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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?

2. 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. [img]/phpBB/images/smiles/icon_smile.gif[/img]

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.

3. 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. [img]/phpBB/images/smiles/icon_smile.gif[/img] 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 . . .

4. Guest
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.

5. 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! [img]/phpBB/images/smiles/icon_smile.gif[/img]

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?"

6. Guest

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>

7. Guest
--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]

8. Guest
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

9. 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.

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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?

11. 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.

12. 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.

13. 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>

14. 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.

15. Guest
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.

16. 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.

17. 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.

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[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?

19. Guest
---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.

20. 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. 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 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>

21. Guest
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>

22. 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.

23. Guest

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.

24. 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.

25. Guest
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.

26. 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.

27. 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.

28. Guest
I was wrong and you were right, again [img]/phpBB/images/smiles/icon_frown.gif[/img]

29. Order of Kilopi
Join Date
May 2004
Posts
6,971
wow a guest...

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

30. Mickal, you should know better than to revive a relativity thread. These things are hard enough to kill as it is!

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