Relativity
Eamples 1. What is the velocity of an electron in a 400 kv transmission electron microscope? What is the velocity in the 6 GeV CESR particle accelerator?. If one million muons enter the atmosphere (10 km high) at a speed of 0.98c, how many will arrive at the surface given that muons have a half-life of 1.56 µs?
Newton s Laws of Motion 1. An object at rest stays at rest and an object in motion stays in motion at constant speed in a straight line unless acted on by an unbalanced force.. Σ F = m dv = ma ΣF = ma dt ΣF = ma 3. F ΣFz= ma = F F1 = F F = F 1 1 F 1y y 1z = Valid in any inertial reference frame, that is, any frame of reference that is not accelerating with respect to any other inertial reference frame. F y z 1 1y 1z
Newtonian or Galilean Relativity * The laws of mechanics must be the same in all inertial reference frames, that is, the laws are invariant. * There is no special or favored frame of reference. * Absolute motion cannot be detected. If an observer in inertial reference frame S measures forces and calculates position and velocity versus time using Newton s Laws, he/she correctly predicts the motion. If another observer in reference frame S moving a constant velocity v in the + direction measures forces and calculates position and velocity versus time using Newton s Laws, he/she also correctly predicts the result. Note that the values each predicts and observes for position and velocity are different.
If frame of reference S ( y z ) is moving in the + direction with speed v compared to frame of reference S (yz), then = - vt y = y z = z t = t d dt d dt d dt u = u - v u y = u y u z = u z d dt d dt d dt a = a a y = a y a z = a z
Eample: Billiard ball collision in two dimensions Cue ball travels at 4.0 m/s and strikes the 8-ball. The 8-ball leaves the collision at an angle of 5 from the original direction the cue ball was travelling. Find the speed of the 8-ball and the direction and speed of the cue ball after the collision. Now repeat in the center of mass frame of reference.
Mawell s Equations r r E da= Q ε r r 0 B da= 0 r r d E ds = Φ dt B r r B ds = I + µ µε 0 0 0 d Φ dt E These equations are not invariant.
Linear Wave Equation: Electromagnetic Waves y = 1 v y t In a region where there is no charge, Q = 0 and I = 0, and E ds = E E B = t dφb dt B = t E = B ds = µε B B t µε E t 0 0 d dt ΦE 0 0 µε E t = 0 0 8 v = 1 / µε = c = 3.00 10 m / s 0 0 µε E = 0 0 t
Does this work for the speed of light? * All waves move in a medium. * The medium for light is called the ether. * Light should travel at c=3.0010 8 m/s only in the rest frame of the ether. It should travel at a different speed in frames of reference which are moving with respect to the ether. c' = c v
Michelson-Morley Eperiment They measured changes in interference fringes caused by differences in the speed of light depending on direction.
They saw no changes!
Einstein s postulates 1. Absolute uniform motion cannot be detected. OR The laws of physics (not just mechanics) must be the same in all inertial reference frames.. The speed of light in a vacuum, c, is independent of the motion of the source. OR All observers measure c for the speed of light in a vacuum.
Consequences - This eplains the Michelson-Morley result. - This removes the parado between electricity and magnetism and wave mechanics. - This abandons the idea that there is such a thing as absolute length or time!
Simultaneity An observer standing on the platform at C sees lightning bolts strike both ends at the same time. Since light from each end traveled the same distance, observer concludes that events were simultaneous.
An observer in the train is sitting at C while train is moving to right. Light from A must arrive before B. Since it came same distance in same time, lightning must have struck the front of the train first, so events were not simultaneous.
Relativity of Simultaneity Events which are simultaneous in one frame of reference will not be simultaneous in another inertial reference frame. Clocks synchronized in one reference frame will not be synchronized in another inertial reference frame.
Galilean transformations = - vt y = y z = z t = t u = u - v u y = u y u z = u z Inverse transformations Switch primed and unprimed and change v to -v = + vt y = y z = z t = t u = u + v u y = u y u z = u z Since light traveling in the + direction in frame S would need to have a speed of u = c - v in frame S, Einstein s Postulates are not consistent with Galilean transformations. Let s assume that the correct transformation to have the form ' = γ ( vt) γ where is a constant that depends on v and c but must approach one as v approaches zero to recover the Galilean transformations. = γ ( ' + vt') To transform from S to S we must have the same form.
Combining these two transformations: = γ ( ' + vt') = γ ( γ ( vt) + vt') = γ γ vt+ γ vt' ( 1 γ ) + γ vt= γ vt' t' = ( 1 γ ) γ v + γ t 1 t = t+ ( γ ) ' γ γ v Time is passing at different rates in the two frames of reference!
