# Time dilation formula?

1. Aug 28, 2009

### Grimble

I know that this is a very basic question but what is the correct formula for time dilation?

In Wikipedia etc. I read ${t^'} = \gamma {t}$ or at least $\Delta{t^'} = \gamma {\Delta{t}}$; yet in this http://en.wikipedia.org/wiki/Twin_p...t_of_differences_in_twins.27_spacetime_paths" 'phase 2' and 'phase 5' imply that the formula is $${t^'} = \frac {t}{\gamma}$$.

Also, if a moving clock is seen to 'go slow' by a stationary observer, then one would expect that less time would be seen to pass in the transformed time, and $${t^'} = \frac {t}{\gamma}$$seems to me to fit that scenario.

I have been looking at this for some time on the internet but, taking heed of the warnings I have been given about believing all I read on there, I have followed the arguments and read the 'derivations' and suchlike, but have a problem:

Whichever way I approach it the formula appears to be the latter viz. $${t^'} = \frac {t}{\gamma}$$ in the same way that $${x^'} = \frac {x}{\gamma}$$ the formula for length contraction.

where:
t is the time on the stationary observer's local clock and
t' is the travelling clock's time, transformed by the Lorentz transformation formulae.

Or are there different formulae applied in different circumstances.

We talk of time dilation - expansion(?) yet also about the moving cock slowing (less time passing)?

Last edited by a moderator: Apr 24, 2017
2. Aug 28, 2009

### dianaj

In a moving system time seems to go slower while objects seem to get longer. As

$$\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} > 1$$

for v > 0, the correct formula for time dilation must be

$$t' = \frac{t}{\gamma}$$

from which follows

$$t' < t$$

e.g. when t minutes has passed in the rest system only t' has passed in the moving system meaning that times moves slower in the moving system.

The length contraction formula must be

$$x' = x \gamma$$

as the length of an object in the moving system appears to be contracted and not dilated.

I hope this sorted out your confusion. Of course this answer is not a derivation of the equations - this is just my line of thought when I forget when to multiply/divide by $$\gamma$$

3. Aug 28, 2009

### diazona

I often go through the same thinking as dianaj. It definitely helps to remember that $\gamma > 1$.

4. Aug 28, 2009

### matheinste

To a "stationary" observer, a clock moving relative to him will appear to run at a slower rate than his. If we call the time between ticks on the "stationary" clock as observed by the "stationary" observer one second, then the time between ticks on the "moving" clock as measured by the "stationary" obsever will be greater than one second, and so in this sense the time in the "moving" frame as observed from the "staionary" frame could be described as expanded. That is, the time between ticks appears to be longer. The term dilated is normally used rather than expanded . So in a certain time as measured by the "stationary" observer on his own clock he oberves a smaller number of ticks on the "moving" observer's clock. More observed time between ticks is taken to mean time passing more slowly, and so the time observed in the "moving" frame by the "stationary" observer can be said to be passing more slowly. I expect you already knew all this but were unhappy with the terminology.

As for time passing more slowly or being dilated, there is no absolute time. For an ideal clock, elapsed time IS the time measured by a comoving observer counting the ticks. Time IS the ticks. To any inertial observer the time elapsed on HIS clock is THE elapsed time, or proper time. For this observer, his own physical time rates never alter.

Matheinste.

Last edited by a moderator: Apr 24, 2017
5. Aug 31, 2009

### Grimble

Thank you, one and all, for your inputs.

It is interesting, Dianaj, and Diazona, that you have the two formulae the opposite way round to Wikipedia, whereas I am inclined to think that you each have one right! Confusing isn't it, a slippery thing to keep one's finger on.

You give a nice summary of the problem, Matheinste: there is no Absolute time – agreed; and time is in the eye of the beholder, if I may paraphrase you.

The big difficulty I see is how to describe 'faster' and 'slower' in time when are we counting the 'clicks' but have no agreement on the size of the clicks.
For instance, one clock may be slower than the other yet record the same number of clicks.
Let me refer you to Einstein's 1920 paper: 'Relativity: The Special and General Theory.'
and in particular to chapter XII. - http://www.bartleby.com/173/12.html" [Broken] where he derives a formula for the time in the stationary system K when ${t^'} = 1$ of
$$t = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$$
in which ${t^'}$ has been replaced by 1 and which we would write:
$$t = \frac{t^'}{\sqrt{1 - \frac{v^2}{c^2}}}$$
He then goes on to state:
'As judged from K, the clock is moving with the velocity v; as judged from this reference-body, the time which elapses between two strokes of the clock is not one second, but
$$\frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$$

seconds, i.e. a somewhat larger time. As a consequence of its motion the clock goes more slowly than when at rest.'

