Physics 2203, Fall 2012 Modern Physics

Transcription

1 Physics 2203, Fall 2012 Modern Physics. Wednesday, Aug. 22, 2012: Ch. 1: Time dila?on, length contrac?on, and transforma?ons. Lorentz Transforma?on In Class exercise Announcements:. Monday s notes posted. We have a tutor: Mohammad Saghayezhian: What?me?

2 Are the laws of physics the same in the two iner3al frames of reference S and S? Yes, according to Newtonian Mechanics. We did this in 2101 many 3mes If we pick an iner?al frame moving with v 0, than the object falls straight done. F=ma

3 u ' Take a careful look at Newton s laws in S and S. u = u '+ v First Law: Law of iner?a, F=0 then velocity is constant. In S the velocity is u = u '+ v In S the velocity is u ' = u v OK! Second Law: F=ma and F =m a Calibrate force measurement (spring): F=F Balance for m: m=m a = d u dt and a ' = d u ' dt ' Assump3on of Newtonian Mechanics t=t

4 Maxwell s equa?on for electromagne?sm state that light in a vacuum moves with the speed of light c=2.99,792,458 m/s (3x10 8 m/s) in every direc3on. BUT, the simple picture below shows that in a moving frame S the speed on light is c+v coming from the right and c v coming from the lea. This means that Maxwell s equa?ons have to be different in the two frames of references. It was accepted by the end of the 19 th century that there was only one ether frame, where the speed of light was the same in all direc?ons. Michelson set out to measure the speed of light in the ether frame. Lorentz developed a transforma?on that made Maxwell s equa?on work in S.

5 Assume arm lengths are equal =L. t 1 = L c + v + t 2 = Δt = t 1 t 2 = L c v = 2L v 2 2Lc v 2 2Lc v 2 2L v 2 Everyone assumed that light traveled in some medium called ether. Michelson and Morely set out to measure the change in the speed of light in a moving reference frame (wrt) ether c ± v. c=3 x 10 8 m/s v ~ 3 x 10 4 m/s velocity of earth wrt sun Define β= v c Δt = 2L c 1 1 β β 2 Binomial Expansion ( 1+ x) n 1+ nx x 1 Δt = 2L c 1+ β 2 1 β 2 = Lβ 2 2 c

6 They did not know which direc?on the Ether wind was blowing so they rotated the interferometer by 90 o Rotation: λ=590 nm L=11 m ΔN 2Lβ 2 λ The?me difference for light going down leg 1 compared to 2 is Δt = Lβ 2 c The period of the light is T=λ/c. If Δt=(n+1/2)T waves out of phase, Int.=0 If Δt=nT waves in phase, Int.=4 As Δt changes the eye sees bright dark modula?ons Number of Fringes N 0.4 N= Δt T = cδt λ Lβ 2 λ They could detect 0.1 of a fringe: Saw Nothing No ether This is a hard experiment 0.1 fringes ~6x10 9 m Thermal expansion 11x10 6 /K What is ΔΤ? Come with your answer Friday

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8 Does the Michelson Morley experiment show that the either does not exist of that it is merely unnecessary?

9 If S is an iner?al frame and if a second frame S moves with constant velocity rela?ve to S, then S is also an iner?al frame. OR the laws of physics are invariant as we change from one reference frame to another iner?al frame. In all iner?al frames, light travels through the vacuum with the same speed, c=299,792,458 m/s in any direc?on. OR the universality of the speed of light. Consistent with Michelson Morley experiment and with Maxwell s equa?ons.

