0
$\begingroup$

Today I was curiously calculating/comparing the times of moving observers and the time recorded by a corresponding stationary observer using Einstein's time dilation equation as detailed in Special Relativity. I was just trying to see the relation between the time recorded by the moving observer compared to the stationary one at various fractions of the speed of light. I noticed that the dilation experienced by the stationary observer when 1 second passes for the moving observer are the same values I get when I divide the time passed for the stationary observer by the time passed for the moving observer at different spans of time (i.e 1 hour and 1 year passed for the moving observer)

I'll post a pic of my results below.

Picture of Excel table for clarification and e.g values

Can anyone tell me the connection between the ratio of time dilation of the moving observer and the stationary observer and the values I'm getting for the time dilation of 1 sec?

$\endgroup$
0
$\begingroup$

The ratio of $\Delta t_s$ to $\Delta t_r$ is always $$ \frac{ \Delta t_s }{ \Delta t_r } = \frac{1}{\sqrt{1-\frac{v^2}{c^2} } } = \gamma $$ This is completely independent of what $\Delta t_s$ or $\Delta t_r$ you choose. Also, if you choose $\Delta t_{r_1} = 1$, then $\Delta t_{s_1} = \gamma$. We must therefore have $$ \Delta t_{s_1} = \frac{ \Delta t_{s_2} }{ \Delta t_{r_2} } = \frac{ \Delta t_{s_3} }{ \Delta t_{r_3} } = \frac{1}{\sqrt{1-\frac{v^2}{c^2} } } = \gamma $$ as you are finding.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.