freely he
will not feel his own weight. This simple
thought made a deep impression on me.
It impelled me toward a theory of
gravitation.”
-Albert Einstein
Principle of Equivalence
• There is no local experiment that can be
done to distinguish between the effects of
a uniform gravitational field in a nonaccelerating inertial frame and the effects
of a uniformly accelerating reference
frame.
Principle of Equivalence
• Free space view is the same as the free
fall view
General Relativity
• The presence of
matter causes
spacetime to warp or
curve.
• Gµv = 8G/c4•Tµv
Schwarzschild’s Solution
• Found the first exact solution of one of
Einstein’s equations
• The metric (distance relation) is:
• ds2 = -c2(1-2MG/c2r)dt2 + dr2/(1-2MG/c2r)
+ r2(dq2+sin2qdf2).
• Describes space-time geometry around a
spherical object of mass M
• Basis of tests of general relativity
Testing the
General Theory of Relativity
• Deflection of
Starlight
• First tested during
total solar eclipse
in 1919
Slowing Down of Clocks by Gravity
• Clock closer to the
mass measures a
shorter elapsed
proper time than a
clock that is further
out.
• Rc=2MG/c2, where Rc
is the Schwarzschild
radius.