Wrong. Perhaps we have different interpretations of what "paradox" means. As I understand it the twin paradox comes about because both twins see the other's clock as moving slowly during the periods between accelerations. Why then is the space-faring twin younger? The difference is that he undergoes accelerations the earth-bound twin does not. Contrary to what you're claiming, what happens during those accelerations is crucial to understanding why the spaceborne twin is younger upon return. You can calculate an answer with SR alone, but you don't know why that answer is correct until you invoke GR. Until you can explain what happens during the accelerations your understanding of the situation is incomplete.

If you look at the GR you'll find that the greater distance between the clocks also perfectly explains why the one who went farther is still younger on return.

There is a difference between computing the result, which I've already acknowledged can be done with the SR time dilation relations only, and understanding why that result is correct, which requires GR. Simply saying that the acceleration periods are unimportant is not an answer. They are what distinguishes the twins (or triplets) from each other and what happens during them is key to understanding the situation. Again I ask you to look into the textbooks. If you do you'll find that every one of them invokes GR as the ultimate explanation of the twin paradox.

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"If it was so, it might be, and if it were so, it would be, but as it isn't, it ain't. That's logic!" - Tweedledee

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It's not geocentricity, it's Relativity. Relativity says any inertial frame can be considered to be "at rest" - There is no difference between "Earth is moving towards the distant star" and "the distant star is moving towards the Earth." Therefore, Relativity predicts the Doppler Effect would be the same in either case.

Since we are on the Earth, we perceive our own inertial frame as being at rest and the star's inertial frame as being in motion. In that case, we'd see a Doppler Effect for the reasons I described, and so we do.

If we were at the distant star, we'd perceive the star's inertial frame as being at rest and the Earth as moving. In that case, we'd expect Earth to measure a Doppler Effect as I described above, and so it does.

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SeanF

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Not entirely true. During the first part of the space-faring twin's journey - when he's moving away from the Earthbound twin - each twin sees the other's clock as moving slowly. But when the space-farer turns for home, he immediately sees the other's clock running fast. However, the Earthbound twin continues to see the spacefarer's clock running slowly. As far as he's aware, the spacefarer hasn't yet turned for home. So the two situations are not symmetrical, even before we start to discuss acceleration.

You're right that acceleration is the key difference which ultimately explains why one twin ages less. It's the acceleration which gets him from one inertial frame into another. But very little of the difference in age occurs during the accelerations. And if the accelerations are instantaneous, then no age difference will accrue during them.

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Okay.

Clock A is sitting in space. Clock B is moving relative to A at a rate of 0.6c. Clock C is also moving relative to A at a rate of 0.6c, but in the opposite direction. From A's motionless point of view, events occur like this:

None of the clocks ever accelerate in any form during this experiment.

Observer A says that Clocks A & B both said "12:00:00" when they passed.
Observer A says that Clocks B & C both said "12:04:00" when they passed.
Observer A says that Clock C said "12:08:00" and Clock A said "12:10:00" when they passed.
Observer A says that Clocks B & C were both running slower than Clock A at all times.

Observer B says that Clocks A & B both said "12:00:00" when they passed.
Observer B says that Clocks B & C both said "12:04:00" when they passed.
Observer B says that Clock C said "12:08:00" and Clock A said "12:10:00" when they passed.
Observer B says that Clocks A & C were both running slower than Clock B at all times.

Observer C says that Clocks A & B both said "12:00:00" when they passed.
Observer C says that Clocks B & C both said "12:04:00" when they passed.
Observer C says that Clock C said "12:08:00" and Clock A said "12:10:00" when they passed.
Observer C says that Clocks A & B were both running slower than Clock C at all times.

Everybody agrees on this. There's no paradox. Even though the "other clocks" are always running slower, the reciprocity is broken by the fact that B&C are in different reference frames. You don't need GR for this.

The "Twin Paradox" is this exact same thing with the "stay-at-home" twin as A and the "traveling" twin as B in the first half and C in the second. Since you don't need GR for this, why would you need it for the "Twin Paradox"?

