1. On 2001-10-24 21:59, Rosen1 wrote:
With regard to forces, what is the same for each inertial frame is the restrictions on forces as defined by SR. SR is useless without forces. Without forces, the twin paradox IS a paradox.
I guess we'll have to ask Bill Clinton just exactly what a paradox is, but, barring that, where do you see the force in this description of the twin paradox?

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That's pretty cool. Good link. (n/t) [img]/phpBB/images/smiles/icon_biggrin.gif[/img]

Ben

3. Thanks Ben

I guess the good ol' "n/t" ain't quite as effective, here.

I'm still looking for additional comments on my twin paradox redux page, btw. And I might as well throw in the general relativity version.

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On 2001-10-24 23:03, GrapesOfWrath wrote:

I guess we'll have to ask Bill Clinton just exactly what a paradox is . . .

Its a Pair-a-Ducks [img]/phpBB/images/smiles/icon_lol.gif[/img]
I'm gonna go back in my hole now.

Hauteden

Having trouble with the imaging

<font size=-1>[ This Message was edited by: Hauteden on 2001-10-25 00:27 ]</font>

5. It's Pair-a-dox, not pair-a-dux. [img]/phpBB/images/smiles/icon_smile.gif[/img]

6. Guest
...,where do you see the force in this description of the twin paradox?

Sorry. The quote in my previous post didn't get through.
---The confusion arises not because there are two equally valid inertial rest frames, but (here's the tricky part) because there are three.---
Here is where the force got in. It's sneaky. However, he chose three inertial frames rather than two only because the bserver in the space ship was slapped to
the ceiling and floor. The definition of nertial frame uses the definition of stationary frame, and the definition of stationary frame says that the simplest laws (which in 1905 was only Newton's mechanics and Maxwell's equations) hold true. The "simple" laws, including quantum mechanics, all have forces in some form.

---A lot of explanations of the twin paradox have claimed that it is necessary to include a treatment of accelerations, or involve General Relativity. Not so.---
I think that he just contradicted himself. A lot of experts have said this sort of thing, but they are saying it wrong.
What they mean is that one can't use the pure kinematic definition of acceleration. If you use a purely kinematic definition, one based solely on geometry, the problem is symmetrical and there IS a paradox. One has to define acceleration as "F/m."

Most people who try to "disprove" SR assume a priori that there are only two inertial frames. If there was no force involved, they would be right.

<font size=-1>[ This Message was edited by: Rosen1 on 2001-10-25 08:21 ]</font>

<font size=-1>[ This Message was edited by: Rosen1 on 2001-10-28 20:42 ]</font>

7. Rosen1

When you refer to "he" ("he chose three inertial frames rather than two" and "I think that he just contradicted himself"), I guess you are talking about me.

I didn't choose three intertial reference frames just to hide the force. I did it to point out that the force can be made negligible, but the analysis still works. It is very similar to that in our common foil's favorite article, Einstein's On the Electrodynamics of Moving Bodies. I expounded on that in the Twins Redux article.

8. Hey, Grapes, I didn't realize that was you! [img]/phpBB/images/smiles/icon_smile.gif[/img]

In Section I.1, Einstein establishes what he means by simultaneity, and the synchronization of clocks in an inertial reference frame. Basically, the clocks must be calibrated and synchronized so that a light pulse that leaves one clock takes the same amount of time to reach the second clock, as it does to return to the first clock. Any number of clocks may be synchronized in this manner, and if one clock is synchronized with two others, then those two will also be synchronized.
I'm going to try again. Consider the following partial description of baseball:

"In baseball, it is important to know which players are responsible for fielding the ball. Consider a situation in which the batter hits the ball over the first baseman's head. The right fielder is responsible for fielding the ball, because he plays behind the first baseman."

Now consider if someone else decided to write a description of baseball based on reading the above, and he wrote:

"In baseball, it is important to know which players are responsible for fielding the ball. Basically, the player who is responsible for fielding the ball is the one who plays behind the first baseman."

The second writer took a specific example from the first description and turned it into a generality. That's what (IMHO) you've done on your page.

