# Time dilation . . . again!

1. Jul 17, 2006

### OS Richert

I'm sorry, I am one of those idiots who just doesn't get it. I see this question has been asked in about three threads below, and I still am having trouble with it. Please be patient with me and allow me to ask it again in my own words.

I am using Modern Physics by Taylor and Zalfiratos to self study the subject. They begin the subject of time dilation with the example of a man on a moving train, who has a light flash from the bottom, reflect off a mirror on the ceiling, and beep when it hits the ground again. The light travels a distance of twice the height of the train (in his reference). TWO observers on the ground see the light travel the two edges of a triangle with height h and base related to the speed of the train. Since the light is measured at the same speed in both reference frames, time must have dilated in the reference frame of the moving train. I will now quote the authors for a second, "This asymmetry may seem surprising, and even to violate the postulates of relativity, since it suggests a special role for the frame S'. . . . in our experiment the frame S' is special, since it is the unique inertial frame where the two events occured AT THE SAME PLACE. "

That sort of makes sense to me. But then the authors go on to describe the 1971 plane experiment where sycrhonized atomic clocks where put on planes and observed to experience time dilation with the US Naval Observatory in Washington, D.C. A 273 +/- 7ns difference. Now to complicate matters, this must also take into account the affect of general relativity with the gravity of the earth, which the author's mention in a note, but they are useing it to prove the portion due to normal SR time dilation.

Okay, finnally my question. Why did the plane time dilate with reference to the ground time? In the rest frame of the plane, the clock is ticking away AT THE SAME PLACE. BUT(!!!), if you ask the pilot and he looks down as he flies over Washington, he could just as well say the earth is moving past him at (whatever) speed it was, and the clock in Washington is ticking AT THE SAME PLACE. So he would measure the Washington clock as dilating. But when they got back together; the plane clocks was definitly behind.

Now, IF the answer to this question has something to do with the acceleration and deceleration of the plane; wouldn't that imply that this IS NOT an example of SR after all, but of GR of some sort. Couldn't you have simply accelerated the plane to its full speed then decelerated immediatly and gotten the exact same result (no need to spend time in the air as part of a seperate inertial frame).

I'm so lost.

2. Jul 17, 2006

### CarlB

Perhaps you need to have a few simple things explained.

First, when Einstein wrote his 1905 paper, the Lorentz transformations and all that were already well known. But they were written in the context of a theory of an aether theory. What Einstein did was to cut loose from the aether assumption. He recognized that it wasn't logically needed, and he removed it.

If you analyze the time dilation from the old aether point of view, then you make the assumption that one of the two frames of reference has clocks that are "correct". The other frame of reference has clocks that are "slow". You can analyze the problem with the plane and the earth in this manner. You will find that the calculations give you the same results as the usual SR theory, because the formulas are the same.

But afterwards, you may realize that the equations are symmetric. That is, you really don't know where the aether frame of reference is, and that actually, both the earth and the aircraft are not on that frame of reference but instead are moving (at unknown speeds) with reference to it.

So both the plane and the earth are moving with respect to this unknown reference frame. What does that have to say to us? It means that since the true reference frame of the world (i.e. the aether) cannot be found, then the equations that relate different choices of reference frame must be symmetric. That is, the reference frame of the rock I throw must have the same relation to the reference frame of the earth as vice versa.

With that perspective, perhaps the fact that the conversion shows that time is slowed, going both ways, will seem more natural.

And like a bunch of people are about to explain to you, the real secret is that events that are simultaneous in one reference frame are not necessarily simultaneous in another. This makes measurement of "time" kind of difficult.

Carl

3. Jul 17, 2006

### Ich

Another simple thing: neither the plane nor Washington are moving inertially. Both are moving on a circle around the center of earth.
SR states the equivalence of inertial frames only.
If you set plane speed=rotation speed of earth and loo at the experiment from an inertial reference frame (eg the one where the center of earth is at rest), you see:
-Washington moving with speed v on a circle
-one of the planes (the one moving to the west) standing still
-one of the planes movin with speed 2v on a circle (making two revolutions)
In this frame, when you read the clocks after the experiment, you see one of the planes without time dilation, Wahington with dilation according to v, the other plane according to 2v.
The result holds in every inertial frame. That means, you can´t transform away non-inertial movement with a single Lorentz transformation. The paths of the three observers are fundamentally different, and SR doesn´t change that fact.

4. Jul 17, 2006

### OS Richert

Thank you. I actually already understand the relative motion concept and that time will slow in both frames. That it why I ask my question. Why if time <appears> or <does> slow in both frames (according to an observer in each), do we learn at the end of the experiment that the plane's time ACTUALLY slowed with respect to the earth?

