According to theory of relativity time dilation $$t=t_0 \gamma \text{ where }\gamma= \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$ suppose time for a person $A$ is $t$

and time for a person $B$ is $t_0$

Both are moving with reference to each other according to theory of relativity there is no such thing like inertial frame of reference magnitude of their velocity is $v$.

for $B$ . $A$ is moving with velocity $v$ magnitude

So time dilation for $B$ is $t_0=t\gamma$

for $A$ . $B$ is moving with velocity $v$ magnitude

So time dilation for $A$ is $t=t_0\gamma$

for $B$,$A$ is traveling in future

for $A$, $B$ is traveling in future

I am confused who is actually traveling in future. I am beganner so kindly explain in simple way. I have this question for long time it always confuses me .


1 Answer 1


The formula you wrote

$$ t = t_0 \gamma $$

is actually incorrect, the right one is with time intervals rather than absolute time

$$ \Delta t = \gamma \Delta t_0 $$

where you wrote correctly that $\gamma = \frac{1}{\sqrt{1-\beta^2}}$ and $\beta = \frac{v}{c}$. This means that if a clock is moving with some speed with respect to our frame, then we would see that clock moving at a different speed when confronting it with our clock (stationary in our frame) and since $\gamma\geq 1$ the time interval is always dilated, namely clocks moving with some speed always tick slower.

This is true for both $A$ and $B$ traveling with a relative speed of $v$. On one end $A$ observes $B$ traveling at $v$, on the other hand $B$ observes $A$ traveling at $v$. Both see the other's clock moving slower.

This is of course counterintuitive, the "mistery" is resolved noticing that special relativity only works for inertial frames. If at some point $A$ and $B$ wanted to stop and check their time, at least one of them (say $A$) would have to accelerate (to either brake or catch up to B) and the special relativity description would no longer apply to $A$ since its reference frame wouldn't be inertial anymore. Then only the description of $B$ (who stayed in an inertial frame all the time) would be the correct one. Take a look at the twin paradox for details.

  • $\begingroup$ Well explained thanks 😊 a little mistake $\beta=\frac{c}{v}$ $\endgroup$
    – Naman
    Apr 29, 2022 at 5:32
  • $\begingroup$ thanks, I corrected the formula. $\endgroup$
    – Andrea
    Apr 29, 2022 at 13:49
  • $\begingroup$ "special relativity would stop working"?!??!!!?? $\endgroup$
    – WillO
    May 3, 2022 at 4:54
  • $\begingroup$ see if you like the new phrasing $\endgroup$
    – Andrea
    May 3, 2022 at 10:55

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