相対性理論の再検証

相対性理論
x^2+y^2+z^2+(ct)^2=0
x'^2+y'^2+z'^2+(ct')^2=0
x^2+y^2+z^2=-(ct)^2
x'^2+y'^2+z'^2=-(ct')^2

t'=At+Bz
x'=x
y'=y
z'=δt+γz

座標系Σ'
(x'、y'、z') = (0、0、0)

座標系Σ
(x、y、z) = (0、0、vt)

z=(v/c)t
0=δt+γz
z=-(δ/γ)t
z=vt=-(δ/γ)t
v/c=-δ/γ
δ=-γv

(x'、y'、z')=(0、0、-vt')
z'=δt=δt'/A=-vt
t'=At+B*0
δt=δt'/A=-vt'=-vAt
δ=-vA
δ=-γv
A=γ

t'=γt+Bz
z'=γ(-vt+z)

(x'^2+y'^2+z'^2)-(z^2+y^2+z^2)=-(ct')^2+(ct)^2
[γ(-vt+z)]^2-z^2=c^2[-(γt+Bz)^2+t^2]

γ^2(v^2t^2-2vtz+z^2)-z^2=
c^2[-(γ^2t^2+2Bγzt+B^2z^2)+t^2]

(γ^2v^2)t^2-2v(γ^2)tz+(γ^2-1)z^2=
c^2(-γ^2+1)t^2+2c^2Bγtz+c^2B^2z^2

γ^2v^2=c^2(-γ^2+1)
(-2v)(γ^2)=2c^2Bγ
γ^2-1=c^2B^2
(c^2+v^2)γ^2=c^2
γ^2=(c^2)/(c^2+v^2)
γ=√[(c^2)/(c^2+v^2)]
γ=√[1/{1+(v/c)^2)}]
B=-v(γ^2)/(c^2γ)
B=-(v/c^2)γ

t'=γ[t-(v/c^2)z]
z'=γ(z-vt)

x+y+z=t
x'+y'+z'=t'
t’=At+B(x+y+z)
x'=γx
y'=γy
z'=δt+γz

座標系Σ'
(x'、y'、z')=(0、0、0)

座標系∑
(x、y、z)=[0、0、-(v/c)t]
z=-(v/c)t
0=δt+γz
z=-(δ/γ)t
z=-(v/c)t=-(δ/γ)t
v/c=δ/γ
δ=γv/c

z'=[0、0、(v/c)t']
z'=δt=δt'/A=(v/c)t'
δt=δt'/A=(v/c)t'=(v/c)At
δ=(v/c)A
A=γ

t'=γt+B(x+y+z)
x'=γx
y'=γy
z'=γ{(v/c)t+z}={z+(v/c)t}

(x'+y'+z')-(z+y+z)=t-t
[γx-x]+[γy-y]+[γ{z+(v/c)t}-z]=
[{γt+B(x+y+z)}-t]

(γ-1)(x+y+z)+γ(v/c)t=(γ-1)t+B(x+y+z)

B=γ-1
γ(v/c)=γ-1
1=γ{1-(v/c)}
γ=1/{1-(v/c)}
γ=c/(c-v)

t'=γt+(γ-1)(x+y+z)
x'=γx
y'=γy
z'=γ{z+(v/c)t}
γ=1/{1-v/c)}
γ=c/(c-v)

ローレンツ変換改定版
t'=γt+(γ-1)(x+y+z)
x'=γx
y'=γy
z'=γ{z+(v/c)t}