Time Dilation

1. Aug 2, 2003

Rockazella

You and a friend synchronies your watches. Your friend gets in a rocket ship and blasts off. He manages to get the ship going to 90% speed of light. You use a telescope to check out his watch. To you it looks like it’s running slow. When he lands back on earth you check both the watches and see that his is behind yours.

We’ve probably all seen a version of this story before. We’ve also probably all heard of the experiment that proved this effect (jet with the atomic clocks).

The only problem I have with it would be if you looked at it from the ‘friends’ point of view. Wouldn’t it be true that if your friend on the spaceship decided to look at your clock with his telescope, he would see yours running slow? If he kept watching it, he would see it become really behind his. Yet when he lands, somehow his is the one that’s all of a sudden behind?

What am I missing?

2. Aug 2, 2003

The problem here comes when he goes to turn around and come back. He must accelerate (that is, change his velocity to come back). In special relativity, velocity must be constant, and that means no changes in it. When you deal with a situation like this, you must use general relativity which permits accelerations, and when you do the calculations, you would wind up getting an answer as to how much the times differ.

3. Aug 2, 2003

Hurkyl

Staff Emeritus
For the sake of being (closer to) 100% accurate, Brad means that the basic formulas of special relativity are only valid in inertial reference frames. (aka ones that don't accelerate)

Special relativity can certainly handle accelerating reference frames, but the formulas must be modified appropriately, which introduces some additional effects. (the same is true in nonrelativistic physics; for example in the reference frame of the passengers of an accelerating vehicle, there is a force pushing people into their seats that one would not ordinarily observe)

4. Aug 2, 2003

zoobyshoe

I asked a different version of
this same question and no one
bothered to try to answer it for
me.

You are correct that this effect
is supposed to create a situation
where each percieves the others
watch to be going slower. This
is actually an illusion, though,
created by the means used to
check each other's time.

I was trying to figure out how the
illusion comes about but no one
bothered to try and explain.

The twin paradox turns out to be
a secondary illusion based on a
misunderstanding of the first
illusion. A cesuium clock sent into orbit came back having lost
time. Cesium clocks normally keep
super-accurate time. This was
misunderstood to be a proof of
a mistatement of Einstein's clocks
Smeone took him to be saying that
the faster an object is moving the
slower time moves for that object.

In fact, it turned out cesium clocks run slower in zero gravity.
It lost time, not because of it's
speed, but simply from being in
zero gravity. To complicate the
illusion astronauts who spend time
away from the sag-creating down-
ward pull of gravity tend to come
back looking a bit younger
than when they left.
-zooby

5. Aug 2, 2003

correct. I wasn't sure if I should mention inertial frames or not. Generally when learning of this paradox people should be aware of its resolution, and I figured it better to err on the side of safety and not invoke the differences of SR and GR in terms of inertial reference frames.

And then I am confused by something here Hurkyl. It has always been my understanding that SR is called "Special Relativity" because it deals with the special case of constant velocity. In utilizing GR to derive the equations, you would in fact derive a version of SR that incorperates acceleration as you stated, is that not true?

6. Aug 2, 2003

so you are saying relativity is wrong?

The clocks in orbit do infact lose time because of their velocity (more accurately also from the angular acceleration).

I should also point out that technically the clocks are still in a gravitational field, they are just merely in free fall.

7. Aug 2, 2003

zoobyshoe

Where are you deriving any impli-
cation I am saying relativity is
wrong from what I said?

-zoob

8. Aug 2, 2003

right there mainly the part about it being a mistatement that objects with faster velocities have slower clocks. They do infact have slower clocks, as measured by an observer in another inertial frame.

9. Aug 2, 2003

Rockazella

Ok, gotcha.
I thought it might have something to do with the accelerations. Does the direction of the acceleration matter, or does it only have to do with the changing speed?

10. Aug 2, 2003

zoobyshoe

You said:

" They do in fact have slower clocks, as measured by an observer in another inertial frame."

