michel123456

Well understood now. Thx. Thx, that is now clear as crystal water. now the return trip: Please explain the bold part, does it come from the dilation formula you posted above?

Oh, I got it: you mean 1.25 hour means 1h and a quarter? That's a bad mix of notation. You are considering as if the travel was 1 hour long, but the travel was 1.25 long Still not understanding this: if it read 12:00 when it left you, you will see it arriving at the second clock reading 12:45.

I am confused. first of all, hours have 60 min, not 100. So 60/0.8=75 or 1h15min (and not 1h 25 min.). Correct? Generally I am confused with the mix of 1.25 & 1.15 in the text above.

You are considering as if the travel was 1 hour long, but the travel was 1.25 long. And you have explained to the community how to travel 1 LightHour in 45 minutes.

Why 01:15 and not 01:25? ! Moderator Note (this is referencing https://www.scienceforums.net/topic/105185-time-dilation-dependence-on-direction/) This thread is now the only place where michel123456 may discuss topics related to time and relativity

Yes.(1) Because you are thinking that A knows the answer right from the beginning. But in reality A makes an observation: a clock is observed ticking at unusual rate. Then he must figure the reason why. In fact, A never observes a clock. A is observing planets, stars & galaxies. When the object gets away, for whatever reason, observer A must figure that a clock on it is ticking slower. When the object gets closer, the clock on it is observed ticking faster.(1) And when the object is receding faster than SOL (I wrote: receding) what is the clock showing I don't know. (1) IIRC this effect is not taken into consideration in the twins paradox.

Sorry for the late reply, I had better things to do... If you don't get the question, figure some examples with regular velocities (lower than SOL) and then increase until you reach near SOL. For example v=1/10 SOL At departure time 10.00 both clocks at A. Point B (arrival) lies 1 LightHour away Clock B needs 10 hours to reach point B (v=1/10 SOL) When B arrives, traveler B reads on its own clock 20.00 (because for the traveler, clock B works as usual) As observed from A, clock B seems to arrive at point B at time 21.00 as shown by clock A (the image of the clock reading 20.00 reaches A at 21.00). It means that during the travel, A has observed the hands of the clock executing 10 rounds in 11 hours :i.e. the clock has been observed by A as running slow. Is that correct so far?

Is that it? A would see B disappear? IMHO A would continue to see B as "frozen". It seems unlikely to me that B could disappear from sight.

One by one lineage is easy: you get a diagram with gene numbering on the horizontal axis and percentage on the vertical. You may need to rearrange the position of the genes in order to get a curve of percentages. Then the same for the second lineage until the 12th. Then you may supperpose all the 12 diagrams in a 3D diagram. You will get something like this https://www.originlab.com/www/products/GraphGallery.aspx?GID=258

I can't remember exactly. This info comes from a book from physician Hans Rosling founder of the Gapminder Foundation. It is a very interesting read, google for gapminder & take time there, it may change your world view. In short, following worldwide data (not speculations), the world is maybe bad, but it is getting better, not worse. Title of the book: Factfulness. Small book cheap. But plenty of info free from gapminder.

. Yes. And what happens in the first figure case?: Speed of missile = SOL A----------(1LH)----------B At point A, time on traveling clock, as seen both by A (stationary) & B (the traveler), reads 10.00 (departure time) For each ticking of the clock (1 sec.) the clock gets away 1 lightsec. The 2 measurements cancel each other. Right?

There is no needfor restriction on population. The models show Earth's population to reach the top on 11 billion inhabitants in around 2100 . see UN model https://population.un.org/wpp/Download/Probabilistic/Population/

The reading 11.00 will show at A at 12.00, 1 hour later. Correct?

Because the missile travels at SOL. The delay is equal to the time traveled. It starts at 10.00, arrives at 11.00, but for observer A, the delay is 1 hour, thus 11.00h minus 1.0h0 = 10.00h Departure 10.00 Arrival time 11.00 delay 1.00h observation from A = arrival time - delay

Where am I trying to go: where Relativity does not allow me to go. Galileo could have thought: lets make a missile that goes at the speed of light, what would A observe? Continuing the previous experiment At 10.00 exactly the missile with clock B is send away. At 11.00 (1 hour later) the missile reaches point B (1 hour-light away). The reading on clock B (as observed by A) is ...10.00 o'clock. IOW A has observed clock B, during its travel, without clicking at all. As if clock B, because it moved at SOL, had stopped clicking (although for observer B, moving together with clock B, has observed clock B ticking as usual). A has observed the traveling clock B "frozen in time" AND worse, Galileo could have thought (because he knew nothing about Relativity), what happens when clock B goes faster than the speed of light? Let's consider this below At twice faster than SOL, the traveling clock reaches point C in 1 hour. Point C lies 2 light-hours away from A A----------(1 LH)----------B---------(1LH)----------C At point A, time on traveling clock, as seen by B (the traveler), reads 10.00 (departure time) At point C, time on the traveling clock, as seen by B (the traveler), reads 11.00 (arrival time) At point B, as seen by B, time on the traveling clock reads 10.00 (time delay 1 hour from arrival time) What is the reading from A? It cannot be 9.00 (because the clock never showed 9.00 o'clock as seen by the traveler, A cannot observe something that never happened) So, what is the reading from A?

