The Theory of Relativity

Transcription

1 At end of 20th century, scientists knew from Maxwell s E/M equations that light traveled as a wave. What medium does light travel through? There can be no doubt that the interplanetary and interstellar spaces are not empty, but are occupied by a material substance or body, which is certainly the largest, and probably the most uniform body of which we have any knowledge? - Maxwell Light (in Maxwell s equations) moves at c=3 x 10 8 m/s. Recall that if you take the constants: ε0 : Permittivity of free space (Electostatic force) μ0 : Permeability of free space (Magnetic force) James Clerk Maxwell ( ) Their product is: ε0μ0= 1/c 2

2 Light moves at 3 x 10 8 m/s. If a car (in inertial frame S ) drives by you (in inertial frame S) at u=0.5 x c, how fast you would observe the speed of the light coming from its headlights? inertial frame S inertial frame S Inertial frames : frames of reference with no net acceleration. Observer in frame S says light moves at c. Does observer in frame S say that light moves at c + u = c + 0.5c = 1.5 c?

3 Michelson-Morley Experiment (1887) The most famous failed experiment. Designed to measure the motion of the Earth with respect to the Luminiferous Ether (the mysterious substance through which light waves propagate). Albert Michelson ( ) Edward Morley ( ) Jointly they won the Nobel prize in 1917 for this work.

4 Michelson-Morley Experiment Designed to measure the motion of the Earth with respect to the Luminiferous Ether (the mysterious substance through which light waves propagate). Motion of Ether with respect to Earth (at ~30 km/s) Experiment was consistent with a motion of ZERO through the ether.

5 Light Propagation Direction Coherent Light From Single Slit Destructive Interference Barrier with Double Slits Screen Constructive Interference Intensity Distribution of Fringes

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7 Light moves at 3 x 10 8 m/s. If a car (in inertial frame S ) by you (in inertial frame S) at u=0.5 x c, how fast you would observe its light? inertial frame S inertial frame S Observer in frame S says light moves at c. Does observer in frame S say that light moves at c + u?

8 Light moves at 3 x 10 8 m/s. If a car (in inertial frame S ) by you (in inertial frame S) at u=0.5 x c, how fast you would observe its light? inertial frame S inertial frame S Observer in frame S says light moves at c. Does observer in frame S say that light moves at c + u? NO says result from Michelson-Morley experiment. Observer in frame S measures that light moves at c!

9 Enter Albert Einstein ( ) Asked simple questions like, If you were traveling at the speed of light and you look in a mirror, would you see your reflection? Einstein finally rejected the idea of the ether and stated the following. The principle of relativity: The laws of physics are the same in all inertial reference frames. The constancy of the speed of light: Light moves through a vacuum (no ether!) at a constant speed c that is independent of the motion of the light source. This is Einstein s Theory of Special Relativity. It s special because it applies to Inertial frames only.

10 Coordinates between two Inertial frames are related by the Lorentz Transformations. Consider frame S with coordinates (x, y, z, t) and frame S with coordinates (x, y, z, t ), relationship between them is: x = a11 x + a12 y + a13 z + a14 t y = a21 x + a22 y + a23 z + a24 t z = a31 x + a32 y + a33 z + a34 t t = a41 x + a42 y + a43 z + a44 t Hendrik A. Lorentz ( ) For the mathematically savy, this is a linear algebra matrix equation: X = A X, where X and X are vectors and A is the transformation matrix.

11 Consider the case: inertial frame S inertial frame S Start t=t at O=O. Then Lorentz transformations become at position O : x = 0 = a11(x - ut) y = y z = z t = a41 x + a44 t

12 We have the following, which is the same as Galileo if a11 = a44 = 1 and a41=0. x = 0 = a11(x - ut) y = y z = z t = a41 x + a44 t Next, incorporate Einstein s 2nd postulate that c is constant for all inertial frames. If a flashbulb goes off at time t=t =0 at O=O, then in frame S at a time t later the light will have traveled a distance (ct) from O. Similarly, in frame S at a time t the light will have traveled a distance (ct ) from O. Gives two equations: x 2 + y 2 + z 2 = (ct) 2 x 2 + y 2 + z 2 = (ct ) 2

13 Combining equations gives the following relations: x = x - ut 1 - u 2 /c 2 = γ ( x - ut) x = γ ( x + ut) y = y z = z OR y = y z = z t = t - ux/c u 2 /c 2 = γ ( t - ux/c 2 ) t = γ ( t + ux /c 2 ) Lorentz Factor is defined as γ = u 2 /c 2

14 Lorentz Factor versus u/c γ γ = u 2 /c 2 1 (always) u / c

15 Weird Implications 1: Proper Time and Time Dilation inertial frame S inertial frame S Strobe light moving at speed u in inertial frame S. Light occurs with time interval Δt =t2 -t1. What is the time interval for observer in inertial frame S? t1= γ ( t1 + u x1 /c 2 ) t2= γ ( t2 + u x2 /c 2 ) Δt=t2-t1 = γ( t1 + u x1 /c 2 - t2 - u x2 /c 2 ) = γ[ (t1 - t2 ) + u (x1 - x2 )/c 2 ] But, in S frame (x1 - x2 )=0 because light is moving with frame at speed u relative to frame S. Δt = γ( Δt ) or Δtmoving = γ( Δtrest) Weird implication 1: γ is always >1, so Δtmoving always > Δtrest!

