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Skladani1
  • v=20 km⋅h−1v = 20 \, \text{km} \cdot \text{h}^{-1} - rychlost vagónu
  • u′=5 km⋅h−1u' = 5 \, \text{km} \cdot \text{h}^{-1} - rychlost uživatele vagónu ve vagónu
Klasická fyzika: u=v+u′=20+5=25 km⋅h−1u = v + u' = 20 + 5 = 25 \, \text{km} \cdot \text{h}^{-1}
  • v=23c,u′=23c,u=43cv = \frac{2}{3}c, u' = \frac{2}{3}c, u = \frac{4}{3}c
  • v=c,u′=c,u=2cv = c, u' = c, u = 2c
→ too bad
Spešl fyzika:
Skladani2
  • v S′S':
    • od t1′t_1' do t2′t_2' a od x1′x_1' do x2′x_2'
    • u′=x2′−x1′t2′−t1′u' = \frac{x_2' - x_1'}{t_2' - t_1'}
  • v SS:
    • t1t_1… t2t_2, x1x_1… x2x_2
    • u=x2−x1t2−t1u = \frac{x_2 - x_1}{t_2 - t_1}
    • u=x2−x1t2−t1=(x2′+vt2′)−(x1′+vt1′)(t2′+vc2x2′)−(t1′+vc2x1′)=(x2′−x1′)+v(t2′−t1′)(t2′−t1′)+vc2(x2′−x1′)=u′+v1+u′vc2u = \frac{x_2 - x_1}{t_2 - t_1} = \frac{(x_2' + vt_2') - (x_1' + vt_1')}{(t_2' + \frac{v}{c^2}x_2')-(t_1' + \frac{v}{c^2}x_1')} = \frac{(x_2' - x_1')+v(t_2'-t_1')}{(t_2'-t_1')+\frac{v}{c^2}(x_2'-x_1')} = \frac{u' + v}{1 + \frac{u'v}{c^2}}
u=u′+v1+vc2u′\boldsymbol{u = \frac{u' + v}{1 + \frac{v}{c^2}u'}} v→0v \rightarrow 0: jmenovatel bude blízko 11, a bude platit u=u′+vu = u' + v
  • v=23c,u′=23c,u=43c1+49c2c2=1213cv = \frac{2}{3}c, u' = \frac{2}{3}c, u = \frac{\frac{4}{3}c}{1 + \frac{\frac{4}{9}c^2}{c^2}} = \frac{12}{13}c
  • v=c,u′=c,u=2c1+c2c2=cv = c, u' = c, u = \frac{2c}{1 + \frac{c^2}{c^2}} = c