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Physics 334 Modern Physics Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on the textbook.

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Presentation on theme: "Physics 334 Modern Physics Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on the textbook."— Presentation transcript:



3 4-1 Foundations of Special Relativity The Experimental Basis of Relativity Einstein’s Postulates 4-2 Relationship between Space and Time The Lorentz Transformation Time Dilation and Length Contraction The Doppler Effect The Twin Paradox and Other Surprises Special Theory of Relativity 1 CHAPTER 4 Special Theory of Relativity 1 It was found that there was no displacement of the interference fringes, so that the result of the experiment was negative and would, therefore, show that there is still a difficulty in the theory itself… - Albert Michelson, 1907 Albert Michelson ( )

4 Newtonian (Classical) Relativity Newton’s laws of motion must be implemented with respect to (relative to) some reference frame. A reference frame is called an inertial frame if Newton’s laws are valid in that frame. Such a frame is established when a body, not subjected to net external forces, moves in rectilinear motion at constant velocity. x z y x’x’ z’z’ y’y’

5 Difference Between Inertial and Non-Inertial Reference Frame


7 The Galilean Transformation For a point P: In one frame S: P = (x, y, z, t) In another frame S ’ : P = (x ’, y ’, z ’, t ’ ) 1. Parallel axes 2. S’ has a constant relative velocity (here in the x-direction) with respect to S. 3. Time (t) for all observers is a Fundamental invariant, i.e., it’s the same for all inertial observers. The Inverse Relations


9 Michelson-Morley experiment Michelson and Morley realized that the earth could not always be stationary with respect to the aether. And light would have a different path length and phase shift depending on whether it propagated parallel and anti- parallel or perpendicular to the aether. Supposed velocity of earth through the aether Parallel and anti-parallel propagation Perpendicular propagation

10 Michelson-Morley Experimental Analysis The Earth’s orbital speed is: v = 3 × 10 4 m/s, and the interferometer size is: L = 1.2 m, So the time difference becomes: 8 × 10 −17 s, which, for visible light, is a phase shift of: 0.2 rad = 0.03 periods Recall that the phase shift is  times this relative delay: or: Exercise 4-1: Show that the time difference between path differences after 90 ° rotation is given by: The Michelson interferometer should’ve revealed a fringe shift as it was rotated with respect to the aether velocity. MM expected 0.4 of the width of a fringe, and could only see 0.01 equal to the uncertainty in the measurement. Interference fringes showed no change as the interferometer was rotated. Thus, aether seems not to exist!




14 The constancy of the speed of light

15 Lorentz Transformation Exercise 4-2: The equations for a spherical wavefronts in S is x 2 +y 2 +z 2 =c 2 t 2, Show that the equation for the spherical wavefronts in S’ cannot be x’ 2 +y’ 2 +z’ 2 =c 2 t’ 2 in the Galilean transformation. Exercise 4-3: Show that x ’ =  (x – vt) so that x = ’ (x ’ + vt ’ ), yields the  factoid  and that for small velocities

16 Lorentz Transformation Exercise 4-4: Use x ’ =  (x – vt) and x = ’ (x ’ + vt ’ ), to find t ’ =  (t – v x /c 2 ) Exercise 4-5: Use x ’ =  (x – vt) and t ’ =  (t – v x /c 2 ) to show that the equations for spherical wave fronts in S and S ’ are the same.

17 Lorentz Transformation Equations A more symmetrical form:

18 Properties of  Recall that  = v / c < 1 for all observers.  equals 1 only when v = 0. In general: Graph of  vs.  : (note v < c )

19 The complete Lorentz Transformation If v << c, i.e., β ≈ 0 and  ≈ 1, yielding the familiar Galilean transformation. Space and time are now linked, and the frame velocity cannot exceed c.


21 The Inverse Lorentz Velocity Transformations If we know the shuttle’s velocity in the rest frame, we can calculate it with respect to the space ship. This is the Lorentz velocity transformation for u ’ x, u ’ y, and u ’ z. This is done by switching primed and unprimed and changing v to –v :

22 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 Lorentz velocity transformation 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?


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

25 Gedanken (Thought) experiments It was impossible to achieve the kinds of speeds necessary to test his ideas (especially while working in the patent office…), so Einstein used Gedanken experiments or Thought experiments. Young Einstein


27 Time Dilation and Length Contraction Time Dilation: Clocks in S ’ run slowly with respect to stationary clocks in S. Length Contraction: Lengths in S ’ contract with respect to the same lengths in stationary S. More very interesting consequences of the Lorentz Transformation:



30 Time Dilation and Proper Time If Mary and Melinda are careful to time and compare their measurements, what duration will they observe? Frank’s clock is stationary in S where two explosions occur. Mary, in moving S ’, is there for the first, but not the second. Fortunately, Melinda, also in S ’, is there for the second. S’ MaryMelinda S Frank Mary and Melinda are doing the best measurement that can be done. Each is at the right place at the right time.