In the baseball example, the right fielder will always meet the stated criteria ("plays behind the first baseman"), regardless of whether or not he is responsible for fielding the ball. On your page, two clocks both motionless relative to an observer will always meet the stated criteria ("a light pulse that leaves one clock takes the same amount of time to reach the second clock, as it does to return to the first clock"), regardless of whether or not the clocks are synchronized.

In Einstein's paper, he talks about three distinct time readings. Let's call them a1, b, and a2. a1 is the time showing on Clock A when the pulse leaves. b is the time showing on Clock B when the pulse arrives. a2 is the time showing on Clock A when the returned pulse arrives.

Keep in mind that these are time readouts on two different clocks. It is not necessarily true to say that b - a1 is the amount of time it took for the pulse to travel from Clock A to Clock B, nor is it necessarily true to say that a2 - b is the amount of time it took for the signal to get back. Agreed?

Einstein describes a specific example in which b - a1 = a2 - b (b is exactly half way between a1 and a2). Then he says that these clocks are synchronized for a stationary observer in this situation because the pulse takes the same amount of time on both trips for stationary clocks.

If the clocks are moving relative to the observer, however, then the light pulse will not take the same amount of time to make both trips. However, moving clocks could still be synchronized if the time read-outs are such that b - a1 is equal to the time the pulse takes to get from Clock A to Clock B and a2 - b is equal to the time the pulse takes to get back.

Let's look at two specific clocks (I'm going to ignore Lorentz contraction and time dilation in this experiment -- it's still valid, just easier to follow). The signal leaves Clock A when it shows ta=0. The signal reaches Clock B when it shows tb=6. The return signal reaches Clock A when it shows ta=12.

A stationary observer says the light takes the same amount of time to go from A to B as it takes to go from B to A. Since A ticked off 12 seconds during the whole trip, it would have ticked off 6 seconds during the out trip and 6 seconds during the return trip. Therefore Clock A showed ta = 6 when the pulse hit Clock B, which was when tb = 6. Since ta = tb, the clocks are synchronized.

An observer who sees the clocks moving (with B "chasing" A), however, might say the light took only half as long to get from A to B as it did from B to A (since B was moving into the light pulse and A was moving away from the return pulse). This observer, therefore, says that Clock A showed ta = 4 when the signal hit Clock B, which was when tb = 6. Since ta <> tb, the clocks are not synchronized.

However, if the signal hits Clock B when it is showing tb = 4, then the moving observer says the clocks are synchronized (ta = tb) while the stationary observer says they're not (ta <> tb).

This is important, here. For the moving observer, the light pulse took 4 seconds to get from A to B and 8 seconds to get from B back to A, yet the clocks are synchronized. For the stationary observer, the light pulse still took 6 seconds on each trip, yet the clocks are not synchronized.

See where I'm coming from? It's not always true that the clocks are synchronized if the pulse takes the same amount of time on both trips (the pulse taking the same amount of time only means that the observer is stationary relative to the clocks, just like the player being behind the first baseman only means that he's the right fielder). . .

9. Guest
When you do the "ships passing in the night" version of the paradox, there are still forces involved. Even though a "signal" is passed instead of a "turn around." The so called "synchronization" involves the use of force. Someone has to at least push the stopwatch button, after all.
The "accelerations" involved in a "signal" are usually microscopic. However, if you want to ignore the detailed microscopic forces, just dump them in "the signal" and "the observer." If you like the grungy microscopic details, then go ahead.
There is no way to "synchronize" two clocks without discussing some type of force between the two clocks.

10. Sorry, SeanF, I should have introduced myself. You are mentioned on that page.

On 2001-10-25 08:38, SeanF wrote:
Keep in mind that these are time readouts on two different clocks. It is not necessarily true to say that b - a1 is the amount of time it took for the pulse to travel from Clock A to Clock B, nor is it necessarily true to say that a2 - b is the amount of time it took for the signal to get back. Agreed?
I'm going to have to disagree, in this context. Both my first sentence that you quoted, and Einstein's first sentence of that section, make it clear that it is an inertial reference frame--Einstein uses the term "stationary system," and the clocks are at points of space.