I do suspect this is the secret for most paradoxes. For instance, the 100m snake and the mischevious boy who wishes to scare the snake by hitting two knives 100m apart. But I am not sure how this applies to the clocks, since we are not talking about simanteous things. The events happen over time in the same place.

But either way, why is it after the plane was brought to rest in the ground frame, did ITS clock show it had slowed, and not the other way, or either way. Was it that fact that we had needed to accelerate the plane to move it to the other FRAME? If we had accelerated the ground into the planes rest from of reference, would the ground then have the slower clock?

I recogonized this as well. Doesn't this go to demonstrate my original point, that the fact that the plane's clock slowed down has nothing to do with SR in the first place.

5. Jul 17, 2006

### Ich

It is still SR. You can calculate the effect easily when you choose an inertial frame in which you do the calculations. It doesn´t matter which frame you choose, but it has to be inertial. Einstein gave the relevant equations in his 1905 paper.
It´s just that you can´t say that non-inertial frames are equivalent. They are not.
And the resolution has nothing to do with initial acceleration. It´s as simple as this: subtract any inertial movement you like (that´s what you´re allowed to do; you mustn´t subtract arbirary movement!), and one plane will be at rest all the time while Washington still makes its circle. So in Washington clocks will show less elapsed time.

6. Jul 17, 2006

### OS Richert

The plot thickens.

I just consulted a seperate Modern Physics book (Bernstein, Fishbane, and Gasiorowicz) which gave the example of the twin paradox; which is not a paradox because the two are actually not symetric. One had to accelerate and decelerate. Now if B accelerates away from A, then turns around, he knows right away he has turned around, but A doesn't realize until the distance between them divided by the speed of light has past (assuming they are radiating light). So B calculates they he is moving away and returning for equal amounts of time; while A calculates that B is traveling outword for longer. After some messy calculations, it is shown that B's time has dilated.

So since there is no absolute zero frame of reference, then it must be the changing of reference frame that causes time dilation. So let's say I am at rest on the earth, and I am manupulating two particles, both at 0.9c. I then deacelerate one particle (B) to 0.6c, wait awile and reacelerate it to 0.9c, just in time so they are together again and can talk to each other. I might make the mistake to think, according to a reference frame attached to particle (A), its time is slower because it was moving "faster" then particle B for a time. But that assumes some sort of absolute reference frame (in this case, the earth frame). But if I was a scientist on particle A; according to my universe, I have always been at rest, and a friend of mine (particle B), also a rest, suddenly acelerates to 0.3c, speeds around for awhile, and slows back down to 0. And now if we compare clocks, particle B has experienced time dilation. Am I correct about this?

7. Jul 17, 2006

### MeJennifer

99% of the confusion is due to bad teaching IMHO.

(1) Seeing a change in a moving object's clock rate, length and (relativistic) mass does not mean that the object in question's clock rate, length or mass is actually changed!

(2) Clock rate, length and mass never change for the object in question in an inertial frame while for a non-inertial frame it depends on how you define it.

(3) Basically making a measurement on something that is moving relative to us distorts our view of it's clock rate, length and mass. For the object measured there is not one single change!

(4) The notion that time dilates is simply a distorted view of what really happens in the twin traveling experiment. Time does not dilate, nothing is going any slower!

(5) That one of the twins has aged less is not the result of time slowing down for him but is because he traveled a different path in space-time. He hapened to travel on a path that took him less time than for the one that stayed behind.

(6) To remove the "paradox" one simply has to let go of the notion of absolute time.

Again, I think that 99% of the confusion is due to bad teaching.

Last edited: Jul 17, 2006
8. Jul 17, 2006

### OS Richert

Wait a second! I thought in the actual 1971 experiment the plane showed less elapsed time?

9. Jul 17, 2006

### OS Richert

But the plane's atomic clock had shown less elapsed time when brought back to rest in the reference of the earth when compared to the US navel clock in Washington, correct?

10. Jul 17, 2006

### MeJennifer

See item (5) to see why.

11. Jul 17, 2006

### OS Richert

MeJennifer,

Thank you for your effort. I see that your original post changes every minute or so! I'm not sure it fundlementally changes my question to consider it in terms of shorter paths of space time from time dilation. But I think I can some up my confusion in one question.

If two particles are in the same frame, and then travel in different frames for a time, and then return to the same frame; is it always true that the particle that actually acelerated and decelerated to change frames will have traveled a shorter path in space-time, and therefor show less elapsed time?

Thanks.

12. Jul 17, 2006

### MeJennifer

Well substitute frame with path in space-time.
But I think you almost got it.
Now is it shorter? No it is the opposite, actually the particle that traveled took a longer path in space-time.

