# Time dilation for Dummies

1. Feb 6, 2008

### osiris774

I've read alot on time dilation and I'm still not grasping how someone traveling close to the speed of light would age slower than someone on let's say earth for instance. Can someone explain in Lamen's terms?

2. Feb 6, 2008

### ZapperZ

Staff Emeritus
3. Feb 6, 2008

### country boy

The guy at rest will see the watch of the moving guy running slow. But each has to age according to his own clock, so the moving guy is seen as aging slower. But how to do the comparison of ages? See the "twin paradox" in ZapperZ's reference.

Now, think about the fact that each guy sees the other as moving, so each sees the other's watch as running slow. But how can they both see the other as aging slower?

4. Feb 9, 2008

### Lojzek

In theory of relativity two events that happen at the same time in one system do not necessarily happen at the same time in another system. For example: if I see that you are much younger than me, it does not mean you see me much older, because the two events (A: me watching and B: you being observed) are not simultaneous in your system.

5. Feb 9, 2008

### rbj

a layman's explanation of the physics, or why time appears to advance more slowly about the guy whizzing past you at (0.9)c .

a fundamental axiom of Special Relativity is that, if you're not being accelerated, your experience and measurement of physical phenomena is the same as the experience and measurement of physical things that some other unaccelerated observer has. even if that other unaccelerated observer has a different constant velocity than you. even if that difference constant velocity is as high as (0.9)c .

you think that you are drifting along in space and that you are "stationary", and it's everything else that is moving past you. but so does the other guy, who has equal claim to be "stationary" and it is you whom is whizzing past him at (0.9)c . Relativity says you're both right. (or neither is right if you are both claiming an absolute degree of being stationary.)

now, suppose you and he are looking at the same beam of light. the laws of physics give an expression (from solving Maxwells Equations) for the speed of light, but the laws of physics have to apply equally to both of you. so you both have to measure the speed of light to be the same. even for the same beam of light.

now, a thought experiment we have for Special Relativity is that of the "light clock", a hypothetical clock that ticks each time a pulse of light bounces off a mirror a meter away (or some other known distance, there is no reason this has to be a meter) and back round trip. the time it takes that light clock to "tick" is the distance of 2 meters divided by c, the speed of light (or 2/299792458 seconds).

now, say you and the other guy have identical light clocks. now let's say that he's whizzing past you, from west to east at (0.9)c with his meter stick and light clock oriented at a right angle to the direction of travel (let's say up and down), as you see his direction of travel. and you appear to him to be whizzing past him from east to west at the same (0.9)c. if he is whizzing past you at (0.9)c, and if his clock (as you would see his clock) was ticking at the same rate as his, you would have to measure the speed of light to be moving faster, because you see the distance that light has to move as farther. it has to move up a meter, but it's also moved east 90 centimeters. it's a right triangle, and because of Pythagorous, between the mirror reflections, that light pulse had to move

$$\sqrt{100^2 + 90^2}$$

or 134.5 cm or 1.345 meter. so, to tick at the same rate that light would have to go faster by 34.5%

but the axiom is that you and he have the same speed of light, because you and he have the same laws of physics. so because you see that his light has to travel a farther distance (than the other guy who is moving along with his meter stick and light clock), and because you measure the speed of light to be the same, if the light has farther to go with his "moving" clock than with your "stationary" clock ("moving" and "stationary" are in the eye of the beholder), you must see his clock as taking a longer time to tick. 34.5% longer. his clock has to be ticking (from your POV) at a rate of 1/1.345 of the rate. you see his the length in time between his ticks to be (2)(1.345)/c or (2)(1.345)/(299792458) seconds.

6. Mar 2, 2011

### jp-london

Can I as a follow-up (dummie) question?

Simply put - If I stand at a point in space and send my friend A East at 60% the speed of light (.6c), and then send my friend B West at .6c - both of these seem feasible for me. What speed are they travelling relative to each other?

Is this even a valid question?