Einstein describes a specific example in which b - a1 = a2 - b (b is exactly half way between a1 and a2). Then he says that these clocks are synchronized for a stationary observer in this situation because the pulse takes the same amount of time on both trips for stationary clocks.
Yes, Einstein says "We have not defined a common 'time' for A and B, for the latter cannot be defined at all unless we establish by definition that the 'time' required by light to travel from A to B equals the 'time' it requires to travel from B to A" (emphasis in the original)

I wasn't so much trying to describe relativity so much, as trying to summarize the logic of that paper.

See where I'm coming from? It's not always true that the clocks are synchronized if the pulse takes the same amount of time on both trips (the pulse taking the same amount of time only means that the observer is stationary relative to the clocks, just like the player being behind the first baseman only means that he's the right fielder). . .
I think it is the difference between necessary conditions and sufficient conditions, but I'm not sure. I'm going to have to tink on this a little while.

[fixed quotation marks]

<font size=-1>[ This Message was edited by: GrapesOfWrath on 2001-10-25 09:38 ]</font>

11. On 2001-10-25 09:24, Rosen1 wrote:
There is no way to "synchronize" two clocks without discussing some type of force between the two clocks.
But, the force can be made negligibly small. I thought Einstein addressed this directly in this article, but a quick search doesn't come up with it. I'll have to look closer.

Anyway, the synchronization of nearly co-located clocks, moving or not, can be done with the tiniest bit of energy or force. Insisting that there is still force there confuses the issue, I think. An understanding of the math is more important.

12. On 2001-10-25 09:36, GrapesOfWrath wrote:

On 2001-10-25 08:38, SeanF wrote:
Keep in mind that these are time readouts on two different clocks. It is not necessarily true to say that b - a1 is the amount of time it took for the pulse to travel from Clock A to Clock B, nor is it necessarily true to say that a2 - b is the amount of time it took for the signal to get back. Agreed?
I'm going to have to disagree, in this context. Both my first sentence that you quoted, and Einstein's first sentence of that section, make it clear that it is an inertial reference frame--Einstein uses the term "stationary system," and the clocks are at points of space.
Yes, but, but, wait a second . . .

Take a simple situation of two clocks which are located such that a light signal takes one time-unit to get from A to B and one time-unit to get from B to A.

Example 1:
If the pulse leaves A when A says 0, hits B when B says 1, and returns to A when A says 2, the clocks are synchronized.

Example 2:
If the pulse leaves A when A says 0, hits B when B says 15, and returns to A when A says 2, the clocks are not synchronized. B is 14 units ahead of A.

In both examples, the pulse took one time-unit to get from A to B and one time-unit to get from B to A, but the clocks are only synchronized in the first example.

In the second example, b - a1 would be 15 - 0, which is 15, not 1. a2 - b would be 2 - 15, which is -13, not 1.

b - a1 does not define the amount of time it takes for the pulse to travel, it merely identifies the difference in the displays of the two clocks at the two times. It will, however, be equal to the amount of time it took the pulse to travel if the clocks are synchronized.

That's how Einstein determines synchronicity. He predefines that the pulse takes the same amount of time for both trips and says the clocks are synchronized if the displays match up with this predefinition. That's not the same thing as saying "the clocks are synchronized if the pulse takes the same amount of time for both trips."

_________________
SeanF

<font size=-1>[ This Message was edited by: SeanF on 2001-10-25 10:03 ]</font>

13. On 2001-10-25 09:51, GrapesOfWrath wrote:
On 2001-10-25 09:24, Rosen1 wrote:
There is no way to "synchronize" two clocks without discussing some type of force between the two clocks.
But, the force can be made negligibly small. I thought Einstein addressed this directly in this article, but a quick search doesn't come up with it. I'll have to look closer.

Anyway, the synchronization of nearly co-located clocks, moving or not, can be done with the tiniest bit of energy or force. Insisting that there is still force there confuses the issue, I think. An understanding of the math is more important.
I've found it interesting reading Rosen's, Grapes', and my own posts on not only this thread but also the other "Ropes and Rockets" thread.

