androstan

I'm pretty sure Mach's principle doesn't require IAAD. You just feel pulled by the rest of the U in a direction indicative of the universe's arrangement some time ago.

If you were in a U with no other matter, and you rotated, you would feel stretching/pulling in your own body as the atoms/molecules on your "left" side stretched away from the ones immediately adjacent ("to the right") and then those stretched away from the next set immediately to the right... etc. In reality, every time you spin around or otherwise accelerate there is a "shock wave" that runs through your body starting somewhere. When you jump off a tall building and land on your feet on the ground, there is a "shock wave" that emanates up from your feet. It meets another "shock wave" that emanates downward, mostly from your hips/thighs.

So "rotating" in a matterless U doesn't seem like rotating. It just feels like an internal strain. Initially it feels like a strain that propagates from your "left" to your "right", but once it's covered your whole body you just feel a general tension across your body, from left to right. When you turn off the propellant you don't "stop rotating" but rather the tension in your body will cease as the bonds holding the atoms and molecules revert back to their "normal" lengths and orientations.

If there is some other bit of matter in the U then your body feels tension in that direction. The tension is asymmetric as the part of your body closest to it is pulled harder than the part of your body further from it. If you turn on the propellant you feel that same tension across your body as the bonds and other interactions between atoms/molecules are torqued a little away from their "normal" state. You see the other body in the U spinning around you. However, you conclude that YOU are spinning, not that the other body is rotating around you, because of the tension you feel across your body when you started the propellant. Similarly, if you turn on the propellant to send you away from the other body, you feel tension from front to back as the atoms/molecules are in one instance pulled toward the other body and in the other instance pushed away from it by the propellant. Again, you conclude that YOU are accelerating. The other body in the U only feels a slight change in tension from front to back as the difference in how hard you are tugging on the front and the back decreases. However, after a time, the fact that you're accelerating is "felt" by the other body and it starts accelerating toward you. If/when you get to the point where it is accelerating toward you as fast as you are accelerating away from it, you both feel as if there is no other matter in the U, i.e. you only feel a tension across your body similar to what you would feel in an empty U in which you suddenly turned on the propellant.

Obviously not that particular experiment, no, but it's a straightforward consequence of special relativity, which has a great deal of experimental support.

 

At first, it seemed like you weren't disagreeing with the physical reality of the situation, just saying that another description was equally valid. Was I mistaken, and are you in fact claiming that SR is wrong?

You can't cite the results of an experiment that has never been done, that's all I'm saying.

You still have the problem of a stationary clock and moving clock not counting photons at the same rate. Whose device is correct?

I don't think the term "correct" really applies here. Nobody is challenging the empirical result that one Cs atom oscillated fewer times than the other. Objectively that's what happened. If one piece of steel rusted more than another and you heard someone asking which hunk of metal was "correct" you'd wonder what they were talking about. Similarly, I don't really know what it means for one Cs atom to be "correct". I think what's more important is to explain why one hunk of metal rusted more, and why one Cs atom oscillated less.

I ventured an explanation for why this phenomenon occurs earlier. Additionally, I am proposing an alternative method of measuring time wherein "time dilation" doesn't occur.

The latter, of course. You disagree that relativity is experimentally well-supported?

We're talking about taking a spring on a relativistic journey and counting oscillations. I have not heard of anyone performing this experiment.

Photon round trips are the same as clocks. And one-way trips or trip segments require communication, and synchronization, and all of the relativistic effects come into play.

When the photon hits the wall and comes back it is reflected downward on a photosensitive plate. The distance between the plate and the reflector is tiny wrt the distance to the wall. Or the person could just count each photon reception. Moving a certain distance at a certain speed only a specific number of photons could go there and back. This is a different kind of clock than i.e. a Cs clock (or an Al/Mg clock).

Acceleration is distance per time per time. You change time, you change acceleration.