I believe Einstein was saying:

They seem to have slower clocks
as measured by an observer in
another inertial frame.

the slowing. Each merely seems slower to the other.

-Zoob

11. Aug 2, 2003

But the equations do indeed show that time does move at a slower rate. Yes it is dependent on being observed, but the equations show a 4-vector. The magnitude of the 4-vector is c. When your components of spatial movement increase, the temporal component must decrease, and vice versa. When you move at light speed in the spatial dimensions, you have no more component for the temporal one, hence time is stopped. The only reason you yourself would not notice this is because you are also moving slower so it seems natural for you.

12. Aug 3, 2003

zoobyshoe

Having been under the impression
or a few years that the cesium clock experiment proved that time
runs slower for faster moving
objects, and then having discover-
ed this particular proof was a mistake, I am very reluctant to
be pressured into believing any
extrordinary sounding claims.

Einstein said:

"The concept true' does not
tally with the assertions of pure
geometry because by the word true' we are eventually in the habit of designating always the
correspondence with a `real' ob-
ject; geometry, however, is not
concerned with the relation of
the ideas involved in it to objects of exerience, but only with the logical connection of
these ideas among themselves"
-Albert Einstein
The Special Theory Of Relativity

Now, even though the equasions
you speak of arrive at the conclusion that time slows down
I don't believe I'm under any
obligation, certainly not from
Einstein, to assume these con-
clusions apply in real life. They
might, they might not. It could
be fact, it could be a cesium
clock illusion. All Einstein cared
of these ideas among themselves."

As I said twice in my original
post in this thread when I started
explain how each clock seems to
be running slower to the other
observer, no one responded.

If you would care to lead me through it slowly, step by step,
I would be willing to learn.

thank you

-zoob

13. Aug 3, 2003

That I shall do.

Before we get there, one thing about the clock experiment. That is not the sole proof. Rather, we also have sub-atomic particles that are detected far lower than they should (example are muons). They should decay very quickly, and thus not reach the ground in the numbers they do. However when we take into account the fact they experience time more slowly than we do, the numbers add up correctly.

Anyways, onto the step by step process (and I did a search for your original thread so I know what to say here, and I am guessing it was the one called What is the time?).

First, we have to get out of the concept of an absolute sense of motion. Let us say you are the ONLY object in space. You are moving at a constant velocity. Is there anyway to determine your velocity? No. There is nothing with which to measure your frame of reference, nor anyway to tell if you are actually moving because there is no acceleration, and so no force.

Now, Let us say there are two people, Bill and Nicole. They are floating in space with constant velocity. Who is moving? From Bill's perspective since he feels no force, he can claim Nicole is the one moving. But Nicole can say the same about Bill. There is no way to determine who is actually moving. They both could, or only one could. IF however, we toss in a third observer, they could determine who is moving where, but then again, the same problem exists only with three people (is the observer moving?).

Now that we have defined an important concept. Velocity is relative. Both Bill and Nicole are equally valid in claiming the other is moving.

Now let us move to the next part. Light. Light travels at a constant velocity as per the Maxwell equations on electromagnetism. But a mathematical result is not enough. This was verified in the famous experiment to detect an ether by its effects that failed to produce them (and subsequent effects). Light seems to be constant no matter which direction and how fast we move. Einstein as a young lad of 16 is supposed to have asked himself what a lightwave would look like if one could catch up with it. He eventually figured out he couldn't but it set in motion his thoughts.

According to the above mentioned items, the speed of light is always the same. No matter who is moving how or when every observer will agree the speed of light is 299,792,458 meters per second. Thus both Bill and Nicole will agree light travels at c. What does this mean?

Well for one it means length must contract in the direction of motion, a property called Lorentz contraction, but the more important one you are questioning is about time.

