# Calculate Time travel with Time Dilation [closed]

Does the following hold:

If we have:

• person A moving at a speed of $0$;
• person B moving at a speed of $xC$ (where $C$ is speed of light, $x$ a fraction)

And if time passes for $m$ minutes, does it hold that after those $m$ minutes

• time passed $m \gamma$ minutes for person A;
• time passed $m$ minutes for person B?

Where $\gamma$ is obtained using the Lorentz Transformation:

$1/(\sqrt{1-\frac{v^2}{C^2}})$

in which $v = xC$

This should imply that person B has travelled in time as a consequence of having a higher speed?

## closed as unclear what you're asking by ACuriousMind♦, John Rennie, Kyle Oman, Kyle Kanos, yuggibJun 23 '15 at 9:26

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• When you say "has traveled in time" to which time are you referring to ? – Joelafrite Jun 22 '15 at 8:58
• @Joelafrite The time for person B relative to person A. So after those $m$ minutes, how many minutes is person A older than person B (as person B should be the one that is travelling in time as he is moving at greater speed, but for him the time passed was also only $m$ minutes)? – JohnAndrews Jun 22 '15 at 9:04
• That is correct but if you want to talk about time travel (forward time travel) You have to decelerate $B$ until $x \ll 1$ and compare the time passed for $A$ and the time passed for $B$. – Joelafrite Jun 22 '15 at 9:50
• "person A moving at a speed of 0;" Anytime you write "at speed" in relativity without saying (or at least meaning) "relative to ..." you are making a mistake. In this case the mistake is implicitly privileging your own frame of reference above others. That is wrong and misses the whole point. – dmckee Jun 22 '15 at 14:45
• I saw this app passing by: play.google.com/store/apps/… – JohnAndrews Jul 19 '15 at 17:17

John, have a look at the simple inference of time dilation due to relative velocity. If you and I are identical twins, and you take a fast out and back trip, when you come back we agree that you've experienced less time than me. As you pointed out, we can relate this to the Lorentz factor and write: $$\Delta t' = \frac{\Delta t}{\sqrt{1-\frac{v^2}{c^2}}}$$ But there's no time travel involved. The Lorentz factor is just a simple application of Pythagoras's theorem, which "works" because of the wave nature of matter. The hypotenuse of a right-angled triangle represents the light path where c=1 in natural units. The base represents your speed as a fraction of c. The height gives the Lorentz factor $\gamma$, where we use a reciprocal to distinguish time dilation from length contraction.