michel123456

45 minutes ago, swansont said:

They are not mixed.

The text:
Using the formula above we find that you will see it tick 1/3 as fast as your own. It will arrive at the second clock in 1.25 hrs when the second clock reads 01:15. But because of the light travel time lag between you and the second clock, you will not see this until until your clock reads 02:15. This means the you will watch the third clock recede for 2.25 hrs by your clock while seeing it ticking 1/3 as fast and thus if it read 12:00 when it left you, you will see it arriving at the second clock reading 12:45. So in the 1:15 it took to make the trip it only ticked off 45 min. (remember, even though you didn't see the third clock arrive until your clock read 02:15, it had actually arrived an hour earlier.)

 

1.15, for example, is not used. All of the time-of-day readings (as one might see on a digital clock) or elapsed times denoting hours and minutes use a colon, per the ISO 8601 standard (further, all clock readings are denoted as such)

The decimals are used for elapsed time calculations, as one might expect when the values used don't result in a whole number

The two notations are used consistently. Decimal point used for decimal numbers, colon used for hours:minutes

 

 

 

 

 

Well understood now. Thx.

1 hour ago, Janus said:

How is using 1.25 hrs any different than saying 1 and a quarter hrs?  or 1 1/4 hrs.  The fact that the number is followed by "hrs" indicates that the numerical value is applies to the one unit, hours, while with 1:15  the ":" denotes a marker between units, or 1 hr, 15 min. 

I'm sorry if this confuses you, but when working with such problems it is easier to the work the math out in decimal hrs and then convert back to Hr and min afterwards if you need to.

If the clock left you moving at 0.8c to a point 1 light hr away ( as measured by you), while both it and your clock both read 12:00, then it will take, by your determination, 1 hr 15 min to reach that point. You, however, will not see him arrive when your clock reads 1:15, as It will take an another hr for the light from his arrival to reach you, so you will see the image of his arrival 2 hr and 15 min after he left, when your clock reads 2:15. In other words, you will "see" his entire trip as being stretched out over 2 hrs 15 min.

As you watched him recede from you, you will see him via relativistic Doppler shift. This does not only effects the measured frequency of the light you get from him, but also the rate at which you would "see" his clock tick.  The Relativistic Doppler shift rate for 0.8c is 1/3.   Since you see this happening for the entire 2 hr and 15 min, you will see that clock ticking off only 1/3 of that, or 45 min.  Thus when you see the clock arrives at that point 1 light hr away, you will see it as reading 12:45.

 

Thx, that is now clear as crystal water.

now the return trip:

On 4/26/2017 at 6:43 PM, Janus said:

Now let's do a reverse trip. you see the second clock read 01:15 and the third clock reads 12:45 when your clock reads 02:15, however this image of those two clocks left 1 hr ago, so the third clock left the second clock when your clock read 01:15 (in other words, if the third clock turns around and heads back the moment it reaches the second clock, by the time you see this, the third clock is already well along its trip back to you.). The return trip takes the same 1.25 hrs, which means it arrive back at you when your clock reads 02:30. This means that you will see its entire return trip occur during the 15 min between 02:15 and 02:30. At a accelerated tick rate of 3 you will see the third clock tick off 45 min during that period, and read 01:30 upon arrival, when your clock reads 02:30. So again, it ticked off 45 min during the 1.25 hr trip, the same as for the outbound trip and accumulated at total of 1.5 hrs for your 2.5 hrs.

Please explain the bold part, does it come from the dilation formula you posted above?

11 minutes ago, Halc said:

Because one and a quarter hours is an hour and 15 minutes, not an hour and 25 minutes.

Oh, I got it: you mean 1.25 hour means 1h and a quarter? That's a bad mix of notation.

2 hours ago, michel123456 said:

On 4/26/2017 at 6:43 PM, Janus said:

if it read 12:00 when it left you, you will see it arriving at the second clock reading 12:45. So in the 1:25 it took to make the trip it only ticked off 45 min. (remember, even though you didn't see the third clock arrive until your clock read 02:15, it had actually arrived an hour earlier.)

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?

