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C ? & D E A ? % > F< C G & : 6 7 8 C E B @ ? H % A ? < % C < A ; <8 %

A 7 & E C G: 6 7 % I: 6 6 ? & %

A 6 7 GH % ? A : ; ? % 6 : ;8 G ? A 7 = E B @ ? %

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R%

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])4-'K/%N'%;5/'1-O%

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β 1⇒ γ = 1 +O(β2) et

ct = ct + β x

x = x + β ct

A5M350,*51%N'%N'KR%G;%N'%M^M'%)R'%

Λ(α) =

coshα sinh α 0 0sinh α coshα 0 0

0 0 1 00 0 0 1

tanhα = βR/R

X = Λ(α1)X

X = Λ(α2)X X = Λ(α1) Λ(α2) X = Λ(α1 + α2)X

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βR/R = tanh(α1 + α2) =βR/R + βR/R

1 + βR/R βR/R

X =

ctxyz

X = Λ(β)X Λ(β) =

γ γβ 0 0γβ γ 0 00 0 1 00 0 0 1

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ct = γ(ct + β · r)r = γ(r + βct)r⊥ = r⊥

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∆x = xB − x

A = 0 ∆t = tB − tA = 0E1-'/+)(('%N'%12345#467462#`M'0K/.%)K%M^M'%(,'Ka%

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∆t = γ∆t

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t =d

vt = δ +

d

v

e%D5,4,%M51%-'M30%f%g%N,-%()%G5/-K'% τ =d

γ(v) v

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γ(v) v

:1%3'K-%)+5,/% t > t et τ < τ

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v

65K/%_5V%

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TA(aller)

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28/04/11 11:25http://upload.wikimedia.org/wikipedia/commons/7/78/Bondi_diagramme_3.svg

Page 1 sur 1

A B

O

P

k(kT)

T

kT

t

x

T'

G'M30%NK%0,S1)(%.M,0%3)/%<(,4'%j%G%

G'M30%N'%/.4'3*51%3)/%_5V%j%k%G%

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<(,4'%'-%_5V%.4Y)1S'1-%N'0%0,S1)KR%(KM,1'KR%

k =

1 + β

1− β≥ 1

28/04/11 11:25http://upload.wikimedia.org/wikipedia/commons/0/0d/Bondi_diagramme_1.svg

Page 1 sur 1

A B

O

T

kT

T

kT

t

x

G'M30%3/53/'%

N'%_5V% β =v

c=

d

c t=

k2 − 1k2 + 1

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d = ck2T − T

2t =

k2T + T

2

>,)S/)MM'%N'%_51N,%`!hWUa%

?R'M3('%X%"%j%"#U% >K/)1-%(F)(('/%

∆tr =

1 + β

1− β∆te = 3∆te

>K/)1-%('%/'-5K/%

∆tr =

1− β

1 + β∆te =

13

∆te

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6)/*%'1%!hr$H%('%+),00')K%E4)/'%'-%0'0%\K)-/'%M'MV/'0%NF.\K,3)S'%

+5Z)S'%!q%M5,0%N)10%(F'03)4'J%E(%0F.4/)0'%0K/%K1'%MZ0-./,'K0'%

3()1[-'H%'1%(F)1%9hrqo%

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γ =20061.5

≈ 1337 =⇒ 1− β ≈ 2.8 10−7

γ =25× 365

15≈ 608 =⇒ 1− β ≈ 1.3 10−6

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βAC =k2

AC − 1k2

AC + 1=

βAB + βBC

1 + βAB βBC

k<AG%j%k

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<_Ga%

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c

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r c= (1− β cos θ) dte

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#$>`-

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νobs =νp

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ct = γ(ct + βx)x = γ(βct + x)y = y

z = z

wx =dx

dt=

γ (β cdt + dx)γ (cdt + β dx)

=w

x + v

1 + wx v/c2

wy =dy

dt=

dy

γ (cdt + β dx)=

wy

γ (1 + wx v/c2)

w⊥ =w⊥

γ (1 + w · v/c2)w =

w + v

1 + w · v/c2

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=⇒ sin θ =sin θ

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cos θ = βsin θ = 1/γ ∆θ = θ − θ ≈ β

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v⊕c≈ 10−4

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dΩ

dΩ=

1− β2

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tr = te +D(te)

c

vapp =v sin θ dte

dtr= c

β sin θ

1− β cos θ

dtr = dte −r · v dte

r c= (1− β cos θ) dte

vapp maximum quand cos θ = β et alors vapp = γβc

vapp > c si β >1√2

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