Yes, I agree. Exactly. So, as observed by A, will there be an (apparent) change or not? (before making any adjustment) What will A observe in his telescope? (before applying any correction on the basis of any theory).

No. I say that WHILE TRAVELING the rate of clock B looked like running slow. While in motion, the rate of clock B was observed clicking slow as observed by A. The clock in motion was NOT clicking slower. But observer A observed the clock as if it was changing rate. No. they are not in sync. The 2 clocks at rest far away both tick at the same rate. But I am discussing the moving clock. What happens when the clock is observed traveling from one location to the other. Yes, you are correct. it didn't change. But what happen to a clock traveling from A to B? As observed by A the traveling clock is observed reading 10.00 at the beginning and 9.00 at the end (sort of speaking).

Say both clocks at rest show 10.00 o'clock. Clocks A & B are perfectly synchronized. You send clock B at a distance of 1 light-hour away. Say that the travel was made in exactly 1 day. Now clock B stands still 1 light-hour away: the observational delay is 1 hour. In this new situation when observer A reads clock A at 10.00 (the next day) at the same instant he reads clock B showing 9.00 (because the delay is 1 hour). Both clocks A & B are ticking at the same rate. But how did clock B change from 10.00 to 9.00? IMHO the traveling clock (B) was apparently ticking slower than clock A. As observed by A, in 24h the moving clock B apparently ticked only 23 hours. If B was never apparently changing rate, it would be impossible to change from 10.00 to 9.00.

Yes. The clock rate would seem to change as observed by the observer standing still. As observed by the moving duck, the moving clock rate would not change at all. There is no actual change of ticking. But both observers will observe the other's clock change rate. Because it takes 1 second for the light to travel the distance (d = ct).

You are misunderstanding me. I do not put invariance into question. I simply say that the fact that SOL is finite is enough & sufficient to introduce a delay. Since the delay is there, the rest is deduction.

And how would the delay come in when a clock travels? Wouldn't it change rate? If this statement is correct, then all the rest is simple logical deduction. There is no need for C being invariant. Also note that in my scenario, if the missile comes back at origin, the clocks will still be synchronized. The observer on earth will observe the traveling clock ticking faster while getting closer.

Even without Relativity, you can work out the question only with the single axiom that SOL is finite. The axiom says that light takes some amount of time to travel. That means that any information that comes to you through the means of light comes from the past. It means that if you are observing a clock away from you, this clock is in the past. So if he knew that, Galileo Galilei, with some thoughts, could get to this: Say that you have 2 synchronized clocks on your desk. Say that you send one clock in a missile (together with a living duck) away from you. When this missile will be far away enough, the clock will look like beying delayed. Which means that during the trip, while traveling, the clock has ticked slower than the one on your desk. The heart rate of the duck has gone slower also (it is not an effect of some mechanical flaw, it is caused by SOL being finite). For the duck, nothing happened, except that for the duck, it is your heart rate that has beaten slower.And it is your clock that is delayed. say that the missile, by the means of some unknown technology, continues its travel accelerating through space. As the missile gets away, the clock seems to tick slower and slower & the ducks heart rate seems, from your point of view, ticking slower. After some years of traveling, the missile will go so fast that, as seen from you, reaching the SOL, the clock will look like stopping , & the duck's heart rate stopping. On the missile, the duck will live as usual next to his ticking clock. Looking back at you, the duck will see that it is your heart that stopped beating and your clock here on hearth stopped clicking. At no instant will any observer see the clock ticking backward or the duck getting younger. SOL is an asymptote.

A light beam was send from the Earth 100 years ago (1920) into space. The first photons of this beam are today 100LY away. They are still in 1920. Because if they hit the eye of some E.T., the E.T. will see the image of 1920. IOW the photons are traveling in space but not in time.

In the observer's FOR. You are observing a particle getting away from you at near to C velocity. This particle appears time dilated. Theory tells us that if you could observe a particle going at FTL (tachyon), you would observe it going backward in time. So I am suggesting that the flip happens at C (between going normally in time & backward in time). Which means that a photon going away, as observed by you, would look as hovering still in time. I suppose that the same counts for a photon getting close to you.

I was thinking of this : You cannot, but a photon can.