LECTURE # 4 DOPPLER –SPACETIME DIAGRAMS PHYS 420-SPRING 2006 Dennis Papadopoulos

Last week we learned that length and time are not absolute, but instead, depend on the frame of reference of the observer. We derived the relations: Using only geometrical arguments and applying Pythagorean's theorem. time dilationlength contraction d cT/2 in red frame in blue frame REVIEW

Time always runs slower when measured by an observer moving with respect to the clock. The length of an object is always shorter when viewed by an observer who is moving with respect to the object. To an observer at rest in one frame, the light in the other frame travels further because the other frame is moving with respect to the observer. Since c is constant in all frames, to each observer time appears to be moving more slowly and length appears to be contracted in a reference frame moving with respect to the observer. Everything is slowed/contracted by a factor of: in a frame moving with respect to the observer.

The Lorentz transformations! SS’ v x x’ The transformation: To transform from S’ back to S:

Velocity Transformation To find the u as a function of u ’ replace v by -v What happens for u x =c ?

Fig. 1-19, p. 30 Find speed of B wr to A Take S’ attached to A -> v=.75 c. u B =-.85 c Speed of B wr A=u ’ =

Fig. 1-20, p. 30

Fig. 1-21, p. 31 How fast does B recede as seen by A

Fig. 1-22, p. 31 Steps: Go to ref frame S’ and transform velocities from S to S’

CLASSICAL DOPPLER EFFECT Source moving Receiver moving

Relativistic Doppler Shift Qualitatively Different from Classical To observer First signal Second signal Time between emission of signals in source frame  t o Time between emission of signals in observers frame  t g =  t o Time between arrival of signals to the observer  t=  t g (v/c) cos  +  t g = =  t o [1+(v/c) cos  ]  v RELATIVISTIC DOPPLER SHIFT V R =v cos  VRVR

BLUE AND RED SHIFT

x y Let’s look at coordinate systems: thinking classically… You rolls a cat toy across the floor towards her cat. x y The same cat, the same cat toy, different (arbitrary) choice of coordinate systems. By Pythagorean’s theorem, the toy rolls: Easier visualization, easier calculation—You are happy (although her cat probably doesn’t care as long as she gets his toy.) This is an example of rotating your coordinate axes in space. Coordinate systems

Space-time Diagrams

Time, the fourth dimension? “Spacetime” x ct In x,y space the two space dimensions are interchangeable if they have the same units. A similar relationship can be used to express the relationship between space and time in relativity. x=ct Light propagating in one dimension in a spacetime coordinate system as viewed from a frame S. The distance traveled is equal to the speed of light times the time elapsed. c  t=  x 45°

The red and blue spaceships viewed in spacetime coordinates. The red and blue spaceships are at x=0, t=0 when one emits a pulse of light. At that instant one of the spaceships starts to move away from the other with velocity v. The lightcone- the distance light has traveled since x=0, t=0 as a function of time. Who perceives him/herself to be at the center of the lightcone? A passenger on the red spaceship or a passenger on the blue spaceship?

The red spaceship’s reality.The blue spaceship’s reality. They are both at the center of the light cone! This can be achieved by rotating their coordinate axes as you did when playing with the cat, except one thing…notice that one set of axes is not orthogonal! You need to add a Lorentz boost.

The velocity of a particle in spacetime coordinates is: The speed of light is absolute, therefore the axes must be drawn to preserve the speed of light! x ct a light pulse is sent out x=ct in frame S in frame S’ x’ ct’ and x’ axes must be symmetrically placed with respect to the light pulse! ct’ A spaceship is traveling in a reference frame with v<c w/r to S ct ct’ x x’Vt x Newtonianly:

x ct x=ct ct’ x’ The axes aren’t perpendicular but are scaled by some factor. They must be symmetric w/r to the light pulse. What is this factor? (ct,vt)  tan  =Vt/ct=V/c

x ct x=ct ct’ x’ The axes aren’t perpendicular but are scaled by some factor. They must be symmetric w/r to the light pulse. What is this factor?

Fig. 1-23, p. 32

Fig. 1-24, p. 33

ct ct’ x=ct x x’ Spacetime is invariant! True, observers will differ on the length of objects or the time events occurred, however, they will always agree on how far apart two events are in spacetime. E1 E2 To find the space and time coordinates of an event in a specific frame, draw lines from the event parallel to the axis of that frame. ! ` TO PROVE USE LORENTZ TRANFORMATION FROM THE PRIMED TO THE UNPRIMED FRAME AND DO A LITTLE ALGEBRA

ct x x=ctx=-ct AB C O Could an event at O cause A? Yes, because a “messenger” at O would not have to travel at a speed greater than the speed of light to get there. Could an event at O cause B? A light signal sent from O could reach B. Could an event at O cause C? No, the spacetime distance between O and C is greater than could be covered by light. It would require time travel. here,now where light that is here now may have been in the past where light that is here now may go in the future

Fig. 1-25, p. 34

DIFFERENT ROUTES IN SPACETIME RESULT IN DIFFERENT AGES. THE LONGEST TIME TO GO FROM a TO b IS ALONG THE STRAIGHT SEGMENT AND THE TIME GETS SHORTER AS THE SEGMENT abd APPROACHES THE LIGHTCONE

The time measured by a heartbeat or the wristwatch of a person equqls its wordline from a to b. Curved wordlines are shorter than straight wordlines. Thus a person arriving via a bent world line has fewer heartbeats than a person following a straight- line (i.e. moving only through space) Photons never age in free space.

TWINS A AND B FOLLOW DIFFERENT LIFES ON STAYING HOME THE OTHER TRAVELING A LOT. EVERY TIME THEY MEET AT THE SAME SPACETIME POSITION Bs WORDLINE IS SLIGTLY SHORTER SINCE SHE HAS MOVED A LOT THROUGH SPACE, WHILE HER BROTHER MOVED ONLY THROUGH SPACE. HER WORDLINE AT THEIR 70 TH BIRTHDAY IS SLIGHTLY SMALLER AND HER PHYSICAL AGE SLIGHTLY SHORTER (BY MAY BE A MICROSECOND)

tB=tA/tB=tA/ OUTWARD JOURNEY xAxA EXACTLY THE SAME FOR RETURN JOURNEY tB=tA/tB=tA/ GAMMA 50, V=.9998c IF A IS 50 YEARS OLDER B IS ONLY ONE YEAR OLDER