Modern Physics (PC301) Class #4 Moore - Chapter R5 – Proper time Chapter R6 – Coordinate Transformation

Don’t forget: Hand in homework Tomorrow by 10 am (8 am for webassign) and Sim2 on Friday 1) Length Contraction Trip to Alpha Centuri revisited. 2) Drawing Two-Observer Spacetime Diagrams time contraction length contraction Lorentz transformation equations Review with some homework problems Revisit length contraction with the diagram 3) Revisit length contraction and simultaneity with the conveyor belt painters and the barn/pole problem.

Homework Questions Due Wed by 8 am (Tomorrow!) Problem Set #2 SR cool – use visualizations Apply transformation equations R1B.2, R1B.4, R1S.5 (events 125m apart on tracks), R1S.7 (particles flying apart), R1A.1 (elevator down) Due Wed by 8 am (Tomorrow!) Also Sim 2 due on Friday First test in two weeks

The Twin Paradox "If we placed a living organism in a box…one could arrange that the organism, after an arbitrary lengthy flight, could be returned to its original spot in a scarcely altered condition, while corresponding organisms which had remained in their original positions had long since given way to new generations." (Einsteins original statement of paradox - 1911) Previously worked in spacetime - need to use proper time.

Alpha Centauri Worldline Twin Paradox t Alpha Centauri Worldline Earth Worldline Ds=? C 13y B Clocks really are running at different speeds -> correct interpretation is brother assumes his frame is inertial and can apply inverse argument. A 4.3y x DISCUSSION

Proper Time - Shortest Possible Time A B I NI For inertial (I) Clock -> For non-inertial (NI) clock ->

Different Views: Read Them This is FUN Tipler: pages 50-53 Ohanian: pages 57-58 Epstein: 85-86 Feynman (6 ideas): 77-79

Binomial Expansion

Evidence Hafele & Keating (1971)

Flying to Australia 10360 miles =16,672,803.84m 630 m/hr - 282m/s What did I not take into consideration =: Earth Movement 10360 miles =16,672,803.84m 630 m/hr - 282m/s Dt=16,4hrs 2.6*10-8s Computer Clock Precision is about 1ns EXTENSION

NIST ATOMIC CLOCKS (1949) NIST-1 Accurate to one part in 1,0*10-11s (1999) NIST-F1 accurate to one part in 1,7*10-15 s

(which is most probable).  Leptons The leptons are perhaps the simplest of the elementary particles. They appear to be pointlike and seem to be truly elementary. Thus far there has been no plausible suggestion they are formed from some more fundamental particles. There are only six leptons ( displayed in Table 14.3) , plus their six antiparticles. We have already discussed the electron and muon. Each of the charged particles has an associated neutrino, named after its charged partner (for example, muon neutrino). The electron and all the neutrinos are stable. The muon decays into an electron, and the tau can decay into an electron, a muon, or even hadrons (which is most probable).   From p480-481 of Thornton and Rex http://www.youtube.com/watch?v=T3iryBLZCOQ

Good reading for next two weeks Read and summarize (type please) Chapter 1. Read chapter 2 for greater understanding. Located on P drive or borrow the book from me. Essay be Isaac Asimov Speed of Light

Length Definition "In an inertial frame, an objects length is defined as the distance between two simultaneous events that occur at its ends." Frame Dependent Quantity My way - find events that occur on t’ axis (i.e. x’=0):Ds2=t’2=t2-x2=t2-(bt)2 = Derive equation - Dsn2=t’2-x’2. @ x’, t’=nd=Dsn2=t2-x2=t2-(bt)2

Visualizing Length Contraction

Two Observer Spacetime Diagram: Time t=t’ O t’ x t Slope=1/b @ x’=0, t’=nd -> nd = t2 - x2 x=bt My way - find events that occur on t’ axis (i.e. x’=0):Ds2=t’2=t2-x2=t2-(bt)2 = Derive equation - Dsn2=t’2-x’2. @ x’, t’=nd=Dsn2=t2-x2=t2-(bt)2 Moore confusing -> d = separation between marks on t & t’ axis

Two Observer Spacetime Diagram: Length x t t’ Length Contraction Slope=b x’ O

Two Observer Spacetime Diagram: Hyperbola Relationship x t t’ Dt2-Dx2=Ds2 x’ O At ds=1, x-0 -> t varies depending on speed 1/x2

From Moore p.108

Lorentz Transformations Q tQ tQ’ Inverse Lorentz Transformations tPQ=gt’OQ Normal Lorentz Transformations P x’ xQ’ bxOP Generalized Normal Lorentz Transformations O xQ x xOP=gx’OQ btPQ

The Barn and Pole Paradox: Home Frame Pole Rest Length (L0) = 10ns Home Frame: Pole moving at b = 3/5 -> L=8ns Barn Length (L0) = 8ns An instant in time when the pole is entirely in barn with doors shut. Seems to Make Sense

The Barn and Pole Paradox: Other Frame Barn moving at b = -3/5 -> L=6.4ns Pole Length (L0) = 10ns How can the pole 10ns long fit into a 6.4ns barn? The runner with the pole does not observe that the pole is enclosed in the barn. Front of pole reaches end of barn @ -6ns End of pole reaches front of barn @ 0ns when pole has already left

The Barn and Pole Paradox Resolution

Geometric Analogy