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Suppose a traveler is going from point A to point B at close to light speed.

From the point of view of the traveler, what is the difference of travel time using classical mechanics or special relativity?

Can we have a formula giving the difference of travel time experienced by the traveler depending on speed? eg :

$$ CM = f(S) \cdot SR $$

With :

  • S = speed
  • CM = travel time using classical mechanics
  • SR = travel time using special relativity.
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    $\begingroup$ Yes ,$f(S)$ should be the gamma factor $\endgroup$
    – Paul
    Jan 27, 2017 at 9:57
  • $\begingroup$ I edited the question see above ;) $\endgroup$
    – user143096
    Jan 27, 2017 at 10:38
  • $\begingroup$ In your edit you completely changed your question. $\endgroup$
    – Paul
    Jan 27, 2017 at 10:46
  • $\begingroup$ Yes should I ask a new question and cancel the edits? Since someone posted an answer that perfectly answers the first question $\endgroup$
    – user143096
    Jan 27, 2017 at 10:47
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    $\begingroup$ Yes,I think you should cancel the edit $\endgroup$
    – Paul
    Jan 27, 2017 at 10:49

1 Answer 1

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A good illustration of this is the light clock. Imagine you have two observers, one sat on a train with a light clock next to them (see image below) and the other sat watching the train. Light clock

Doing a little bit of Pythagoras on the relativistic case we see: $$ L^{2} = d^{2} + (\frac{u*t_{r}}{2} )^{2} $$ Where $t_{r}$ is the time it takes the light to move from the bottom mirror, to the top and back again. We can also see that $L = \frac{c t_{r}}{2}$ since this is the time it has taken light to traverse this path and, from the stationary observer's case we see $d = \frac{c t_{c}}{2}$ where $t_{c}$ is the time it takes for light to go from bottom to top and back again in the classical case.

If we substitute these into our equation we get:

$$ (\frac{c t_{r}}{2})^{2} = (\frac{c t_{c}}{2})^{2} + (\frac{u*t_{r}}{2} )^{2}$$

Dividing through by $\frac{c t_{r}}{2}$ gives:

$$ 1 = (\frac{t_{c}}{t_{r}})^{2} + (\frac{u}{c})^{2} $$

and if we rearrange this to get an equation for $t_{c}$ in terms of $t_{r}$ we get:

$$ t_{c} = t_{r} \sqrt{1-(\frac{u}{c})^{2}} $$

Where $\sqrt{1-(\frac{u}{c})^{2}} $ is often denoted by $\frac{1}{\gamma}$, this factor is your $f(S)$ proposed in your question.

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  • $\begingroup$ I am really sorry I realized the question I asked was not exactly what I wanted and I edited it.. Maybe I should have asked another one :/ But thank you for your effort !! Should I ask a new question and cancel the edits so that it makes sense for everyone? You indeed answer perfectly to my first question :) $\endgroup$
    – user143096
    Jan 27, 2017 at 10:41
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    $\begingroup$ Alright I rolled back to previous version and accepted your answer ! Thank you for your time $\endgroup$
    – user143096
    Jan 27, 2017 at 10:51

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