It's obvious that Rosen's training and/or interest is on the physical side and mine (and Grapes' as well, I think) is more on the mathematical side . . . [img]/phpBB/images/smiles/icon_smile.gif[/img]

If the rockets are farther apart than the length of the rope, the rope's gotta break, regardless of where the "force" comes from, eh? [img]/phpBB/images/smiles/icon_wink.gif[/img]

14. On 2001-10-25 09:54, SeanF wrote:
That's how Einstein determines synchronicity. He predefines that the pulse takes the same amount of time for both trips and says the clocks are synchronized if the displays match up with this predefinition. That's not the same thing as saying "the clocks are synchronized if the pulse takes the same amount of time for both trips."
But it includes it, certainly. He defines what he means by "the pulse takes the same amount of time for both trips," and doesn't both trips mean all pairs of trips--not just the one pair that happened to make things work right?

Otherwise, I'm trying to think of a counterexample that would support your objection, but I can't.

15. On 2001-10-25 10:10, GrapesOfWrath wrote:
On 2001-10-25 09:54, SeanF wrote:
That's how Einstein determines synchronicity. He predefines that the pulse takes the same amount of time for both trips and says the clocks are synchronized if the displays match up with this predefinition. That's not the same thing as saying "the clocks are synchronized if the pulse takes the same amount of time for both trips."
But it includes it, certainly. He defines what he means by "the pulse takes the same amount of time for both trips," and doesn't both trips mean all pairs of trips--not just the one pair that happened to make things work right?

Otherwise, I'm trying to think of a counterexample that would support your objection, but I can't.
Oh, it includes it, certainly, although clocks could be synchronized for a moving observer in which the trips would not take the same amount of time.

And yes, it means all pairs of trips. With the synchronized clocks, we have something like:

a1 - b - a2
0 - 1 - 2
2 - 3 - 4
4 - 5 - 6

With the non-synchronized clocks, it becomes:

a1 - b - a2
0 - 15 - 2
2 - 17 - 4
4 - 19 - 6

All the trips show clock synchronization in the first example and non-synchronization in the second.

Note that the clocks are both ticking off seconds (or whatever the units are) at the same rate in either case -- in both cases, they would both agree on how much time elapsed between any two events. However, in the non-synchronized case, they would not agree at what times the two events actually occured.

_________________
SeanF

<font size=-1>[ This Message was edited by: SeanF on 2001-10-25 10:18 ]</font>

16. Wouldn't an observer of the second set of clocks say that the time to go from a to b was 15 seconds and -13 seconds? Seems OK, and consistent with Einstein's logic.

Light dawns!

Your distinction is between what can be measured, and what actually occurs! Is that correct? Einstein, of course, says that the measurement is what actually occurs. I should make that point more clear, is that it?

17. On 2001-10-25 10:31, GrapesOfWrath wrote:
Wouldn't an observer of the second set of clocks say that the time to go from a to b was 15 seconds and -13 seconds? Seems OK, and consistent with Einstein's logic.

Light dawns!

Your distinction is between what can be measured, and what actually occurs! Is that correct? Einstein, of course, says that the measurement is what actually occurs. I should make that point more clear, is that it?
No, not quite -- at least, I don't think so. The observer in this case is sitting at Clock A. He can check the time on A directly (I think Einstein talks in his paper about how a local clock can be read directly but a distant clock cannot?).

The observer sees the signal go out at 0 and return at 2, so he knows it took 2 units for the round-trip. Since he knows the pulse took the same amount of time out as it took back (that's predefined, right?), he knows that it's "stopping point" at Clock B would be right in the middle of the total round trip, at 1 unit. He now knows that Clock A read 1 when the signal hit Clock B.

Because of the signal he received from Clock B, he knows that Clock B said 15 when the signal hit it. So, he now knows that Clock A said 1 at the same time Clock B said 15 and the two clocks are not synchronized.

He can't use the 15 from Clock B to determine the travel time of the pulse, because he does not know if that clock is set correctly or if it's ticking time correctly. All he can use that 15 for is to determine what Clock B read at the moment the pulse reached it. He does knows that Clock A is set correctly and ticking time correctly. The only way he knows that the pulse hit Clock B when Clock A read 1 is because of the predefined definition that the pulse takes the same amount of time out and back (which is really just a conclusion of the predefined condition that light travels at a constant velocity).

