Relativity of Count – Spin Counter

That’s very relevant. If that’s your personal justification, it will continue to cause you reject length contraction until it is dealt with. Again, why don’t we discuss your real reasons for your positions, rather than some after-the-fact justification at the end of which you’ll still fall back and say “well, my real reason still stands”?

Because you are merely imagining that argument, possibly to avoid the fact that you have no defense for your own position other than as PhysBang’s, “well those smart guys told me that SR says…

So if you disagree with what I have proposed. SHOW me your logic/math for your disagreement. I need not show you my reason for why I believed it in the first place. All I have to do it show you the errors in yours.

“What will the train see of the station’s light spinner? And why?”

Carleas directly answered this question and gave you the explanation why that did not appeal to SR. He told you about the difference in arc length. You have not addressed this other than to be incredulous. Try working it out.

More of your BS. I and everyone acknowledged the arc. I even drew a pictorial of it. I am asking, repeatedly, “SO WHAT???”

What is the conclusion that is to be drawn from such a thought?

No one is arguing against the idea that there is an arc. But let’s get a move on. Why do we care? What is the next step in reasoning? What is the conclusion from such a thought? Then where does that lead us (relevant to the OP)?

If the arc length is different from one frame to another, and the speed is the same, the time is different. So the spinners read different times. So… I don’t understand what you’re looking for. SR accepts and accounts for what you’re calling “relativity of count”, but it calls it time dilation or relativity of simultaneity.

Those are VERY different concepts. You can have time dilation without any trace of “relativity of simultaneity”.

What does the train see of the station’s spinner? - That is the next step of concern. I don’t accept PhysBang’s answers as yours. He’ll just repeat rhetoric spins (“the repeated meme”).

You haven’t fully expounded what you mean by “relativity of count,” so I’m not sure which best captures it. But your objection seems to be that the clock read differently in different frames, which is involved in both time dilation and relativity of simultaneity.

In some sense yes, but in special relativity they are inextricably related.

I agree with PhysBang:

Also,

I didn’t imagine it. You said:

If, for instance, the set of all whole numbers and the set of all even whole numbers have the same number of elements, then everything that follows from that (including your personal rejection of length contraction) is left unsupported. If this argument prevents you relenting on your assault on SR, it’s something we’ll have to deal with sooner or later. And if it’s why you know, it’s better that we deal with it sooner.

Then you run into the Twin Clocks Paradox.

Either twin A ages more than twin B or visa versa. It can’t be both. Either time is dilated on the train or it is dilated at the station. If it were both, there would be no actual dilation at all. They would always agree in the end.

So try again.

What is “the end” in which they would agree? Let’s say “the end” is a flasher going off at a given time according to the train clock. They certainly wouldn’t agree on what time it went off, since everyone agrees their clocks aren’t synchronized.

Also, the “Twin Paradox” is easily resolved using Special Relativity.

No that is not the end. The end is when the train stops. Or when the twins get back together.

If the train were to stop itself and its clock as soon as they read 4:00, the station would see them read 4:00 and see it’s own clocks read perhaps 4:02. It is that simple. No magic.

Special Relativity misuse is what proposed the paradox in the first place. It is easily resolved using common sense. Resolve to the Twins Paradox

But now back to the question, what will the train see of the station’s clock timing?

You’re getting a little ahead of yourself. We’ve answered that question, and your argument against it can be summed up as, [-(

The Twins Paradox is not a problem for Special Relativity, so saying “then you run into the Twin Clocks Paradox” is not a reason to reject our answer.

What do you think the solution is???

Which twin aged more?

I agree with the solution in the link I posted earlier. And this one and this one (both with nice animations).

I didn’t ask what you agree to. I asked which twin aged more.

The twin at home, as all three explanations indicate.

So if the twin at home ages more, that means that the twin B’s clock was moving slower and to the traveler, the Earth’s clock was moving faster.

So I ask again, what does the train see of the station’s clock?

Take a look at some of Sardin’s papers: arxiv.org/find/grp_physics/1/au: … /0/all/0/1 and do read The Other Meaning of Special Relativity.

In the twin paradox, symmetry is violated. That is not the case here. Both the station and the train see the other’s clock as moving more slowly.

It is violated in exactly the same way. The object that accelerated away from the Frame of Origin (in this case, the still ground) is the one that will be time dilated. So the train’s clock turns slower than the station’s clock.

True, but we haven’t been calculating for acceleration. We’ve just been dealing with the period of constant relative velocity after acceleration. And even with initial acceleration, the observers would see each others clocks as running slowly during the period of constant velocity.

We also don’t have a well defined “end” in this situation. If the train decelerates, then we will have a similar situation to the Twin Paradox. But if the station accelerates to reach the speed of the train (ignoring that it’s practically unlikely), the end result should be that the clocks are still synchronized.