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GrapesOfWrath
2001-Oct-25, 03:03 AM
On 2001-10-24 21:59, Rosen1 wrote (http://www.badastronomy.com/phpBB/viewtopic.php?topic=15&forum=1&start=30&35):
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 (http://mentock.home.mindspring.com/twins.htm) of the twin paradox?

Ben Benoy
2001-Oct-25, 03:22 AM
That's pretty cool. Good link. (n/t) /phpBB/images/smiles/icon_biggrin.gif

Ben

GrapesOfWrath
2001-Oct-25, 03:52 AM
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 (http://mentock.home.mindspring.com/twinrdux.htm), btw. And I might as well throw in the general relativity version (http://mentock.home.mindspring.com/twin2.htm).

Hauteden
2001-Oct-25, 04:13 AM
On 2001-10-24 23:03, GrapesOfWrath wrote:

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

http://www.dusko.net/duck/pictures/010.jpg

Its a Pair-a-Ducks /phpBB/images/smiles/icon_lol.gif
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>

GrapesOfWrath
2001-Oct-25, 11:49 AM
It's Pair-a-dox (http://www.aip.org/history/einstein/ae63.htm), not pair-a-dux. /phpBB/images/smiles/icon_smile.gif

2001-Oct-25, 12:17 PM
...,where do you see the force in this description (http://mentock.home.mindspring.com/twins.htm) 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>

GrapesOfWrath
2001-Oct-25, 12:31 PM
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 (http://www.fourmilab.ch/etexts/einstein/specrel/www/). I expounded on that in the Twins Redux article (http://mentock.home.mindspring.com/twinrdux.htm).

SeanF
2001-Oct-25, 12:38 PM
Hey, Grapes, I didn't realize that was you! /phpBB/images/smiles/icon_smile.gif

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). . .

2001-Oct-25, 01:24 PM
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.

GrapesOfWrath
2001-Oct-25, 01:36 PM
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>

GrapesOfWrath
2001-Oct-25, 01:51 PM
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.

SeanF
2001-Oct-25, 01:54 PM
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>

SeanF
2001-Oct-25, 02:01 PM
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 . . . /phpBB/images/smiles/icon_smile.gif

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? /phpBB/images/smiles/icon_wink.gif

GrapesOfWrath
2001-Oct-25, 02:10 PM
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.

SeanF
2001-Oct-25, 02:16 PM
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>

GrapesOfWrath
2001-Oct-25, 02:31 PM
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?

SeanF
2001-Oct-25, 02:46 PM
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 . . .

SeanF
2001-Oct-25, 02:49 PM
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. /phpBB/images/smiles/icon_frown.gif

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 . . .

GrapesOfWrath
2001-Oct-25, 03:00 PM
At least you haven't deleted it...yet /phpBB/images/smiles/icon_smile.gif

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?

SeanF
2001-Oct-25, 03:32 PM
On 2001-10-25 11:00, GrapesOfWrath wrote:
At least you haven't deleted it...yet /phpBB/images/smiles/icon_smile.gif

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>

2001-Oct-25, 03:42 PM
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.

GrapesOfWrath
2001-Oct-25, 03:47 PM
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.

2001-Oct-25, 03:50 PM
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.

robert_d
2001-Oct-25, 03:56 PM
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.

SeanF
2001-Oct-25, 04:23 PM
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! /phpBB/images/smiles/icon_smile.gif

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 . . .

SeanF
2001-Oct-25, 04:25 PM
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.

SeanF
2001-Oct-25, 04:35 PM
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?

2001-Oct-25, 05:16 PM
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.

SeanF
2001-Oct-25, 05:44 PM
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>

2001-Oct-25, 06:57 PM
[/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 & /phpBB/images/smiles/icon_cool.gif 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?

robert_d
2001-Oct-25, 07:02 PM
Thanks for the reply, SeanF. Would you know if relativity dominates the doppler effect at high % of C? Or are they mathematically the same magnitude? In other words if a clock is coming towards you at 3/5 C, do the two effects cancel out and the clock appear to run normal?

