Answer to Question #134825 in Mechanics | Relativity for sajal kumar

Question

Apply the special theory of relativity to prove that the time interval ( t ) in the

moving reference frame ( S' ) appears to be longer by a factor $\frac{1}{^{\sqrt{1-\frac{v^{2}}{c^{2}}}}}$ observer in the stationary frame ( S ) .

Answer 1

t=$\dfrac{t^`}{1-\dfrac{v}{c^`\scriptscriptstyle k^`\Rightarrow k}}$ = $\dfrac{t^`}{1-\dfrac{v}{c}}$

whereas, the same perceived time interval in k from the perspective of k.is given by,

$t=t^`(\dfrac{t^`}{1-\dfrac{v}{c^`\scriptscriptstyle k^`\Rightarrow k}})=t^`(1+{V \over C})$

$t={t^` \over 1-{V \over C}}=t^`(1+{V \over C})$

leading to v=0 unless t is transformed by a certain factor say $\gamma$ with respect to k, and by other factor say $\beta$ with respect to k'. This leads to;

$t={\beta t^` \over 1-{V \over C}}=\gamma t^`(1+{V \over C})$

which can be satisfied only if $\beta =1 / \gamma$

which results to;

$\gamma =$ $\cfrac{1}{\sqrt{1 - \cfrac{V^2}{C^2}}}$

$\therefore$ $t=\dfrac{t^`}{\gamma(1-\dfrac{v}{c})}$ =$\gamma t^`(1+{V \over C})$

hence the prove that;

$\gamma =$ $\cfrac{1}{\sqrt{1 - \cfrac{V^2}{C^2}}}$