# Time dilation-doubts

1. Sep 4, 2013

### suryanarayan

Can anyone explain about time dilation in a simple way?

I am confused about these facts:

1) If an observer A at rest relative to a moving spacecraft sees an event e1 in the space craft and measures the time interval to be t1 on his clock and t2 on the clock in the spacecraft.... if the observer B in the space craft sees this event and measures it as t3 on his clock and as t4 on A's clock...how are the different times connected?

2) in the case of muons travelling more distance than they are supposed to, the time we measure as their half life on the clock with us is supposed to be the dilated time and their true half life is the one measured by an observer on the muon frame in his clock...is this right??

3) if two very precise clocks are synchronized and one is kept at rest on the earth and the other is taken on a voyage and finally are brought together..Will the times be still different ??if yes ,why does this happen??
Are the clocks relevant in this discussion?
please explain this in a simple way(without world line diagrams)

2. Sep 4, 2013

3. Sep 4, 2013

### Staff: Mentor

I modified the question slightly because an event doesn't have a time interval, it is a point in time idealized as having no duration.

The events for observer A and observer B are related by the Lorentz transform. In this formula the time and position for one observer are t, x, y, and z, and for the other observer they are t', x', y', and z':
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/ltrans.html#c2

Yes, a clock carried with the muon would be time dilated in our frame, and as a result it would display the normal muon half-life at the time 50% of the muons had decayed.

In an inertial frame moving clocks run slow. The amount that they run slow is a factor $dt/d\tau=(1-v^2/c^2)^{-1/2}$. You can use that rule in any inertial frame to determine what they each read when they are brought together.

4. Sep 4, 2013

### Staff: Mentor

In general, yes. The experiment has even been done in real life by putting one of a pair of very sensitive clocks on an airplane.

This situation (separating two clocks, moving them in different ways, then bringing them back together) requires more than just time dilation for a proper explanation - if you search for "Twin Paradox" you'll find much more. Don't do this until after you thoroughly understand the basics of the Lorentz transformations, though.

5. Sep 4, 2013

### suryanarayan

im more confused about the role of clocks...and the connection between these times(t1,t2,t3,t4)

6. Sep 4, 2013

### Staff: Mentor

The times in one reference frame are connected to the times in another reference frame by the Lorentz transform. Did you read the link I provided?

7. Sep 4, 2013

### ghwellsjr

Aren't there just two clocks? A has a clock and B has a clock, correct? So, if the event occurred on the spacecraft when B's clock on the spacecraft displayed t2, then t3 will be equal to t2, correct?

Now as far as A is concerned, the standard way in Special Relativity for A to measure the time according to his own clock at which a remote event occurred is for him to continually send radar signals to the remote location and wait until he sees the reflection of the signal when the event occurs. Then he takes the average of the times according to his clock when the particular radar signal was sent and received and that is what he will call t1.

Now since no identifiable event occurred for A at his time of t1 (he doesn't even have any awareness of the event at that time), there is no similar way for B to assign the time you are calling t4. The only way he could do this is for him to have been sending radar signals reflecting off of A's clock keeping track of the sent and received times and then when A finally does the calculation for t1, B will have to see what time A gets and then he will have to go back to his list and find out which radar signal reflected off A's clock at that time and then he can take the average of the sent and received times to calculate t4.

I could draw you a world line diagram which would make this all very clear but since you seem to have an aversion to them, you will have to be satisfied with my "simple" explanation.

Last edited: Sep 4, 2013
8. Sep 4, 2013

### harrylin

In addition to the other replies: with "their true half life" you probably mean their proper half life - the half life as measured in a frame in which they are in rest. Relativity considers the observations with either inertial frame as equally "true" or "apparent".

9. Sep 4, 2013

### Nickelodeon

I think the actual variation in time is just dependent on what acceleration you subject the voyaging clock to and there will be no time variation once your clock is traveling at a constant speed, even if it traveled for years and years. I'm not sure what happens when you relatively subject it to minus acceleration to return it to its original position next to the clock on earth. I sort of think it will read the same time despite the claims of the quoted plane experiment.

10. Sep 4, 2013

### Staff: Mentor

The acceleration is irrelevant and the amount of time lag between the two clocks will depend on their relative speed and the time they've been traveling.

We can calculate the time experienced on a journey between two points in spacetime by using the formula $\tau = \sqrt{\Delta{t}^2-\Delta{x}^2}$ where $\Delta{t}$ and $\Delta{x}$ are the time and space differences between start and end as measured by any inertial observer. Remarkably, although different observers will see very different values of $\Delta{t}$ and $\Delta{x}$, the value of $\tau$ comes out the same.

For example (assuming that the speed and direction changes are instantaneous, which doesn't change the result but does simplify the calculation - no calculus required!) we have two clocks. One of them stays at rest on the earth and the other one travels at a speed .5c for one year (as measured by us stay-at-home people), then turns around and returns to earth.

Using the measurements of the stay-at-home clock (we choose this one because it is inertial) we have three interesting points in spacetime:
A: (x=0, t=0 years): The two clocks both read the same time, and the traveling clock starts its journey.
B: (x=.5 lightyears, t=1 year): Using the measurements of our earth-based observer, the travelling clock has been traveling outbound for a year, so is one-half light-year away. The traveling clock turns around and start back to earth.
C: (x=0, t=2 years): The two clocks meet again, as the travelling clock completes its two-year round trip.

Use the $\tau = \sqrt{\Delta{t}^2-\Delta{x}^2}$ formula between points A and C and you'll get the time elapsed on the stay-at-home clock: Two years.

Use that formula on the A-B leg and then the B-C leg followed by the traveling clock, and you'll get .75 years for each leg, for a total time elapsed on the travellng clock of 1.5 years. Try this with the clock moving faster, or traveling further for a longer time, and you'll get even bigger lags.

Last edited: Sep 4, 2013