Additional pulses sent back and forth can allow him to verify that Clock B is ticking off units at the same rate as Clock A, but is simply off by a constant 14 units.

As far as the measurement being what actually occurs, that's only true as far as Clock A goes. Because this observer measures the total trip time as 2 units, it is taken as real that the trip took 1 unit out and one unit back.

A moving observer, on the other hand, might say the total trip time was 6 units, with 2 units out and 4 units back. That is taken as real for him, with the conclusion that Clock A is experiencing time dilation and only ticked off 2 units during the "real" 6 unit duration. For this observer, the clocks are not even synchronized in the first example, because he calculates that Clock A would be showing 0.6666 when the pulse hits Clock B instead of 1 . . .

18. Having just read that last post I typed, I am so glad I'm not a teacher. I don't think I can explain things worth s**t. [img]/phpBB/images/smiles/icon_frown.gif[/img]

I think I need to take the time to actually write something out in Notepad or something and go back and reread it and edit it and what-not a few times before I actually post it . . .

19. At least you haven't deleted it...yet [img]/phpBB/images/smiles/icon_smile.gif[/img]

If it weren't for embarassing post hanging out all over the internet, I wouldn't be learning nothing.

OTOH, yours doesn't seem that bad. I don't really find anything to disagree with, though, and that worries me. Why do you think it is inconsistent with what I wrote on that page?

20. On 2001-10-25 11:00, GrapesOfWrath wrote:
At least you haven't deleted it...yet [img]/phpBB/images/smiles/icon_smile.gif[/img]

If it weren't for embarassing post hanging out all over the internet, I wouldn't be learning nothing.

OTOH, yours doesn't seem that bad. I don't really find anything to disagree with, though, and that worries me. Why do you think it is inconsistent with what I wrote on that page?
It just seems to me that the wording on your page suggests that the clocks are synchronized if the pulse takes the same amount of time on both legs of the trip.

You say "Basically, the clocks must be calibrated and synchronized so that a light pulse that leaves one clock takes the same amount of time to reach the second clock, as it does to return to the first clock." In order to achieve the specified result (the light pulse takes the same amount of time), all you would need to do is have the observer stationary relative to the clocks. The two clocks could be nowhere's near to being synchronized, but if they're located 300,000 km apart, it's still going to take one second for the pulse to go one way and one second for it go the other way.

The clocks are synchronized if the two clocks are displaying the same time at the exact instant the pulse reaches Clock B. Clock B will simply identify what time it was displaying when the pulse hit it. Clock A's display needs to be calculated based on the time it was showing when the pulse left and the time it was showing when the pulse returned.

Using the time definitions I mentioned above, the clocks need to be set up and calibrated so that b - a1 = a2 - b (of course, that's only true for a stationary observer, but that much I think we can agree is a given).

_________________
SeanF

<font size=-1>[ This Message was edited by: SeanF on 2001-10-25 11:33 ]</font>

21. Guest
I think what is confusing everybody is the fundamental difference between the "one rocket turns around" problem and "two passing ships" problem. In the "rocket turns around" there is only one observer, and in the "passing ships" there are two observers. Even more important, the clock in the "turn around" problem experiences the same acceleration (F/m), pretty much similar forces, as all the other systems on the ship. In the "ships passing" problem, the signal exerts a force primarily on two clocks. The other systems on the ship are not greatly affected by the signal.

In the case of the twin who turns on his retrorockets and changes direction, his entire body is affected by the forces involved. His body and his clock experience similar stresses. The assumption is that second order effects are ignored, which means that he is not crushed by floor. Under this condition, SR says that if he turns around, the force will prevent him from being older than his twin when he gets back. It isn't only the clocks that show the paradox, it is the other systems on the ship. There is no separation possible between clock and the other systems on the ship. The force of the floor caused by the "turn around" affects all systems on the ship because all the systems experience turn-around force. The measurement of age is real in that the astronauts biological age and the clock agree. They have too. The astronauts body uses electrical forces as does the clock, and they experienced similar stresses from the turn around.

