1. P1X*Dynamics & Relativity : Newton & Einstein Chris Parkes October 2005 Special Relativity Postulates Time Dilation Length Contraction Lorentz Transformation Relativistic Energy http://ppewww.ph.gla.ac.uk/~parkes/teaching/DynRel/DynRel.html Part II - “ When a man sits with a pretty girl for an hour, it seems like a minute. But let him sit on a hot stove for a minute and it's longer than any hour. That's relativity. ” 2005 Centenary Read the textbookPhenomena close to speed of light, high energy

2. Michelson Morley Interferometer Ether was a hypothetical substance through which light travelled The universe was thought to be a stationary frame of reference full of ether. So, if the earth is traveling at v with the ether then light should have one velocity c+v (w.r.t. ether). In the perpendicular direction the velocity would just be c In 1887 Michelson & Morley set up this apparatus in which a light beam is split into two perp. Beams, then recombined. By observing interference between the two beams, they showed that the speed of light was constant Maxwell’s eqns of electromagnetism (1865) Contain a velocity c for light (emag. waves)

3. Einstein’s Postulates First Postulate –Laws of physics are the same in any inertial frame of reference (principle of relativity) See before, co-ordinate axes moving at constant velocity No preferred frame of reference, no frame is more “correct”- cannot tell if you are moving at constant velocity Second Postulate –The speed of light in vacuum is the same in all inertial frames Independent of the motion of the source at rest I measure speed of light c (~3x10 8 ms -1 ) If it is the headlight of a train moving at u Newton: speed =c+uEinstein: speed still =c !!! c v Newton’s laws are the low velocity limit of Einstein’s special relativity

4. Co-ordinate Transformations Revisited 0 ut Frame S Frame S’(x’,y’) x’ y V of S’ wrt S 0 x Now, consider point moving, differentiate Obtain Galilean velocity transformation (as before) i.e. despite looking convincing this is wrong at high velocities. It gives us the problem for our train headlight. We have assumed time is the same for our observer in S (station)and S’ (train) t=t’ No t  t’, and the velocities must be defined as: (and similarly with distances…)

5. Time Intervals: Simultaneous Events Two events simultaneous in one reference frame are not simultaneous in any other inertial frame moving relative to the first. Two bolts seen simultaneously at C Right bolt seen first at C’ Left bolt seen second at C’ Two lightning bolts strike A,B S-frame : simultaneous lighting strikes S’-frame : right bolt hit first

6. Time Dilation Time interval is shortest in a reference frame where a clock is at rest. Moving clocks appear slower to an observer. i.e. an observer looking at a moving clock, measures a longer time on her watch than on the moving clock Lets prove this: v Stationary observerMoving w.r.t. observer light mirror In both frames (S,S’) observers agree light is travelling at c but they disagree on the distance (path) the light has travelled S’S Frame S:Time taken d This is called the proper time (measured in stationary frame) This is the shortest possible time interval, moving observers measure a longer time

7. Time Dilation continued Frame S’ (moving observer): L L vt d Distance travelled by light =2L By Pythagoras, So time taken Now, So,(By substituting for d in expression above) Hence, time measured by moving observer Define velocity relative to light speed And time “correction” factor Hence, can rewrite time as Time appears slower (to stationary observer) for moving system t>t 0 (So,  1 always)

8. Time Dilation Example: b-quark decay An unstable b quark is produced The b quark travels ~ 4mm before decaying It is travelling at 0.99c (  =9) Hence 4mm Image reconstructed by DELPHI particle physics Experiment at CERN However, the average lifetime (at rest) of a b-quark is 1.5 ps (1 pico-s is 1x10 -12 s) So why did we measure 13ps ? Time dilation

9. Length Contraction Lengths are longest in a reference frame where the object is at rest. Moving Objects appear shorter to an observer. Only applies to lengths in direction of travel light mirror L0L0 v vt Stationary observerMoving w.r.t. observerS’S Frame S:Time taken for light to bounce back and forth L L 0 is the proper length (measured in stationary frame) This is the longest possible length, moving observers measure a shorter length Frame S’ (moving observer): Time taken for light to travel from source to mirror = t 1 Corresponding distance travelled Hence,

10. Length Contraction continued Length contraction (remember  >1) Time taken for light to travel from source to mirror = t 2 Corresponding distance travelled Hence, Frame S’ (moving observer) cont.: So, total time But,from time dilation Moving object appears to shrink by a factor  in the direction of travel L vt 2

11. Including S’ origin movement If S’ movement in x direction, then y,z co-ords unaffected Lorentz Transformations Covered special cases –Time dilation, –length contraction General form of how to relate two frames S, S’ vt Frame S Frame S’(x’,y’) x’ y V of S’ wrt S 0 x Transform (x,y,z,t) in S to (x’,y’,z’,t’) of S’ A distance x’ in S’ is seen as x’/  w.r.t. S 0 –S’ moving with velocity v along x axis Spatial Transform Same as Gallilean transforms if  =1, i.e.  =0, low velocity approx.

12. Lorentz Tranformations : time Lorentz Transformations Summary: A distance x in S is seen as x/  w.r.t. S’ Including S origin movement (velocity –v w.r.t. S) but from spatial transform Hence, since Moves from Frame S to a Frame S’ travelling at velocity v along x axis Thus, To move from S’ to S, just reverse sign on v e.g. (t  t’for low velocity)

13. Lorentz Velocity Transform Particle moving at speed v in S Frame S’ moving at speed u wrt S’ What speed (v’) is it moving at in S’ ? Return to our velocity question… Velocity in frame SVelocity in frame S’ Now, from Lorentz transforms Hence, orand Is speed of light a constant? Put v x =c (e.g. c for velocity of headlight of train, u is speed of train) i.e. if it moves with c in one frame it moves with c in all frames  second postulate …and low velocity limit? back to Gallilean transformSo, denominator

14. Spacetime So, for moving systems as measured by stationary system: Time got longer, distance got shorter. But more than this - from Lorentz transforms we know: Space co-ords in one frame depends on both space and time in other frame And time in one frame depends on both space and time in other Space & time have become intertwined - ‘joined’ in spacetime. Space and time mix between frames. We have to consider spacetime 4-vector co-ordinates (x,y,z,t) Spacetime distance is conserved between frames (i.e. a rotation) -the increased time co-ord compensates the reduced spatial co-ord. -Define co-ords as (x,y,z,ct) 3D space distance 2 is: 4D space-time distance 2 is: x ct light Minkowski spacetime

15. That Equation Mass is a form of energy –Can be used to create new particles A->B+C Rest mass m 0 E = mc 2 m =  m 0  m 0 c 2 also since More useful to express in terms of momentum:

16. Relativity Top 5 Laws of physics are the same in any inertial frame of reference The speed of light in vacuum is the same in all inertial frames Moving clocks appear slower to an observer. Time difference is shortest when the clock is at rest in the reference frame Moving Objects appear shorter to an observer Object is longest when the object is at rest in the reference frame space and time get mixed  spacetime –Transform (x,y,z,t) in S to (x’,y’,z’,t’) of S’ where Postulate 1 Postulate 2 Time Dilation Length Contraction Lorentz Transformations