SeanF
2001-Oct-25, 07:16 PM
Rosen, my opinion is that the returning twin would in fact be younger than his earth-bound brother. Whether the "clocks" in question are atomic clocks, quartz crystals, biological entities, or what not doesn't matter -- they are all just denoting passage through the "time" dimension, and it is the rate of that passage that's affected.

IMHO, anyway. /phpBB/images/smiles/icon_smile.gif

One thing about SR is the simultaneity issue on top of the time dilation issue. Time dilation is direction independent, but the skewing of simultaneity is not -- it's skewed "up" in front of you and "down" behind, and it is this change in perspective when the twin turns around that breaks the symmetry of the time dilation.

I should probably point out that I've had no formal education on Relativity or any physics past High School AP . . . this is all just stuff I've picked up from reading about it, etc.

SeanF
2001-Oct-25, 07:17 PM
On 2001-10-25 15:02, robert_d wrote:
Thanks for the reply, SeanF. Would you know if relativity dominates the doppler effect at high % of C? Or are they mathematically the same magnitude? In other words if a clock is coming towards you at 3/5 C, do the two effects cancel out and the clock appear to run normal?

Robert, that is absolutely cool. /phpBB/images/smiles/icon_smile.gif I've never thought about that before, so I don't have an answer ready to give you, but let me work on it.

My initial reaction is that there may be a specific velocity at which the effects would exactly cancel each other out, but I'm not sure . . .

2001-Oct-25, 07:24 PM
One thing about SR is the simultaneity issue on top of the time dilation issue. Time dilation is direction independent, but the skewing of simultaneity is not -- it's skewed "up" in front of you and "down" behind, and it is this change in perspective when the twin turns around that breaks the symmetry of the time dilation.

All right then! What from a mathematical point of view defines "turning around?"

This is a trick question. I don't think that there is any criteria from pure mathematics that determines which twin has turned around. I am saying that from the physical point of view, forces and only forces break that symmetry. Therefore, forces are more reliable "causes" than dilation, contraction, and other kinematic concepts.

SeanF
2001-Oct-25, 07:49 PM
All right then! What from a mathematical point of view defines "turning around?"

This is a trick question. I don't think that there is any criteria from pure mathematics that determines which twin has turned around. I am saying that from the physical point of view, forces and only forces break that symmetry. Therefore, forces are more reliable "causes" than dilation, contraction, and other kinematic concepts.

This is a trick question! /phpBB/images/smiles/icon_smile.gif

The Law of Inertia (rephrased for SR) tells us that an object in an inertial frame will remain in that inertial frame unless a force acts upon it. It is, of course, impossible for either twin to change inertial frames without a force.

However, consider a plain piece of glass through which light rays travel undeflected. Now go and grind that glass into a lens, through which light rays are refracted. I don't think I would agree with any claim that the force behind the grinding is what causes the light rays to bend, would you?

It may require a force to get from one situation to another, but the consequences of being in any given situation are consequences of that situation, not consequences of the force . . .

Hmmm . . . does all that sound like, "Uh, I don't know?"

2001-Oct-25, 11:20 PM

I found another way the "turn-around twin" and the "ships passing problem" are different. The Captain flyby sees the clock his clock reset. Mr. Turnaround never notices his clock reset. The effect of the turn around in terms of time is only noticed by Mr. Flyby. Why is that?

This time, I will claim that it is a fundamental difference. We will consider all time measurements a type of synchronization. In other words, once you detect the other ship and measure the time, you can subtract that time from the time of the next measurement immediately. In your counter example, you chose to calculate the difference later after the entire trip is complete, at the very end, but the results should not be different then for a measurement by measurement calculation.

Scenarios that involve the spaceman doing something are also equivalent. You are merely using the spaceman as an additional component of your clock.

Suppose Mr. Flyby resets his clock the moment he sees the other ship whizz by. The rhodopsin in his eye, a molecule shaped which is shaped like and antenae (the old tree like ones) actually swings around due to the electromagnetic field of the light. To the rhodopsin molecule, and all the nerves responding to it, the force of the light is not negligible. It may be negligible to most of the other neurons in this person, to other people on the ship, to the rocket engines. However, the communication device and recording devices attached to it don't find this force "negligible." The force of the light on the rhodopsin together with the memory part of the brain have been changed, and it is an irreversible reaction.