In the "passing ships" problem, the signal that synchronizes the ships only has to pass between two clocks. The other systems are isolated from the signal, or not affected very much by the signal. By Newton's Laws, or the quantum equivalent (conservation of momentum) for every action there is a reaction. So the two clocks indeed experience the same force. And whatever mechanisms the clocks run on experience a "twin paradox." However, this force only affects the clocks.

Suppose that no roockets are blazing. The man on the spaceship going back may be much older than the man on the spaceship going out when they pass each other. When he gets back, he may still be much older than the twin left behind. There is no way a "signal" can synchronize ages. So this can be considered a pure measurement problem. The clocks disagree as to how much time passed, but not the other inhabitants.

However, the effect was very real for the clocks. One clock had the action, and the other clock had the reaction (or is it vica versa). In this case, everything except the clock is somewhat isolated from the signal.

As far as the "coming" ship is concerned, the ship came from another galaxy far far away, passed a ship going out, they asked what time it is (quickly), set their watches (quickly) and moved on. The only system affected by Newton's Third Law were the clocks. No other system was effected.

The two "twin paradoxes" are completely different. The extent of the force doing the "synchronization" is different. However, in both cases, one needs a force to synchronize the clocks.

22. On 2001-10-25 11:32, SeanF wrote:
It just seems to me that the wording on your page suggests that the clocks are synchronized if the pulse takes the same amount of time on both legs of the trip.

You say "Basically, the clocks must be calibrated and synchronized so that a light pulse that leaves one clock takes the same amount of time to reach the second clock, as it does to return to the first clock."
I think I see what you're driving at, but it's similar to what I suggested before, so I'll try something. How about if the sentence read "Basically, the clocks must be calibrated and synchronized so that, by the clock measurements, a light pulse that leaves one clock takes the same amount of time to reach the second clock, as it does to return to the first clock."

I added the phrase "by the clock measurements". How's that? If that's not it, what would you suggest?

I'm going to change the page to point to BA's new board, too.

23. Guest
But, the force can be made negligibly small. ...Anyway, the synchronization of nearly co-located clocks, moving or not, can be done with the tiniest bit of energy or force. Insisting that there is still force there confuses the issue, I think. An understanding of the math is more important.
[/quote]

No it can't. By negligibly small, it merely means that the effects are restricted to the clocks or even to one of its component parts. The clock is greatly affected. Now, the inhabitants on the ship won't age more slowly just because a signal passed between them. However, they are different observers anyway. There are three people involved in the "ships passing" problem, not two as in the turn around. Furthermore, everything except the clock is isolated from external forces.

The clocks, and the clocks alone, are affected in a BIG way. A negligibly small force only means the forces making the signal is spatially localized on the rocket ship. Every time Einstein says synchronization, I imagine forces. However, they don't always extend pass the clock.

24. Member
Join Date
Oct 2001
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19
The interesting thing is what one observer can actually see at the position of the other. We all note that light takes time to travel. So if a clock moves away, even at slow non-relativistic speed, it must appear to slow down so that it ends up reporting a time exactly late by the time of light travel (equivalent to red shift). The difference with relativity is that this clock would appear to speed up on return (blue shift) so that when it returns its time would coincide with the observer's time. This is what throws me (due to math defficiency) I can't separate the two effects in my mind.

25. On 2001-10-25 11:47, GrapesOfWrath wrote:

I think I see what you're driving at, but it's similar to what I suggested before, so I'll try something. How about if the sentence read "Basically, the clocks must be calibrated and synchronized so that, by the clock measurements, a light pulse that leaves one clock takes the same amount of time to reach the second clock, as it does to return to the first clock."

I added the phrase "by the clock measurements". How's that? If that's not it, what would you suggest?

I'm going to change the page to point to BA's new board, too.
That was my whole problem with trying to "correct" your statement in the first place was that I couldn't think of a straight-forward, simple, way to say what I was thinking! [img]/phpBB/images/smiles/icon_smile.gif[/img]

Adding the "by the clocks' measurements" is a step in the right direction, though, and I certainly can't claim to have anything better to offer . . .

