In the time dilation equation $$t'=t/\sqrt{1-(v/c)^2}$$ where $ t'$ is the time measured by an observer in motion for the same event, where $t$ is time measured by the observer at rest.

Imagine a situation where both of the observes measure a time of 5 sec on their watches. These events (an observer seeing the fifth tick on his watch) are not simultaneous.

Now, my question: is the time taken by the observer in motion given by the equation $$t'=t/\sqrt{1-(v/c)^2}.$$ Does this signify that as time is travelling slower for the observer in motion it takes more time than the observer at rest to measure the time interval of 5 sec?

  • $\begingroup$ Its recommended you type your math out with this code in the future! meta.math.stackexchange.com/questions/5020/… $\endgroup$ – Hritik Narayan May 24 '15 at 5:59
  • $\begingroup$ I think the answer is yes. You are basically asking what the time dilation formula means/how to apply it, I think? $\endgroup$ – innisfree Nov 4 '15 at 12:39
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    $\begingroup$ Your question is very hard to make sense of. You say "both observers measure a time of 5 sec on their watches". Where is each one when they start measuring? You say that the fifth ticks are "not simultaneous". According to whom? There's always some observer who considers them simultaneous. You need to specify exactly who is doing what, when and where. $\endgroup$ – WillO Nov 4 '15 at 15:06
  • $\begingroup$ As WillO said, the question is unclear. Saying the events are "not simultaneous" is meaningless because there is no unique notion of simultaneous, you need to say in which frame they are not simultaneous. $\endgroup$ – ACuriousMind Nov 4 '15 at 18:00
  • $\begingroup$ @acurious that's my bad. I added the comment about simultaneous as answers said if they both see 5s then $\gamma=1$. Which I'm sure isn't what the OP meant. $\endgroup$ – innisfree Nov 4 '15 at 21:49

(Only SR answer)

For both the observers to measure the same time duration between two specific events, (5 secs in this case) we have the condition that $t'=t$.

This is only possible when $1-(v/c)^2$ is equal to $1$. Which implies that the relative velocity $v$ of one of the observers with respect to the other is zero.

So if two observers measure the same time duration between two specific events , they don't move relative to each other! (which rules out the conclusions you made in the last statement!)

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    $\begingroup$ -I think u have misinterprated the question.I wanted to saw that both are measuring a 5 sec during with their own watch(they are not measuring any event in a particular reference frame but they are measuring a time interval of 5 sec according to their local time). $\endgroup$ – SHM May 24 '15 at 17:36
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    $\begingroup$ -You are not understanding.Is it always necessary for two bodies to be at rest to measure a time interval of 5sec? $\endgroup$ – SHM May 25 '15 at 2:28
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    $\begingroup$ -Imagine a situation.you know that time runs slower as speed increases.Now let,observer A with a speed of 0 meter per sec.Another observer B with speed of 0.99c .so,time moves slower for B with respect to A.So,clock moves slower for B than A.Let both measure 5 sec with their watches.That does not mean that they are with rest with each other.My question is if 5 sec is longer B than A with respect to an independent observer who measures B's time with respect to C. $\endgroup$ – SHM May 27 '15 at 14:26
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    $\begingroup$ @soham Who is C? I mean what is its state of motion wrt A or B? $\endgroup$ – Dvij Mankad Nov 4 '15 at 11:13
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    $\begingroup$ This answer is at best misleading, because it talks about "the time duration for an event" (which is, of course, always zero), when it should be talking about the time duration between two events. $\endgroup$ – WillO Nov 4 '15 at 15:04

The question seems a bit unclear. But I am trying to answer it according to what I understand the question is and that is as follows:

Let's say $B$ is moving wrt $A$ in the positive $X$ direction with speed $v$. They are at the same point in space at some instant and both of them start their clock precisely when they are at the same position in space. Now when any of the observers observes $5\ seconds$ elapsed in his own clock fires himself. So will it happen that Observer $A$ will never observe Observer $B$ to fire himself (because $B$'s clock is running slow wrt $A$)?