Consider Mr. Turnaround who is watching his on board clock the entire time. When that rocket turns around, and the flame turns on, the only thing he notices is increased pressure on his pants. He does not see the clock jump. Before the turn around, after the turn around. Both clock and man are under the influence of the thrust, and they are close together. This results in the man and clock being in agreement.

In the "ships passing" problem, consider the Mr. Incoming man. He is also watching his clock. The other ship passes closely, a radio signal passes from one clock to another and the two clocks are set to the same time. The incoming man DOES see the clock jump. Neither he nor the twins can predict how much, or in which direction, since no one knows what the difference between clocks was before they get there. However, Spaceman Incoming does see the reading on his clock change suddenly. By a strange coincidence, at the same time he feels a mysterious pressure on his pants. By mysterious I mean one with no reaction force.

The force that passed from one ship was a radio wave, which moved the electrons in the antenae significantly. The radio wave affected nothing else immediately on the ship. Every other consequence of that radio wave occurs after the two ships have separated. If

Here is another way to put it. In SR, an inertial frame is one where each observers is not under the influence of a nonzero net force. The antennae and the rhodopsin, while in the vicinity of the other spacecraft, are under the influence of a nonzero net force. Either the light or the radio wave. Both exert significant force on some electrons.

<font size=-1>[ This Message was edited by: Rosen1 on 2001-10-25 19:23 ]</font>

2001-Oct-25, 11:59 PM
--However, consider a plain piece of glass through which light rays travel undeflected. Now go and grind that glass into a lens, through which light rays are refracted. I don't think I would agree with any claim that the force behind the grinding is what causes the light rays to bend, would you?--

I was saying that "proximity" in space time can only be defined in terms of forces between two objects. If the two objects interact by means of a known short range force, then the occur simultaneously and in the same place. The grinding occurred earlier in time,probably in a different location.

The bending of light is another matter. It is simultaneous with the light passing through the surface of the lens. Now, just why is it simultaneous and how do we know?

Suppose the lens is in a vacuum. The lens bends the light partly because it has a different index of refraction than the vacuum. The reason that it has a different index of refraction is because of the polarizability of glass is not zero at optical frequencies. The reason that the polarizability is not zero is that the electrons in the glass can be accelerated by the electromagnetic disturbance that is a light wave. Microscopically, the forces do define whether the glass is in an inertial frame.

Even macroscopically, the light exerts radiation pressure on the lens. If we wanted to know when the light hits the surface of the lens, the only way to do it is measure the radiation pressure. The radiation pressure is related to shape of the lens and the index of refraction.

In fact, one of the chapters of Lorentz's book "Theory of the Electron" is devoted to explaining why a piece of glass doesn't develope birefringence at high velocity. One would have thought that the internal forces of the electrons would create stress. One moving electron creates a magnetic field that the other moving electron is forced to pass through. One would think that it would develope birefringence. However, it doesn't. By the calculation of forces, he proved it doesn't.

Another thing. Index of refraction will change at high velocities. There is a transformation for index of refraction. Its related to the change in magnetic permeability of the material. We would call that a relativistic constituent relation because it relates to the macroscpic material properties. However, the these are the result of microscopic forces.

[/quote]

2001-Oct-26, 11:58 AM
The title of Einstein's paper was "On the Electrodynamics of Moving Bodies." It was not "On the Electrokinetics of Moving Bodies." Without discussing fields and forces, SR means nothing.

[quote]
---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?---

Not really. In Einsteins example, both clocks are synchronized by a lightening bolt hitting nearby. However, the incoming man has no way to validate that the other ship was near the lightening bolt, or even that he is near the lightening bolt. Even if the two trains are exactly the same distance apart. Short range forces are still necessary to synchronize the two trains.