16 Weird Implications II: Proper Length and Length Contraction inertial frame S inertial frame S Object with length L =x2 -x1 moving at speed u in inertial frame S. What is the length L for observer in inertial frame S? x1 = γ ( x1 - u t1) x2 = γ ( x2 - u t2) Δx =x2 -x1 = γ( x2 - u t2 - x1 + u t1 ) = γ[ (x2 - x1) - u (t2- t1)] But, in S frame we measure the length at time t2= t1. Therefore, Δx = γ( Δx) or Lrest = γ( Lmoving) Weird implication II: γ is always >1, so ΔLrest always > ΔLmoving!

17 Does this really happen in nature? Yes, it does. Cosmic Rays hit upper atmosphere all the time. Some of these collisions create muons (like electrons but have more mass). Muons decay with an average lifetime of τ = 2.20 x 10-6 s (=2.20 μs). If # of muons at t=0 is N0, then at later time there will be: N(t)=N0 e -t/τ In 1963, Frisch & Smith measured the # of muons at the top of Mt. Washington, 1907 m above sea level: 563 muons / hr with u= c. What will the number of muons be at sea level? Muons travel 1907 m in (1907 m/ c) = 6.39 μs. Therefore, N(t)=N0 e -t/τ = (563 muons/hr) e -(6.39 μs)/(2.20 μs) = 31 muons /hr But, Frisch & Smith actually measured 408 muons / hr at sea level!

18 Does this really happen in nature? Yes, it does. Answer is that the average lifetime of muons is measured in their inertial frame. In our frame it is very different (time dilation): Δtexperiment = γ( Δtmuon) = (2.20 μs) /(1-u 2 /c 2 ) 1/2 Therefore, in the experiment frame, t=2.20 μs / ( ) 1/2 = 22.5 μs N(t)=N0 e -t/τ = (563 muons/hr) e -(6.39 μs)/(22.5 μs) = 424 muons /hr In good agreement with the measured value, 408 muons / hr. What about in the muons frame? They should decay in 2.20 μs?

19 Does this really happen in nature? Yes, it does. What about in the muons frame? They should decay in 2.20 μs? Answer is that they see the length (height) of the mountain contract in their frame. In their frame, they travel a distance of (length contraction): ΔLmuon =( ΔLexperiment) / γ = (1907 m) x (1-u 2 /c 2 ) 1/2 Therefore, in the muons frame, ΔLmuon = 1907 m x ( ) 1/2 = m Thus, they travel this distance in (186.6m / c) = μs, or N(t) =N0 e -t/τ = (563 muons/hr) e -(0.625 μs)/(2.20 μs) = 424 muons /hr. Same answer!

20 Does this really happen in nature? Yes, it does. In Intertial frame of experiment, muon moves with speed u. In Intertial frame of Muon, mountain moves with speed u u u IMPORTANT: Both observers are correct. Their perspective differs but the outcome is the same. It s Relative.

21 Relativistic Doppler Shift Previously, we used the measurement of the wavelength of absorption lines in stars as an indicator of the stars radial velocities. λobs - λrest λrest Δλ = = vr / c λrest Relativistically, consider a source that emits a signal at time t1 and another signal at time t2 = t1 + Δt1 as measured by an observer at rest with respect to this source (they are in the same Inertial frame, S). Consider what this looks like to another observer in an Inertial frame, S, moving relative to frame S with a speed u. S S The time between the emission of light observed in frame S is Δt = γδt. In this time, observer in frame S determines that the distance to the light emitting source has traveled a distance of γuδt cos(θ).

22 Relativistic Doppler Shift The time between the emission of light observed in frame S is Δt = γδt. In this time, observer in frame S determines that the distance to the light emitting source has traveled a distance of γuδt cos(θ). Time interval between the arrival of first and second signals, Δtobs = γδt + (γ Δt cos(θ))/c = γδt [1 + (u/c) cos(θ) ] Take Δt to be the time interval between the emission of light wave crests in inertial frame S, and Δtobs to be the measured time interval between the light wave crests in frame S. Then, the frequencies of the light wave are: ν = 1/Δt in frame S, and νobs = 1/Δtobs in frame S Relativistic Doppler shift is then νobs = νrest 1 - u 2 / c (u/c) cos θ = νrest 1 - u 2 / c vr/c

23 Relativistic Doppler Shift νobs = νrest 1 - u 2 / c (u/c) cos θ = νrest 1 - u 2 / c vr/c For the redshift: z = λobs - λrest λrest = Δλ λrest This becomes: z = ( 1 + v r / c 1 - vr / c ) - 1 For example, the quasar, SDSS has a redshift of z=6.28, measured from the redshift of its Lyman-α hydrogen emission line. The radial velocity from Earth of this quasar is calculated by solving for vr above, vr = c!!