A straight answer is 'Yes'. Since the clock of B runs slower wrt $A$ this has to happen. In other words, $A$ does note in his diary that $B$'s clocks were faulty and its hands took more time to show 5 minutes elapsed. That is what Time Dilation says.

But the most interesting portion is not done yet. The relativity principle suggests that $B$ will also note similar about $A$! i.e. $B$ will observe that his ($B$'s) clock has shown $5\ minutes$ elapsed but $A$'s clocks have not. They are running slow - Taking more time to show $5\ minutes$ elapsed.

Now is it a paradoxical or contradictory phenomenon? Not at all. Becuase the definition of simultaneity and invariance of the speed of light clearly dictates that simultaneity depends on the frame. Events that are simultaneous in one frame need not to be simultaneous in other and thus whatever is happening is perfectly okay and more like the only logical possibility of what can happen.


Yes indeed two observes in different frames of referencez will not necessarily agree on the time nor the positions. However they will all agree on what is called the space-time interval $s$ that follows this definition $$s^2=x^2+y^2+z^2-c^2t^2$$ Where you would recognize $x$, $y$ and $z$ to be the familiar cartesian coordinates. All observers regardless of their speed, will agree on this interval, but they wont necessarily agree on particulary when or where an event occured.

  • $\begingroup$ This is an incredibly bad answer. It suggests that two observers, at rest with respect to each other, will experience relativistic time dilation just because they see themselves as being at different locations. $\endgroup$ – WillO Nov 4 '15 at 15:00
  • $\begingroup$ Hmm I do not see how it implies anything like this! If it is so bad answer yourself! You don't even can explain how it implies anything as ridiculous as you just said. $\endgroup$ – user97166 Nov 4 '15 at 15:25
  • $\begingroup$ What I gave is a simple example just to understand that even without relativity, it is possible for two observers not to agree when something happens. $\endgroup$ – user97166 Nov 4 '15 at 15:29
  • $\begingroup$ But these observers do agree about when everything happens. If their watches are synchronized, they both agree that the light leaves the star at time $T$, that it reaches observer A at time T+1, and that it reaches observer B at time T+100,000. There are no disagreements (and can't be, because the observers' relative velocity is zero). $\endgroup$ – WillO Nov 4 '15 at 15:32
  • $\begingroup$ Indeed but no observers can know when the light came out of the star, they can only tell when they have measured it. It is not a thought experiment here! It is an example that I give without any relativity and it does not break anything since it is not the same king of disagreement indeed. It is not a real difference in time but it can seem to, I am not reinventing relativity, but just giving an example where a similar kind of disagreement follows a similar rule in terms of this interval I talked about. The example I gave ideed has nothing to do with relativity and I mentionned it! $\endgroup$ – user97166 Nov 4 '15 at 15:38

Is the time taken by the observer in motion given by the equation $$t'=t / \sqrt{1-(v/c)^2}$$

Yes. The important point to appreciate is that the Lorentz factor is derived very simply from Pythagoras's theorem. Take a look at the simple inference of time dilation due to relative velocity on Wikipedia:

enter image description here Public domain image by Mdd4696, see Wikipedia

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, and it's $\sqrt{1 - v²}$. You can write v²/c² because c=1, or you can write that as (v/c)², and we use a reciprocal to distinguish time dilation from length contraction.

Does this signify that as time is travelling slower for the observer in motion?

Kind of. But note that time isn't "travelling", and motion is relative. If you and I are two observers passing one another in space, we could each claim that the other guy is in motion, and that his clock is the one ticking slower. To appreciate what's going on here imagine we each have a parallel-mirror light clock. You see your clock light going like this ||, and my clock light going like this /\, and I do too. This is the "twins paradox" but it's nothing special. When we're separate by distance I look smaller than you and you look smaller than me. But we know about persepctive, so we don't shout woo, paradox! Nor should we when we're separated by relative motion and my time looks slower than yours and your time looks slower than mine. It's just another type of perspective.

it takes more time than the observer at rest to measure the time interval of 5 sec?

Yes. But note that you can only deem one of the observers to be at rest if the other one does and out-and-back trip.


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