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SeanF

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SeanF, as I said earlier, since none of your earlier simpler thought experiments worked out, you have to keep making more and more complex thought experiments, and you hope that people will just eventually get lost while trying to figure out the more complex ones.

But there is absolutely no legitimate reason why you should deviate from the original simpler thought experiment from the 1905 paper itself, except that it leads to a paradox and you want to try to avoid the paradox.

You said:

“Observer A says that Clocks A & B both said "12:00:00" when they passed.
Observer B says that Clocks A & B both said "12:00:00" when they passed.
Observer C says that Clocks A & B both said "12:00:00" when they passed.

Observer A says that Clocks B & C both said "12:04:00" when they passed.
Observer B says that Clocks B & C both said "12:04:00" when they passed.
Observer C says that Clocks B & C both said "12:04:00" when they passed.”

According to SR theory, your second set of observations is not correct, since observer A would see the B and C clocks time dilating in the ratio of 1 : √ 1 – (v^2/c^2), because observer A would see himself as being “stationary” and he would see the B and C clocks as “moving”. AND, observers B and C would see themselves as being “stationary”, while they would see each other and A as “moving”. So, observer A could not see the same time on the B and C clocks that the B and C observers see on their own clocks when B and C pass. According to SR theory, observer A would see a “time dilated” time on the B and C clocks when B and C pass. B would see dilated time on the C clock but not on his own, and C would see dilated time on the B clock but not on his own. So observers B and C would disagree about what time they see on each other’s and their own clocks.

According to SR theory, B would not see the same time on the C clock as on his own clock since he would see himself as “stationary” and the C clock as “moving”, so B would “see” the C clock as time dilated. And C would not see the same time on his own clock as he sees on the B clock, since C would see himself as “stationary” and the B clock as “moving”, so C would see the B clock as being time dilated.

In fact, in this thought experiment of yours, each observer, A, B, and C, would “see” his own clock as ticking normally, while he would “see” the other two clocks as ticking more slowly.

Sam5, nobody can disagree as to what B and C said at the moment B and C pass each other (and I do mean nobody - regardless of relative motion of inertial frame, everybody who can see that B and C passed each other will agree that they both said 12:04:00 at the moment it happened). This is a single point in space-time, at which Clocks B and C are co-located; that is, they are at the same point in space and the same point in time. It's a single "event," and thus must be agreed upon by all observers.

Surely you understand that two clocks which are ticking at different rates could still end up showing the same time simultaneously at some point, don't you?

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SeanF

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Yes, but, B and C were already moving relatively, so B “saw” C’s clock time dilating, while C “saw” B’s clock time dilating, while neither saw their own clock time dilating. So they would not have the same readings unless you reset their timers exactly when they meet.

In your thought experiment, you’ve got several errors and misconceptions. In the first place, you say that B and C are moving relative to A at .6c, but you don’t point out that C and B are moving relative to each other at a higher rate.

Notice down at the bottom of your chart that you have C further away from B than A is. So you have A and B separating at one rate, while you have C and B separating at a faster rate. If B were separating from A at the same rate as B separates from C, then C and A would not separate at the bottom of your chart. They would move relative to B, but they would not separate relative to each other.

What you have in your chart is B and C converging and then separating from A at the same rate, but you’ve got B and C converging and separating from each other at a faster rate. This is because you have A as “stationary” in your chart, relative to us, the chart viewers, and also relative to our computer screens.

Another problem with your thought experiment is that you say:

“Observer A says that Clocks B & C both said "12:04:00" when they passed.
Observer B says that Clocks B & C both said "12:04:00" when they passed.”

So what you have here is only A’s “opinion” about what B is “supposed to say” he would “see” on his clock. But you can’t do that in SR theory. You can only say what A “sees” on his own and the B and C clocks. Then, if you want to get B’s opinion, you have to see the thought experiment from B’s perspective and he will tell you what he “sees” on his own and A’s clock, and what he “sees” on the C clock too. And of course, what B will actually “see” is his own clock ticking normally, while he “sees” the A clock as time dilated and the C clock time dilated more than the A clock.