24 Relativisitic Velocity Transformations Relativistic formula for Velocities comes from taking the differential version of the Lorentz relations with respect to time (replace all x,y,z,t with dx,dy,dz,dt): dx = γ ( dx - u dt) dy = dy dz = dz vx = dx dt = = γ ( dx - u dt) γ ( dt - u dx/c 2 ) dx/dt - u ( 1 - u [dx/dt]/c 2 ) dt = γ ( dt - u dx/c 2 ) vx = vx - u ( 1 - u vx/c 2 ) Similarly: vy vy = γ ( 1 - u vx/c 2 vz = ) vz γ ( 1 - u vx/c 2 )

25 Relativisitic Velocity Transformations vx - u vy vx = ( 1 - u vx/c 2 vy = ) γ ( 1 - u vx/c 2 vz = ) vz γ ( 1 - u vx/c 2 ) inertial frame S inertial frame S sin θ = vy / v = γ -1 View from S View from S

26 Enter Albert Einstein ( ) Asked simple questions like, If you were traveling at the speed of light and you look in a mirror, would you see your reflection? Einstein finally rejected the idea of the ether and stated the following. The principle of relativity: The laws of physics are the same in all inertial reference frames. The constancy of the speed of light: Light moves through a vacuum (no ether!) at a constant speed c that is independent of the motion of the light source. This is Einstein s Theory of Special Relativity. It s special because it applies to Inertial frames only.

27 Relativistic Momentum Consider the Fact : Conservation of Momentum must occur in all inertial frames (true because one can rotate the previous figures by 180 o and must get same answer)! p = γmv Important: Some books talk about the relativistic mass, where m = γm, and m is the rest mass. Your textbook uses the mass, m, as a Lorentz Invariant. It does not change with Inertial Frame, but it intrinsic to each object. The reality is that while the mass does not increase, the force required to change the momentum of a particle does, by Newton s 2nd Law: F = dp/dt = d/dt (γmv).

28 Relativistic Energy Change in Kinetic Energy using change in momentum: dp dt dx dt K = F dx = dx = dp = v dp xi xf xi xf pi pf pi pf p = γmv where pi and pf are the initial and final momenta (measured at positions xi and xf). Integrate by parts, and convenience set pi = 0 vf K = pf vf - p dv 0 vf = γmvf 2 - m v dv v 2 /c 2 = γmvf 2 + mc 2 ( 1 - vf 2 /c 2-1) Dropping the f subscript and generalizing: K= γ mv 2 + mc 2 / γ - mc 2 K = γmv 2 + γmc 2 / γ 2 - mc 2 = γ[mv 2 + mc 2 (1-v 2 /c 2 )] - mc 2 = γ(mv 2 + mc 2 - mv 2 ) - mc 2 K = mc 2 ( γ - 1) Then the Relativistic Energy: E + Erest = mc 2 ( γ - 1); where E = γmc 2 and Erest = mc 2. Particle s kinetic energy is the total Relativistic energy minus the rest energy. One can show that E 2 = p 2 c 2 + m 2 c 4, holds even for massless particles (i.e., photons carry momentum and energy but have no mass)

29 Twin Paradox Relativity is very, very weird. Consider the following (attributed to Paul Langevin in 1911). Consider two twins. One twin undertakes a long space journey in a spaceship. She travels at almost the speed of light to a distant star, while the other twin remains on Earth. The traveling twin finally returns to Earth from the distant star, traveling again at almost the speed of light. The Earth-bound twin says, I observe that your velocity is very large relative to mine, so I observe your clock running much slower than mine. Therefore, when you return to Earth, you will be younger than me. The traveling twin says, Well, in my frame, your velocity is very large relative to mine. I observe that your clock on Earth is running much slower than mine. Therefore, when I return to Earth, you will be younger than me. Who s correct?

30 Barn-and-Ladder Paradox Consider a barn of length L. The barn has both doors open. Jane is standing in the barn observing her friend Emma. Emma is running toward the barn at a speed very close to the speed of light. Emma is carrying a ladder of length L in its rest frame. Jane wants to trap Emma in the barn, so she says, Because of length contraction, Emma s ladder appears shorter than the length of the barn. So, I will slam the door on the far end closed. Then, as soon as the back end of her ladder enters the barn, I will slam the front door closed and trap her. But, Emma says, Well, in my inertial frame, the length of the barn contracts, and so my ladder is longer than it. Therefore, when Jane slams the back door, it will hit my ladder and I won t be trapped. Ladder (at rest) Moving Ladder u Ladder in Barn Frame of Barn Barn (at rest) Barn (at rest) u Who s correct? Frame of Ladder Ladder (at rest) Ladder (at rest) u Barn (at rest) u Moving Barn Ladder in Barn