Suppose a policeman had solve the mystery of whether the two trains really were close to each other when the messages were exchanged. I assume that the messages could travel a large distance, so for all either knows the trains could have been far apart. would be the following. He would interview Mr. Incoming, and ask what happened. Mr. Incoming will say, "I saw a flash, I smelled the ozone, and the thunder rattled me. Then I received a message from Mr. Outgoing telling me the time" The policeman takes an airplane and gets to Mr. Outgoing destination and asks him what happened."I saw a flash, I smelled the ozone, and the thunder rattled me. Then I received a message from Mr. Incoming telling me the time." The policeman says, "Well, they both saw the flash, they both smelled the ozone, the thunder shook them. These all represent forces with a very limited range. They must have been close together when they communicated!"

---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.
---How would either observer know that? The only way that they can determine that the two pulses traveled for the same amount of time, in the stationary frame, is to compare notes. Or ask someone in the traveling frame. In any case, forces are involved.

By static force, I mean one that does not change with velocity.

As I said before, in the passing train situation, the effect of the forces is extremely local at the turn around point. Only the communication and detection devices are immediately effected. The recording devices (including human memory) may work later on. A man in the ship will see that the other ships time has been received. In the turnaround case, the forces affect everything in the rocket. Mr. Turnaround has set up the clock not to react directly to the Newtonian thrust or vibration. He himself is cushioned against the thrust. However, the forces that reset the clock are neither the Newtonian thrust nor vibration.

Maybe part of the problem is that we usually don't consider the force fundamental. In Newtonian, there is a mathematical definition of foce. Then the Third Law of Motion (for every action there is a reaction) then breaks the circular logic of the definitions. I claim that a modified version of the Third Law breaks the circular logic of SR.

T

SeanF
2001-Oct-26, 12:34 PM
On 2001-10-26 07:58, Rosen1 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?---

Not really. In Einsteins example, both clocks are synchronized by a lightening bolt hitting nearby. However, the incoming man has no way to validate that the other ship was near the lightening bolt, or even that he is near the lightening bolt. Even if the two trains are exactly the same distance apart. Short range forces are still necessary to synchronize the two trains.

"Not really?" What exactly are you disagreeing with? All I said in that paragraph is that if a signal leaves Clock A when Clock A reads a certain time, and that signal arrives at Clock B when Clock B reads a certain time, you cannot simply subtract those two time readings to calculate the amount of time the signal took.

If you have two clocks sitting a mere inches apart, with one set on CST and the other set on EST, subtracting the times would seem to tell you it took an hour for the light pulse to get from one clock to the other. It did not.

How can you disagree with that?

---How would either observer know that? The only way that they can determine that the two pulses traveled for the same amount of time, in the stationary frame, is to compare notes. Or ask someone in the traveling frame. In any case, forces are involved.

Not true. They "determine" that the two pulses traveled for the same amount of time in the stationary frame because "time", at least within Einstein's paper, is predefined that way.

If you have two objects that are not moving relative to each other, it is taken as a given that a light-speed pulse will take the same amount of time to get from object one to object two as it takes to get from object two to object one. That is the starting definition of "time," and it comes from the presupposition that light speed is a constant.

robert_d
2001-Oct-26, 01:20 PM
Even that quick summary of the "twins" paradox bothers me. It says that "Bob lets 4 years pass" - So they must have agreed Bob would return in 8 years time. (Thinking at that point that time was unchanging). His clock is slowed by dilation so that he turns around when Ann's clock says 5 years have passed. My problem arises (and the reason I brought up the doppler effect) is that Ann's clock must appear to slow down to Bob (if he could see it instantaneously) But he CANNOT. So what would Ann see and what would Bob see when Bob's clock said four years?

SeanF
2001-Oct-26, 01:28 PM
On 2001-10-25 15:02, robert_d wrote:
Thanks for the reply, SeanF. Would you know if relativity dominates the doppler effect at high % of C? Or are they mathematically the same magnitude? In other words if a clock is coming towards you at 3/5 C, do the two effects cancel out and the clock appear to run normal?

Robert,

On checking, this doesn't ever happen . . . the blueshift effect increases more quickly than the time dilation effect does, so you're always receiving the signals faster and faster even though the accelerating source is transmitting them more slowly.

SeanF
2001-Oct-26, 01:36 PM
Rosen,

In my post up there where I denoted all the individual events, I started out saying we could ignore the "setting" of the clocks and just assume that they happen to be match up when they passed; I also provided a scenario where we simply look at the elapsed time.