We can actually use an equation for it, and I oversimplify it a bit, basically it is like this:

$$ds^2=t^2 - dx^2$$

Which means that the space-time distance between two events (since it is space + time we speak of events and not of locations) is related to the elapsed time (t) minus the spatial distance (x).

Now because the two particles started being together (which you call in the same frame) and ended up being together it means that their $$ds^$$ must be exactly the same. Now the one that accelerated covered more space right, so that means that its time must be less.

Let's say for the sake of argument that the space-time distance for both of them is 25 so we get:

$$25=t^2 - dx^2$$

Now assume that the traveling particle covered a distance (x) of 4, so the elapsed time (t) for that particle will be: 3

Now the particle that stayed behind covered a distance (x) of 0, so the elapsed time (t) for the particle that stayed behind will be 5.

See that there is no slowing down of time involved here.

Note: the complete equation is:

$$ds^2=c^2dt^2 - dx^2 - dy^2 - dz^2$$

Which includes all 3 space dimensions instead of 1 in my example and includes the speed of light. The speed of light can be eliminated if we express distances in terms of lightspeed.

Last edited: Jul 17, 2006
13. Jul 17, 2006

### OS Richert

Thank you. I think that clears it up for now.

14. Jul 17, 2006

### Hans de Vries

Twin Brothers in an "absolute reference frame"

Twin Brothers in an "absolute reference frame"

A good way to understand these seemingly paradoxical aspects of time
dilation is to study it from an “absolute reference frame”. That is IF there
exist such a frame and IF we will one day be able to locate it.

Now we don’t actually need to find such an “absolute reference frame”
since in SR every frame could equally well be the absolute reference
frame. We might take an arbitrary frame and just presume that it’s the
special one.

Aging goes fastest in absolute frame and everything what moves
progresses slower. The faster it moves, the slower it progresses. Now
look at the two cases of the Twin “paradox”

1) The first twin stays at rest all the time in the absolute frame. The
other starts his voyage into some direction and turns back halfway to
go back to his twin-brother. It’s evidently that the second one ages
less, OK, but now the other way around.

2) It starts the same: The first twin stays at rest and the second one
starts his voyage. In this case however the second twin doesn’t turn
back but continuous to travel in the same direction. The first twin now
goes after him and does so at a higher speed to overtake his brother at
some point. At this point they can compare their ages.

It turns out that the situation is reversed now. It’s the first of the twins
who has aged less. (by exactly the same amount). This is because of
the higher speed the first twin needs to travel to overtake his brother.
Aging slows down so much (in this shorter period) that at the end he
has aged less as his brother who continued to travel always at the
same speed.

Now look at case 2 from the eyes of the second brother: He first sees
his brother getting farther and farther away and then, halfway, he sees
his brother changing speed to get back to him. The second brother may
presume that he’s at rest all the time and that it’s the first twin brother
who went on a voyage and came back, and, therefore the first brother
should have aged less, which is indeed exactly what they find out!

What you see here is just one of the reasons why we can’t locate this
“absolute reference frame” and formally can not even be sure it exist.
Formally SR doesn’t need it. At the other hand one can also say that
each and every frame could be the “absolute reference frame”

We have not been able to locate it. Even with the really extremely
of SR.

Regards, Hans

Last edited: Jul 17, 2006
15. Jul 17, 2006

### MeJennifer

So I hope it is clear to everyone that the explanation that Hans de Vries gives here is diametrically opposed to the explanation that I gave.

While he holds the idea that time in some way can be seen as slowing down I hold the opposite opinion. The age of an object depends on the path taken in space-time not on the effects of relativistic measurements.

16. Jul 17, 2006

### Hans de Vries

Wrong. There is actually no contradiction at all.

17. Jul 17, 2006

### MeJennifer

Well that is your opinion, I simply think it is incorrect to explain that one of the twins in younger because his time slowed down. He is younger because less time has elapsed not becuase his time went slower.

The idea of time slowing down and using absolute time in this example will only cause more confusion for those who try to understand it.

Last edited: Jul 17, 2006
18. Jul 17, 2006

### Hans de Vries

Let me reformulate more clearer then:

Aging goes fastest in the absolute frame and everything what moves
progresses slower. The faster it moves, the slower it progresses.

That is, the amount of distance covered is always subtracted from
the amount of time covered. Thus, the formula you use is the same old
standard SR one I use.

You do say however: "Less time has elapsed" OK, but this is in the
other reference frame: The reference frame of the traveler.
In the rest frame time, which is what I am talking about, the aging
of the traveler goes slower.

