special relativity remember the michelson and morley experiment in 1887? experiment

Special relativity

Remember the Michelson and Morley experiment in 1887?

• Michleson and Morley thought it was a failure because it could not detect the aether wind, but the most important conclusion of this experiment it:

• The speed of the light is not influenced by the motion of earth (that moves at a considerable speed of ~100,500 kmh= 27,916 m/sec around the sun)

Hendrik Lorentz

• Hendrik Antoon Lorentz (18 July 1853 – 4 February 1928) was a Dutch physicist who won the 1902 Nobel Prize in Physics 

• Hendrik Lorentz was born in Arnhem,  (The Netherlands. He studied physics and mathematics at the University of Leiden, In 1875 Lorentz earned a doctoral degree.   In 1878, only 24 years of age, Lorentz was appointed to the newly established chair in theoretical physics at the University of Leiden.

• In 1895, with the attempt to explain the Michelson-Morley experiment, Lorentz proposed that moving bodies contract in the direction of motion. Lorentz worked on describing electromagnetic phenomena (the propagation of light) in reference frames that moved relative to each other. He discovered that the transition from one to another reference frame could be simplified by using a new time variable which he called local time. Lorentz's publications made use of the term “local time” without giving a detailed interpretation of its physical relevance.

That was the beginning of the theory of relativity …

•

• At the time of Newton and Galileo, the laws of motion were simple. ..

• First law (or law of inertia). Every object continues in its state of rest, or of uniform motion in a straight line, unless compelled to change that state by external forces acted upon it.

• This law does not change if motion occurs in aframe of reference that moves with constant velocity (that is, no acceleration). Such frame is called inertial

V’p=VT+Vp

• If the train moves with constant velocity vT , and a passenger inside the train walks with velocity vp (say, in the direction of the motion of the train) the conductor on the platform will see the passenger speeding by him at velocity v’p= vT+vp.

Δ’s= vT Δt + Δs The displacement of the

passenger inside the train in the interval of time Δt is, (according to the conductor)

Δ’s= v’p Δt = (vT+vp) Δt

So if vp Δt = Δs is the displacement of the passenger inside the train, the relation between Δ’s and Δs is

Δ’s= vT Δt + Δs

Example: a train travels with a constant velocity of 100 km/h. A passenger walks inside the train with velocity of 1 km/h in the direction of the motion of the train. What is the displacement of the passenger according to an observer on the platform of the train in 1 min? How would your answer change if the passenger moves in the opposite direction of the train?

• Answer. According to the conductor, the velocity of the passenger relative to the sidewalk is 101 km/h. So, if Δt= 1 minute= 1/60 hr, the displacement is 1.68 km.

• If the passenger moves in the opposite direction of the train, its velocity relative to the sidewalk is 99 km/h. So his displacement is 99x 1/60= 1.65 km

Example: a train stops in the station, and after loading the passengers, it starts moving with initial velocity of 6 km/h. A passenger is still talking to a friend standing on the sidewalk, and when the train moves, she walks in the train with velocity 2 km/h so to keep talking to her friend. How fast and in which direction should the friend on the sidewalk move so to keep talking to the passenger on the train?

Answer: The problem does not say in which direction the passenger on the train moves, but since she wants to talk to a friend in the station, she has to go toward the station! So her friend will see her moving with velocity v’p= 6-2= 4km/h away from him, and if he wants to keep talking to her, he should run at 4 km/ h in the direction of the train.

To summarize:

• The velocity of a passenger in a train with respect to the sidewalk is

v’p = vT + vp Where vT is the velocity of the train and vp is the velocity

of the passenger in the train.

. The displacement of a passenger with respect to the sidewalk is

Δ’s = vT Δt + Δs.

The same equation are valid also in 2 or 3 dimension. We only need to use vectors instead of scalars .

• In these problems we assume that the time elapsed in the train is the same as the time elapsed in the station. That is,

Δt= time elapsed in the train= Δt’= time elapsed in the station.