In neither case is the incoming clock changed at the moment it passes the outgoing clock. Doesn't matter.

The "force" of the signal hitting the receiving system or the observer's eye may not be negligible, but it is irrelevent to the effects of SR. In the context of the twin paradox, it just doesn't matter, and neither does any "force" imparted by the acceleration of the twin turning around.

Okay, let me say this -- GR makes predictions about effects of acceleration. Those effects would come into play with a twin turning around and coming back. However, there are still specific predictions made by SR that the returning twin would be younger.

Consider if both twins are sent off in spaceships with equal acceleration programs, but one travels farther before stopping and turning back around. In this case, any GR effects of the accelerations would be absolutely equal for the two twins (in facts, any effects of the acceleration at all would be equal). However, SR still predicts that the twin who travels farther would end up younger because he spent more time in simple relative motion.

SeanF
2001-Oct-26, 01:45 PM
On 2001-10-26 09:20, robert_d wrote:
Even that quick summary of the "twins" paradox bothers me. It says that "Bob lets 4 years pass" - So they must have agreed Bob would return in 8 years time. (Thinking at that point that time was unchanging). His clock is slowed by dilation so that he turns around when Ann's clock says 5 years have passed. My problem arises (and the reason I brought up the doppler effect) is that Ann's clock must appear to slow down to Bob (if he could see it instantaneously) But he CANNOT. So what would Ann see and what would Bob see when Bob's clock said four years?

Robert, this is where you need to consider the simultaneity issue of SR.

Bob says his clock says 4 and Ann's clock says 3.2 at the same time. Ann says Bob's clock says 4 and her clock says 5 at the same time.

When Bob turns around and comes back the other way, the direction of the simultaneity changes. If he changes direction instantaneously when his clock says 4, he would go from a reference frame in which Ann's clock is 3.2 "right now" to a reference frame in which Ann's clock is 6.8 "right now." Hence, to Bob, Ann's clock ticks 3.2 years during the trip out, and 3.2 years during the trip back, but is still at 10 when he gets back (and his clock says 8).

As to what he would actually "see" if he were receiving light signals from Ann's clock during the whole trip, he would still receive signals for the entire 0-10 ticking of Ann's clock. That is, if he was recording the signals, the recording would not suddenly jump from 3.2 to 6.8. The Doppler Effect (red shift and blue shift) combined with the SR time dilation would produce some really weird visual timings of Ann's clock.

It's very complicated, and very difficult to visualize . . .

_________________
SeanF

<font size=-1>[ This Message was edited by: SeanF on 2001-10-26 09:45 ]</font>

GrapesOfWrath
2001-Oct-26, 02:02 PM
So what would Ann see and what would Bob see when Bob's clock said four years?

I think Sean has already addressed this, but I want to point out that any inertial frame of reference is valid--they just don't "see" things at the same time, of course. If you accept that the equations are consistent (that is, they aren't inherently self-contradictory), then the thing to do is choose a single reference frame that makes your calculations the simplest.

When two observers are side by side, they will see the same thing, essentially.

2001-Oct-26, 11:14 PM
In my post up there where I denoted all the individual events, I started out saying we could ignore the "setting" of the clocks and just assume that they happen to be match up when they passed; I also provided a scenario where we simply look at the elapsed time.

I just don't think that this is a valid assumption. In order to start counting, the two rocket men have to communicate to know that the two settings are the same. To do this, a force must exist between specific devices on the outgoing and incoming rockets. This is the physical reason. It has to be the force.

SeanF
2001-Oct-27, 01:47 AM
On 2001-10-26 19:14, Rosen1 wrote:

In my post up there where I denoted all the individual events, I started out saying we could ignore the "setting" of the clocks and just assume that they happen to be match up when they passed; I also provided a scenario where we simply look at the elapsed time.

I just don't think that this is a valid assumption. In order to start counting, the two rocket men have to communicate to know that the two settings are the same. To do this, a force must exist between specific devices on the outgoing and incoming rockets. This is the physical reason. It has to be the force.