10 We have to be very careful and define how we will measure?me in reference frame S. Here is one solu?on. Posi?on clocks (or people with clocks) everywhere on or x,y,z grid. This will enable us to determine (x,y,z,&t) for any event in S. Always make sure you have a proper?me measurement as a reference:

11 S frame of reference: Δt=distance/velocity Δt =2h/c S frame of reference: Δt=distance/velocity Δt ' Δt Δt ' Δt γ = Δt= 1 1 β 2 Δt ' 1 β 2 cδt 2 2 Solving for Δt Δt= 2h c = h 2 + vδt β 2 β= v c This is not smoking mirrors but real Iden?cal clocks one in space Half life of radioac?ve par?cles Twin Paradox

12 Δt= Δt ' 1 β 2 f=2.3786x10 10 Hz 1) Accurate atomic clocks In 1971 four Atomic clocks were flown around the world in jet planes. The predic?on of the?me difference using the equa?on above was 275±21 ns. The measurement gave 273±7 ns. 2) Radio ac?ve decay N(t) = N 0 e λt Half-life N(T 1/2 )= N 0 2 T 1/2 = ln 2 λ In 1976, experiments with muons were conducted at CERN in Geneva. A muon has the charge of an electron but a mass 207?mes that of the electron. The muons were accelerated to v=0.9994c and the decay rate was measured and compared to the decay when the muons were at rest. T 1/2 = T ' 1/2 1 β 2 T 1/2 = 28.9T ' 1/2

13 N(t) = N 0 e λt Half-life N(T 1/2 )= N 0 2 T 1/2 = ln 2 λ How many of the original N 0 are lea aaer a?me t? N(t) = N 0 e λt write t=nt 1/2 N(t) = N 0 e λnt 1/2 = N ( 0 e λt ) n 1/2 N(t) = N 0 2 n

14 Δt= Δt ' 1 β 2 According to Frank s understanding of modern physics, Mary s biological clock?cks slower that his, so he claims that Mary will return to earth younger than he. Mary on the other hand says Frank is moving away from her frame of reference and when she returns he will be younger. Symmetry would argue that they will be the same age when she returns. What is the correct answer????

15 Δt= Δt ' Einstein s original statement of the twins paradox in 1911: If we 1 β 2 placed a living organism in a box... One could arrange that the organism, aaer an arbitrary lengthy flight, could be returned to its original spot in a scarcely altered condi?on, while corresponding organisms which had remained in their original posi?ons had long since given way to a new genera?ons. Answer to Twin Paradox: There two cases are not symmetric. Frank is in an iner?al frame of reference but Mary is not. She must accelerate to get going and then decelerate and turn around. Mary will be younger!

16 Problem Muons are elementary par?cles with a (proper) life?me of 2.2 µs. They are produced with a very high speeds in the upper atmosphere when cosmic rays collide with air molecules. Take the height L 0 of the atmosphere to be 100 Km in the reference frame of the Earth, and find the minimum speed that enables the muons to survive the journey to the surface of the Earth? Define the birth and decay of a muon as the?cks of a clock in the frame of reference of the muon. Proper measurement of 3me. The clock is moving with respect to the frame of reference of the earth. If the muon is moving with a speed ~c, then the?me it takes to reach the earth is Δt = L 0 c = 100km 3x10 8 m / s = 333µs Δt = Δt 0 1 u2 = 2.2µs 1 u2 = 333µs Solution: u = c

17 Problem #1.26 A group of π mesons (pions) is observed traveling at a speed of 0.8 c in the physics laboratory. a) What is the factor γ for the pions? b) If the pions proper half life is 1.8x10 8 s, what is their half life as observed in the lab frame? c) If there were ini?ally 32,000 pions, how many will be lea aaer they have traveled 36m? d) What would be the answer to (c) if one ignored?me dila?on? γ = a) 1 1 β 2 β=0.8 b) t 1/2 (lab) = γt 1/2 (rest) = 5 3 x1.8x10 8 = 3.0x10 8 s γ = 5 3 N(t) = N 0 e λt Half-life N(T 1/2 )= N 0 2 T 1/2 = ln2 λ c) T (lab) = d 0.8c = c = 1.5x10 7 s T (lab) = 5t 1/2 (lab) Number left N= N = 1000 d) T = 8.33t 1/2 N = N 0 =

18 Lets measure the length of the railroad car in S and S. The car is moving with velocity v. The observer in S (Q) watches the train go by and measures the?me Δt. So he says the car has a length L=vΔt Length Contraction L = L '( 1 β 2 ) = L ' γ L L' L' is proper length = L 0 L = L ( 0 1 β 2 ) = L 0 γ L L 0 The observer in S watches Q move to the lea with velocity v. He measures the?me it takes for Q to go from the right to lea Δt. So L =vδt But we know the equa?on for?me dila?on, which gives v = L Δt = L ' Δt ' : Δt ' = Δt 1 β 2 = γδt Always make sure you know the proper Length!