You’ve never thought that there is any difference between the time measured on the surface of earth and the time measured inside a moving vehicle, right?

But Michelson and Morrey discovered that the speed of light c is the same if measured in the direction of the motion of earth and in the direction opposite to the motion of earth …

Think about it …

• The earth is like a train carrying a passenger (a beam of light). The speed of light measured in the direction of the motion of the earth should be c+ vE, and in the direction perpendicular to the motion of the earth should be c. But instead it is c both times.

Not quite convinced yet?

The observer on the ground should measure the speed of this wave as c + 15 m/s

But both observers actually measure the speed of this wave as c!

It would be like seeing a passenger in the train moving with the same velocity, regardless of whether the train

leaves the station or comes into the station …

…But if the passenger in the train could run at the speed of light, then the speed of the train won’t make any difference. You will still see her moving at the speed of light regardless of the train’s velocity.

• A better example are the headlight and the tail light of a train in motion . You will see the light travelling at the same speed, regardless if the train comes toward you or goes away from you.

Einstein’s Postulates (1905)• First Postulate

– The laws of physics are the same in any inertial frame of reference (principle of relativity)

• Second Postulate– The speed of light in vacuum is the same in all

inertial frames• That is, the speed of light is c (~3x108 m/s) and is

independent of the motion of the source• For example, If the light comes from the headlight of

a train moving at with velocity v

Newton: speed =c+v Einstein: speed still =c !!!

c

v

• From now on, Δ’t and Δ’s are the time and the displacement of a moving object P on the train (i.e., an inertial frame that moves with constant speed vT ) measured by an observer in the station, (i.e., from a stationary inertial frame)

• and Δt and Δs are the time and the displacement of a moving object measured on the train (i.e., within the moving inertial frame)

• For small velocities,

v’p = vT + vp

that is

Δ’s/ Δ’t= vT + Δs/ Δt

But if P moves at the speed of light c, it should be

Δ’s/ Δ’t = v’p =vT + c =c Δs/ Δt= Δ’s/ Δ’t =c

So, either Δ’t has to change, or Δ’s has to change, or

both have to change.

Time dilation: Δ’t > Δt

That is, time in the station appears to pass slower than on the train, or: an observer at the station that looks at a moving clock on the train measures a longer time on her watch than an observer on the train.

Let’s prove this:

v

observer in the train Observer at the station

light

mirror

both observers agree that light is travelling at speed=cbut they disagree on the distance (path) the light has travelled

S’S

Observer S:

Time taken c

dt

2

speed

distance

d

Observer S’

L L

Vt’

d

The train travels with velocity v. The distance travelled by light is 2LSo: the time taken by the light to come back to the mirror(measured in S’) is

By Pythagoras,

From these equations2

22

2

222

2

'

2

'

2

'

2'

vtd

ctL

vtdL

c

Lt

In S,

2222

222222

2

2'2

2'2

22

)1('

''

2

2

tctc

tvtctc

d

cv

vtctct

Solving both equations for d^2 and equating

22

22

2

ctdd

ct

c

dt

Vt’

d=c/(2t)

Hence, time measured by an observer in S’ (the station)

So, t’ > t because the “ correction factor”

cv

21

1

Hence, we can rewrite time as tt '

Is always >1

2

2

1'

cv

tt

L=c/(2t’)

Example: b-quark decay• The b-quark is an unstable sub-atomic particle.• The b quark travels ~ 4mm at 0.99c so, (=9) before

decaying• When the particle travels, its life span (i.e., as observed

in S’)

4mm

Image reconstructed byDELPHI particle physics

Experiment at CERN

pssms

m

v

st 13

11103.11810399.0

004.0'

(1 pico-s is 1x10-12 s)

However, the average lifetime at rest (i.e., in S)

of a b-quark is t=1.5 psThe discrepancy is explained by the time dilation

pspstt 5.135.19'

How do we measure the length of an object?