In dealing with thought experiments in SR, there are always assumptions. One might say, "there are two clocks, located some distance apart, that are not moving relative to each other." Now, from a literal standpoint, there would need to be communications and forces going back and forth in order for an observer to verify that the two clocks are, in fact, motionless relative to each other.

However, these forces are irrelevent to the thought experiment, the purpose of which is simply to demonstrate what would happen in the given case. It doesn't really matter if any observer within the experiment itself knows for a verifiable fact that the clocks are relatively motionless; all that matters is that they are.

The same is true in this thought experiment. It doesn't matter that anybody on the rockets or the planet could actually observe the clocks at the appropriate times. All that matters is that they are synchronized when they pass.

Besides, the outgoing ship could easily be broadcasting it's clock time. An observer on the incoming ship could just as easily use that signal to calculate how quickly the outgoing ship is approaching, how it's clock is running, and what time the outgoing ship clock will be showing when they pass. He can then set his own clock in order to have them synchronized when they pass.

He can do this minutes, hours, even years ahead of time, if the duration of the experiment is long enough.

Yes, Rosen, there would be forces involved in the experiment, but the predictions of SR are separate from and independent of those forces.

GrapesOfWrath
2001-Oct-27, 06:03 AM
On 2001-10-26 19:14, Rosen1 wrote:
I just don't think that this is a valid assumption. In order to start counting, the two rocket men have to communicate to know that the two settings are the same. To do this, a force must exist between specific devices on the outgoing and incoming rockets. This is the physical reason. It has to be the force.

I like this.

What if, in my shipping lanes setup, I had some folk stationed between the shipping lanes, and only they read the clocks? They don't leave the "stationary" reference frame, and after they're done, they compare their readings of the clocks. That way the two ships never communicate directly with each other, and the readings are purely passive, but the results of the experiment will still be the same.

Lusion
2001-Oct-27, 04:40 PM
[quote]On 2001-10-25 11:50, Rosen1 wrote:
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.

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.

I'm confused. Why would the forces of synchronizing clocks play into time differences?

Gallileo reasoned that if two bricks side by side fall at the same speed, then both bricks should still fall at that same speed if they were cemented together. From this he concluded that all things being equall, all masses fall at the same speed. (Of course, he did confirm this with a bunch of experiments).

It would seem to me, then, that if it takes some particular force to synchronize clocks to a clock on a particular ship that's travelling side by side with another, very massive ship, then it would still take that same amount of force if the ships were cemented together, right?

Also, it seems to me that if synchronizing once actually causes differences in time frames, then synchronizing fifty times would cause fifty times the difference.

Could you explain this to me? Or, if I'm misinterpreting you, could you clarify?

2001-Oct-28, 12:22 AM
---If you accept that the equations are consistent (that is, they aren't inherently self-contradictory), then the thing to do is choose a single reference frame that makes your calculations the simplest.---

This is true, and used all the time
when the "field equations" are well known.
For instance, we know that in a reference frame a charged particle has an electric field and a magnetic field associated with it. One uses the magnetic field to make
the electric field "fit" the postulates of SR. This is obvious since they teach one about electricity and magnetism long before they teach you about SR. However, in the frame moving with the charge, the magnetic field generated by the charge doesn't exist. One doesn't need SR particularly to predict the law of magnetism from measurements of an electric field, since magnetism was already well known.

However, what if either the "field equations" are either totally unknown or too complicated to solve directly? What about that "rope" between the preprogrammed space ships which broke even though in the stationary frame it didn't "actually" change length?

Someone asked "why" it broke. Almost everyone agreed that in the frame of the ships, the two spacecraft moved apart and stretched the rope. However, in the stationary frame, the rope it didn't experience a "contraction" until it broke. Someone said that it did contract. However, the only thing that contracted, if anything, was the hypothetical length of the unstressed rope. The Lorentz contraction wasn't a physical contraction in this case, it was the contraction of an abstraction.
The rope is too complicated (all those atoms!) to analyze using the electromagnetic field equations, even aside from the additional complexities of quantum mechanics. The forces on the rope have to calculated at a purely macroscopic level.