31 Time Dilation Mary and Melinda measure the times for the two explosions in system S ’ as t ’ 1 and t ’ 2. By the Lorentz transformation: This is the time interval as measured in the frame S ’. This is not proper time due to the motion of S ’ :. Frank, on the other hand, records x 2 – x 1 = 0 in S with a (proper) time:  = t 2 – t 1, so we have:


33 Mirror FrankMary Time Dilation Example: Reflection Fred L S’S’ v S c∆t/2 v∆t/2 Exercise 4-7: Show that the event in its rest frame (S ’ ) occurs faster than in the frame that’s moving compared to it (S).

34 Time stops for a light wave Because: And, when v approaches c : For anything traveling at the speed of light: In other words, any finite interval at rest appears infinitely long at the speed of light.


36 Length Contraction Frank Sr., at rest in system S, measures the length of his somewhat bulging waist: L p = x r  x ℓ Now, Mary and Melinda S’, measure it, too, making simultaneous measurements ( t’ l = t’ r ) of the left, x ’ l, and the right x’ r endpoints Frank Sr.’s measurement in terms of Mary’s and Melinda’s: ← Proper length where Mary’s and Melinda’s measured length is: Moving objects appear thinner! Lp Frank Sr.

37 Length contraction is also reciprocal. So Mary and Melinda see Frank Sr. as thinner than he is in his own frame. But, since the Lorentz transformation is symmetrical, the effect is reciprocal: Frank Sr. sees Mary and Melinda as thinner by a factor of  also. Length contraction is also known as Lorentz contraction. Also, Lorentz contraction does not occur for the transverse directions, y and z.

38 Lorentz Contraction A fast- moving plane at different speeds. v = 10% c v = 80% c v = 99.9% c v = 99% c

39 Experimental Verification of Time Dilation Cosmic Ray Muons: Muons are produced in the upper atmosphere in collisions between ultra-high energy particles and air-molecule nuclei. But they decay (lifetime = 1.52  s) on their way to the earth’s surface: No relativistic correction With relativistic correction Top of the atmosphere Now time dilation says that muons will live longer in the earth’s frame, that is,  will increase if v is large. And their average velocity is 0.98c!

40 Detecting muons to see time dilation At 9000 m it takes muons (9000/0.998c ~ 30 μs) about 15 lifetimes to reach earth. If N o = 10 8 and t = 15τ, N ~ 31 muons should reach earth. From relativistic approach, the distance traveled is only 600m at that speed in 1 lifetime (2 μs) and therefore N = 3.68 x 10 7 Experiments have confirmed this relativistic prediction

41 Space-time Invariants The quantity Δs 2 between two events is invariant (the same) in any inertial frame. Δs is known as the space-time interval between two events. There are three possibilities for Δs 2 : Δs 2 = 0: Δx 2 = c 2 Δt 2, and the two events can be connected only by a light signal. The events are said to have a light-like separation. Δs 2 > 0: Δx 2 > c 2 Δt 2, and no signal can travel fast enough to connect the two events. The events are not causally connected and are said to have a space-like separation. Δs 2 < 0: Δx 2 < c 2 Δt 2, and the two events can be causally connected. The interval is said to be time-like. (∆s) 2 = (c 2 ∆t 2 ) - [∆ x 2 + ∆y 2 + ∆z 2 ] This is a quantity that is invariant under Lorentz transformation. It is defined in the following way;





46 Waves from a source at rest Viewers at rest everywhere see the waves with their appropriate frequency and wavelength.


48 The Relativistic Doppler Effect So what happens when we throw in Relativity? Exercise 4-8: Consider a source of light (for example, a star) in system S ’ receding from a receiver (an astronomer) in system S with a relative velocity v. Show that the frequency can be obtained from Where f 0 is the proper frequency Exercise 4-9: What would be the frequency if the source was approaching? Exercise 4-10: Use the results from exercise 8 and 9 to deduce the expressions for non-relativistic velocities. c∆tv∆t

49 Using the Doppler shift to sense rotation The Doppler shift has a zillion uses.

50 Using the Doppler shift to sense rotation Example: The Sun rotates at the equator once in about 25.4 days. The Sun’s radius is 7.0x10 8 m. Compute the Doppler effect that you would expect to observe at the left and right limbs (edges) of the Sun near the equator for the light of wavelength = 550 nm = 550x10 -9 m (yellow light). Is this a redshift or a blueshift?







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