User talk:84.15.186.187

March 2016

Hello, I'm Dcirovic. I wanted to let you know that one or more of your recent contributions to Louis van Gaal has been undone because it did not appear constructive. If you would like to experiment, please use the sandbox. If you think I made a mistake, or if you have any questions, you can leave me a message on my talk page. Dcirovic (talk) 11:46, 12 March 2016 (UTC)

If this is a shared IP address, and you did not make the edits, consider creating an account for yourself so you can avoid further irrelevant notices.

January 2020

Hello, I'm DVdm. Your recent edit to Time dilation appears to have added incorrect information, so I removed it for now. If you believe the information was correct, please cite a reliable source or discuss your change on the article's talk page. If you think I made a mistake, or if you have any questions, you can leave me a message on my talk page. Thank you. DVdm (talk) 20:11, 22 January 2020 (UTC)

DVdm,

the article claims the following:

$\Delta t'={\frac {\Delta t}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}},$ which expresses the fact that the moving observer's period of the clock $\Delta t'$ is longer than the period $\Delta t$ in the frame of the clock itself.

Please mote, that $\Delta t'$ is $\gamma$ times greater. Hence, moving observer's clock measures greater time interval $\Delta t'$ contrary to predictions, that moving clock slows down. Because the author confuses, who is at rest and who is moving. Single clock is moving from point A to point B and would measure $\gamma$ times shorter time interval, than two spatially separated and synchronized clock in points A and B. These synchronized clocks represent rest frame of observer. Accordingly, two spatially separated and synchronized clock would measure $\gamma$ times greater interval. Hence, single clock is moving relatively to this rest frame, which consists of two (at least) synchronized clocks . It makes no sense to say, that rest frame of observer is moving relatively to a single clock; because single clock occupies infinitesimal area of space. --84.15.190.237 (talk) 12:19, 23 January 2020 (UTC)

Left: Observer at rest measures time 2L/c between co-local events of light signal generation at A and arrival at A.
Right: Events according to an observer moving to the left of the setup: bottom mirror A when signal is generated at time t'=0, top mirror B when signal gets reflected at time t'=D/c, bottom mirror A when signal returns at time t'=2D/c

This is about two events on a clock. Δt is the time between the events as measured in the clock's frame, in the diagram on the left. Δt' is the time as measured between the same events as measured in the moving frame, in the diagram on the right.

The expression Δt' > Δt means that for the ones for whom the clock is moving, it takes more time (Δt', read on their local watches) between the events, than it takes time (Δt, read on their local watches) for the ones who fly with the aparatus. So, for example the ones who stay with the clock say that it takes Δt = 1 nanosecod between the events, whereas the ones for whom the clock is moving say, NONONO, we measure Δt' = 3 nanoseconds between your events, so your clock, which is moving, runs slow as compared to ours.

Hope this helps - DVdm (talk) 12:45, 23 January 2020 (UTC)

No DVdm, it doesn’t. Who is finally moving?

„This is about two events on a clock. Dt is the time between the events as measured in the clock's frame. Dt' is the time as measured between the same events as measured in the moving frame.“

First, I don’t see any unprimed clock’s frame. I see only primed frame. The article considers dilation of unprimed clock but claims that primed frame is moving!!!. But I see that unprimed single clock is moving (or changing its spatial coordinate) in the primed frame, but not vice versa (as the article claims). Just eliminate mentally everything around - your computer screen, surrounding of your office etc.; so how can primed frame be moving relative to nothing?

The article changes opinion who is moving in every sentence. You started with a single clock which measured interval $\Delta t$ and told that this clock was at rest. Then you assumed that a certain frame was moving relative to this clock and measured interval $\Delta t'$. Then you apparently thought like this: “what's the difference? relativity will forgive everything” and assumed that since that frame was moving relative to the clock, we can also say that the clock was moving relative to the frame, hence $\Delta t$ is shorter.

So, it is 100% clear, that:

1) You have denoted interval $\Delta t'$ as an interval that was measured by a moving observer.

2) This interval was not shorter, but longer than $\Delta t$

By the way - in relativity observers rarely move. Here is definition who is an observer in relativity

It is well-known, that moving clock measures shorter interval of time, but not longer. You can tell me whatever you want, but, according to the article, moving observer’s clock measured longer time interval. At least It is very unfortunate wording.

I don't understand how readers can sort this nonsense out.

Time dilation is purely about the way how one compares time intervals; one must denote correctly denote who is “moving” and who is “at rest”. An observer who is “at rest” cannot measure time interval by means of a single clock, because events (departure and arrival of moving clocks) are spatially separated. Let's say that a train is moving from London to Glasgow. Suppose there is a passenger with a clock in this train. Since moving inertial clock passes by and never returns, an observer on Earth (who is “at rest”) needs at least two Einstein – synchronized clocks so as to measure interval of time; one in London and another in Glasgow.

An observer in the train (moving observer, he changes his spatial position in the Earth frame) can do fine using just one clock; he compares readings of his clock with that in London (when the train passes by London) and with another one in Glasgow (when the train passes by Glasgow). Measured by means of a single clock interval of time is shorter, than measured by means of two Einstein – synchronized clocks.

