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Theory of Relativity Albert Einstein Physics 100 Chapt 18
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watching a light flash go by
v c The man on earth sees c = (& agrees with Maxwell) 2kk
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watching a light flash go by
v c If the man on the rocket sees c-v, he disagrees with Maxwell
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Do Maxwell’s Eqns only work in one reference frame?
If so, this would be the rest frame of the luminiferous Aether.
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If so, the speed of light should change throughout the year
upstream, light moves slower downstream, light moves faster “Aether wind”
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Michelson-Morley No aether wind detected: 1907 Nobel Prize
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Einstein’s hypotheses:
1. The laws of nature are equally valid in every inertial reference frame. Including Maxwell’s eqns 2. The speed of light in empty space is same for all inertial observers, regard- less of their velocity or the velocity of the source of light.
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All observers see light flashes go by them with the same speed
No matter how fast the guy on the rocket is moving!! c Both guys see the light flash travel with velocity = c
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Even when the light flash is traveling in an opposite direction
Both guys see the light flash travel past with velocity = c
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Gunfight viewed by observer at rest
He sees both shots fired simultaneously Bang! Bang!
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Viewed by a moving observer
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Viewed by a moving observer
He sees cowboy shoot 1st & cowgirl shoot later Bang! Bang!
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Viewed by an observer in the opposite direction
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Viewed by a moving observer
He sees cowgirl shoot 1st & cowboy shoot later Bang! Bang!
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Time depends of state of motion of the observer!!
Events that occur simultaneously according to one observer can occur at different times for other observers
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Light clock
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Seen from the ground
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Events y (x2,t2) (x1,t1) x x x1 t x2 x
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Same events, different observers
Prior to Einstein, everyone agreed the distance between events depends upon the observer, but not the time. Same events, different observers y’ y’ y (x2,t2) (x1,t1) x x (x1’,t1’) (x2’,t2’) t’ t’ x1’ x1’ x2’ dist’ x’ x’ x1 t x2 x dist
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Time is the 4th dimension
Einstein discovered that there is no “absolute” time, it too depends upon the state of motion of the observer Einstein Space-Time Newton Space & Time completely different concepts 2 different aspects of the same thing
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How are the times seen by 2 different observers related?
We can figure this out with simple HS-level math ( + a little effort)
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Catch ball on a rocket ship
Event 2: girl catches the ball wt v= =4m/s w=4m t=1s Event 1: boy throws the ball
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Seen from earth Location of the 2 events is different Elapsed time is
V0=3m/s V0=3m/s Location of the 2 events is different Elapsed time is the same The ball appears to travel faster d=(3m)2+(4m)2 =5m w=4m v0t=3m dt v= = 5m/s t=1s
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Flash a light on a rocket ship
Event 2: light flash reaches the girl wt0 c= w t0 Event 1: boy flashes the light
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Seen from earth Speed has to Be the same Dist is longer Time must be
V V Speed has to Be the same Dist is longer Time must be longer d=(vt)2+w2 w vt c= = dt (vt)2+w2 t t=?