On 4/26/2017 at 6:43 PM, Janus said:

You looking at this clock when your clock reads 12:00 will visual see the clock as reading 11:00. (since it took the light carrying that image 1 hr to travel from the other clock, to get to you when your clock reads 12:00, it had to leave the other clock when it read 11:00)

 

A third clock travels from your position to the second clock at 0.8c

Using the formula above we find that you will see it tick 1/3 as fast as your own. It will arrive at the second clock in 1.25 hrs when the second clock reads 01:15. But because of the light travel time lag between you and the second clock, you will not see this until until your clock reads 02:15. This means the you will watch the third clock recede for 2.25 hrs by your clock while seeing it ticking 1/3 as fast and thus if it read 12:00 when it left you, you will see it arriving at the second clock reading 12:45. So in the 1:15 it took to make the trip it only ticked off 45 min. (remember, even though you didn't see the third clock arrive until your clock read 02:15, it had actually arrived an hour earlier.)

Generally I am confused with the mix of 1.25 & 1.15 in the text above.

On 4/26/2017 at 6:43 PM, Janus said:

So let's work out a couple of examples:

Assume you are situated 1 light hr from a stationary(to you) clock that is synchronized to your own clock (when your clock reads 12:00, it reads 12:00)

You looking at this clock when your clock reads 12:00 will visual see the clock as reading 11:00. (since it took the light carrying that image 1 hr to travel from the other clock, to get to you when your clock reads 12:00, it had to leave the other clock when it read 11:00)

 

A third clock travels from your position to the second clock at 0.8c

Using the formula above we find that you will see it tick 1/3 as fast as your own. It will arrive at the second clock in 1.25 hrs when the second clock reads 01:25. But because of the light travel time lag between you and the second clock, you will not see this until until your clock reads 02:25. This means the you will watch the third clock recede for 2.25 hrs by your clock while seeing it ticking 1/3 as fast and thus if it read 12:00 when it left you, you will see it arriving at the second clock reading 12:45. So in the 1:25 it took to make the trip it only ticked off 45 min. (remember, even though you didn't see the third clock arrive until your clock read 02:15, it had actually arrived an hour earlier.)

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.

On 4/26/2017 at 11:43 AM, Janus said:

A third clock travels from your position to the second clock at 0.8c

Using the formula above we find that you will see it tick 1/3 as fast as your own. It will arrive at the second clock in 1.25 hrs when the second clock reads 01:15

Why 01:15 and not 01:25?

1 hour ago, Eise said:

And now turn the direction. B travels back to A. Now the B's clock ticks faster than A's, from A's viewpoint.

Yes.(1)

1 hour ago, swansont said:

That’s what A observes, but if A understands the propagation delay, this is not correct. A does not observe the clock to be running slow, because the time is the combination of the reading and the delay.

IOW, if A does the measurement properly, this us not true. However, if A does the measurement incorrectly, A will get the wrong answer. One possibility being they assume the clock ran slow. They might also conclude that a dragon ate an hour of time. There are a large number of wrong answers one could apply.

 

What’s the point of asking the same question over and over again? Are you hoping someone will eventually give you a different answer?

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.

On 9/12/2020 at 3:37 PM, Halc said:

Don't get the question.  The only measurements you describe are both of them in each other's presence at the time of departure.  They both measure 10:00 on both clocks at that time.  Not sure what it would mean for those identical observations to 'cancel'.

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?

15 hours ago, Janus said:

Thus, if B were traveling away from A at c, its light emitted back towards A would be not moving at all relative to A. If B started at any non-zero distance from A, A would see light coming from it right up to the moment B started moving away at c, and then would see no image of B at all.

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

1 hour ago, Airbrush said:

Could you please explain this model?  How does this tell you that 11 billion is the max?  What about fresh water shortages?  How about wars over resources?  I always thought that unless people intentionally restrict population growth, there would be a big, accidental war that eliminates half the population in the aftermath of a nuclear exchange.  Oops! 😲

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.

.

12 minutes ago, Halc said:

It depends on what theory Galileo uses.  Using the ballistic theory of light, the missile is immediately invisible at A because it is going faster than light, and thus light emitted by it cannot propagate towards A.  It can only fall behind.

Yes.

12 minutes ago, Halc said:

Using a pre-relativity ether theory (Newtonian), and assuming A is stationary in that ether, the missile clock reads 11:00 when it passes C and it takes 2 hours for that light to go back and reach A, so A sees the missile clock ticking at 1/3 rate.