 

e.g.: Say that under relativistic time compression, the observer on the train experiences 1/10th the amount of time as the observer on the platform. A spring on the train, stretched and released, will oscillate 1/10th as many times per trip as a spring stretched with the same force and released on the platform.

 

In androstanian physics, however, we have declared that both observers actually experience the same amount of time, and it just all "clocks" that are distorted. Hence the spring on the train, stretched to the same amount, undergoes 1/10th the acceleration. So you either have to say that f=ma is not true, or say that fundamental forces change in magnitude.

Away from earth the spring displacement is measured when the train is accelerating to get the acceleration-dependent component of force.

Back on earth the spring-displacement is measured outside the train to get the vertical component of force.

Now, on the moving train, the string and spring hang at an angle. The measured resultant spring-displacement squared equals the sum of the squares of the two measured components, assuming Euclidean axioms. If the train accelerates more or less, the angle changes and the vertical component decreases or increases respectively to compensate. The resultant force (spring displacement) is always the same.

But you're not asking about such a "static" situation (before I thought you were), but rather about a dynamic situation. A situation wherein the spring oscillates. Your claim is that one person will leave the station, go to the wall (or around the world), and upon arrival he will tell his stationary friend that his spring oscillated fewer times. The stationary friend will agree and say his spring oscillated more times than the moving person. Is this an experiment you did or are you referencing someone else's experimental results?

The result of that experiment is the same. However, f=ma no longer holds, so you'll have to redefine what you're measuring (since that is the definition of force).

Justification?

Can't have a round trip under those conditions.

Well, the person can't. But we can just as well switch things around and talk about photon round trips. We could also talk about portions of a one-way trip for the person.

I think you are ignoring the effect of the acceleration of the train.

We were talking about frames in uniform relative motion.

But either way, the results of the experiments are still the same. The spring displacement required to break the string is the same on an accelerating train as on an inertial train.

So you just have to say that f=ma is not true, then. Since the same net force, applied to the same mass, will accelerate it different amounts depending on whether or not its on train.

The results of the two experiments are identical, the spring displacement required to break the string is the same in any inertial frame.

It's not that simple, because it isn't just C that changes. Everything has to change along with it. One example: pulling on a string with enough force to break it, measured by the acceleration of a mass. Except that if your unit of time is "trip," then you get a different values for acceleration, hence force, hence tensile strength, hence intermolecular forces, hence all electromagnetism. Depending whether you're pulling on the string on the train or not.

Someone on the train has a box full of blocks of a wide variety of sizes. S/he attaches each to a spring that is attached to the roof of the train and measures the spring's displacement for each block, recording each displacement. Then s/he attaches a string to the roof and ties a block to it to see if it'll break. Repeat until it does. Take an identical string and use a block with a slightly smaller spring displacement to see if the string will break. Repeat until you have identified the block with the smallest spring displacement that breaks the string.

I think that if this person gets off the train and does the experiment on the ground, they will get the same result. Or if s/he gets on a train going a different speed, s/he will get the same result.

Someone on the train has a box full of blocks of a wide variety of sizes. S/he attaches each to a spring that is attached to the roof of the train and measures the spring's displacement for each block, recording each displacement. Then s/he attaches a string to the roof and ties a block to it to see if it'll break. Repeat until it does. Take an identical string and use a block with a slightly smaller spring displacement to see if the string will break. Repeat until you have identified the block with the smallest spring displacement that breaks the string.

 

I think that if this person gets off the train and does the experiment on the ground, they will get the same result. Or if s/he gets on a train going a different speed, s/he will get the same result.

The experiment would be easier if you just tie the string to the roof and tie the other end to a spring. Pull on the spring until the string breaks, noting the spring's displacement from equilibrium. It will be the same whether you are on the moving train or not.

I was wondering, with time dilation. If it occurs when an object is moving, no matter how fast, wouldn't that mean that as we ourselves are moving including other objects, everything is constantly being/getting jumbled around in time, That nothing is in phase..

