The formula for calculating how much time passes for a still observer is T=TO/(1-v^2/c^2)^1/2. How do I calculate what TO is?
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3$$\Delta t' = \frac{\Delta t}{\sqrt{1-v^2}}$$ In natural units, where $t^{'}$ is the time observed by a stationary object, and $t$ is the time as measured by the object in motion. – Mar 09 '14 at 17:44
1 Answers
The equation for time dilation is
$$\Delta t =\frac{\Delta t_{0}}{\sqrt{1-\frac{v^2}{c^2}}}=\Delta t_{0}\gamma$$
Where $\Delta t$ is the time that passes for the moving observer (according to the stationary observer), $\Delta t_{0}$ is the time that passes for the stationary observer, $v$ is the relative velocity between the two observers.
There is also length contraction:
$$ l =l_{0}\sqrt{1-\frac{v^2}{c^2}}=\frac{l_{0}}{\gamma}$$
Where $l$ is the objects length according the stationary observer and $l_{0}$ is the original length (or length of object according to moving observer). There is also relativistic energy
$$E=\gamma mc^2$$
Where this expression accounts for both rest mass energy and kinetic energy. The factor $\gamma$ can be anything from 1 to infinity (when $v=c$).

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