But, consider just what he is saying here: the time t is the time in system K, the stationary system and is, therefore, 1 second proper time, but the time ${t^'}$ is the time from the system ${K^'}$ transformed into co-ordinate time (as we refer to it).

So where Einstein says that the clock slows one could just as easily say, that the time of system ${K^'}$ has been 'shrunk' or 'contracted' by the transformation such that one second proper time, system K, is now equal to
$$\frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$$
seconds co-ordinate time, which is greater than one.

So 1 second proper time in system ${K^'}$ (measured in system ${K^'}$ it is in an inertial frame of reference) upon transformation becomes only $$\frac{1}{\frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}}$$ seconds.
So one might say that it has slowed down as the same duration now measures less elapsed time, or, that the units of time have shrunk and that time now passes faster!
It all depends on what one is comparing, number of units or size of units.

My personal preference would be to say that when transformed by the Lorentz equations, the units of time become smaller.

Last edited by a moderator: May 4, 2017
6. Aug 31, 2009

### matheinste

When you compare your clock to a moving clock, the moving clocks units, that is time between ticks, as measured by you, appears to you to be longer than those on your clock. Because number of ticks and length between ticks are inversely proportional, the moving clock will appear to you to have experienced less ticks compared to your clock.

Time dilation refers to the lengthening of the distance between ticks, compared with your clock, of a clock moving relative to you as observed by you. So comparative length between ticks in the frame moving relative to you, appears to you to be longer, or dilated, That is time dilation. But the comparative number of ticks in the frame moving relative to you as observed by you appears to be decreased, that is smaller, not dilated, when compared to your clock. This is true for any observer when observing another clock moving relative to his.

Of course if you interchange the roles of the observers and let the obvserver who was considered to be moving, now consider himself, quite legitimately for the present pupose, to be at rest, then he will consider the other clock to be slower than his. All clocks moving relative to any observer will be observed by him to run slower than his own.

Matheinste.

7. Sep 1, 2009

### Grimble

Let us take an example of time dilation; if we say that for a clock moving at 0.8c, for each second that passes and is indicated on that clock, how much time, (time dilated), will the stationary observer see pass?