Since the relative velocity of A and B is different from the relative velocity of B and C, according to your chart, with the relative velocity between B and C being greater than the velocity between B and A, then B would “see” the A clock time dilate a little, while he would “see” the C clock time dilate more during the relative motion. So, by the time B and C meet, B would “see” the A clock dilated a little and the C clock dilated more. So B can’t “see” the C clock reading the same as his own clock, unless you did some thought experiment manipulation and had the C clock start out with an advanced reading, or if you reset the B and C clocks when they meet. But even if you did that, the A observer and the B observer would disagree about what time they both “see” on the C clock, since C is moving at one velocity relative to A and a different velocity relative to B.

The more complicated you make these thought experiments, the more time it takes me to figure them out and find your errors, but I’m willing to continue when I have the spare time to work on them.

Sure they can - if their clocks are set right to begin with.

Neglecting to point out something that is so painfully obvious as that is not an error.

I still don't see what the problem is. B's relative motion to C is greater than his relative motion to A, yes. Of course. If B and C weren't moving relative to each other, they would never pass each other. Why is this a problem?
No, no, no. Perhaps that was bad terminology on my point. It's not Observer A's opinion of Observer B's opinion of Clock B, it's Observer A's opinion of Clock B. How about, "Observer A says that Clocks B & C were both at '12:04:00' when they passed." Is that better? Adjust all the other statements accordingly.

I never said anything about how the clocks were initially set up (again, pardon me for not explicitly stating the obvious). When one clock is running slower than the other, that does not mean the first clock is always behind the other. If the slower clock started out ahead of the faster clock, it will stay ahead (but less and less ahead) until the faster clock "catches" it. At that moment, they'll both show the same time (how about that?). From then on out, the slower clock will be behind the faster, lagging farther and farther behind as we go on.

Hmm. First you complain about things I've left out, then you complain that I'm making things unnecessarily complicated. The "simple" version of this experiment was the first three sentences:

Everything I added after that was just specifics - the actual times displaying on the clocks at specific points, etc.

(I also feel compelled to point out that this particular thought experiment was intended to show Eta C how I see the Twin Paradox as strictly SR. As such, whether it's simpler than others I've described to you never came into my mind.)

At any rate, I keep trying to make the experiments simpler until I can find one you understand, and then move ahead from there. :-?

(Speaking of which, what happened with my post with the four statements and asking you where the logic breaks down in it? You never answered...)

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SeanF

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Ah Ha! See? I knew you knew something that you weren’t telling us. Now, why don’t you tell us the start-out time reading on all the clocks, as seen by their own “stationary” observers, at the very start of your thought experiment?

You know what this reminds me of? That little disk shaped plastic toy that has the four buttons around the edges, and a button in the center. So I hit the center button and I get a “beep”. Then I have to hit the button that lit up when I heard the “beep”. Then I hit the center button again and I get a “beep” and a “bop”. Then I have to hit those two buttons in order to keep the game going. But it keeps getting more complicated. By the time I get up to a “beep”, “bop”, “boop”, “beep”, “beep”, “bonk”, “bop”, I’m just about bonkers, and finally I hit the wrong button and I get an offensive “buzzzzzz”.

My analysis of your last thought experiment is correct, but if you add just one more moving observer to your thought experiment, I might go bonkers.

No, I’m not complaining at all. I find this very relaxing and enjoyable. I’m just pointing out what you did, so some of the others can understand what you are doing.

Sam5,
You are the only one here who does not understand. The rest uf us are waiting for you to "get it". There may be some quibbles with some finer points, but you are the only one here who is disagreeing with relativity (either SR or GR).

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Sam5, I shouldn't have to tell you anything else about the clocks. If you understand SR, you already know (or at least you know how to calculate it). So let's take a little quiz.

