Joonis 8 Keha m liigub ajas.
Tavaruum K nihkus K´ suhtes s1,2. Keha M nihkus K suhtes s3,5, kuid K´ suhtes aga s3,4. Tavaruum K hyperruumis K´ - K( x6´,0,0,t4 ). Keha M tavaruumis K – M( xf,0,0,t4 ), kuid K´ M( xg´,0,0,t4 ).
Tavaruumis K:
Hyperruumis K`:
M( xg´,0,0,t4 )
K( x6´,0,0,t4 )
m( x´,0,0,t )
M( xf,0,0,t4 )
m( 0,0,0,0 ) = m( 0 )
Hyperruumis K´:
Aeg:
( t # t2 # t3 # t4 )
Hyperruumis K´:
I
II
III
IV
M( x1´,0,0,t ) # M( xb´,0,0,t2 ) # M( x4´,0,0,t3 ) # M( xg´,0,0,t4 )
K( x2´,0,0,t ) # K( x3´,0,0,t2 ) # K( x5´,0,0,t3 ) # K( x6´,0,0,t4 )
I
IV
m( x´,0,0,t ) = m( x´,0,0,t )
Tavaruumis K:
I
II
III
IV
M( x1,0,0,t ) # M( xb,0,0,t2 ) # M( xd,0,0,t3 ) # M( xf,0,0,t4 )
m( x,0,0,t ) # m( xa,0,0,t2 ) # m( x,0,0,t3 ) # m( 0,0,0,0 )
Keha m nihkus K´ suhtes s2,5, kuid K suhtes s = 0. Keha m tavaruumis K – m( 0,0,0,0 ), kuid
K´ suhtes aga – m( x´,0,0,t ). Keha m kaugust ( nihet ( s ) ) kirjeldabki aeg ( t ).
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