26. On 2001-10-25 11:56, robert_d wrote:
The interesting thing is what one observer can actually see at the position of the other. We all note that light takes time to travel. So if a clock moves away, even at slow non-relativistic speed, it must appear to slow down so that it ends up reporting a time exactly late by the time of light travel (equivalent to red shift). The difference with relativity is that this clock would appear to speed up on return (blue shift) so that when it returns its time would coincide with the observer's time. This is what throws me (due to math defficiency) I can't separate the two effects in my mind.
Robert,

The thing you've got to keep in mind is that the Relativity-predicted time dilation is in addition to the red-shift and blue-shift you're describing.

SR basically says that even if the stationary observer "corrects" for the red-shifted and blue-shifted distortions, the moving clock will still be ticking more slowly, both out and back.

27. Rosen,

Is there really a "fundamental difference" between those two thought experiments (one rocket turns around vs. two rockets already going in opposite directions)? I mean, granted, if you were to actually take a ship out and stop it and turn it around, you'd have to deal with the accelerative effects predicted by GR on top of the SR effects, but in terms of SR itself, there really isn't any difference, is there?

In the two-ships example, we've got three distinct events:

A. Outgoing clock passes stationary clock
B. Outgoing clock passes incoming clock
C. Incoming clock passes stationary clock

If the stationary clock and the outgoing clock are synchronized at A, and the outgoing clock and incoming clock are synchronized at B, then the incoming clock will be behind the stationary clock at C (the word "synchronized" is an adjective, not a verb, in this sentence).

So if the outgoing clock simply stops and turns around at B, why would we need to add or consider some additional force to explain why it's behind when it gets back?

28. Guest
On 2001-10-25 12:35, SeanF wrote:
Rosen,

---Is there really a "fundamental difference" between those two thought experiments (one rocket turns around vs. two rockets already going in opposite directions)?
---

I think so. Well, not in terms of the mathematics of SR. However, physically it makes a great deal of difference to the twin in the rocket that is moving out. Lets look at it in terms of biological age, in addition to the clocks. The question is how does the biological age differ from the reading on the "clocks."

The twin in the "passing ships" experiment is never coming home again. He doesn't turn on his rockets, he is going nowhere. He will never meet his twin brother again. The question of what there biological age is at "simultaneous" times is moot. There is no way that can be important to him, and the answer depends on inertial frame anyway.

The fellow coming in never met the twin on earth before. In fact, he can't truly be an identical sibling to the twin on earth. He was never on earth before. He receives a message from the brother in space when they pass each other by. Now, he isn't affected at all by a "turn around." However, when he meets the brother, their difference in biological ages won't mean anything. He has never been on earth, he doesn't know what the brothers biological age "should" be. The biological age has nothing to do with the setting of the clock, and is totally unaffected by it.

However, I repeat. Some component of the clock on board the "coming" ship was greatly affected during the passing. The crystal, or atom, or mainspring had to make a large change to "synchronize." As far as the "main spring" goes, it has suffered a trauma at least as large as the "turn around" ship. One can imagine a hand, reaching from the other ship and twisting. It would be very hard on the hands, no? The clock is much more sensitive to forces from the other ship than the other components, even if the other components transmit the information.

The twin who turns back is in a different category. When he comes home, he will actually see that he is younger than his brother. The returning brother will see that less clock time has passed for him than for his earth brother. He knows that they were born at the same time, which actually means that they were forced out of the uterus at the same time. A force has synchronized everything on both ships while on earth. There was no other synchronization. Therefore, he concludes that something happened to him on the way out. When turning around, a force affected both the mainspring of the clock and him.

I suppose variations on the twin paradox are asking: does the clock time agree with the time measured by some other device (like our biochemistry). The answer is yes, for the twin who turns around. The answer is no for the man coming in who was never on earth before.

The "ships passing" problem has a third character who never met the twins. He can never have been part of the "Paradox Triplets." There are only the "Paradox Twins." The third character has never experience a force (at least, not in our solar system). The forces between objects on his ship and objects on the other ship is extremely localized. Therefore, I think the problems have a significant physical difference if not a mathematical one.