Did you notice that for each "passing" (A & B, B & C, and A & C), I told you how all three observers would "read" the two clocks that pass each other, but said nothing about the distant clock? Specific example: When B passes C, I told you what everybody says about Clocks B & C, but I didn't say anything about what anybody says about Clock A. So, please answer these three questions:

What time does Observer A conclude was displaying on Clock A at the moment B passed C?
What time does Observer B conclude was displaying on Clock A at the moment B passed C?
What time does Observer C conclude was displaying on Clock A at the moment B passed C?

You know Special Relativity, right? So what does it predict for those three answers?

“Beep”, “bop”, “boop”, “beep”, “beep”, “bonk”, “bop”

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SeanF

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Kaptain K, as far as GR is concerned, there is no irresolvable clock paradox, since only one of two clocks “really” slows down. This is due to an acceleration force being experienced by one of the clocks, the one that slows down. But with SR there is an irresolvable clock paradox, since both of two relatively moving observers “see” each other’s clock slow down at the same rate, by means of light signals, while they do not “see” their own clock slow down at all Therefore, one of the two relatively moving clocks in SR theory can not “really” be slowed down while the other is not, not based solely on “relative motion”.

I’m still waiting for you to answer my earlier questions: Please tell us how you came up with the 36 seconds to advance the A clock at the beginning of one of your earlier thought experiments. In the first half of the experiment you had A and C set at 00:00 at the beginning, and in the second half you had A set at 00:36 while you had C set at 00:00. They weren’t moving relative to each other in either half of the experiment. This advance for the A clock gave a false reading at the end of your experiment and made it appear that observer B saw A’s clock ticking rapidly, whereas, according to SR theory, he actually saw the A clock ticking more slowly.

And also, I asked you to tell us the start times on all your clocks, A, B, and C, as seen by their own observers, at the beginning of your last thought experiment. And I would like to know the speed of B relative to C.

Some of the readers of this thread might not be as familiar with the SR numbers as we are, so when you artificially set your clocks to different times at the beginning of your thought experiments, so that you can get their times to work out the way you want them too later in the thought experiments, I think you need to tell everyone what those different beginning set-times are and why you used those particular numbers. Otherwise you might leave people thinking all the start times are 00:00 or 12:00:00, and I’m sure you wouldn’t want to give anyone a false impression or a misconception about any of your clock start times or your own thought experiments.

And when you have two different relative velocities in your thought experiments, please let us know what both of them are so we can run them through the Lorentz Transformation equation to find out what A, B, and C actually “see” on their own clocks and the other clocks.

It takes all the fun out of it if you don't do it yourself, Sam5, but since you insist...

"Light is always propogated in empty space with a definite velocity c which is independent of the state of motion of the emitting body." - Albert Einstein

We have an emitter/receiver with a mirror. A pulse of light is sent from the emitter, bounces off the mirror, and returns to the receiver. A clock on the emitter/receiver times the duration of the light's round-trip.

Since the velocity of light is c regardless of the motion of the emitting body, the velocity of light relative to the emitter & mirror will be c-v and c+v, where v is the velocity of the emitter assembly relative to the observer.

So, t1 = d/(c-v) and t2 = d/(c+v), where t1 is the time from the emitter to the mirror, t2 is the time from the mirror back to the receiver, and d is the distance between the emitter/receiver and the mirror. t1+t2=t, of course, where t is the total round-trip time.

Now, if v=0, meaning the emitter assembly is motionless relative to the observer, then both c-v and c+v are equal to c. This means t2=d/c and t1=d/c, so t1=t2. Therefore, t=t1+t2 becomes t=t1+t1, then t=t1*2, and finally t1=t*0.5. So, the light pulse spends half the total round-trip time on the way out and half on the way back.

Now, if v=0.6c, then c-v=0.4c and c+v=1.6c. This means t2=d/1.6c and t1=d/0.4c. From that last equation, d=t1*0.4c. Plugging that in for d in the next-to-last equation gives us t2=(t1*0.4c)/1.6c, which simplifies to t2=t1*0.25. Therefore, t=t1+t2 becomes t=t1+t1*0.25, or t=t1*1.25, or t1=t*0.8. So, the light pulse spends four-fifths (80%) the total round-trip time on the way out and only one-fifth (20%) on the way back.