Single clock cannot be "at rest” because it has no reference frame attached. As soon as you attach reference frame (lattice of synchronized clocks) to an observer you turn this observer into “stationary” one; all other moving bodies change spatial positions in his frame and their clocks measure shorter interval of time.

In the article the single clock (unprimed) changes spatial position in the primed frame; hence in this exact case unprimed clock is moving and primed is “at rest”, that’s why primed interval is longer. It is not good to change opinion in every sentence; it is necessary to denote unprimed clock as “moving’ and primed frame as “at rest”.

Let’s say there are two relatively moving frames: S and S’. Two spatially separated and Einstein – synchronized clocks of frame S measure shorter interval than single clock of frame S' and vice versa.

Please look at this animation in the same article; it makes everything clear. There is a sequence of synchronized clocks and single moving clock which measures shorter time intervals. Or maybe you think that on this picture a sequence of clocks should be depicted as "moving" and single clock as "at rest"?

Transversal time dilation. The blue dots represent a pulse of light. Each pair of dots with light "bouncing" between them is a clock. For each group of clocks, the other group appears to be ticking more slowly, because the moving clock's light pulse has to travel a larger distance than the stationary clock's light pulse. That is so, even though the clocks are identical and their relative motion is perfectly reciprocal.

84.15.190.237 (talk) 15:29, 23 January 2020 (UTC)

Your question "Who is finally moving?" is meaningless. They are moving with respect to one another. Two events taking place on one clock (in this case on the t-clock) are considered, and the times between these events for two different observers are calculated, based on the essential assumption that light speed is the same for both. As seen on the t'-clock, the t'-time between two ticks on the moving t-clock is longer than the t-time between these events on the t-clock itself. It takes 3 of their nano-t'-seconds to complete a cycle of something that takes just 1 nano-t-second on the moving clock. So the t'-observers say that the moving t-clock is "running slower" than their own t'-clock.

It's a pretty standard situation, and pretty basic, supported by tons of sources in the literature, four of which to be found in the relevant section Time dilation#Simple inference of velocity time dilation. - DVdm (talk) 16:20, 23 January 2020 (UTC)

This is perfectly correct, DVdm, because t clock, as you say, is moving.

But, according to the article: "which expresses the fact that the moving observer's period of the clock $\Delta t'$ is longer than the period $\Delta t$ in the frame of the clock itself."

Hence, according to the article while moving observer's t' clock measures interval between two events $\Delta t'$, clock at rest t measures interval $\Delta t$, i.e. moving clock t' measures longer interval of time, or is ticking faster than clock t.

I say again that wording is not good to say the least, because the article confuses who is moving and picture is wrong, they say that t' is moving relative to t, while t should be moving relative to t'.

Again, what do yo mean saying "As seen on the t'-clock,"? What does it mean "as seen"? How does an observer "see"? Does he stare at his own clock? He must some way COMPARE readings between his own clock(s) and moving one, isn't it really clear? Have you ever heard about synchronization of clocks in the observer's frame? For what purpose an observer synchronizes clock, have you ever thought about that?

Read it here in Spacetime - Mutual time dilation

It turns out that in mutually observing the duration between ticks of clocks, each moving in the respective frame, different sets of clocks must be involved. In order to measure in frame S the tick duration of a moving clock W' (at rest in S'), one uses two additional, synchronized clocks W1 and W2 at rest in two arbitrarily fixed points in S with the spatial distance d. 90.135.153.131 (talk) 19:08, 23 January 2020 (UTC)

This is an observer in his rest frame on the picture - do you see how many clocks he has? Not just one, the whole space is filled with his synchronized clocks. And moving clocks change spatial position relative to his clocks and run slower, not vice versa - the whole frame is moving somewhere relative to a single clock. It is wrong to say, that rest frame of an observer is moving relative to a single pointlike clock. Vice versa.

Figure 1-1. Each location in spacetime is marked by four numbers defined by a frame of reference: the position in space, and the time (which can be visualized as the reading of a clock located at each position in space). The 'observer' synchronizes the clocks according to their own reference frame.

90.135.153.131 (talk) 19:16, 23 January 2020 (UTC)Cordially Yours (talk) 19:38, 23 January 2020 (UTC)

Of course "as seen" means "as measured" or better, "as calculated" or even better, "as deduced" or "as reasoned". No eyes are needed here. Think only in terms of the two specific events on the t-clock, and read what I and the cited sources say until you understand it. Note that some sources might swap the primed and unprimed coordinates. That is of course entirely and utterly irrelevant. If none of what I and the sources say help you out of your confusion, then I'm afraid I can't help you, and... perhaps you might consider another hobby. You are not the only amateur who struggles with this. Sorry about that - DVdm (talk) 19:52, 23 January 2020 (UTC)

No problem, DVdm ! I just wanted to improve the article.Cordially Yours (talk) 20:03, 23 January 2020 (UTC) It's a pity that you don't want to accept my valuable notes ! The article claims that moving clock is ticking faster, that is wrong. Cordially Yours (talk) 20:05, 23 January 2020 (UTC)

Please excuse me DVdm for expressing doubts about your competency; I have looked at your user page. It turns out you are THE scientist and experienced specialist. Just at first, according to your reasoning, I thought that you were some kind of a donkey and absolutely do not understand anything. Please excuse me again. Cordially Yours (talk) 20:26, 23 January 2020 (UTC)

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