7. Mar 2, 2011

### Staff: Mentor

It's a perfectly valid question. You need to use the relativistic addition of velocity formula, explained here: How Do You Add Velocities in Special Relativity? (And here: Einstein Velocity Addition)

Their speed relative to each other is about 0.88c.

(Note that this thread is over 3 years old!)

Last edited: Mar 2, 2011
8. Mar 3, 2011

### Gabe21

i guess im still not getting it, but if 2 objects depart in opposite directions both objects are moving at .6c from origin; how can there only be a distance of .88 light years between them after one year of travel?

if you only recorded the distance of one of those objects after a years time would it still have traveled .6 lightyears?

9. Mar 3, 2011

### Staff: Mentor

Whenever you mention distance or time you must specify the frame in which the measurements were made. In this case the frame is that of one of the moving ships. According to ship A, after 1 year ship B will have traveled 0.88 light years with respect to ship A.
If you measure their travel with respect to the original frame (the frame in which the ships travel at 0.6c), then according to that frame the ships will each travel 0.6 light years in 1 year.

But different frames measure length and time differently.

10. Mar 7, 2011

### jp-london

Thanks for the follow up and link to calculating speed in special relativity

I suppose the lay-persons challenge is getting over practical experience where relative speeds simply add up.

So time dilates for all cases of any velocity between 2 points, just that in our experience the dilation is so small it is not noticible until you get to significant relative speeds.

Indeed, most primers on relativity start by putting us into our known mindset and we think 'relative' speed means your speed 'on the train' plus the speed of 'the ball you throw in it'.

This keeps us thinking you can add them up.

Am I correct that, in fact, what we need to consider is that the speed between you and the ball you throw is different from your view than it is from the observer outside the train.

So if the train goes at a speed X, and you throw a ball at Y, you see the ball going away from you at Y, but the outsider not on the train does NOT see the ball moving away from you at Y, but at a value (slighlty) less than Y because you and the ball are both moving relative to the person outside the train (and time dilates some for you and the ball relative to the observer). Indeed if your train moves at .8c relative to me, and you throw at .8c relative to you, I see the ball moving away from you at much less than .8c (if I calculate correctly, it would be seen to be moving away from you at 0.175c)... yeh?

Thanks

11. Mar 7, 2011

### Staff: Mentor

Right. According to the outsider's observations, the ball separates from the thrower at a rate of 0.175c.

12. Mar 10, 2011

### snuz2001

Suppose we have two trains moving at 0.6c relative to each other, along 2 long tracks 1 m apart, and the passengers in each train can see the other train's clocks. What time will passengers in train A see in train B at say, 1 PM in train's A time? at 2 PM? 3 PM? and vice versa?
Thanks

13. Mar 23, 2011

### snuz2001

Suppose we have a long road. every 150,000 km (93,000 miles) along the road there is a clock, and above it a high resolution camera. The clocks are numbered, 1, 2, 3....100. All clocks are synchronized.

A car is riding along the road, at half the speed of light (150,000 km/sec) relative to the road. On its roof a big clock, and a high resolution camera.
Every time the car pass by a clock, the cameras in the car and on the road takes a picture of both clocks: the one on the road and the one in the car.
The question is: If at clock #1 on the road, both pictures from the cameras, the one on the road and the one on the car, will show exactly 09.00.0000000, what will they show in clock 2, and 3?

If you think that there is time dilation, remember that relative to the car, the road is the one which is moving, and since unlike the twin paradox, no acceleration is involved, which clock will slow down?

14. Mar 23, 2011

### Staff: Mentor

The clocks are synchronized in the frame of the road, but not according to the frame of the moving car. And the distance between those clocks is less according to the frame of the car.

OK: Both car clock and road clock #1 read exactly 9am when they pass each other.

According to the road frame, it takes the car D/v = .5c-s/.5c = 1 sec to reach clock #2, so when the car passes clock #2, clock #2 will read 1 sec after 9am.

