Can clock transport tell us anything important

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1. Nov 19, 2014

dbkooper

Clock transport can be used to compare one clock with another in an absolute sense. All we have to do is to transport a clock between two A-frame clocks that have been synchronized per Einstein's definition. Let's call the transported clock "T" and the left-hand and right-hand E-synch'd clocks "E1" and "E2" respectively. We will need to transport T twice - at different speeds wrt A - in order to have a comparison.

In both cases, T and E1 read zero at the start. It is most important to bear in mind the simple fact that the temporal relationship between E1 and E2 is constant. (This means "constant within the A frame," not "constant as seen by any outside observers.") (In other makes them actual experiments in anyone's view.)

In the first transport case, each frame* sees the other moving at 40/41c.
(*The T-clock frame and the E-clocks frame, aka, Frame A)

-------------[0]-->

---------<--[0]------Frame A------[?]

-------------------------[0.30]-->

[?]------Frame A-----[1.10] difference = 0.80

In the 2nd case, each frame sees the other moving at 35/37c.

-------------[0]-->

---------<--[0]------Frame A------[?]

-------------------------[0.34]-->

[?]------Frame A-----[1.06] difference = 0.74

If T and E1 always tick at the same physical rate (as special relativity demands), and given the fact that E1 and E2 always have the same temporal relationship, then T must faithfully carry or transport this relationship to E2.

In case one, we see that this transported relationship shows up as E2 reading 0.80 different from E1's transported time of 0.30.

But in case two, we see that T now differs from E2 by only 0.74.

Only if T read .26 at the end (of the 2nd case) would it have physically ticked at the same rate as E1 (because the latter is always - per the 1st case - 0.8 different from E2).

We see that simple clock transport tells us the very important fact that clocks do indeed run at physically different rates even with no acceleration or gravity.

This is important because it means that a clock's absolute motion affects its rate.

And this is important because it is evidence of absolute motion.

2. Nov 19, 2014

Staff: Mentor

Is this because E1 and E2 remain at rest relative to each other all the time, while T is transported? I'll assume that's the case in what follows.

Given that, what does "temporal relationship" mean? Does it mean the relationship between readings on clocks E1 and E2 that happen "at the same time" according to some inertial frame? I'll also assume that's the case in what follows.

If "temporal relationship" means what I assumed it does above, and E1 and E2 are always at rest relative to each other as I assumed above, then the temporal relationship between E1 and E2 will change depending on which frame we use, but once we've chosen a particular frame, the relationship is always constant within that frame--any inertial frame.

Whether the relationship is "constant as seen by outside observers" depends on what you mean by "seen". Do you mean what they actually see with their eyes? (As in, what light signals they receive from each clock?) Or do you mean what they "see" as in what is true in whatever inertial frame they are at rest in?

As you can see, what you appear to think is a "simple fact" actually hides a lot of subtleties. I would advise re-thinking your position in the light of that.

SR demands no such thing; in fact, it demands the opposite: if T and E1 are in relative motion, they tick at different rates. If the relative motion is very slow, the difference in rates is very small; but it's always there. You can't handwave it away. You can ignore it as a reasonable approximation in some scenarios (such as our everyday lives here on Earth, where all relative velocities are much less than the speed of light), but you can't ignore it if you are trying to draw general conclusions.

I won't even comment on the rest of your post because I can't make sense of it, given the obvious misstatement I've just pointed out. As I said above, you need to re-think your position.

3. Nov 20, 2014

harrylin

I'm not sure what you try to show there; perhaps you simply argue that a clock that is transported at constant high speed from one "stationary" clock to another should be found to be behind when compared with the second clock? That is certainly correct.
However, as Peterdonis already remarked, what may be a key sentence in your discourse is certainly wrong - apparently some publication misinformed you about special relativity. It could be useful if you mention that source (but, may I remind you: don't give a link to a "bad" site!).

Last edited by a moderator: May 7, 2017
4. Nov 20, 2014

dbkooper

To PeterDonis:

To be more precise, we can replace clocks with people.

Your statement that special relativity says that inertially-moving clocks run at physically different rates then becomes the claim that SR says that people in different inertial frames age differently.

Please tell us how and where SR says this. It would help to have a specific example.

5. Nov 20, 2014

harrylin

I don't know if he will be so kind; in any case, there is no need to replace clocks by people and this is what SR predicts:

Suppose at two points A and B of the stationary system two clocks are given which are synchronous in the sense explained in § 3 when viewed from the stationary system. Suppose the clock at A to be set in motion in the line joining it with B, then after the arrival of the clock at B, they will no longer be found synchronous, but the clock which was set in motion from A will lag behind the clock which had been all along at B by an amount , where t is the time required for the journey.
- https://en.wikisource.org/wiki/On_the_Electrodynamics_of_Moving_Bodies_(1920_edition)

Last edited: Nov 20, 2014
6. Nov 20, 2014

dbkooper

Hi, harrylin, thanks for the Einsteinian example. Note that it matches my examples, given that my boo-boo's are corrected. (My case 1 clock times at the end should have been .225 and 1.025.) [BTW, Einstein's formula was only approx. - the real deal is t(1-sqr(1-v^2)) if c = 1.]