If the round-trip time takes 120 seconds, then an observer stationary to the system says the light took 60 seconds out and 60 seconds back. An observer who sees the system moving at 0.6c relative to himself says the light took 96 seconds out and 24 seconds back. 96-60 = 36 = 60-24. Therefore the "moving" observer concludes the clock is 36 seconds later when the pulse hits the mirror than the "stationary" observer does.

Specifically, if a clock at the emitter reads 11:59:00 when the pulse leaves and 12:01:00 when the pulse returns, the "stationary" observer would conclude the light hit the mirror when that clock read 12:00:00. However, an observer moving relatively at 0.6c says the light would hit the mirror when that clock read 12:00:36.

If a clock located at the mirror says 12:00:00 when the light hits the mirror, the "stationary" observer would conclude the two clocks are synchronized. The "moving" observer, however, would conclude that the emitter clock is 36 seconds ahead of the mirror clock.

So from this we can conclude that two clocks located one light-minute apart (which would be the required distance for the pulse to take two minutes round trip) which are synchronized in the "stationary" frame would be off from each other by 36 seconds in the relatively "moving" frame.

The fact that 36 seconds is exactly what I needed to set the clocks off by to make the final numbers work out is not coincidence or something I "fudged." It is a result of the fact that Special Relativity is internally consistent and so it will all work out in the end.

When adding relativistic velocities, v = ( v1 + v2 ) / ( 1 - ( v1 * v2) / ( c * c ) ), where v is the total velocity, v1 and v2 are the separate original velocities, and c, of course, is the speed of light. Plugging 0.6c in for both v1 and v2 results in approximately 0.88c (unless I did my math wrong), which should be the velocity of B relative to C and vice-versa.

You've got to know that equation (or know where to look it up ) if you're going to deal with SR...
The explanation I gave above for 36 seconds is the basis for the simultaneity issue in SR. At the point when A & B pass each other in this experiment (which we can consider to be the "start of the experiment"), C is quite some distance away, and A, B, and C are all three in different inertial frames. Therefore, while it's easy and universal to say that A & B were both at 12:00:00 when they passed each other, the time displayed on C at that same moment (in other words, what C-clock "tick" event is simultaneous with the A-B-passing event) will depend on which of the three inertial frames you're looking from.

This will also be true for the question of what time is displayed on A when B & C pass each other, and also for the question of what time is displayed on B when A & C pass each other.

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Sam5I cannot speak for all readers of this thread, but of all the posters, you are the one who is least familiar with SR. You are arguing about thought experiments, which are just an attempt to put into words the mathematics of SR. The word descriptions are not the math!

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Kaptain K, as far as I know, the math of the theory is fine. It’s some of the concepts that are in error. For example, Einstein has his light-speed c-regulator as being fixed with the frame he calls “stationary”. But when we switch to the other frame and call it “stationary”, then the c-regulator becomes fixed with that frame.

The theory seems to work out as long as we think from the point of view of only one “stationary” frame per thought experiment. But when we realize that both frames are equal, inertial, and moving relatively, then the observer in the so-called “moving” frame will see his frame as the “stationary” one and the other frame as the one that is “moving”. When we switch frame and observer points of view and alternate each frame as being “stationary”, which we must do, since all motion in the theory is relative, then that’s when a major error of the theory is revealed and the clock paradox results.

Sam5, you're embarrassing yourself.

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<3

Please, get this straight:

The math is the theory!
No more, no less. If "the math of the theory is fine", the theory is also (by definition) "fine". Again, you are arguing about the word decription of the theory, not the theory itself. If you understood the theory (the math), you would not be picking nits over the words.

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Sam5:

http://math.ucr.edu/home/baez/RelWWW/wrong.html

Sorry Kaptain K, but the math is only part of the Special Relativity theory. If you want to say that a separate individual light-propagating medium is stationary with every observer in every inertial frame, and then when there is relative motion between inertial frames, the universe splits up into multiple parallel universes so that all the observers in all the different inertial frame see different sets of things happen, then you still don’t have a valid theory that is based on reality, but you’ve got a fine plot for a science fiction movie.