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How is t related to t0? t= time on Earth clock
t0 = time on moving clock wt0 c = (vt)2+w2 t c = ct = (vt)2+w2 ct0 = w (ct)2 = (vt)2+w2 (ct)2 = (vt)2+(ct0)2 (ct)2-(vt)2= (ct0)2 (c2-v2)t2= c2t02 1 1 – v2/c2 c2 c2 – v2 t2 = t02 t2 = t02 1 1 – v2/c2 t = t0 t = g t0 this is called g
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Properties of g = 1 1 – v2/c2 Suppose v = 0.01c (i.e. 1% of c) g = =
1 – (0.01c)2/c2 1 1 – (0.01)2c2/c2 g = = 1 1 – 1 1 – (0.01)2 1 0.9999 = = g = g =
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Properties of g = (cont’d)
1 1 – v2/c2 Suppose v = 0.1c (i.e. 10% of c) 1 1 – (0.1c)2/c2 1 1 – (0.1)2c2/c2 g = = 1 1 – 0.01 1 1 – (0.1)2 1 0.99 = = g = g =
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Let’s make a chart v g =1/(1-v2/c2) 0.01 c 0.1 c 1.005
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Other values of g = 1 1 – v2/c2 Suppose v = 0.5c (i.e. 50% of c) g =
1 – (0.5c)2/c2 1 1 – (0.5)2c2/c2 g = = 1 1 – (0.25) 1 1 – (0.5)2 1 0.75 = = g = g = 1.15
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Enter into chart v g =1/(1-v2/c2) 0.01 c 1.00005 0.1 c 1.005 0.5c
1.15
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Other values of g = 1 1 – v2/c2 Suppose v = 0.6c (i.e. 60% of c) g =
1 – (0.6c)2/c2 1 1 – (0.6)2c2/c2 g = = 1 1 – (0.36) 1 1 – (0.6)2 1 0.64 = = g = g = 1.25
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Back to the chart v g =1/(1-v2/c2) 0.01 c 1.00005 0.1 c 1.005 0.5c
1.15 0.6c 1.25
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Other values of g = 1 1 – v2/c2 Suppose v = 0.8c (i.e. 80% of c) g =
1 – (0.8c)2/c2 1 1 – (0.8)2c2/c2 g = = 1 1 – (0.64) 1 1 – (0.8)2 1 0.36 = = g = g = 1.67
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Enter into the chart v g =1/(1-v2/c2) 0.01 c 1.00005 0.1 c 1.005 0.5c
1.15 0.6c 1.25 0.8c 1.67
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Other values of g = 1 1 – v2/c2 Suppose v = 0.9c (i.e.90% of c) g = =
1 – (0.9c)2/c2 1 1 – (0.9)2c2/c2 g = = 1 1 – 0.81 1 1 – (0.9)2 1 0.19 = = g = g = 2.29
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update chart v g =1/(1-v2/c2) 0.01 c 1.00005 0.1 c 1.005 0.5c 1.15
1.25 0.8c 1.67 0.9c 2.29
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Other values of g = 1 1 – v2/c2 Suppose v = 0.99c (i.e.99% of c) g =
1 – (0.99c)2/c2 1 1 – (0.99)2c2/c2 g = = 1 1 – 0.98 1 1 – (0.99)2 1 0.02 = = g = g = 7.07
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Enter into chart v g =1/(1-v2/c2) 0.01 c 1.00005 0.1 c 1.005 0.5c
1.15 0.6c 1.25 0.8c 1.67 0.9c 2.29 0.99c 7.07
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Other values of g = 1 1 – v2/c2 Suppose v = c g = = = = g = g =
1 – (c)2/c2 1 1 – c2/c2 g = = 1 0 1 1 – 12 1 = = g = g = Infinity!!!
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update chart v g =1/(1-v2/c2) 0.01 c 1.00005 0.1 c 1.005 0.5c 1.15
1.25 0.8c 1.67 0.9c 2.29 0.99c 7.07 1.00c
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Other values of g = -0.21 1 1 – v2/c2 Suppose v = 1.1c g = = = = g =
1 – (1.1c)2/c2 1 1 – (1.1)2c2/c2 g = = 1 1-1.21 1 1 – (1.1)2 1 -0.21 = = g = g = ??? Imaginary number!!!
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Complete the chart v g =1/(1-v2/c2) 0.01 c 1.00005 0.1 c 1.005 0.5c
1.15 0.6c 1.25 0.8c 1.67 0.9c 2.29 0.99c 7.07 1.00c Larger than c Imaginary number
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Plot results: 1 1 – v2/c2 g = Never-never land x x x x x v=c
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Moving clocks run slower
t0 1 1 – v2/c2 t = t0 t t = g t0 g >1 t > t0
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Length contraction v time=t L0 = vt Shorter! man on rocket =vt/g =L0/g
Time = t0 =t/g Length = vt0 =vt/g =L0/g
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Moving objects appear shorter
Length measured when object is at rest L = L0/g g >1 L < L0 V=0.9999c V=0.1c V=0.86c V=0.99c
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Length contraction
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mass: m = g m0 time F=m0a = m0 t0 a time=t0 Ft0 m0 Ft0 m0 = g Ft0 Ft
change in v time F=m0a = m0 t0 a time=t0 m0 Ft0 m0 change in v = Ft0 change in v m0 = mass increases!! g Ft0 change in v Ft change in v = g m0 m = = m = g m0 t=gt0 by a factor g
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Relativistic mass increase
m0 = mass of an object when it is at rest “rest mass” mass of a moving object increases g as vc, m m = g m0 as an object moves faster, it gets harder & harder to accelerate by the g factor v=c
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summary By a factor of g Moving clocks run slow
Moving objects appear shorter Moving object’s mass increases By a factor of g
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Plot results: 1 1 – v2/c2 g = Never-never land x x x x x v=c
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Twin paradox a-centauri 4.3 light years Twin brother & sister
She will travel to a-centauri (a near- by star on a special rocket ship v = 0.9c He will stay home & study Phys 100
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Light year distance light travels in 1 year dist = v x time = c yr
1cyr = 3x108m/s x 3.2x107 s = 9.6 x 1015 m We will just use cyr units & not worry about meters
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Time on the boy’s clock v=0.9c v=0.9c d0 v tout = = d0 v tback = =
d0=4.3 cyr v=0.9c v=0.9c According to the boy & his clock on Earth: d0 v 4.3 cyr 0.9c tout = = = 4.8 yrs d0 v 4.3 cyr 0.9c tback = = = 4.8 yrs ttotal = tout+tback = 9.6yrs
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What does the boy see on her clock?