Now if we put the appropriate figures into the formula for time dilation $t = \gamma {t^'}$
where:
t = the proper time for the inertial observer.
v = 0.8c
t' = co-ordinate time in the moving clock's frame
$$\gamma = \frac {1}{\sqrt{1 - {\frac{v^2}{c^2}}}} = 1.67[/itex] then $t = 1.67 {t^'} So 1.67 seconds on the moving clock is equivalent to 1 second on the observer's clock, i.e. less time is seen to pass by the observer, the clock slows. But if the moving clock has 'ticked' (one second ticks) 1.67 times for each of the observer's seconds then is it not ticking faster in the observer's frame of reference? And, in the infamous Twin Paradox, would the travelling twin not have aged 16.7 years in his frame of reference, while only 10 had passed for his sedentary sibling? And would that sibling not, therefore, see his brother aging faster than himself? 8. Sep 1, 2009 ### dianaj You have [tex]t' \cdot 1.67 = t$$ and so $$t' = 1s \Rightarrow t = 1.67 s$$ The time passed describes an event: the 'hand' on the clock moving from on place to another, one second passing. According to the observer this event event takes 1.67 sec's. But according to the moving clock it only takes 1 sec. Therefore time must go slower for the moving clock (after, say, 100 of these events, the moving clock will have aged 100 secs while the observer will have aged 167 secs). It's a good idea to think of $$t$$ as $$\Delta t$$: as a time interval describing an event. Usually solves my problems, when I get confused. 9. Sep 1, 2009 ### matheinste In every case you need to make clear who is making the observation and which clock they are observing. "for each second that passes and is indicated on that clock" as observed by who? " how much time, (time dilated), will the stationary observer see pass?" on which clock? Matheinste. 10. Sep 1, 2009 ### Grimble Yes but, t' has been transformed and is in co-ordinate units and t, being in the inertial observer's frame of reference is in proper units. So what we are saying in your first equation is that 1.67 co-ordinate seconds are equal in duration to 1 second of Proper time at the current velocity. So really the proper time seconds (as seen by an observer travelling with the clock) have shrunk when transformed at 0.8c such that it takes 1.67 of them to have the same duration as 1 second proper time as measured from the observer's inertial frame of reference. OK, my friend, I will restate it: Let us take an example of time dilation; we will take a clock in an inertial reference frame, moving at 0.8c relative to an observer in another inertial reference frame. Then for each second that passes in the moving clock's reference frame, how much time, (which will be time dilated), will the remote observer see pass from his reference frame? As would be seen by a local observer in the clock's reference frame. A very good question, for the moving clock observed in its own reference frame (i.e. by a local observer) will be displaying 1 sec Proper time. And the remote observer's clock (relative to which the moving clock is travelling at 0.8c) will also be shewing 1 sec (and it will also be displaying Proper time). The dilated time is the moving clock's proper time transformed into co-ordinate time in the remote observer's frame of reference. But his clock will be displaying HIS local (proper) time. So you seem to have highlighted another conundrum, on which clock could the transformed time be displayed? One might think that it would be the moving clock as observed by the remote observer, but, if it were a clock with hands (as is often supposed) then how could the remote observer read anything but the position of the hands? It is after all the units of time that have changed (from proper time to coordinate time) not the hands of the clock. Maybe the observer has another clock set to co-ordinate time? Or is the difference in time exactly that: a difference in the units of time and the clocks would all read the same but be measuring different units of time. That is to say both observer's read 1 second on the moving clock, but the 1 second the observer moving with the clock reads (proper time), is equal in duration to 1.67 of the second the remote observer reads (co-ordinate time). 11. Sep 1, 2009 ### matheinste First, remeber that there is no "moving" or "stationary" clock, just two clocks moving relative to each other. As they are at rest in inertial frames they can each, for the sake of simplicity consider themsemselves to be at rest and the other moving. Next, you must take into account the relativity of simultaneity. Let the two clocks be colocated at the origin and there be set to both read zero (as your transformation equations imply). They will of course at this point read zero simultaneouly for both observers. However, the time when the clock of the observer who considers himself at rest reads one second IS NOT simultaneous , in his own frame, with the time when he observes the other clock read 1 second. The time of greater than 1 second shown on the clock of the observer at rest IS simultaneous, in his frame, with the time when he observes the moving clock reading 1 second. That is, to the observer at rest the other clock appears to be running slow. Of course the reciprocal case also applies. Matheinste. 12. Sep 4, 2009 ### Grimble Yes, exactly, that confirms my understanding. Thank you. And this too is exactly my understanding, for the other clock's time has been transformed. This last part is what Einstein described in the following section of his paper: “Relativity: The Special and the General Theory”. In chapter XII. “The Behaviour of Measuring-Rods and Clocks in Motion” he writes: But if t is the time shewn on the observer's clock then it is in the (proper?) seconds of an inertial frame of reference. Whereas the is surely the equivalent time in transformed (co-ordinate?) seconds. So one second on the observer's clock would be simultaneous with transformed seconds on the other clock. And the transformed seconds are smaller (contracted) and consequently they pass quicker and so the other clock will be seen to be speeded up, not slowed?:surprised And is not what is shewn on Minkowski diagrams? And thank you once again for all your help, Grimble. 13. Sep 4, 2009 ### matheinste Hello Grimble. The last part should read -----So one second on the observer's clock would be simultaneous with //www.bartleby.com/173/M5.GIF[/PLAIN] transformed seconds on the other clock. And the transformed seconds are LONGER. ----- If you interchange observers you get exaclty the same result. We have [itex]{t}^{'} = \gamma t$ and so the transformed seconds are greater in length than the "stationary" observer's seconds. Then for the inverse transform cosidering the frame in which the other observer is "stationary" and ${t}^{'}$ is his time intrerval we have ${t}= \gamma {t}^{'}$

Where

$$\gamma =$$
and is $\geq 1$

Matheinste

14. Sep 26, 2009

### Grimble

Hello again matheinste and apologies for the delay.
I'm sorry, but how can the transformed seconds be longer, when one second on the observer's clock would be simultaneous with transformed seconds on the other clock?
For is >1
Agreed.
But surely, the formula that Einstein has derived:
when ${t^'} = 1$ gives us ${t} = \gamma {t^'}$
No, they must be shorter in length.
I'm sorry and mean no disrespect to you, but I think that while changing labels to shew reciprocality is fine if those labels are arbitrary, it is inadvisable to do so where the labels have been given particular meanings.
In this case $t^{'}$ was appropriated by Einstein to denote the transformed co-ordinates. But that is purely my own view.