If you take the theory down to its most fundamental level, in Einstein’s own clock paradox thought experiment he winds up with observer A and observer B disagreeing about what they “see” at the end of the experiment. Observer A “sees” observer B’s clock running slow and lagging behind, while observer B “sees” observer A’s clock running slow and lagging behind. So we wind up with two sets of clock face readings when A and B unite, a total of 4 readings in all for only 2 clocks.

We have what A sees on his own clock, and what A sees on B’s clock, and they are different readings from what B sees on his own clock and what B sees on A’s clock. This does not conform to reality. In General Theory, atomic clocks slow down when they experience acceleration forces. That is based on reality and conforms to observation. So there is no paradox in the GR theory. If you think there is no paradox in the SR theory, then you just don’t understand the SR theory.

And by the way, pendulum clocks speed up when they experience acceleration while atomic clocks slow down, so it is not “all of time that slows down” when clocks experience acceleration, since different clocks react to acceleration in different ways.

There is no light-propagating medium. I told you before that one second and 186,000 miles are the same thing, and that's why light always travels 186000 miles in one second regardless of who's looking at it.

{Edited to Add}: K's right, the math is the theory. The "Twin Paradox" and such thought experiments are merely demonstrations or examples.
No, we don't. Observer's A and B both agree on what Clocks A and B read when they meet. They disagree on what the clocks read while they are separated (SR predicts these disagreements), but when they meet, all is well.
No, in the General Theory of Relativity time flows differently in gravity wells and during acceleration. It is not a physical response of the clocks, it is a fundamental property of space and time itself.

There are some physical responses - a pendulum clock will change its rate if you grab hold of the pendulum, for pete's sake - but those aren't part of Relativity (Special or General). General Relativity predicts changes in time (as does Special Relativity), and the clocks merely reflect that.

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SeanF

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SeanF, in your previous thought experiment, on page 3 of this thread, you had clocks A and C and they weren’t moving relatively. They were in the same frame. And you said the start times for both clocks was 00:00, then later you said their start times were 00:36 for the A clock and 00:00 for the C clock.

So you wound up with two different start times for the A clock.

What you did was remove the “paradox” from the end of your thought experiment and place it at the beginning of your thought experiment. So you didn’t resolve the paradox, you just moved it from one end of the thought experiment to the other.

Nope, it's not a paradox. A and C remain physically separated for that entire experiment, and so it does not make one whit of difference what A reads when C reads "12:00:00".

When B and C pass each other, they tell each other what they think A reads at that moment. B says "A reads 12:00:36 right now, and when it gets to me (because it's moving) it'll read 12:01:40." while C says "No, A reads 12:00:00 right now and when you get to it (because you're moving) it'll read 12:01:40."

A isn't there, it's off in the distance. They're both calculating the "current value" of A based on light that left A some time ago, so it does not cause any kind of "paradox" that they disagree.

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SeanF

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Your statement is misleading. There is a light-speed-regulator that travels with every large astronomical body. And because of that, light slows down when it pass near the sun or other massive bodies. The stronger the gravitational field, the slower light travels as it passes through it.

I’m not sure if we should call this a “propagation medium” or what. But this is why we on earth see light traveling at c, here locally at the surface of the earth, while a distant high-c galaxy sees light traveling at c relative to that galaxy.

When the light is emitted by that distant galaxy it is traveling at c relative to the galaxy but less than c relative to the earth. So you can’t say that, “light always travels 186000 miles in one second regardless of who's looking at it.” You can say that light travels at 186000 miles in one second while traveling near the surface of the earth, but you can’t say that light travels inside a distant high-c galaxy at 186,000 miles in one second both through that galaxy and relative to the earth at the same time.

This is one of the most fundamental and basic concepts of nature, and I don’t understand why you can’t understand it.

