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Richard Hachel
Apr 16, 2013, 7:59:04 AM4/16/13
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Richard Hachel a ï¿œcrit :
On a donc dans R' la nouvelle coordonnï¿œe y':
$$\LARGE y'=\frac{y-V_o(t-\frac{\sqrt{x^2+y^2+z^2}}{c})}{\sqrt{1-\frac{(V_o)^2}{c^2}}}$$
La transformation rï¿œciproque devenant logiquement dans R:
$$\LARGE y=\frac{y'+V_o(t'-\frac{\sqrt{x'^2+y'^2+z'^2}}{c})}{\sqrt{1-\frac{(V_o)^2}{c^2}}}$$
R.H.
$$\LARGE t'=~\frac{[~t-\frac{\sqrt{x^2+y^2+z^2}}{c}~]-~\frac{y~.V_o}{c^2}~}{\sqrt{~1-\frac{(V_o)^2}{C^2}}}~+~\frac{\sqrt{~x'^2+y'^2+z'^2}}{c} $$
La transformation rï¿œciproque devenant:
$$\LARGE t=~\frac{[~t'-\frac{\sqrt{x'^2+y'^2+z'^2}}{c}~]+~\frac{y'~.V_o}{c^2}~}{\sqrt{~1-\frac{(V_o)^2}{C^2}}}~+~\frac{\sqrt{~x^2+y^2+z^2}}{c}$$
Test fr.sci.divers
R.H.
Richard Hachel a ï¿œcrit :
On a donc dans R' la nouvelle coordonnï¿œe y':
$$\LARGE y'=\frac{y-V_o(t-\frac{\sqrt{x^2+y^2+z^2}}{c})}{\sqrt{1-\frac{(V_o)^2}{c^2}}}$$
La transformation rï¿œciproque devenant logiquement dans R:
$$\LARGE y=\frac{y'+V_o(t'-\frac{\sqrt{x'^2+y'^2+z'^2}}{c})}{\sqrt{1-\frac{(V_o)^2}{c^2}}}$$
R.H.
$$\LARGE t'=~\frac{[~t-\frac{\sqrt{x^2+y^2+z^2}}{c}~]-~\frac{y~.V_o}{c^2}~}{\sqrt{~1-\frac{(V_o)^2}{C^2}}}~+~\frac{\sqrt{~x'^2+y'^2+z'^2}}{c} $$
La transformation rï¿œciproque devenant:
$$\LARGE t=~\frac{[~t'-\frac{\sqrt{x'^2+y'^2+z'^2}}{c}~]+~\frac{y'~.V_o}{c^2}~}{\sqrt{~1-\frac{(V_o)^2}{C^2}}}~+~\frac{\sqrt{~x^2+y^2+z^2}}{c}$$
Test fr.sci.divers
R.H.
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