The other consequence is that time slows down as you move faster and faster. A simple way to illustrate this is to use something called a light clock. Simply put, you have two mirrors facing each other with a photon bouncing between them. Each tick we can say represents for the sake of simplicity, one second (a tick being defined as making a cycle from one mirror to the other and back again). We are sitting at rest in our setting observing a clock also at rest. Next, we have a second light clock slide past us. As we observe it the path the photon has to take is diagonal since the whole system moves as the photon is in transit. However, as was stated earlier, in the Bill/Nicole part, the clock has equal footing to say it is at rest, and so it will still seem as though it is at rest and merely traveling up and down. We measure it to have taken longer than the other clock (say 2 seconds), but it measures itself to have taken 1 second to complete it, and that we are the ones moving more slowly. Why can it say we are moving slowly? Because at the velocity it would have to travel to attain a time dialation such as that, it would take that long for light from us, and thus information about us, to catch up with it and tell it we are doing that. Hence it is because of the relativity of observers in motion we detect time to pass more slowly for objects in motion because each has a valid claim that they are at rest (remember SR deals ONLY with constant velocity).

Now, what back to the effect on length. Say Nicole is traveling in a spaceship. Before taking off on a test course that is 2 million meters long, Bill measures the spaceship to be 10 meters long. Nicole enters the track and for the entire duration that she is on the 'track' she travels at a constant velocity. Bill has sophisticated equipment he can use to also measure the length of the spacecraft. He starts a timer at the instant the front of the ship passes a point and stops it when the tail passes the same point. Since this is all constant velocity, he can use simple d=vt. BUT using the new knowledge we have from the time bit, we know that Nicole can claim she is at rest and that Bill is moving. And since she observes Bill to be moving however fast she was, she will note his clock running slow, and thus his measurement of her craft using the slower time rate will result in a shorter length. He has a watch that will only elapse a shorter time in other words. Hence the spacecraft will have contracted.

So what does this all have to do with spacetime? Well let us use this as an example. Say at the start it always takes Nicole exactly 10 seconds to traverse the 2 million meters. However, she starts to run the distance at a slant. Bill measures a longer time for her to have completed the run. At first he might think maybe the ship was no longer traveling at the pre-determined velocity as always. But then Nicole tells him she traveled at a slight diagonal. We can then clearly see what has happened. If before she was traveling solely in the x direction, then her velocity vector was totally in that direction and hence it took 10 seconds. But as she went in a diagonal in the y direction, some of her constant velocity now had to be split. Since only the x direction would wind up at the end of the track, she has less velocity in that direction, but the magnitude of the resulting vector will be her pre-determined velocity (simple s2 = x2+y2). The same thing happens in spacetime as I stated above. If we are at perfect rest in space, then our velocity in time will be the most it can be, c. The most we can be in space is c, and perfectly at rest in time. But remember, they are not seperate, but rather united in spacetime. Thus our spacetime velocity is always c. To put it more mathematically, s2 = x2+y2+z2-c2t2. The -c2t2 part is a conversion of the temporal dimension to a spatial one so that it can be added, and it is negative due to the treatment of time as having an imaginary parameter, since it is not exactly the same as a spatial dimension. One can see that as the sum of the spatial velocity vectors' magnitudes approaches c2 (thus an actual velocity in spacetime of c) the temporal one, t, must become zero. This can be stated more effectively in calculus with rates of change.

ds2 = dx2+dy2+dz2-c2dt2

We know that the first derivative of position vectors is the velocity vector and hence we see the change in spacetime is equal to the sum of the velocities of space and time. As dx, dy, and dz increase, dt must decrease and vice versa.

Hope that helps, and any other PF people may expand or clairify points as needed. Also thanks to Brian Greene for some assistance in some of the explination via [bold]The Elegant Universe[/bold].

14. Aug 3, 2003

zoobyshoe

First off let me say how generous
it was of you to take the time to
compose that response. Thank you.

I follow with ease up to the photon clocks. From there on I
wander in and out of comprehen-
sion. I read the whole thing over
many times as carefully as I could.

