Relativity of Count – Spin Counter

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.

I’m afraid you would have to prove that one. The math won’t workout (well… bend enough rules and you can make anything seem to workout).

Running the same speed, but no longer in sync. Remember, the twin A actually does age more and twin B less. Time dilation isn’t merely a perception issue.

So I take it you want to defend the stance that both station and train see the other clock moving slower than its own?

I just showed that: a clock traveling at constant velocity will be observed to run slowly by a stationary observer, because that observer will see a point on the edge of the clock traveling at the speed of line trace a longer path than will a similar particle tracing the edge of a clock in his own frame.

Right, but twin A is the only twin that accelerates in that problem. If twin A were to accelerate to constant velocity, and then twin B accelerated to the same constant velocity in the same direction (such that they are in the same frame again), the clocks would be synchronized again.

Not particularly. There are dozens of more reputable sources around the internet that do a much better and more thorough job of saying exactly what PhyBang, Farsight, and I have been saying.

This is what you have claimed;

And you haven’t supported that notion at all. Nor can you if you do your math right, because it isn’t true.

No, this is what you said;

No, merely running the same speed. The aging doesn’t reverse. The clocks would be offset.

I wouldn’t be including Farsight in your crew. And you really need to stick just to “Carleas”.

So far, you still haven’t raised any point that I can see as relevant to the OP.

I supported that notion in my last post:

Because both observers see the other observer’s clock as moving, both see the other’s clock as running slowly.

To borrow a refrain: show the math. I honestly don’t know how the acceleration equations would work out, but I know that 1) in the Twin paradox, only one twin accelerates, 2) it is that acceleration that makes their clocks read differently, 3) I’m not taking your word for it that the math works out, because as far as I can tell neither of us have an equation that calculates the effect of constant acceleration on time.

Wrong, by your own sources.

“More aging” == “clocks running FASTER”.

a = v/t
a’ = v’/t’
(falsely assuming v’ = v);
a’ = v/t’ =>> v = a’t’
a = v/t =>> v = a
t

since t’ = L(t - vx/c^2);
a’ = v /((L(t - vx/c^2))
or
v’ = a’ * L(t - vx/c^2)
v = a * L(t’ - vx/c^2)

It is just arithmetic/algebra.

Except that acceleration doesn’t involve a constant v, so it’s not just plug and play. As I said in one of these threads, acceleration has to require an integral over time to account for the changing velocity.

You’re conflating the two situations we’re talking about. In the spin counter example, both observers see the other’s clock as running slowly. In the Twin Paradox, one twin ages more than the other. Both still see the other’s clock as running slowly during the periods of constant velocity, but they ultimately undergo asymmetric experiences. The Twin Paradox also requires Relativity of Simultaneity to solve :-"