 

And for my next question, which may be a little philosophical (for this section). If such things as ourselves, 'conscious' beings, perceive "time" and its effects within the context of existence. Could that mean relativity is only accountable for things such as ourselves, since, as 'time' is conveyed through us, in terms of the river of time, its effects can only be interpreted via the system for which it is perceived..

STR, like the other poster said, just says what you observe. What you observe are clock ticks. One person counts fewer clock ticks than another. This doesn't necessarily mean that one person undergoes more/fewer of the deleterious effects of aging (senescence). Some people take the "strong stance" i.e. if we had 2 identical copies of you and we sent one on a relativistic journey and left you at home, the journeying twin could come back long after you'd died. The journeying twin's motion actually slowed down all biological processes, including physical decay. Other people take the "weak stance" i.e. that the journeying twin only counts fewer clock ticks but returns to find the twin at home has senesced by the same amount.

But no, things don't become "out of phase" or "jumbled". Some things just count fewer clock ticks, or their internal machinery goes slower.

To say the "trip" is the same quantity of time for both observers, they would have to be measuring different values of C (as well as differences in pretty much every other imaginable clock, from radioactive half lives to rate of thought), and consequently of all other fundamental constants. You would end up with totally different physics for each observer.

Well, you have to pick a reference. For trips they will get different values of c. Let A and B approach their work place and bounce a radar signal off it. Let A approach at 1/2 and B approach at 1/4.

The time (in work trips) that A will meet back up with his/her pulse is given by:

T/2 = 1 - T

T = 2/3 of a trip

For B:

T/4 = 1 - T

T = 4/5 of a trip

A calculates c to be 1/2 and B calculates it to be 2/3.

So you can set your trip-distance to be 1 and use it as the standard, or you could set the speed of light to 1 and use that as your standard.

(emphasis added)

 

And that's the deal-breaker.

Well, one method requires you to quantitatively reinterpret your measurements to do useful things with them but is generally applicable to observations so far. The other method doesn't require quantitative reinterpretation , but isn't as general. So it's arguable. One is more practical, the other contains fewer assumptions.

No, it isn't consistent. That's the point. "It always takes me exactly the same amount of time to get to work: one trip."

Well when you compare different results you have to use one standard, one clock. If you're going to talk about 10 trips to work, you have to pick one of those trips as your standard to compare the others to. Unfortunately you can't because they're in the past, so you'll have to pick another standard. You could pick earth rotations, Cs clock oscillations, etc. In practice we use whatever is most convenient and easy, whatever's practical.

Let's say you go to work and back by the same exact path over and over, counting each time. Someone else goes there and back by a different path, also counting each time, and you both leave your starting points simultaneously starting from rest. Eventually you'll both arrive at work simultaneously, maybe on the 500th trip or the 10th trip or the 10⁸⁰ trip. If you took N trips while your friend took M trips, then by the "trip standard" all you can objectively say is that you took N/M times more/less trips than your friend. Whether this is because you moved faster or the path you took was shorter isn't known. You could both go back and lay down rulers to measure the path lengths, then you would know how much faster/slower you were moving than him.

You could both carry clocks with you and compare those results too. Should you believe the trip standard or the clock standard? Well, the trip standard will work fine if all you want to predict is when the next time you'll both meet at work again, assuming neither of you changes your habits. The trip standard works fine whether you and him are moving at the same speed or not. However, a change in habit (walk faster, take a different path) would constitute a new reference standard, and everything must be compared to a common reference standard.

The clock standard, on the other hand, does not work if you are not both moving at the same speed. You both notice that your clocks do not agree. You two will not be able to decide when you'll meet at work again since you don't know whose clock to believe. However, you reason that the clock ticks every time light is emitted. If the vector components of the light are perpendicular to the clock's velocity vector then the resultant will be the vector difference, and the time it takes for light to propagate into and back out of the clock will be increased. By using simple vector algebra you arrive at a transform that corrects your clock for differences in speed between you and your friend. Now you can both not only predict when you'll meet the other (according to either clock), you can also correct for changes in velocity.