You located the correct thread"What's the time?" You may
recall I mentioned having been
confused by an encyclopedia entry
that contained an obvious mistake.

I do not have a scanner which is
unfortunate because I would very
much like to be able to present
the text and graph here for some-
one to clear up for me.

Let me present the problem and
see what you have to say:

A and B are moving relative to
each other such that a point ar-
rives where they are close enough
to synchronize their clocks. Then
they separate.

When A's clock reads 1 second he
sends a light signal toward B.

When B recieves the signal he
notes his clock reads 2 seconds.

The light signal is instantan-
eously reflected back to A.

What time will A's clock read when
the signal gets back to him?

-zoob

15. Aug 3, 2003

Ok, first off, let me make sure this is not a misread. When you say instaneously reflected back, does that mean that right upon hitting B the signal is instantly back at A, or does it mean it is reflected back to A without any delay, but still has travel time?

If it is the first case, there is the problem. Instaneous transmission of signals is not possible.

If not, then both people will be valid in claiming the other's clock is moving slower. Why? A and B are moving apart. Each has the valid claim that they are at rest since it is constant velocity. However, to us, the observer at rest to both of them (note: it is to both of them) we know that the light has to catch up to either one of them. In other words, if A and B are going at 0.95c, it will take some time for light traveling at c to convey the time of A to B, and then time again for the reflected signal from B to reach A since the distance between them is increasing. In this way, thanks to the constant speed of light under any inertial frame, each and every observer can claim they are at rest and it is the other one who is moving, and thus experiences a time dialation.

The problem arises if you modify this. Say upon reciving the signal from A, person B turns around and catches up with A to compare clocks. SR says both must regard the other as ticking slower because after all, each can claim they were at rest, hence the symmetry of relativity. BUT, notice what B did. He had to accelerate dramatically to catch up with A. In this sense, B now knows for sure he was moving because he felt a force, and thus the symmetry is broken. B's clock will be definitively slower than A's clock. If they were traveling at around 99.5% the speed of light at first, and B turns around after 3 years and accelerates to catch up with A. According to his clock, 6 years will have passed (remember either one could have claimed to be at rest and hence if B turns around and goes to catch up with A again at the same velocity, it will take an additional 3 years as per his clock). However, A, having continued to move at a constant velocity will have measured 60 years to have gone by. That is what acceleration (and hence forces) do to your inertial frame. You essentially jump between various inertial frames as you accelerate so that you only experience 6 years, where as A, remaining in a constant inertial frame will experience 60 years.

16. Aug 3, 2003

zoobyshoe

Whoa! Take it easy there. You are
biting off more than I can chew.

First : When I say instan-
taneous reflection I mean it
does not take any time for the
light signal to reverse direction.
It's velocity on it's way back to
A is plain old C.

Given this problem and its stip-
ulations, what time will A's
clock read when the signal gets
back to him?

-zoob

17. Aug 3, 2003

Janus

Staff Emeritus
Since you give no value for the relative velocity between A and B, a value answer can't be given. But a formula answer can.

given that v = the relative velocity between A and B and t1 is the time between the clocks passing each other and the emission of light by A (as measured by A). then T , the time showing on A's clock upon return of the light signal is:

T = 2vt1/(c-v) + t1

In your example, t1 = 1 sec, so it becomes

T = 2v/(c-v) + 1 secs

18. Aug 4, 2003

zoobyshoe

Janus,

Thank you. That answer was very
much to the point.

I found this site:

Relativity Tutorial
Address:http://www.astro.ucla.edu/~wright/relatvty.htm Changed:9:29 PM on Saturday, September 21, 2002

which has both text and diagrams
and provides a common reference
I can point people to with my
questions. The encyclopedia was
definitely a dead end in that regard.

I'm going to have to mull this
site over a while to figure out
what it is I don't understand.

In the mean time thanks to all