According to the car frame, it takes clock #2 (D/γ)/v = (.5c-s/1.155)/.5c = 0.866 sec to reach him, so when clock #2 passes him, his car clock will read 0.866 sec after 9am. (Note that according to the car, the distance between the road clocks is reduced by a factor of γ = 1.155 due to length contraction.)

(Similar reasoning will tell you the clock readings when the car passes clock #3.)

Of course there is time dilation. Each frame views the other's clocks as running slow. The car sees the road clocks as running slow and the road frame observers see the car clock as running slow. What's the problem?

15. Mar 23, 2011

### Naty1

We explain, as above, not WHY something like this happens...because it makes no common sense....but we explain WHAT happens...WHAT IS experimentally observed.....that is, the theory is experimentally confirmed and verified.....

"Why do positive and negative charges attract each other?"

Or
"Why does elapsed time vary with gravitational potential?"

We can describe WHAT happens thanks to keen insights by predecessors, but not WHY.

Now that would be a real accomplishment!!

16. Mar 23, 2011

### snuz2001

The question was: "what will the pictures from both cameras, the one in the car, and the one on the road will show?"
An answer could be: At clock 1, the picture from the camera above the clock will show 090001 at the road's clock, and 09000086 at car's clock. and at clock 1, the picture from the car will show 0900001 at the car's clock and 09000086 at the road's clock.
try to answer it, and see if you do not run into a problem.

17. Mar 23, 2011

### Gabe21

the road camera will show the car clock as running slow and the car cam will show the road clock running slow. but you cant look at it from both positions at the same time so their is nothing wrong.

18. Mar 23, 2011

### Gabe21

if you r in the car its the road clock that is doing the moving, and if you are on the road its the car clock that's doing the moving.

19. Mar 23, 2011

### Staff: Mentor

At clock number n the clock on the road will show 09:00 plus $$n\,1\,s$$

At clock number n the clock on the car will show 09:00 plus $$n\,\frac{\sqrt{3}}{2}\,s$$

Last edited: Mar 23, 2011
20. Mar 24, 2011

### ghwellsjr

Your question has been answered but maybe you should have also asked how that answer adheres to time dilation (and length contraction). Although this was pointed out, it didn't "click" with you so let me try to explain it in more detail.

First off, you defined a hundred clocks/cameras along the road and one clock/camera on the car. As was pointed out, in the frame of the road, it takes one second for the car to traverse the distance from one clock to the next, so the camera on the car will show 100 pictures of different clocks with one-second difference between them.

But there is only one clock on the car and since it is time dilated, according to the rest frame of the road, it will be running slow by a factor of 0.866 which is the time difference shown on each of the one hundread cameras along the road. These cameras are showing the times on the one clock on the car and have only one picture each in them.

Now you are probably thinking that this is only a one-way time dilation and "where's the symmetry", correct?

Well, as was pointed out, from the rest frame of the car, the clocks along the road are time-dilated by a factor of 1.155, meaning that it takes 1.155 seconds according to the car for them to tick each second. But they are closer together by a factor of 0.866. These two factors cancel out so that when the car passes each clock, they read one second apart and, of course, the time on the car's clock has only progressed 0.866 seconds because they are closer together by that factor.

So from the car's FOR, it takes one hundred pictures of the road clocks that are one second apart and its own clock has advanced by 0.866 seconds between each of these shapshots and this is the time interval that shows up on each of the one hundred road cameras.

Can you see now how each observer sees the other one as experiencing time dilation?

By the way, the reason why your scenario may not seem symmetrical is because you are comparing the times taken by one camera of a hundred clocks with the times taken by a hundred cameras of a single clock. Usually, when showing the symmetry of time dilation, there is just one camera/clock for each observer and they take pictures of each others' clocks at one-second intervals according to their own clock. Because of the finite light propagation time, both cameras would end of with a series of pictures that had the same times on them (separated by 1.732 seconds when moving away from each other or 0.577 seconds when moving towards each other). This is called the Relativistic Doppler Effect.

Last edited: Mar 24, 2011