Ein Laser
x
→
L
a
s
e
r
{\displaystyle {\vec {x}}_{Laser}}
x
→
P
h
o
t
o
n
{\displaystyle {\vec {x}}_{Photon}}
c
=
299.792.458
m
s
{\displaystyle c=299.792.458{\frac {m}{s}}}
S
{\displaystyle S}
v
{\displaystyle v}
a
=
0
m
s
2
{\displaystyle a=0{\frac {m}{s}}^{2}}
S
′
{\displaystyle S'}
x
→
B
e
o
b
a
c
h
t
e
r
{\displaystyle {\vec {x}}_{Beobachter}}
S
0
{\displaystyle S_{0}}
v
=
0
,
a
=
0
{\displaystyle v=0,a=0}
Das Photon
x
→
P
h
o
t
o
n
{\displaystyle {\vec {x}}_{Photon}}
c
t
{\displaystyle ct}
x
→
L
a
s
e
r
{\displaystyle {\vec {x}}_{Laser}}
v
t
′
{\displaystyle vt'}
c
t
{\displaystyle ct}
v
t
′
{\displaystyle vt'}
c
t
⊥
v
t
′
{\displaystyle ct\perp vt'}
x
→
B
e
o
b
a
c
h
t
e
r
{\displaystyle {\vec {x}}_{Beobachter}}
x
→
P
h
o
t
o
n
{\displaystyle {\vec {x}}_{Photon}}
c
t
′
{\displaystyle ct'}
c
t
′
>
c
t
{\displaystyle ct'>ct}
S
{\displaystyle S}
S
′
{\displaystyle S'}
t
≠
t
′
{\displaystyle t\neq t'}
t
{\displaystyle t}
t
′
{\displaystyle t'}
t
∼
t
′
{\displaystyle t\sim t'}
γ
{\displaystyle \gamma }
Durch die Beobachtung entnehmen wir, dass
t
′
=
t
γ
{\displaystyle t'=t\gamma }
und
(
c
t
)
2
=
(
c
t
′
)
2
−
(
v
t
′
)
2
{\displaystyle (ct)^{2}=(ct')^{2}-(vt')^{2}}
Nun können wir ausklammern
c
2
t
2
=
c
2
t
′
2
−
v
2
t
′
2
{\displaystyle c^{2}t^{2}=c^{2}t'^{2}-v^{2}t'^{2}}
und mithilfe des Distributivgesetz vereinfachen
c
2
t
2
=
t
′
2
(
c
2
−
v
2
)
.
{\displaystyle c^{2}t^{2}=t'^{2}(c^{2}-v^{2}).}
Dies können wir nun mit
c
2
t
′
2
{\displaystyle c^{2}t'^{2}}
t
2
t
′
2
=
1
−
v
2
c
2
{\displaystyle {\frac {t^{2}}{t'^{2}}}=1-{\frac {v^{2}}{c^{2}}}}
hieraus können wir dann die Quadratwurzel ziehen
t
t
′
=
1
−
v
2
c
2
{\displaystyle {\frac {t}{t'}}={\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}
Wenden wir das von oben an kommt heraus
γ
=
1
1
−
v
2
c
2
{\displaystyle \gamma ={\frac {1}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}}
Da
a
=
0
m
s
2
{\displaystyle a=0{\frac {m}{s^{2}}}}
v
=
c
o
n
s
t
.
{\displaystyle v=const.}
l
′
=
l
1
−
v
2
c
2
{\displaystyle l'=l{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}
und
t
′
=
t
1
−
v
2
c
2
{\displaystyle t'={\frac {t}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}}
Wie schon im erseten Versuch schießt ein Laser
x
→
L
a
s
e
r
{\displaystyle {\vec {x}}_{Laser}}
x
→
P
h
o
t
o
n
{\displaystyle {\vec {x}}_{Photon}}
c
=
299.792.458
m
s
{\displaystyle c=299.792.458{\frac {m}{s}}}
S
{\displaystyle S}
x
→
B
e
o
b
a
c
h
t
e
r
1
{\displaystyle {\vec {x}}_{Beobachter1}}
E
k
i
n
,
B
e
o
b
a
c
h
t
e
r
1
{\displaystyle E_{kin,Beobachter1}}
x
→
B
e
o
b
a
c
h
t
e
r
2
{\displaystyle {\vec {x}}_{Beobachter2}}
E
k
i
n
,
B
e
o
b
a
c
h
t
e
r
2
=
0
{\displaystyle E_{kin,Beobachter2}=0}
Beobachter
x
→
B
e
o
b
a
c
h
t
e
r
1
{\displaystyle {\vec {x}}_{Beobachter1}}
x
→
P
h
o
t
o
n
{\displaystyle {\vec {x}}_{Photon}}