d=4.3 cyr v=0.9c v=0.9c According to the boy her clock runs slower tout g 4.8 yrs 2.3 tout = = = 2.1 yrs tback g 4.8 yr 2.3 tback = = = 2.1 yrs ttotal = tout+tback = 4.2yrs
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So, according to the boy:
d=4.3 cyr v=0.9c v=0.9c his clock her clock out: 4.8yrs yrs back: 4.8yrs yrs She ages less total: 9.6yrs yrs
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But, according to the girl, the boy’s clock is moving &, so, it must be running slower
v=0.9c According to her, the boy’s clock on Earth says: tout g 2.1 yrs 2.3 tout = = = 0.9 yrs tback g 2.1 yrs 2.3 tback = = = 0.9 yrs v=0.9c ttotal = tout+tback = 1.8yrs
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Her clock advances 4.2 yrs & she sees his clock advance only 1.8 yrs,
A contradiction?? She should think he has aged less than her!!
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Events in the boy’s life:
As seen by her As seen by him She leaves 0.9 yrs 4.8 yrs She arrives & starts turn short time ???? Finishes turn & heads home 0.9 yrs 4.8 yrs 9.6+ yrs 1.8 + ??? yrs She returns
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turning around as seen by her
According to her, these 2 events occur very,very far apart from each other He sees her finish turning He sees her start to turn Time interval between 2 events depends on the state of motion of the observer
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Gunfight viewed by observer at rest
He sees both shots fired simultaneously Bang! Bang!
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Viewed by a moving observer
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Viewed by a moving observer
He sees cowboy shoot 1st & cowgirl shoot later Bang! Bang!
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In fact, ???? = 7.8+ years 0.9 yrs 4.8 yrs 7.8+ yrs short time ???
as seen by him as seen by her She leaves 0.9 yrs 4.8 yrs She arrives & starts turn 7.8+ yrs short time ??? Finishes turn & heads home 0.9 yrs 4.8 yrs 9.6+ yrs 1.8 + ???yrs 9.6+ yrs She returns
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No paradox: both twins agree
The twin that “turned around” is younger
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Ladder & Barn Door paradox
Stan & Ollie puzzle over how to get a 2m long ladder thru a 1m wide barn door ??? 1m 2m ladder
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Ollie remembers Phys 100 & the theory of relativity
Stan, pick up the ladder & run very fast 1m tree 2m ladder
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View from Ollie’s ref. frame
Push, Stan! 1m 2m/g V=0.9c (g=2.3) Stan Ollie
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View from Stan’s ref. frame
But it doesn’t fit, Ollie!! 1m/g V=0.9c (g=2.3) 2m Ollie Stan
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If Stan pushes both ends of the ladder simultaneously, Ollie sees the two ends move at different times: Too late Stan! Too soon Stan! 1m clunk clank V=0.9c (g=2.3) Stan Ollie Stan
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Fermilab proton accelerator
V= c g=1000 2km
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Stanford electron accelerator
v= c 3km g=100,000
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status Einstein’s theory of “special relativity” has been carefully tested in many very precise experiments and found to be valid. Time is truly the 4th dimension of space & time.
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test g=29.3
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