What about velocities? ' = γ ( vt) d' = γ ( d vdt) t' = γ t+ ( 1 γ ) dt' = γ dt + γ v ( 1 γ ) γ v d u' = d' dt' u' = 1+ = γ( d vdt) d vdt = ( 1 γ ) dt v d ( 1 γ ) γ + dt + γ γ v d d v dt u v = ( 1 γ ) d ( 1 γ ) 1+ u γ v dt γ v
dy u' = dy' dy = dt y = dt' 1 dt v d 1 + ( γ ) ( γ ) γ γ 1 + γ γ v u' y u' z = γ 1+ = γ 1+ u ( 1 γ ) γ v u u y z ( 1 γ ) γ v u d dt
Since the speed of light should be c in all reference frames, if, then. u = c u ' = c u'= u v 1 1+ ( γ ) γ v u c= c v 1 1+ ( γ ) γ v c γ = 1 1 v c γ = 1 1 = 1 β 1 c v
Now that we know what gamma is, we can simplify some equations: 1- v c γ γ β = = t t 1 v t v c ' ( ) = + = γ γ γ γ u'= u v 1 1 v u u v 1 vu c + = ( ) γ γ vt ' ( ) = γ u' u 1 1 v u u 1 vu c y y y = + = γ γ γ γ ( ) y y ' =
Time Dilation c t' = D c t v t = + D 1 t = t' = γ t' 1 v c
Using the Lorentz transformation to go from the moving frame to the fied frame we get the same result. t t'+ v' = γ c t t t '+ v ' t'+ v ' 1 1= γ γ 1 c c γ t t t ' t ' + v ( ' ' ) 1= γ ( 1 ) 1 c t t = γ ( t ' t ') 1 1 t= γ t'
How about using the Lorentz transformation to go from the fied frame to the moving frame? v t' = γ t c v t' = γ t γ c v t' = γ t γ c v t v t' t t 1 v = γ γ = γ t c c t' = t/ γ
Length Contraction The length of an object measured in its own reference frame is, called its proper length. L p In any other reference frame, the object will appear shorter. A train car of proper length L p is passing a pole at speed v. When the back of the train is at the pole, the front is a distance ahead. L The observer on the ground sees the front of the train ahead by only L = / γ L p. To the observer outside, objects contract in the direction they are moving.
Again, using the Lorentz transformation we get the same result. ( ) ' = γ vt ( ) γ ( ) ' ' = γ vt vt 1 1 1 ' = γ γv t ' = γ L' = L = γ L p
Twin Parado v = 08. c γ = 5 / 3 Homer watches Ulysses travel at ( ) to a planet 4 light-years away, turn around, and return at the same speed while Homer stays on earth. Homer ages 10 years while Ulysses is gone. t = d / v = ( 4 ly)/( 08. c) = 10 yr Ulysses ages only 6 years! tp = t / γ = ( 10yr)/( 5 / 3) = 6yr Ulysses says he traveled 4.8 light-yrs in 6 years (v=0.8c). L = L / γ = ( 4 ly)/( 5/ 3) = 4. ly p t = L/ v = ( 4. ly)/( 08. c) = 6 yr The parado is: Why couldn t Homer be considered the one moving and Ulysses the one at rest? Then Homer would age less than Ulysses. So which one ages more?
Spacetime Diagrams
Construct a space-time diagram for Homer and Ulysses. 1 11 10 9 8 ct (light-years) 7 6 5 4 3 1 0 0 1 3 4 5 6 (light-years)
1 11 10 ct'=6 9 ct'=5 8 Ulysses ct (light-years) 7 6 5 Homer ct'=4 turn around ct'=3 light 4 ct'= 3 ct'=1 1 0 0 1 3 4 5 6 (light-years)
Invariance of the Space-time Interval Both lengths and times are different in different frames of reference. That is, the Lorentz transformation changes both space and time coordinates. Is anything invariant under the Lorentz transformation? The space-time interval, s, between two events is invariant: ( ) ( ) ( s ) = ( c t) ( r) = ( c t) ( ) + ( y) + ( z) ( s' ) = ( c t' ) ( r' ) = ( c t' ) ( ' ) + ( y' ) + ( z' ) ( s ) = ( s' ) s The space-time interval,, may be positive, zero, or negative. If its positive, the interval is time-like. There is a reference frame where these events occur in the same place. If its zero, the interval is light-like. If its negative, the interval is space-like. There is a reference frame where these events occur at the same time. Is the space time interval for Ulysses outward trip the same for both frames of reference? How about our train/platform simultaneity problem?
For sound waves: f ' = Doppler Effect v v ± v v where v O is the speed of the observer and v S is the speed of the source, both with respect to the air. O S f Motion of the source and the observer change the frequency by different amounts. Einstein s postulates assert that this can t be true for electromagnetic waves.
Space-time Diagram for a Stationary Source 1 11 B ct A 10 9 8 7 ct 6 5 4 3 1 0-6 -5-4 -3 - -1 0 1 3 4 5 6
Space-time Diagram for a Moving Source (w/o Relativity) 1 11 B ct β=0.3 ct' A 10 9 8 7 ct 6 5 4 3 1 0-6 -5-4 -3 - -1 0 1 3 4 5 6
Moving Observer (w/o Relativity) B 1 11 10 ct A 9 8 7 ct 6 5 4 3 1 0-6 -5-4 -3 - -1 0 1 3 4 5 6
Relativistic Doppler Effect We can leave out the motion of the observer since we will be calculating the frequency in his/her rest frame. But we have to account for the fact that the clocks are running differently in the frames of the source and the observer. For EM waves: f f f c f 1 f = = 0 c v 0 γ 1β γ = = S 1-β 1β 1± β 1 β f 0 f = ( 1+ β)( 1 ) 1β β 0 0 f where we use the top sign when the source is approaching and the bottom sign when its receding.
Space-time Diagram for a Moving Source (w/ Relativity) 1 11 B ct β=0.3 ct' A 10 9 8 7 ct 6 5 4 3 1 0-6 -5-4 -3 - -1 0 1 3 4 5 6