In the absence of a theory about light propagation the "trip clock" is superior because it is consistent. Both persons count their trips and agree on the ratio of trips. They agree on when they'll meet again. It has the flaw of being ungeneral, i.e. a new standard has to be used every time there is a change in the physical situation. With a theory about light propagation clocks become superior because now we can reference everything to a single standard, the net propagation rate of signals ©.

So in the twin paradox, if I said the traveling twin aged 1 year while the stay at home twin aged 50 years, you would disagree and say that they both aged exactly one "traveling twin's journey?" Do you see the problem?

Again, if they come together and their atomic clocks disagree, it would seem better to use the "number of journeys" clock, since it is consistent.

This is comparing a clock against itself, which is not a good way to evaluate a clock. All real clock measurements use two or (preferably) more different clocks.

I mentioned comparing "train passings" with each person's atomic clock.

The point at issue between the two persons seems to be how to explain why cesium atoms emit light less frequently when in motion. The Cs atom absorbs radiation from its surroundings, is excited in some way, there is some internal processing, and the radiation is re emitted as the atom "settles back down".

If the atom moves as a longitudinal wave and the light moves as a transverse wave, they have perpendicular components. When the light encounters the atom they blend into a wave that has both longitudinal and transverse components, the resultant velocity of which still has to be c. So it would take the light signal more time to propagate through the atom and back out (re emitted) if the atom were moving due to the net propagation rate being the vector difference between the longitudinal (v) and transverse © components. So, the ratio between light's propagation speed "in space" (only transverse components) and its propagation speed through the longitudinal wave of the moving atom would be c/(c²-v²)^1/2, which would explain the particular form of the Lorentz factor.

What if the two people decide that "number of train station passings" is their chosen unit of time? Reasoning that the train is undergoing uniform periodic motion, they both reasonably conclude that each cycle of the train is a good clock. After 400 cycles they compare results and find they agree exactly on the elapsed time, even though they were in relative motion wrt each other.

Put another way, perhaps they both use atomic clocks too. After coming together they find that their atomic clocks do not agree, but that their "train passing" clocks do. They agree that the train passing clock is a superior timekeeper.

This would contradict the particular physical interpretation of STR, that time itself dilates. If the theory only applies to particular clocks and not to others, then it isn't really a theory about time but rather it is a theory about those particular clocks. If the electromagnetic radiation emitted by an atom is redshifted when in motion then one person may call it "time dilation". Another person may simply call it a velocity-dependent redshift.

With respect to length contraction, that has to do with how length is measured for nonlocal inertial frames. If I'm holding a ruler, then it is my local frame and I can measure the length of any other colocal object by simply comparing them directly. For an object moving away from me I cannot simply hold up my ruler to it. I have to measure the length of the departing object remotely. The most sensible way to do this is with radar. Let's say I take two identical rulers and I send one (ruler B) hurtling off away from me while I keep mine (ruler A) at home. I wish to know the length of ruler B compared to ruler A. I need to measure the length of ruler A in a way that is comparable to ruler B. I send a light signal along ruler A and I count how many clock ticks it takes for the signal to traverse the ruler. I then send a signal to ruler B and count how many of B's clock ticks it takes for the signal to traverse the ruler. I count that it takes fewer of B's clock ticks to traverse ruler B than it takes A's clock ticks to traverse ruler A. Since the measured speed of light is constant, fewer clock ticks (smaller time) means smaller length/distance.

Looked at another way, if you (A) are standing stationary on the embankment and someone else (B) is running at +v, person B will necessarily count fewer clock ticks than A as s/he sends a light signal to the end of the train and back. This is perhaps more obvious than the previous example.