The problem with a single example like Einstein's is that it does not necessarily show intrinsic (or physical) clock slowing due to the use of asynchronous clocks in the "stationary" frame. (See PS below.)

The only way to use clock transport to prove real clock slowing is by using two examples and comparing them, as I did.

If we replace clocks with people, then we can eliminate all extraneous "stuff" so we can see the real story. I used clock transport just to see if it could be done. It, admittedly, is not the best way. The ideal way is to use Triplets.

By using Triplets, we can eliminate all vestiges of acceleration, E-synch., an outsider's view, odometers, triangles, the optical Doppler, the lack of symmetry, the lack of history, and of course gravitation.

But I stand by my claim that comparing two examples of clock transport does indeed show real or physical clock slowing. This is because, once set, the E-synch'd clocks in a given frame do not change their temporal relationship. If they did, then they would no longer be E-synch'd.

Since the transported clock matched the left-hand E-clock at the start, it should match it at the end ---- unless, of course, the transported clock's physical tick rate changes from that of the E-clocks.

Both the Triplet case and the above-given transport cases tell us that the physical times of both clocks and people vary with absolute motion through space.

If one still insists that special relativity claims that clocks really slow or that people really age differently, then one must cope with the fact that this conflicts with SR's claim that absolute motion is meaningless.

PS
"Different observers at rest in their respective frames disagree over the time interval between two events because they calculate the difference in the readings of two clocks at rest relative to themselves. THE LACK OF AN ABSOLUTE SYNCHRONIZATION for these clocks causes the variation in delta-t from observer to observer."

[_introduction to the theory of relativity_ by Sears & Brehme, Addison-Wesley, p. 87]

7. Nov 21, 2014

PAllen

No, it shows the transported clock ticked slow compared to a sequence of E-synched clocks. If you did the same experiment with clocks synched in a frame in which the transported clock is at rest during transport, you would find that each each of the original frame's E-synch clocks have the same rate, but are out of synch with each other by an amount proportional to their separation.

8. Nov 21, 2014

Staff: Mentor

Sure.

No, because to compare the "ages" of two people who are spatially separated (which they must be if they are both at rest in different inertial frames, except possibly for one unique event where their worldlines cross), you need more than just their clock rates; you need a simultaneity convention, so you know which events on each one's worldline happen "at the same time", in order to compare their respective ages at those events. But simultaneity is relative. So what SR actually says is that there is no invariant way to compare the ages of two people who are at rest in different inertial frames.

9. Nov 21, 2014

dbkooper

Do you agree that the E-synch'd clocks do not change during the experiments? Do you agree that the transported T-clock must faithfully carry E1's time to E2 IFF both T and E1 always have the same physical rate ?

10. Nov 21, 2014

dbkooper

But let's take a look-see at the Triplets Case:

Bob passes Ann (he's moving to the right, she's going left) when they are the same age.

Bob meets Bill when they are both the same age.

Bill (or 'Copy-Bob') goes on to catch up with Ann, and their ages are different.

No simultaneity convention is involved, because no clocks are involved.

All we have are people in different frames who age differently (physically).

Why do people in different (inertial) frames age differently? Fairy dust? ;)

11. Nov 21, 2014

PAllen

Yes, No. That is, whether E1 and E2 are in synch is frame dependant. You've got to think about more than rates. When comparing clocks, you ask two questions:

1) is a minute for both the same?
2) is 2:00 the same?

These are completely different questions. All observers agree that if E1 and E2 are mutually motionless, then the answer to (1) is the same. However, any observer for which E1 and E2 are moving, will disagree about (2) compared to someone for whome E1 and E2 are motionless.

Last edited: Nov 21, 2014
12. Nov 21, 2014

PAllen

I find this example nonsensical. Let's start with 'same age'. Each person's age is indpendent of who is observing whom. Given this, your whole scenario makes no sense to me.

13. Nov 21, 2014

Staff: Mentor

Huh? First, they are all in every frame. Frames are infinite in space and time, they are not little boxes that things can be in or out of. Second, this doesn't show which twin aged faster or slower (you can always find a frame where any given triplet ages the fastest and the other two age slower)

14. Nov 21, 2014

Staff: Mentor

Ok.

Ok.

Yes, because by specifying that Bob and Ann were the same age when they passed, and Bob and Bill were the same age when they passed, you have also specified that Bill and Ann must be different ages when they pass (Bill will be younger than Ann).

Wrong: no simultaneity convention is involved because all of the age comparisons are done when the people are at the same spatial location. (Clocks are involved because the biological processes that cause aging are effectively clocks.) But, for example, when Bob and Bill pass, Ann is spatially separated from them, so specifying how old Ann is "when" Bob and Bill pass requires a simultaneity convention, and different conventions will give different answers (and no answer is more "right" than any other; there is no invariant fact of the matter).

