τ = 10 seconds . In a non-relativistic N 1 = N The muon survival is given by the law of radioactive decay N(t)=N exp /.

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Muons on the moon Time ilation using ot prouts Time ilation using Lorentz boosts Cheking the etor formula Relatiisti aition of eloities Why you an t eee the spee of light by suessie boosts Doppler shifts Cop Raar Classifiation of interals Time-like,spae-like, light-one Proper time Getting to the proper frame Spae-like interals Spae-like interals an ausality Speial Relatiity

What will we o in this hapter? In this hapter we will work out some of the onsequenes of Speial Relatiity. Our first eample esribes an eperiment to measure the time ilation of osmi ray muons. We o this using ot prout inariants as well as Lorentz boosts. We net obtain a formula for the relatie eloity between two referene frames whih iffers from the Newtonian form. This formula shows that one annot eee the spee of light by by suessie boosts. We isuss the relatiisti Doppler shift for eletromagneti waes an isuss how op raar works. We onlue by isussing the lassifiation of spae-time interals base on the interal ot prout. We show that if the square interal is greater than zero, there eists a referene frame where the two eents hae no spatial separation an the minimum time separation obserable in any referene frame. This minimum time is alle the proper time. We esribe how to jump to this frame. Suh interals are known as time-like interals. If the square interal is negatie, we say that the interal is spae-like. We show that there eists a frame where the two eents our simultaneously an with the minimum spatial separation. If two eents hae a spae-like interal, they annot be ausally linke in the sense that one eent annot ause the other eent. Eents that hae time-like interals, howeer, are not neessarily ausally linke, of ourse.

Muons on the mountain The muon is a subatomi partile whih an be thought of as a heay eletron. It is the priniple omponent to osmi rays an has an aerage lifetime of. miroseons. This lifetime an be easily measure by stopping the muon in plasti an looking for its eay with sintillation ounters. osmi μ The trigger bo (left) requires the upper pale to fire an that the lower to not fire. The bo forms the sope trigger. The time between the muon arrial an its eay light is measure on sope. The muon surial is gien by the law of raioatie eay N(t)=N ep /. 6 to. 0 seons 0 ( t τ ) The stoppe muon time elays are well fit τ = Mt MKinley 600 m The typial osmi ray mu moes at about 98% of the spee of light. To trael from the top of Mt. MKinley to sea leel woul take: 600m 6 Δ t = 0 seons 8 0.98 3 0 m/ s or about 0 ( ) τ μ. In a non-relatiisti N = N 000 worl sea leel ep( 0) Mt MKinley But this isn t een lose to being true. There is only a fator of 5 ifferene in the osmi rate on the top an bottom of a mountain. 3

#: ( 0 h) #: ( (h/) 0) Lets analyze this relatiistially We show two eents # when the muon passes the mountain top an # when the muon hits sea leel. h = 600 m How muh time has elapse in the muon frame? We will sole this problem seeral ways. The best way is to iretly use inariane of the interal square length. h r = ( h/ h) ri r = h ν In the muon frame r = Δ 0 sine ( tμ ) the two eents take plae at the muon position. Thus r ir μ ( t ) = Δ μ μ μ Thus ( tμ ) h Δ = h ν ( tμ ) h h h Δ = = ν ν h or Δ tμ = or Δ t =Δt ν 8 ( ) μ earth The time between two eents loate at the same position in some frame is alle the proper time. We will show that it is always the minimum time measure in any frame. Plugging in the numbers we get 600 Δ t μ = = 0.98 3 0 6 0.98 4. 0 se This is only lifetimes rather than 0! Hene many more muons at the top of Mt. MKinley surie their trip to sea leel before 4 eaying than Newton woul preit.

( / - ) r = h h From last leture, the boost for illustrate ase was t' γ γ t = ' γ γ Muon on Mountain w/ Boosts h = 600 m ' We nee to reerse the sign for our ase t ' γ γ h / = = ' γ γ -h γ γ / h / γh / γh / = γ / γ -h γh γh h h h t ' = γ = = We note that ' =0 whih is a goo hek that we hae inee booste to the μ's rest frame We finally hek the result with the etor form. r i t ' = γ t ; r = h ˆ, = ˆ, t = h / h h h t ' = γ t ' = γ h t' = t' is the proper time sine '=0 To the muon, the mountain is traeling upwars with a eloity. It takes the h mountain a time of Δt μ = to pass from top to sea leel. The muon thus measures the mountain height as: h' = Δ t = h μ h' is Lorentz ontrate 5

Relatiisti eloity aition formula Consier a partile moing with eloity u in a frame moing with eloity wrt the lab frame. We gie the interal between # an #. I # # = ( Δt, uδt) ' Δt' γ γ Δt γδt+uγδt = = Δ' γ γ uδt γδt+ γ uδt We use a positie sign in the matri sine when objet is traeling along boost ais it appears with a larger eloity in booste (prime) frame. Δ ' γδt+ γuδt ' = = ( Δt')/ γδt+u γδt / ' + u = + u / u ( ) This formula is learly ifferent from the Galilean eloity formula = u +. It has the effet of preenting one from surpassing the spee of light by suessie boosts. Here is a plot of ersus for arious u. You only hae approahing but neer eeeing. / 0.8 0.6 0.4 0. 0 / u = 0 u = 0.5 u = 0.50 u = 0.75 0 0. 0.4 0.6 0.8 u = 0.9 = 0.9 0.9 0.9 ' = + = 0.994 + 0.9*0.9 Close, but no igar! 6

