Theory of Relativity Albert Einstein Physics 100 Chapt 18

watching a light flash go by v c The man on earth sees c = (& agrees with Maxwell) 2kk

watching a light flash go by v c If the man on the rocket sees c-v, he disagrees with Maxwell

Do Maxwell’s Eqns only work in one reference frame? If so, this would be the rest frame of the luminiferous Aether.

If so, the speed of light should change throughout the year upstream, light moves slower downstream, light moves faster “Aether wind”

Michelson-Morley No aether wind detected: 1907 Nobel Prize

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.

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

Even when the light flash is traveling in an opposite direction Both guys see the light flash travel past with velocity = c

Gunfight viewed by observer at rest He sees both shots fired simultaneously Bang! Bang!

Viewed by a moving observer

Viewed by a moving observer He sees cowboy shoot 1st & cowgirl shoot later Bang! Bang!

Viewed by an observer in the opposite direction

Viewed by a moving observer He sees cowgirl shoot 1st & cowboy shoot later Bang! Bang!

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

Light clock

Seen from the ground

Events y (x2,t2) (x1,t1) x x x1 t x2 x

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

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

How are the times seen by 2 different observers related? We can figure this out with simple HS-level math ( + a little effort)

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

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

Flash a light on a rocket ship Event 2: light flash reaches the girl wt0 c= w t0 Event 1: boy flashes the light

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=?

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

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 – 0.0001 1 1 – (0.01)2 1 0.9999 = = g = g = 1.00005

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 = 1.005

Let’s make a chart v g =1/(1-v2/c2) 0.01 c 1.00005 0.1 c 1.005

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

Enter into chart v g =1/(1-v2/c2) 0.01 c 1.00005 0.1 c 1.005 0.5c 1.15

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

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

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

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

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

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

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

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

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!!!

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 

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!!!

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

Plot results:  1 1 – v2/c2 g = Never-never land x x x x x v=c

Moving clocks run slower t0 1 1 – v2/c2 t = t0 t t = g t0 g >1  t > t0

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

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

Length contraction

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

Relativistic mass increase m0 = mass of an object when it is at rest  “rest mass” mass of a moving object increases g as vc, m m = g m0 as an object moves faster, it gets harder & harder to accelerate by the g factor v=c

summary By a factor of g Moving clocks run slow Moving objects appear shorter Moving object’s mass increases By a factor of g

Plot results:  1 1 – v2/c2 g = Never-never land x x x x x v=c

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

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

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

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

So, according to the boy: d=4.3 cyr v=0.9c v=0.9c his clock her clock out: 4.8yrs 2.1yrs back: 4.8yrs 2.1yrs She ages less total: 9.6yrs 4.2yrs

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

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!!

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

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

Gunfight viewed by observer at rest He sees both shots fired simultaneously Bang! Bang!

Viewed by a moving observer

Viewed by a moving observer He sees cowboy shoot 1st & cowgirl shoot later Bang! Bang!

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

No paradox: both twins agree The twin that “turned around” is younger

Ladder & Barn Door paradox Stan & Ollie puzzle over how to get a 2m long ladder thru a 1m wide barn door ??? 1m 2m ladder

Ollie remembers Phys 100 & the theory of relativity Stan, pick up the ladder & run very fast 1m tree 2m ladder

View from Ollie’s ref. frame Push, Stan! 1m 2m/g V=0.9c (g=2.3) Stan Ollie

View from Stan’s ref. frame But it doesn’t fit, Ollie!! 1m/g V=0.9c (g=2.3) 2m Ollie Stan

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

Fermilab proton accelerator V=0.9999995c g=1000 2km

Stanford electron accelerator v=0.99999999995 c 3km g=100,000

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.

test g=29.3