If the object is at rest, we just use a ruler …

but if the object appears moving, we should figure out how to measure its leftmost and rightmost points at the same time. For example, we should ask someone else to stop by and be there to help out…

• So, Tom and Mary should run after the moving object with a ruler, and place it on top of the object while still running … but when Tom places the “0” of the ruler to the leftmost point of the object, yells at Mary “I got it”, and she tries to line up the ruler with the rightmost point of the object … the object in question has moved some more … and Mary’s reading won’t be accurate …

So, the best way of measuring the length of a moving object can’t have anything to do with John or Mary…

• We place a light and the mirror on the leftmost point of the object, and another mirror on the rightmost point of the object, and we measure the time that the light takes to bounce back from the source, to the mirror, and back to the other mirror … and we use the relation:

• length’= ½ (time’ x c) (remember, the “primes” mean for us that the length and the time are measured from the station) to estimate the length.

• Once again, time and distances are intertwined … and the consequences are …

• Length’< actual length• i.e: Object in a train appears shorter to an observer S’ at the station than to an observer S in the

train

Note: It Only applies to lengths in direction of travel

lightmirror

Lv

vt’

Observer on the train observer at the stationS’:S:For S:

The time taken for light to bounce back and forth is

c

Lt

2

speed

distance

L’

L is the length measured on the train

lightmirror

Lv

Vt’

S’:S:

L’

For S’ (the observer in the station): the speed of light is still c.

Time taken for light to travel from source to mirror = t1,

Corresponding distance travelled:Hence, solving for t1,

111 ' ctvtLd

vcLtLvtct '

111 '

Length contraction(remember >1)

Time taken for light to travel from mirror to source = t2

Corresponding distance travelledHence, solving for t2

222 ctvtLd

vcLtLvtct '

222 '

So, total time vcL

vcLttt ''

21'

2'2

)1(

'2

'2))((

)(')('

2

2

22

cL

c

L

vccL

vcvcvcLvcL

c

v

But,from time dilation cL

cLtt 22'2'

L

L 'Object in S appears

to shrink by a factor in the direction of travel

to an observer in S’

L

vt2

• Exercise : A rod measures 3 m on a spaceship that travels with a speed of .4c. What is its measure according to an observer on Earth?

• Solution; the observer on Earth sees the length of the rod L contracted by a factor

• 09.1

)4(.1

12

That it, L’= length measured on Earth= L/(1.09), where L is the length of the rod on the spaceship. So, L’=2.74 m.

Lorentz Tranformations

A distance x on S (train) is seen from S’ (station) as x’=x/ . Thus, instead of x’ = x+vt’ , we have

)'('

'/'

vtxx

vtxx

)'(' vtxx )(' 2c

xvtt

'

'

zz

yy

e.g.

•We can also solve this equation in terms of t. We get the following set of transformations

Suppose a shuttle takes off quickly from a space ship already traveling very fast (both in the x direction). Imagine that the space ship’s speed is v (as measured by the sheep’s speedometer) , and the shuttle’s speed relative to the space ship is u. What will the shuttle’s velocity u’ be in the rest frame S’?

Relativistic Velocity Transformation

Answer: According to Galileo, the velocity of the shuttle is u’= v+u. But according to Einstein, things are not so simple …

T he velocity of the shuttle is Δx’/ Δt’. If we us the Lorentz transformations,

)(' 2cxvtt )'(' vtxx

We obtain

)1('

2cuvvuu

You can verify that if u=c, then u’ =c as well.

Cool, uh?

• Example: If one fires a bullet that travels with speed of .3c (relative to the gun) from a vehicle that moves with speed of .5 c (according to the vehicle’s own speedometer) what is the velocity of the bullet as observed from Earth?

