Special relativity Part II

Recall Einstein’s Postulates (1905) First Postulate –The laws of physics are the same in any inertial frame of reference (principle of relativity)

Recall Einstein’s Postulates (1905) Second Postulate The speed of light in vacuum is the same in all inertial frames That is, the speed of light is c (~3x10 8 m/s) and is independent of the motion of the source

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 v T ) 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)

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=c but they disagree on the distance (path) the light has travelled S’S Observer S: Time taken d

Observer S’ L L Vt’ d The train travels with velocity v. The distance travelled by light is 2L So: the time taken by the light to come back to the mirror (measured in S’) is By Pythagoras, From these equations In S, Solving both equations for d^ 2 and equating

Vt’ d=c/(2t) Hence, time measured by an observer in S’ (the station) So, t’ > t because the “ correction factor” Hence, we can rewrite time as Is always >1 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 by DELPHI particle physics Experiment at CERN (1977) (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 ps The discrepancy is explained by the time dilation

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 go from the source, to the mirror, and bounce 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’< 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 light mirror L v vt’ Observer on the train observer at the station S’:S: For S: The time taken for light to bounce back and forth is L’ L is the length measured on the train

light mirror L v 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 t 1,

Length co n traction (remember  >1) Time taken for light to travel from mirror to source = t 2 Corresponding distance travelled Hence, solving for t 2 So, total time But,from time dilation Object in S appears to shrink by a factor  in the direction of travel to an observer in S’ L vt 2

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 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/ . To summarize: we have proved that the time and the distance in a frame that moves with constants velocity v =(v, 0, 0) are related to the time and the distance measured from the the stationary frame by the relation: ∆t’= ɤ ∆t and ∆x’= ∆x/ ɤ. So, if we are measuring the segment 0x=x in a moving inertial frame from the origin of a stationary inertial frame, Galileo would obtain x’= x+vt’. (t’ is the time as measured by the observer in the stationary frame!). Lorentz instead gets e.g.

It is natural to assume that also the time t and t’ are related through a linear expression that involve also the distance x. That is, we assume that t’= ɤ t + a x, where a is a constant to be determined…

How to determine the constant a? Since the speed of light is constant in every inertial frame, the measure of the speed of a ray of light that shoots from the origin of the moving frame is the same in the stationary frame C= x/t= x’/t’

Using this identity, we can evaluate the constant a and derive the transformation:

From which follows that:

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, And we let u= Δx / Δt, we obtain You can verify that if u=c, then u’ =c as well.

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

V’ pg = velocity of police relative to ground v bp = velocity of bullet relative to police V’ og = velocity of outlaws relative to ground V pg = 1/2c V og = 3/4cV bp = 1/3c police outlaws bullet 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 Galileo’s addition of velocities V b ’= 1/3 c+ ½ c = 5/6 c > ¾ c. Justice served!

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

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

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?

12.5 years each way.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 each way. There and back makes the tripThere and back makes the trip 12.5 x 2 or 25 years x 2 or 25 years.

The gamma factor: Velocity (v) is v = 0.8c therefore… β = v/c= 0.8 γ =γ = β² 1 - β² = = 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 view Jerry 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.