Now, we can take your wording and make it “work out” by saying this: “people on earth see light traveling locally on earth at c”, and “people in high-c galaxies see light traveling locally in their galaxy at c”, but we can’t say that their light is traveling at c relative to the earth while it is traveling through their galaxy.

It does make a difference when you’ve already said the A clock starts out at 00:00. So you can’t change that in the same thought experiment. If you do change it, then you’ve got the paradox of A “seeing” his own clock read both 00:00 and 00:36 at the same time, and that is impossible.

Do you understand that in reality a single observer holding a single clock can not see two different readings on the single clock at the same time?

But B doesn’t “see” clock A reading :36 seconds at the beginning. You already said that in the beginning all clocks read 00:00 and you said that you were using instantaneous light signals, so all observers know that all clocks are reading 00:00 in the beginning.

In Einstein’s thought experiment, he has all clocks as being “stationary” in the beginning so that all observers can synchronize all of them and make them synchronous. We can do the very same thing while B is stationary at C. So all observers know that all clocks are reading 00:00 at the beginning. Then the relative motion begins. And when A and B unite, observers A and B will disagree about what their own and each other's clock readings are.

No, A sees A at 12:00:00 when B passes C. C, in the same inertial frame as A, sees A at 12:00:00 when B passes C. B, in a different inertial frame as A and C, sees A at 12:00:36 when B passes C. No single Observer disagrees with himself at any point.

I never said anything regarding "instantaneous light signals," and I never said that all observers saw the same thing on all three clocks at the same time. You better go back and read that again.

Instantaneous communication is prohibited under the rules of SR (not arbitrarily - the logic flowing from the constant speed of light concludes it), and it's a good thing, too, because there would be paradoxes without it. That's just another part of how SR is internally consistent. The simultaneity difference in relative motion would result in paradoxes if the observers in relative motion could communicate instantaneously. But they can't.

In my thought experiment, B is never stationary at C, so we can't "do the very same thing." In Einstein's thought experiment, the observer who changes his relative motion sees the distant clock differently after the change than before. When comparing Einstein's thought experiment to mine, My C is Einstein's B before the relative motion starts, and my B is Einstein's B after the relative motion starts.

Crap. Does Einstein's experiment say "We move B to A" or "We move A to B"? I'm assuming it's the former. If it's the latter, than my A is Einstein's B; my C is Einstein's A before the relative motion starts; and my B is Einstein's A after the relative motion starts. Either way, the result is the same.

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SeanF

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No, its a physical response of the clocks, depending on the type of clocks used. An atomic clock slows down in a “gravity well”, but a pendulum clock speeds up in the same gravity well. If we set an atomic clock and a pendulum clock to tick out an accurate astronomical “year”, then if we put them in a gravity well, the atomic clock will run slow and the pendulum clock will run fast, while the astronomical year hasn’t changed at all. We can be in a gravity well here on earth and then move out into space in a spacecraft, and the astronomical year won’t change, whereas an atomic clock in space will speed up while the pendulum clock in space stops running altogether.

These kinds of clock-related “time” dilations were discovered centuries ago. The knowledge that gravity wells affected the tick rate of pendulum clocks was so well known in the 19th Century, the gravitational potential at different elevations on the earth’s surface was measured with pendulum clocks.

In 1877 Maxwell wrote about the force of gravity, “Its intensity, as measured by pendulum experiments, decreases as we ascend”. He also wrote this about time, “We thus see the theoretical possibility of comparing intervals of time however distant, though it is hardly necessary to remark that the method cannot be put in practice in the neighborhood of the earth, or any other large mass of gravitating matter.” So he knew very well that gravity wells can alter the tick rates of clocks. By experiment we have learned that gravity wells slow down atomic clocks and they speed up pendulum clocks. These are simply the laws of nature. But relative motion can’t possibly alter the tick rate of any clock, since the clock experiences no “force” as a result of the “relative motion”.

LOL. “Boop”, “bop”, “beep”, “bip”. Was it a “bip” before a “bop” or a “boop” before a “beep”? I know how you feel. So we won't go bonkers simultaneously, in our own frames, maybe we need to take a break.