# Doppler shift # We hae a light emitter at rest in the prime frame whih is moing at a eloity with respet to the unprime frame where there is a light etetor at =. Eent # is sening of the first wae at = = t = t =0. Eent # is the sening of the seon wae at = 0, t = τ. We are going to fin the oorinates of eent an eent in the reeier (unprime) frame using a Lorentz boost. We will then stay in the unprime frame an ompute the time when eah of the two waes are reeie. The elay between the when the two waes are reeie is the perio of the Doppler shifte light. ' ' t γ γ 0 0 = = γ γ 0 0 t γ γ ' τ γ ' τ = = γ γ 0 γ τ' # ( ) # 0,0 ( γ τ ', γτ ') ( re) T# = + t = ( re) γτ ' T# = + t = + γτ ' ( re) T# = + γτ ' ( ) Δ T = τ = + γτ ' = γτ ' ( ) ( ) 7

More Doppler Shifts re sen ( ) ( ) From last slie τ = γτ'. In terms of frequenies: fre = / τ an fsen = / τ ' fsen = γ ( ) fre = f f γ fre = fsen = fsen + f = f f > f re sen re sen ( ) ( )( + ) ( ) We reeie an inrease in frequeny if the soure approahes the reeier as shown aboe. The -ratio woul inert if soure was retreating from reeier an we woul reeie a lower frequeny. Lets apply this to op raar with a ouble Doppler shift. The raar striking the retreating ar is shifte own. The ar ats as an eho soure of what it reeies an the op reeies a shifte own ersion of the ar eho. f = f an f = f + re sen sen re ar op ar ar sen op sen op 9-7 ( 0 )( ) 8 re sen sen sen fop = far = fop = fop + + + + Δf f re f fop = = = f + + For a 00 km/hr ar / -7 0 if f= GHz Δf 0 00 Hz Beat freq

Classifiation of interals Consier the interal between eents r = ( Δt r). Components of repen on the frame they are iewe in, but rr i is the same in all frames an an thus be use to lassify the interal. rr i > 0 is alle "time-like" rr i = 0 is alle "light one" rr i < 0 is alle "spae-like" If two eents are separate by a "time-like" interal, there is a frame where the two eents take plae at ifferent times but at the same loation. In fat it is ery easy to fin suh a frame by a Lorentz boost inline along the separation etor between the two eents. # ' ( t ) Δt' γ γ t = 0 γ γ 0 = γ + ( t t) γ ( ) ( ) r or ( ) ( ) = = t t t t # But if >, then γ=/ - will be imaginary. Hene we only hae suh a frame r r Δt > r or if < or = < ( t t) Δt ( Δt) rir = r ir > 0 ( Time-like interal) 9

When we hae a time-like interal, the boost to eliminate Δ is simply the apparent eloity etor between the two eents! In this frame there is no spatial separation an a (proper) time separation of Proper time 6 5 4 3 Δt Δ t = proper In any other frame, Δt' > Δt whih ( Δt' ) r' ir' = r ir = ( Δt proper ) ( proper ) rr i proper is ery easy to show using inariants Δt'= Δ t + r' i r' >Δt proper Boosting to arbitrary frames mies Δt' an r ' an moes us along a hyperbola of form Δt' r ' r i ' = onstant. ( ) 0 Δt proper 0 3 4 5 6 The aboe is plot of the time an spae oorinates for a time-like interal iewe in many frames. In the proper frame there is no spatial separation an a minimal time interal. As one boosts away from this frame, the time interals an spae interals both get large an the interal approah an r asymptote gien by Δ t = r The asymptote is the relationship one woul get for light. For a fie spatial separation, the time interal is always larger than that for light. 0

( t' ) Spae-like interals For the ase where rr i < 0, we an no longer efine a proper time. But an show that there is a minimal separation between the eents ( Δt' ) r' ir' = r ir r' = Δ + r > r ir By boosting from frame to frame we moe along the re hyperbola shown below for the ase where r ' = 5 4 3 0 Δt lightone time-like min spae-like r 0 3 4 5 # ' If two eents are linke by a spae-like interal, they annot be ausally linke in the sense that eent # annot hae ause eent #. This is beause there is a referene frame (ompute below) where eent # an eent # our simultaneously but separate. Fin for this frame. lab Chek w/ Lorentz boost # ( ˆ ) ; ηlab γ ( ˆ ) ηlab ir lab γ ( i ˆ ) r = Δ t r r = Δ t ˆ η = = Δt r r = 0 Δt Δt Δt = = min rrˆ ˆ = i osθr ˆ ˆ r r Possible if Δ t < rr i & not unique ( ) ( t t) γ ( ) t ( t) min ( ) ( t ) 0 γ γ t = Δ ' γ γ 0 = γ = =