19 This just had to happen so that we can reconcile the changes in half life in different moving coordinate systems. Muon Decay. In the frame of the moving Muon the halflife is the true half life, 2.2µs. But the height of the mountain in the Earth s frame has been contracted to 650 m. In the Earth s frame of reference the height of the mountain is the proper height, 4700 m. But the half life has been dilated to 16µs.

20 Ques?on on Home work Explain in your own words the terms 3me dila3on and length contrac3on.

21 L You do this. Prove that L=L. Hint Δt=Δt =0 L γ = 1 1 v 2 / = 3.2 Example: A triangle shaped spaceship flies by an observe on the earth at 0.950c. When the ship s size is measured in S (moving at v) the height y is 25. m and the length L=50m. What is the shape of the ship seen by the observer?? Answer: y =y=25m. L contracts. L(S)=50/γ = 15.6m

22 You are in the rail road car (S ) moving to the right with velocity v. You have a He/Ne Laser whose wave length is 594 nm. I will set on the ground in frame S and measure the frequency and wavelength of the light from your laser. Since the laser is at rest in your coordinate system you can make a proper measurement of the period T =T 0 and of the wavelength λ =λ 0 =594 nm. What is Wrong? Tell me Friday. We know that I will see?me dila?on so the period in S is. T = γ T 0 We know that I will see length contrac?on so in S,. λ = λ 0 / γ Everybody agree??? Let us calculate the speed of the wave in S. Velocity = λ T = λ 0 γ 2 = cγ 2 = c1 v2 T 0

23 Lets do this using what we know about rela?vity. Assume a small explosion occurs in the car at P, marking the wall. In S we know that this occurred at a posi?on x, y, z, t. In the S system this explosion occurred at x, y, z, t, i.e. at P, t. The x value is the distance from the end of the car to P. In S this is contracted. x vt = x' γ x' = γ ( x vt) How do we transform what we know in S (x, y, z, t ) into S (x, y, z, t)? Classically this is known as a Galilean Transforma3on. This is what we did in x' = x vt y' = y z' = z t ' = t x = x'+ vt ' y = y' z = z' t = t ' x' = γ ( x vt) y' = y z' = z t ' =?

24 How do we transform t into t? x' = γ ( x vt) y' = y z' = z t ' =? x = γ ( x'+ vt ') y = y' z = z' t =? Subs?tute x in first box into x equa?on in box #2 x γ = γ x vt ( ) + vt ' ( ) t ' = γ t + x v ( 1 γ 2 ) γ 2 1 γ 2 ( ) γ 2 = β 2 x' = γ ( x vt) y' = y z' = z t ' = γ t vx x = γ ( x'+ vt ') y = y' z = z' t = γ t '+ vx'

25 x' = γ ( x vt) y' = y z' = z t ' = γ u' x = u' y = u' z t vx u x v 1 vu x u y γ 1 vu x u z γ 1 vu x Δx' = γ ( Δx vδt) Δy' = Δy Δz' = Δz Δt ' = γ u x = u y = u z Δt vδx u ' x + v 1 + vu ' x u ' y γ 1 + vu ' x u ' z γ 1 + vu ' x u ' x = Δx' Δt ' = u ' y = Δy' Δt ' = u ' z = Δz' Δt ' = ( Δx vδt ) γ γ Really Strange! Δt vδx Δy Δt vδx Δz Δt vδx