Regards, Hans

Last edited: Jul 18, 2006
19. Jul 17, 2006

### MeJennifer

Well hopefully we can agree to disagree.

Relativistic measurements are like lenses, they simply distort the measurements on the objects you want to measure.

So when someone claims that a clock rate, lenght or mass changes it is due to the distortions of the measurement, it has to do with the "funny" behaviour of the constancy of the speed of light, it is not actually changing for the object in question.
For the object in question there is not change whatsoever. For an accelerated frame it is more complicated because the measuring frame itself is no longer flat due to those same distortions, but in princple it applies there as well.

Ultimately, who could reasonably argue that the clock rate, length or mass changes at the source because it is measured by something that is in relative motion to it while knowing that these effects can explained by the constancy of the speed light?
That is the same a persisting that a doppler shifted sound frequency is actually the frequency at the source because it is measured as such.

Really, we know why we measure those changes.

Special relativity is simple compared to general relativity. But the same relativistic measurement distortions exist there and it becomes increasingly harder to understand if you consider those distortions actual properties of the objects measured.

Would you try to insist that a free falling rock approaching the earth is accelerating, because we measure so from earth? That a photon approaching the event horizon of a black hole never actually reaches it or let alone crosses it because it is measured so from someone away from the black hole?

Last edited: Jul 17, 2006
20. Jul 17, 2006

### OS Richert

At risk of backtracking on my understanding and saying something stupid, can I jump back in after listening to you two converse.

It would seem that the issue is who moves farther through space time; that person will have their clock have less elapsed time when the two bodies come back together within the same reference frame. Is it always true that the body that had to acelerate then decelerate (or vice versa) to get back to the beginning reference frame (as opposed to the one that was always at rest) will move father through space time? Indeed, I wish to ask it again with my ealier example.

In this case, I would say the particle that stayed at 0.9c traveled "farther" with respect to earth. But from a reference frame attached to it, it was always at rest, while the other particled acelerated then decelerated; so the second particle traveled father through space-time and must have elapsed LESS time. If this is true once they are back in the same reference frame, would we then expect if the particles are identicle unstable subatomic particles with a short half life, that particle A will actually <blink> out of exsistance before particle B?

21. Jul 17, 2006

### MeJennifer

What is essential with regards to who accelerates and who does not is where the force is applied.

In general relativity an object moving according to the gravitational force (in case you prefer to call it a force) is not accelerating, it is traveling on a so called geodesic, and one of the characteristics of a geodesic is that time is maximized, it is only accelerating if a non-gravitational force is applied to it, for instance an electro-magnetic force. This causes an object to divert from traveling on a geodesic and hence time is no longer maximized.

22. Jul 17, 2006

### OS Richert

So, let me try to put this into my own words.

Two bodies sit next to each other at rest in their frame of reference.
Later, the two bodies find themselves in motion relative to themselves.
Later still, the two bodies find themselves next to each other at rest in the same reference frame again.
In conclusion, which ever body experienced some non-gravitational force to cause the aceleration into motion (at which time, it found itself in relative motion to the other body) and then another force to somehow reverse this acceleration to bring it back to rest in the original frame, THAT body left its geodesic and therefor did not have its time maximized and therefor experienced less elapsed time.

23. Jul 17, 2006

### MeJennifer

Yes, that is correct!

24. Jul 17, 2006

### pervect

Staff Emeritus
Yes, with one important pre-condition. Space-time must be flat for this to work.

In flat space-time, the body that moves inertially maximizes its proper time.

In flat space-time, there is only one geodesic path between any two points, just as in Euclidean plane geometry there is only one straight line between any two points. And the particle that follows this geodesic path between two points in flat space-time will be the particle with the longest elapsed time on its clock. This particle will also experience no inertial forces.

25. Jul 17, 2006

### Hans de Vries

There are no distortions in the measurements.

The life time of the muon becomes 10 or 100 times longer depending on
how close it is to the speed of light, it ages slower. Chemical reactions
go slower. et-cetera...

Do not jump to conclusions. Many things look strange and non consistent
when studying Special Relativity, but believe me, there are all good
physical explanations and with time comes understanding.

Point is that step by step one has to understand in detail what is really
going on. How Lorentz contraction is caused by non simultaneity on
one hand, but also how it arises physically from Lienard Wiechert for
EM fields or Klein Gordon for quantum mechanical fields. Then you start
learning what physics actually causes the speed of light to be constant
and what causes the actual laws of Special Relativity. What causes
the impression of non simultaneity et-cetera.

I will have written a small illustrated booklet in a few months or so just
for this purpose, sparing the reader for example from the need to study
complicated graduate text on Relativistic Quantum Mechanics.

Regards, Hans