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Re: équation relativiste inverse (y->y';y'->y).

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Richard Hachel

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Apr 14, 2013, 4:24:27 PM4/14/13
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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.


Richard Hachel

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Apr 15, 2013, 4:00:33 PM4/15/13
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$$\LARGE t'=\frac{t-[~\frac{y~.V_o}{c^2} +\frac {d}{c}~]}{\sqrt{~1-\frac{(V_o)^2}{C^2}}}+\frac{\sqrt{~x'^2+y'^2+z'^2}}{c} $$


Richard Hachel

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Apr 15, 2013, 4:12:21 PM4/15/13
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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:

$$\LARGE y=\frac{y'+V_o(t'-\frac{\sqrt{x'^2+y'^2+z'^2}}{c})}{\sqrt{1-\frac{(V_o)^2}{c^2}}}$$





$$\LARGE t'=~\frac{t-[~\frac{y~.V_o}{c^2} +\frac {\sqrt{x^2+y^2+z^2}}{c}~]}{\sqrt{~1-\frac{(V_o)^2}{C^2}}}~+~\frac{\sqrt{~x'^2+y'^2+z'^2}}{c} $$

$$\LARGE Wￜￜￜￜￜￜￜￜￜￜￜￜￜￜￜￜￜￜￜￜￜￜￜￜￜￜￜￜￜￜￜￜￜￜￜￜￜￜￜￜￜￜ~~!!!$$


R.H.


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