2. Special Theory of Relativity 2.1 Classical Relativity An event seen by two observers. Position, velocity and acceleration 𝑥! = 𝑥 − 𝑉𝑡 2.2 Michelson-Morely

Experiment

𝑡!"! =2𝐿𝑐

1

1 − 𝑢! 𝑐!

𝑡!"! =2𝐿𝑐

1

1 − 𝑢! 𝑐!

• Rotating the experiment →

phase changes at the eyepiece

• No motion of the fringes was observed.

• The ether does not exist. Therefore, there is no preferred inertial frame.

2. We need a new theory that reflects the constancy of the speed of light.

Special Theory of Relativity is based on two postulates:

1) The principle of relativity: The laws of physics are the same in all inertial reference frames.

2) The principle of the constancy of the speed of light: The

speed of light in free space has the same value c in all inertial reference frames.

The second postulate explains the failure of the Michelson-Morely experiment to observe a “phase change” (i.e., the motion of the fringes due to the directional change of the ether). The first postulate doesn’t allow for a preferred frame of reference (i.e., all inertial frames are equivalent). 3. The Lorentz Transformation Imagine having a stationary frame S, and a frame S’ moving along the 𝑥, 𝑥′ direction with velocity V w.r.t. S. A light pulse is emitted when the two origins cross and a spherical light pulse is observed in both inertial frames 𝑆, 𝑆′ . Let’s try to correlate the space-time events 𝑥, 𝑡 → 𝑥!, 𝑡′ 𝑥! − 𝑐𝑡! = 𝜆 𝑥 − 𝑐𝑡 Location of the pulse on the right side

𝑥! + 𝑐𝑡′ = 𝜇(𝑥 + 𝑐𝑡) Location of the pulse on the left side

such that 𝜆 and 𝜇 only depend on 𝑉, and c = constant.

Example: Compare the space-time coordinates between 𝑆 and 𝑆! along the 𝑥, 𝑥′ axes for the wave front moving in both the positive and negative directions. Compare the space-time coordinates after 10.0 ns has elapsed in the 𝑆 frame. Example: A meter stick is at rest in the 𝑆! frame as it is moving with 𝑉 = 4 5 𝑐 w.r.t. the 𝑆 frame. What's the length of the meter stick as observed in the 𝑆 frame. Example: A clock is at rest in the 𝑆! frame as it is moving with 𝑉 = 4 5 𝑐 w.r.t. the 𝑆 frame. How long does it take to "tick off" one second as observed with the clocks in our 𝑆 frame? Calculate the velocity equations: 𝑣!! , 𝑣!! , and 𝑣!!. Example: The 𝑆! frame is moving at 0.900 𝑐 and a flashlight is at rest in that frame shining a beam in the +𝑥′ direction. How fast does the wave front appear to be moving in the 𝑆 frame? Example from Chapter 1: A pion is moving through the laboratory at a speed of 0.931 𝑐. The pion decays into another particle, called a muon, which is emitted in the forward direction (the direction of the pion's velocity) with a speed of 0.271 𝑐 relative to the pion. You don't need to know the Lorentz Transformation in three dimensions, but since you asked:

~r 0 = ~r + ~��

�

� + 1~� · ~r � ct

�

4. Relativistic Doppler Shift What does a wave look like when you go from 𝑆 → 𝑆′. That is, a traveling wave

𝐸! 𝑠𝑖𝑛 𝑘 𝑥 − 𝑣𝑡 looks like what (??) in the 𝑆′ frame? Find the wavelength 𝜆′ and frequency 𝑓′ as observed in the 𝑆′ frame. Red Shift

𝑟 =1 + 𝛽1 − 𝛽

𝑧 ≡𝜆!"# − 𝜆!"

𝜆!"

𝑧 = 𝑟 − 1 (the red-shift)

𝑧 = 0 (no shift) 𝑧 > 0 (red-shift) 𝑧 < 0 (blue-shift) Redshift: 𝜆! = 𝑟 𝜆!" Blueshift: 𝜆! = 𝜆!" 𝑟 Compare this to the definition of red-shift using the Hubble constant. 𝑧 = !!

! 𝑑 where 𝐻! ≅ 68 (𝑘𝑚/𝑠)/𝑀𝑝𝑐

Example (2.8) A distant galaxy is moving away from the Earth at such high speed that the blue hydrogen line at a wavelength of 434nm is recorded at 699 nm, in the red range of the spectrum. What is the speed of the galaxy? What is the red-shift, 𝑧?

Space-Time Invariant Quantities (4 vectors) 𝑥! = (𝑐𝑡, 𝑥, 𝑦, 𝑧) a 4-dimensional space-time vector Event 1 𝑥!

! = (𝑐𝑡!, 𝑥!, 𝑦!, 𝑧!) Event 2 𝑥!

! = (𝑐𝑡!, 𝑥!, 𝑦!, 𝑧!) The Metric Equation: Δ𝑠! = 𝑥!

! − 𝑥!! ∙ 𝑥!,! − 𝑥!,! = 𝑐! Δ𝑡! − Δ𝑥! − Δ𝑦! − Δ𝑧!

• This equation specifies the frame-independent space-time

interval Δ𝑠 between two events, given their coordinate separations Δ𝑡, Δ𝑥, Δ𝑦, and Δ𝑧 in any given inertial frame.

• This equation applies only in an inertial reference frame.

• This equation is to space-time what the pythagorean theorem is to Euclidean space.

Δ𝑠! > 0 (time like) "Causal" Δ𝑠! < 0 (space like) Δ𝑠! = 0 (light like) on the "light cone" The "space-time interval squared" is a relativistic invariant. It is the same in all inertial frames.

Δs ! = Δs′ !

𝑐! Δ𝑡! − Δ𝑟! = 𝑐! Δ𝑡′ ! − Δ𝑟′ ! Example: In the solar system frame, two events are measured to occur 3.0 h apart in time and 1.5 h apart in space. Observers in an alien spaceship measure the two events to be separated by only

0.5 h in space. What is the time separation between the events in the alien's frame?

5. Space Time Diagrams

Slope = 1/𝛽 The new axes for 𝑥′ and 𝑐𝑡′ are found from the Lorentz Transformation Along the 𝑐𝑡′ axis, 𝑥′ = 0 so that 𝑐𝑡 = 𝑥 ⁄ 𝛽 Along the 𝑥′ axis, 𝑐𝑡′ = 0 so that 𝑥 = 𝑐𝑡 ⁄ 𝛽 where 𝑡𝑎𝑛 𝛼 = 𝛽

Space Time

Diagrams (cont'd) The curved path in 𝑥 − 𝑦 space is longer because its differential path length is determined by:

𝑑𝑠! = 𝑑𝑥! + 𝑑𝑦! similar to the Pythagorean theorem.

The curved path in 𝑥 − 𝑐𝑡 space is shorter because its differential path length is determined by: 𝑐 𝑑𝑡′ ! = 𝑐 𝑑𝑡 ! − 𝑑𝑥 !

(i.e., the "space-time interval squared" is a relativistic invariant). More precisely,

Δs ! = Δs′ !

and 𝑑𝑥′ ! = 0 as long as you’re on the curved path.