29. On 2001-10-25 13:16, Rosen1 wrote:

Some component of the clock on board the "coming" ship was greatly affected during the passing. The crystal, or atom, or mainspring had to make a large change to "synchronize." As far as the "main spring" goes, it has suffered a trauma at least as large as the "turn around" ship. One can imagine a hand, reaching from the other ship and twisting. It would be very hard on the hands, no? The clock is much more sensitive to forces from the other ship than the other components, even if the other components transmit the information.
Ah, no -- from a purely mathematical standpoint, this is not necessary for the thought experiment. The "incoming" clock doesn't need to be "set" or "adjusted" at the moment of passing; it just "is" set the same.

If that's too much coincidence, just consider it this way:

Event A: Outgoing clock passes stationary clock
Event B: Outgoing clock passes incoming clock
Event C: Incoming clock passes stationary clock

All observers will agree on these observations:

1: Time showed on stationary clock at Event A
2: Time showed on stationary clock at Event C
3: (Corollary to 1 & 2) Amount of time passed for stationary clock between Events A and C
4: Time showed on outgoing clock at Event A
5: Time showed on outgoing clock at Event B
6: (Corollary to 4 & 5) Amount of time passed for outgoing clock between Events A and B
7: Time showed on incoming clock at Event B
8: Time showed on incoming clock at Event C
9: (Corollary to 7 & 8) Amount of time passed for incoming clock between Events B and C

If there were no time dilation, then the sum of (6) and (9) would be equal to (3), but it's not. It's less. Any and all observers would agree on this, even one in an inertial frame other than the three directly involved in this thought experiment.

Less combined time passed between A & C for the two moving clocks than for the stationary clock, even though both moving observers would say the stationary clock was always ticking more slowly than their own.

And the exact same thing happens if a single clock just turns around at Event B.

_________________
SeanF

<font size=-1>[ This Message was edited by: SeanF on 2001-10-25 13:44 ]</font>

30. Guest
[/quote]

>---Ah, no -- from a purely mathematical standpoint, this is not necessary for the thought experiment. The "incoming" clock doesn't need to be "set" or "adjusted" at the moment of passing; it just "is" set the same.---
From the purely physical viewpoint, the incoming space ship has to record the time that the other space ship passed it. Other wise, there is no way to tell "how much time passed."
Maybe its not the main spring that is affect. The outgoing space ship at blast off looked at his clock, and when another ship was detected via radio looked at the time again. Radios work because an electromagnetic field from the radio wave pushes free electrons through a wire. The outgoing ship sent a signal that was recorded, telling how much time elapsed. The radio and tape recorder were certainly greatly affected byelectromagnetic forces on the ship. At that moment, getting a radio signal from the outgoing, looked at his clock. When he landed on earth, he added the differences up and compared it to the earth twins passage of time. However, this couldn't be done unless the radio antennae had electrons loose enough to be moved by a radio wave. It was only the radio antennae, and devices attached to it, that was effected by the passage of the other ship. The force from the other ship wasn't really "insignificant," it was merely "highly localized."

---1: Time showed on stationary clock at Event A
2: Time showed on stationary clock at Event C
3: (Corollary to 1 & 2) Amount of time passed for stationary clock between Events A and C
4: Time showed on outgoing clock at Event A
5: Time showed on outgoing clock at Event B
6: (Corollary to 4 & 5) Amount of time passed for outgoing clock between Events A and B
7: Time showed on incoming clock at Event B
8: Time showed on incoming clock at Event C
9: (Corollary to 7 & [img]/phpBB/images/smiles/icon_cool.gif[/img] Amount of time passed for incoming clock between Events B and C---
As I said before, the ships must interact via forces in order to take those differences.

Biological processes are controlled by forces that are also constrained by SR. Therefore, I believe that the hypothetical, untested "assymmetry" with a turn around twin could be real. Alot of these so called "solutions" to the paradox claim that the twins would be the same age. I just don't believe it.

Mesons showing a time dilation shouldn't be different from a free radical in the human body. If the mesons show a time dilation, so would the human body. Am I wrong?

What do you believe? Would biological twins, in the "twin paradox" experiment, show an assymmetry in biological aging or not? If so, why? If not, how does the math fit the symmetry of the situation situation?

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