This is where SR would really come in useful! All physicists know the low velocity (v<<c) approximation of force equations for a rope. They can measure a few properties of the rope, they can predict and measure the speed of sound and transverse waves on the rope, the distance between rockets, and the length of the unstretched rope. However, they can not predict the tension on the rope without SR. If they tried, they would get the answer that the tension doesn't change. Therefore, the rope doesn't break in the stationary frame.

This "Lorentz contraction of an abstraction" could bother some people. After all, there is no clue as to what physical property changed in the rope. It broke spontaneously, as if by magic. Although the problem is supposedly solved (the unstretched length of the rope does contract), nonscientists can sense that the logic is incomplete. An abstraction like "the normal rope length" can not break a rope.

SR enables one to predict the forces at velocities close to the speed of light. In the frame of the rockets, the rockets are moving apart. The unstretched length of the rope remains the same. The tension is easily calculated by Hookes Law.However, in the stationary frame, the distance between rockets front to back remains constant.

SR says that even if you fix your viewpoint on in the stationary frame, you can not assume that the equations of force on the rope are not velocity dependent. The universe, according to SR, only has "Lorentz invariant forces" regardless of the complexity of the system. The equations of tension measured at low velocities was merely an approximation of the actual equations of motion. In the stationary frame, the tension of the rope has to consist of one nearly constant component plus another force that is proportional (roughly) to velocity. SR enables one to calculate and predict the velocity dependence of the tension.

One nice thing about SR is that one can predict forces at high velocity from low velocity measurements. Magnetism is one known before SR.

The weak nuclear forces, that make the meson decay, are forces studied after SR was conceived. There was no way before SR to predict that the meson would decay slower at high velocities. They measured the decay time at low velocities. However, there was no way to tell that there was a "velocity dependent" weak force that slowed down the decay. The fact that time slowed down is true, but very abstract. What is more real to some of us is that forces that turns on at high velocities slowed everything, including the fast moving meson, down.

It is impossible to determine what frame is inertial and what is a noninertial frame without forces because the lack of forces on the observer are implied in the definition of inertial frame.

GrapesOfWrath
2001-Oct-28, 10:41 PM
On 2001-10-27 02:03, GrapesOfWrath wrote:
What if, in my shipping lanes setup, I had some folk stationed between the shipping lanes, and only they read the clocks?

OK, I have changed the Twins Paradox Redux page (http://mentock.home.mindspring.com/twinrdux.htm). I changed it to include that, and also Sean's clarification in the discussion of Section 1.1. And I changed that attribution section. Thanks.

The old version (http://mentock.home.mindspring.com/twinrold.htm) is still around, for comparison.

_________________
rocks
<font size=-1>[Fixed old version url]</font>

<font size=-1>[ This Message was edited by: GrapesOfWrath on 2001-10-28 17:43 ]</font>

2001-Oct-29, 12:12 AM
Dear "Grapes of Wrath." I have looked at our new Webpage.It looks very nice. I have one teeny suggestion as an addition.

[quote from Grapes]
---Unfortunately, when a clock is turned around, it no longer remains in an inertial reference frame, and the principles cannot be applied so simply.---[end quote from Grapes]

You should explain why the laws of mechanics don't apply when the clock is turned around. My own explanation, which you of course don't have to use, is that the laws of mechanics have to hold for each part of the spaceship including the Third Law of motion. At turn around, the third law is violated if we restrict our POV to the cabin of the spaceship which contains the clocks. For every action there is an equal and opposite reaction. Restricting yourself to the spaceship, and ignoring both the thrust and the other space ship, the third law does not apply. There is an action (some particles in the spaceship accelerate) and no reaction corresponding particles in the same spaceship accelerate in the exact opposite direction as defined in the first and second laws).

" An example of particles accelerating are the electrons in whatever apparatus comprise the receiver of the the signal. Clearly, some electrons in receiver don't stay in the same "inertial frame." In the ships passing example, only a few electrons "change inertial frames" during passover and in the "reverse direction" case all the electrons in the cabin change inertial frames."

Maybe there is a way you can condense it, somehow. Or, maybe, add an appendix?