• Answ. We just use the formula

where v= .5 c and c= .3 c. All the c’s at the denominator simplify nicely. We get

u’ = c (.3+.5)/(1+.15)=

.69c

)1('

2cuvvuu

V’pg

= velocity of police relative to ground

vbp

= velocity of bullet relative to police

V’og

= velocity of outlaws relative to ground

Vpg

= 1/2c Vog

= 3/4cVbp

= 1/3c

police outlawsbullet

Example: As the outlaws escape in their really fast getaway ship at 3/4c, the police follow in their pursuit car at a mere 1/2c, firing a bullet, whose speed relative to the gun is 1/3c. Question: does the bullet reach its target a) according to Galileo, b) according to Einstein?

In order to find out whether justice is met, we need to compute the bullet's velocity relative to the ground and compare that with the outlaw's velocity relative to the ground.

In the Galilean transformation, the velocity of the bullet relative to the ground is simply the sum of the bullet’s velocity and the police car’s velocity. Since

51 13 2 6

5 36 4

v v v vbg bp pg bg c c c

c c

Therefore, justice is served!

Galileo’s addition of velocities

51 13 2 6

5 36 4

v v v vbg bp pg bg c c c

c c

Therefore, justice is served!Vb’= 1/3 c+ ½ c = 5/6 c > ¾ c. Justice served!

Einstein’s addition of velocitiesDue to the high speeds involved, we really must relativistically add the police ship’s and bullet’s velocities: (sorry, v

5 37 4 !c c justice is not served

1 13 2 5

721 13 2

v1 /bg

c cc

c c c

What is a light year?

What is a light year?

• A light year is the distance travelled by the light in a year (approx.  9.4 × 10^(15) meters) . For example, The nearest known star (other than the Sun), Proxima Centauri, is about 4.22 light-years away.

The twin paradox

The twin paradox

• Tom and Jerry are twin brothers. Tom stays on Earth, and Jerry travels a distance of 10 light years at 80% of the speed of light to get to a nearby star. If Tom and Jerry are 30 when Jerry leaves for his space trip, how old are they when Jerry returns to earth?

• Solution: if the light travels 10 years to cover the distance covered by Jerry on his space ship, him, who travels at 80% of the speed of light, will take 10/.8= 12.5 years 12.5 years each way.each way.

• There and back makes the trip There and back makes the trip

• 12.5 x 2 or 25 years.12.5 x 2 or 25 years.

The gamma factor:

• Velocity (v) is v = 0.8c therefore…

• β = v/c= 0.8

γ = 1

1 - β²1 - β² = 1 - 0.64 = 0.36 and the √ of 0.36 is 0.6 !!

(β² = 0.8² = 0.64)

γ = 1 ÷ 0.6 = 1²/³ = 5/3

Tom’s point of view …

• Tom sees Jerry’s clock is running slow by

γ = 5/3

• Therefore Jerry’s clock reads 25 years ÷ γ

• 25 ÷ 5/3 = 15 years! So When Jerry comes back to Earth, Tom, is 55 and Jerry is 45.

Jerry’s point of viewJerry does not see his clock running slow, but he sees the distance contracted by γ = 5/3 (because, according to point of view, he is “in the station” and his brother is “on the train” !)

So, according to Jerry’s , he travels a distance of 10 ÷ 5/3 = 6 light years each way; 12 years is the time that the light would take to complete the journey, so Jerry takes t = x/v = 12/0.8 = 15 years.

• Exercise. Two twin brothers travel on separate space ships. One travels 10 light years with velocity of .5 c; the other travels the same amount with velocity of .7 c. If the brothers are 30 when they begin their journey, what is their age when they arrive?

• Solution: compute the factor gamma for both trips.

• The first gamma is …

• The second gamma is …

• Then you should compute how long both trips last; the first trip lasts …

• And the second trip lasts …

• Then you have to measure how much the time contracts for each traveler by dividing the duration of each trip by gamma.

• You get …

• Which is how much each twin has aged during the trip.