15. Nov 22, 2014

harrylin

Hi db,

The triplets example is well known, it was recently discussed here: https://www.physicsforums.com/threads/acceleration-free-twin-paradox.781369/ See also my post #17 there. Indirectly I referred to Langevin's explanation that the change of velocity breaks the symmetry. He used that "twin" example to make the same argument that you make here.
See p.47 + next starting from p.50 of https://en.wikisource.org/wiki/Translation:The_Evolution_of_Space_and_Time

As you can infer, he was a big promoter of SR, and he claimed the opposite of what you say that "SR claims". In contrast, Einstein did make such claims - but at another time he agreed with Langevin, and at again another time he apparently agreed with Minkowski. In reality SR is on this topic interpretation-neutral.

However, the reference that you cite is not wrong, just a bit myopic: while ignoring the physical cause of the lack of an absolute synchronization for those clocks, it is certainly correct! If you don't find that result then it is due to a little calculation error. People on this forum will be glad to be of help to find the glitch in your calculation, if you present it in full detail.

And once more: on this forum the messenger will be shot if he or she does not heed the rules. As I could see where you were heading, I gave you a link in my first reply which you evidently ignored. You are probably going to argue for what that link calls "LET", and it appears that PAllen may be headed for an argument in favour of the "block universe", or for a third opinion that I'm not aware of. It doesn't matter: this forum does not appreciate such discussions without end (it's a matter of belief systems).
For this forum it is however useful to inform about these different views of reality as recorded in the literature in order to understand the different ways in which people describe the predictions. Evidently some description somewhere by someone made you think that that person's interpretation of SR is SR. It isn't.

Last edited: Nov 22, 2014
16. Nov 22, 2014

PAllen

I didn't intend any such argument. I only mentioned simultaneity convention, which operationally defined irrespective of interpretation, and what type of quantities are observer independent, which is also not related to interpretation.

17. Nov 22, 2014

harrylin

Thanks for the precision! That is what jokingly is called the "shut up and calculate explanation". ;)

18. Nov 22, 2014

stevendaryl

Staff Emeritus
Are you saying that this experiment can reveal which clocks are absolutely at rest? Because that is certainly not the case.

It's interesting to look at "slow clock transport" from two different frames.

In frame $F_1$, you have two clocks $E_1$ and $E_2$ at rest that are synchronized. You have a third clock,$T$. that is initially synchronized with $E_1$, and is slowly (at a speed much less than $c$, as measured in frame $F_1$) carried to clock $E_2$. Then in the limit of very slow transport, $T$ will agree with $E_2$ when it gets to $E_2$. In the frame $F_1$, all three clocks are synchronized at all times.

Now, look at the exact same events from the point of view of a different frame, $F_2$. According to this frame:
1. $E_1$ and $E_2$ are moving at some velocity $v$ in the direction from $E_1$ to $E_2$.
2. $T$ is moving at a slightly greater velocity than the other two clocks.
3. All three clocks are time-dilated (they run slower than a stationary clock).
4. $E_1$ and $E_2$ run at the same rate, while $T$ runs slightly slower.
5. Initially, the three clocks are NOT synchronized, according to this frame. $E_1$ and $T$ are initially synchronized, but $E_2$ is behind (set to an earlier time).
6. When $T$ moves to $E_2$, it loses time compared to $E_2$, so even though $T$ starts off ahead of $E_2$, they show the same time when they get together.
So both frames agree that $E_1$ and $T$ are initially synchronized. Both frames agree that at the end, $E_2$ and $T$ are synchronized. The two frames disagree, though, about whether $E_1$ and $E_2$ are synchronized.

There is no way to resolve this disagreement and figure out who is "really" right. There is no way for slow clock transport to discover any absolute speed.

19. Nov 22, 2014

stevendaryl

Staff Emeritus
Here's an analogous "paradox" of Euclidean geometry. There are three cities arranged in a triangle: Alphaville, Boomtown and Carson. Alphaville an Carson lie on highway $A$ that runs west-to-east. Boomtown lies halfway between Alphaville and Carson, to the north of highway $A$. There is also a highway $B$ running from Alphaville to Boomtown, and a highway $C$ running from Boomtown to Carson.

Now, along each of the highways is a mileage marker, telling how far you've traveled down that highway. Let's assume that
1. At the intersection of $A$ and $B$ at Alphaville, both highways show the same mileage (say 50 miles).
2. At the intersection of $B$ and $C$ at Boomtown, both highways show the same mileage (say 100 miles).
3. At the intersection of $C$ and $A$ at Carson, highway $A$ shows 130 miles, while highway $C$ shows 150 miles.
By your reasoning, this shows that highways that travel east (highway $A$) have mileage markers that are farther apart than highways that run northwest (highway $B$) or highways that run southwest (highway $C$). We know this because highway $A$ was "synchronized" with highway $B$ at Alphaville, and highway $B$ was synchronized with $C$ at Boomtown, but $C$ was ahead of $A$ by the time they got to Carson.

20. Nov 22, 2014

dbkooper

Can you justify this claim experimentally?