<font size=-1>[ This Message was edited by: Rosen1 on 2001-10-28 19:15 ]</font>

<font size=-1>[ This Message was edited by: Rosen1 on 2001-10-28 20:14 ]</font>

<font size=-1>[ This Message was edited by: Rosen1 on 2001-10-28 20:23 ]</font>

<font size=-1>[ This Message was edited by: Rosen1 on 2001-10-28 20:39 ]</font>

GrapesOfWrath
2001-Oct-29, 11:29 AM
On 2001-10-28 19:12, Rosen1 wrote:
You should explain why the laws of mechanics don't apply when the clock is turned around.

The first sentence of the second paragraph is "The principle of relativity is applied to 'all frames of reference for which the equations of mechanics hold good,'" though I guess the reader would have to make some sort of inference. In that page, I wasn't trying to explain special relativity, per se, I was more trying to step through the logic of that 1905 paper of Einstein's. For an old friend of ours, who did not want his name associated with it, unfortunately.

Maybe there is a way you can condense it, somehow. Or, maybe, add an appendix?

I want to think about that for awhile. I've added your name to the end--if that's OK.

2001-Oct-29, 12:58 PM

Fine. Furthermore, you don't have to use all of it.

Maybe you could just state the idea that the clocks or some equivalent procedure is significantly affected by a force between the two ships, and leave it at that.

GrapesOfWrath
2001-Oct-29, 02:27 PM
On 2001-10-29 07:58, Rosen1 wrote:
Maybe you could just state the idea that the clocks or some equivalent procedure is significantly affected by a force between the two ships, and leave it at that.

I'm glad you rephrased it that way, 'cause I disagree with it. There doesn't seem to be any significant force or "force" effect at all. The context is special relativity, but still, the effect can be demonstrated with pure mathematics--no force involved. The spacetime is changed under transformation, that's all.

2001-Oct-30, 12:15 AM
I'm glad you rephrased it that way, 'cause I disagree with it. There doesn't seem to be any significant force or "force" effect at all. The context is special relativity, but still, the effect can be demonstrated with pure mathematics--no force involved. The spacetime is changed under transformation, that's all.

[/quote]

You were right and I was wrong. (Ouch) At least about the ships passing problem. That is pure mathematics.

However, a separate explanation should be give in the turnaround problem. I still feel that the the turnaround problem is NOT pure mathematics since there is onyl one observer who shifts from one to another frame. There has to be a force to explain that shift of that one observer.

I guess what I meant to say is that the agreement between these two different scenarios may be logically linked to the definition of stationary frame. At this point, this is all my speculation.

GrapesOfWrath
2001-Oct-30, 11:55 AM
I still have the t-shirt from the old BA board that says "You are right, I was wrong." I wear it all the time. It comes in very handy.

I used to insist that a force must be involved, in the twin paradox, but I now realize that even the forces can be relativized. A person falling on earth is under the influence of gravity, yet in their own reference frame, spacetime is "flat." That was the great insight. I think we still have much to appreciate and learn from that.

SeanF
2001-Oct-30, 01:21 PM
On 2001-10-29 19:15, Rosen1 wrote:
However, a separate explanation should be give in the turnaround problem. I still feel that the the turnaround problem is NOT pure mathematics since there is only one observer who shifts from one to another frame. There has to be a force to explain that shift of that one observer.
his is all my speculation.

Rosen, you're absolutely right that there needs to be a force that causes the shift in observation, but it's simply the observation that's important.

You can talk about how the air is thinner at the top of the mountain than at the bottom without talking about the force necessary to get a person from the bottom to the top . . . and you can talk about how the perception of time on the distant world is different going towards it than going away from it without talking about the force necessary to turn the person around.

2001-Oct-31, 01:15 PM
I was wrong and you were right, again /phpBB/images/smiles/icon_frown.gif

mickal555
2005-Jun-29, 04:26 PM
wow a guest...

sorry guy's- I'm going to bed...

Eroica
2005-Jun-29, 04:46 PM
Mickal, you should know better than to revive a relativity thread. These things are hard enough to kill as it is! :D

Normandy6644
2005-Jun-30, 12:33 AM
Yeah seriously, it's bad enough that they happen in the first place! :D :wink: