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The equation for time dilation is typically given as $t'=\frac {t} {\sqrt {1-\frac {u^2} {c^2}}}$ - however, in the Feynman lecture notes, it is said that if a man moving at velocity $u$ records 1 second elapsed, then a stationary man will observe $\frac {1} {\sqrt {1-\frac {u^2} {c^2}}}$ seconds elapsed. Simply plugging in $t'=1$ and $t=\frac {1} {\sqrt {1-\frac {u^2} {c^2}}}$ creates problems - am I flipping the values for $t'$ and $t$? If so, why would the moving system's time be taken as $t$ and the stationary one's be taken as $t'$?

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  • $\begingroup$ $t$ is always the time of the observer that is at rest with respect to the event being measured. $\endgroup$ – bapowell Feb 9 '18 at 0:38
  • $\begingroup$ Is that not how I had it originally? $\endgroup$ – Andi Gu Feb 9 '18 at 0:55
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In SR (special relativity) textbooks the standard configuration presents two reference frames $S$ and $S'$ with aligned spatial coordinates and $S'$ moving in the $+x$ direction relative to $S$ with speed $v$. In that configuration the time dilation is given by
$\Delta t = \gamma \Delta t'$
where:
$t$ time as measured by the stationary observer $S$
$t'$ time as measured by the moving frame $S'$, that is proper time
$\gamma = 1 / \sqrt{1 - v^2/c^2}$ Lorentz factor
$v$ velocity of the moving frame
Then to emphasize the proper time, that is the time measured by a clock at rest in a reference frame, $t'$ is stated as $\tau$, so usually the time dilation is written as
$\Delta t = \gamma \Delta \tau$

However what is important to stress is that in SR two IRF's (inertial reference frames) in relative motion are symmetric, that is they experience the same description of physical events. The moving frame measures a time dilation of the rest observer as well.

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  • $\begingroup$ Great, this clears everything up - this equation matches up with my understanding of it. $\endgroup$ – Andi Gu Feb 9 '18 at 19:56
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it is said that if a man moving at velocity u records 1 second elapsed, then a stationary man will observe $\frac{1}{\sqrt{1−\frac{u^2}{c^2}}}$ seconds elapsed.

Isn't this consistent with moving clocks run slower? By stipulation, the (relatively) moving man finds that 1 second has elapsed according to his wrist-watch. This is the proper time, i.e., all inertial observers agree that this man's wristwatch shows an elapsed time of 1 second.

But, in any inertial reference frame in which this man is (uniformly) moving with speed $u$, the elapsed time in this frame is greater (the moving clock runs slower) by a factor of $\gamma = \frac{1}{\sqrt{1−\frac{u^2}{c^2}}} \gt 1.$

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  • $\begingroup$ Intuitively, I understand it - it makes sense that moving clocks run slower than stationary ones. However, the essence of my question is, what does $t'$ represent and $t$ represent? Because I have a feeling I'm getting them mixed up. $\endgroup$ – Andi Gu Feb 9 '18 at 2:05
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    $\begingroup$ @AndiGu asks "what does $t'$ represent and $t$ represent?" - It really should be $\Delta t$ and $\Delta t'$. In this SR context (observers in uniform relative motion etc.), it is clear that $t$ in your post is the proper time $\Delta \tau$ (equal to the elapsed coordinate time $\Delta t$ in the frame in which the two events are co-located) and $t'$ is the (elapsed) coordinate time in a relatively moving reference frame (in which the two events are not co-located). Just remember this: the elapsed proper time between two events is the smallest elapsed time. $\endgroup$ – Alfred Centauri Feb 9 '18 at 2:40
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The man moving at speed $u$ carries a clock, which measures the time $t \equiv t_0$. In the man's reference frame (where there is the clock), the clock is at rest. For this reason is less confusing to call this measured time as $t_0$.

Then, a person at rest, who is watching that the previous man is moving at $u$, will measure $t= \gamma t_0$ or, in your notation, $t' = \gamma t$.

For example: muons have a mean lifetime of $\tau_0 = 2 \times 10^{-6}$ seconds at rest (i.e. in their own reference frame). However, we observe that, when they enter in the atmosphere at high velocity, they travel more distance than what we expected for this mean lifetime. This is because, in our reference frame, their mean lifetime is bigger (even 50 times bigger) due to time dilation $\tau= \gamma \tau_0$.

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  • $\begingroup$ So in the case of your muon example (and using the notation I used in my question), and from a stationary frame of reference. $t'$ is the time elapsed from a stationary observer, and $t$ is the time elapsed from the muon's (moving) frame of reference? $\endgroup$ – Andi Gu Feb 9 '18 at 2:18
  • $\begingroup$ Using your notation, $t$ is the time measured for the muon in its own frame of reference (an object is at rest in its own frame, obviously); and $t'$ is the time measured from a stationary observer who is watching the muon moving at speed $u$. So the stationary observer measures $t' = \gamma t$, which is a bigger value. $\endgroup$ – GaussJordan Feb 10 '18 at 13:49
  • $\begingroup$ Great - however, this doesn't really match up with the notation used for space coordinates (i.e. $x$), since $x'=\gamma x$, and in this case, $x'$ represents coordinates as measured from the moving observer's frame of reference. $\endgroup$ – Andi Gu Feb 10 '18 at 22:55
  • $\begingroup$ For this reason I recommend you to use the followig notation: $L=L_0 /\gamma$ and $t=\gamma t_0$ for distance contraction and time dilation, where the sub0 índex refers to measurements in the frame which is moving (and at rest from this point of view), while the no sub0 index refers to changing the frame to another with a stationary observer who is watching the other frame. $\endgroup$ – GaussJordan Feb 12 '18 at 10:50

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