Grimble.

15. Sep 26, 2009

### matheinste

Hello Grimble.

I really have nothing to add.

Matheinste

16. Sep 26, 2009

### Staff: Mentor

Grimble:

Try writing down the full coordinates for each event of interest, in each frame. Einstein didn't do that in his book (at least I don't remember him doing so), but it might help, since it will show explicitly how the formulas work out.

We have a clock at the origin of the system K'. Two successive ticks of that clock have coordinates, in K', of:

Tick 1: x' = 0, t' = 0.

Tick 2: x' = 0, t' = 1.

Now transform into the system K. The two events now have coordinates:

Tick 1: x = 0, t = 0 (by definition; this is where the origins of the two systems cross).

Tick 2:
$$x = \gamma \left( x' + v t' \right) = \gamma v$$,
$$t = \gamma \left( t' + v x' \right) = \gamma$$.

So a time interval that "looks like" 1 in system K', "looks like" $\gamma$ in system K. We can interpret this as saying that the clock at rest in K' is "running slow" with respect to K, because viewed from K, the time between two successive ticks of the clock at rest in K' is $\gamma$ instead of 1.

17. Sep 26, 2009

### Bob S

Here is a very specific real-world test. BNL (Brookhaven Nat. Lab.) physicists stored muons with γ=29.4 in a circular ring. The muon's lifetime at rest is about 2.2 microseconds. In the ring, their lifetime was about 65 microsecons in the lab reference frame.
Bob S

18. Sep 26, 2009

### Rasalhague

Be aware that many (most?) texts take the word dilation in the opposite sense to Matheinste’s “Time dilation refers to the lengthening of the distance between ticks, compared with your clock, of a clock moving relative to you as observed by you.” For example, in Spacetime Physics (p. 66, problem 10), Taylor and Wheeler explicitly state that by time dilation they mean an increase in the number of seconds: “This time lapse is more than one meter of light-travel time. Such lengthening is called time dilation. To dilate means to stretch.” Presumably everyone who presents the relation in the form

$$\Delta t' = \gamma \Delta t$$

and calls this “time dilation” is going by Taylor and Wheeler’s interpretation. Of course, this is just a matter of words.

You identify t as "the proper time of the inertial observer" and t' as "co-ordinate time in the moving clock's frame", but our input--the information we actually have (this one second)--is the time between the clock's ticks in the clock's rest frame.

The time between the clock's ticks in the clock's rest frame is the proper time between these events. That's the co-ordinate time between them in the clock's rest frame. Proper time is co-ordinate time between events in a frame where they happen in the same place.

The value we want to calculate (our output) is the co-ordinate time between ticks with respect to an inertial frame in which the clock is moving at 0.8c, so we need to multiply one second by gamma to find the (longer) amount of time that will have passed between ticks in the frame where the clock is moving, namely 5/3 = 1.667 seconds for every tick of the moving clock. The proper time between two events is always shorter than the co-ordinate time between them in a frame where they don't happen in the same place.

Equivalently, we could refer to the time we're trying to calculate as the proper time between two events, one of which is simultaneous, in the frame where the clock is moving at 0.8c, with one tick of the clock, and the other of which is simultaneous, in the frame where the clock is moving at 0.8c, with the next tick of the clock. There's no paradox because in the clock's rest frame (the frame where it isn't moving)--where these two events happen in different places--even if we arrange for the first event to be simultaneous with a tick of the clock, the other won't be simultaneous with the next tick of the clock but rather will still lie in the future when the clock shows one second.

Last edited: Sep 26, 2009
19. Sep 26, 2009

### matheinste

I am sure that ALL authors agree on their use of the term "time dilation". If I appear to use it in the opposite sense then it is wrong of me to do so and my explanation of what I believe them to be saying is flawed. Perhaps one source of some confusion for you may be my use of the words "distance between ticks". This does not refer to the spatial distance travelled by light between ticks.

Matheinste.

20. Sep 26, 2009

### Rasalhague

I gathered that you meant the temporal distance (amount of time) between ticks.

Incidentally, Taylor and Wheeler refer to "light-travel time" in that quote just to clarify the significance of their use of meters (rather than seconds) as a unit of time.