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?

21 hours ago, Airbrush said:

Rather than colonizing Mars, the Moon, asteroids, or a giant solar-system cruiser, we should colonize Earth!  We could build bunkers that resist the worst natural disasters.  People living near the ocean should be prepared with a tsunami and hurricane shelter.

Establish self-sufficient settlements in areas previously uninhabited, such as deserts that are near the ocean, Northern Canada, or Antarctica.  Such independent settlements would use solar and wind power, and enough renewable biofuels (those in warm climates) to satisfy energy required for a growing settlement, and pump sea water inland to desalinate.  Such settlements could produce food on farms and use surplus fresh water to plant trees where trees and other vegetation had not grown before.

Sooner or later governments will place restrictions on population.  Like China's one child per couple policy, but maybe not that strict.  Maybe a 2-child limit.  Maybe medicine would progress to the point that people could know in advance it they should have children.   Maybe they learn their children would have genetic problems, so they decide to not have children.

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/

1 hour ago, swansont said:

Right, arrival time is 11:00, so the clock reads 11:00

You have not established any reason why the clock would not tick during the trip. It's just something you've asserted with no justification.

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

49 minutes ago, swansont said:

Why is it 10:00?

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

15 hours ago, Janus said:

He will see whatever time that was on clock B when the light left it. So what?  Where are you trying to go with this?

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?

21 minutes ago, swansont said:

the clock will always display the same time as the other clock, and the difference in the signal will simply be due to the delay time of the signal.

Yes, I agree.

22 minutes ago, swansont said:

If you send a signal at an intermediate point, it will read whatever time it was, offset by the signal travel time.

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).

4 hours ago, swansont said:

 

No. You are ignoring the delay in the signal travel time.

Let’s say that the clock really was slowed by an hour in that trip - the clock ran 23/24 as fast. It would lose an hour. At 9:00, it reads 8:00. It sends a signal to A saying it’s 8:00 and it gets there an hour later. A gets the signal at 10:00, and it says “it’s 8:00”

That’s not what you say is happening, though.

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.

3 hours ago, Eise said:

Two clocks, 1 lighthour distance, A and B, not moving relative to each other At A it is ten o'clock. A looks at the clock of B and also sees ten o'clock. Does that mean that the clocks are in sync?

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. 

4 hours ago, swansont said:

It didn’t “change”. At 10:00, it reads 10:00 to an observer with it. Observer A gets a signal from B saying 9:00, with the knowledge that there is a 1 hour delay

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).

21 hours ago, swansont said:

No, as observed by both observers, the clock rate would not change.

If you think otherwise, you need to show this, not just assert it.

 

That doesn't give a change in the clock rate. That gives an (apparent) change in the clock phase (time), but this can be adjusted for, as we do with actual clocks in use. You subtract out t = d/c (if the clocks are at rest with respect to each other)

You are confusing the clock's time with the signal that is being received. That's only a problem if we were trying to reconstruct the time from the signal frequency. You could just broadcast a "at the tone the time will be..." signal and then reconstruct the time from the signal delay. Then there is no frequency shift of the signal to worry about.

 

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.

52 minutes ago, swansont said:

That would seem to simply be propagation speed changes and/or doppler shift. The signal would change, but the clock's rate would not.

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.

52 minutes ago, swansont said:

Under a scenario where c is not invariant (i.e. SR does not hold): If you move away at some v and go to a place 1 light-second away and come to rest, and the clock signals differ by 1 light second, that's because it takes 1 second for the light to travel the distance (d = ct). The clocks actually read the same, which means there was no change in rate.

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).

30 minutes ago, Eise said:

This is ridiculous. First you should know better, there are more than enough explanations of the basics of special relativity, and the invariance of the speed of light is essential in all explanations. Second, this invariance is a fact. So try the following:

Take a muon: we know its half-life from laboratory experiments. Now we know that muons are produced by cosmic rays in the upper atmosphere. However given their half-lives, next to none should ever reach the earth's surface. But they do. This is explained by the time dilation we observe for the muons. Now take your explanation. Just assume the muon has a wristwatch that we can see. Now explain why its time seems slower for us, than for the muon itself.

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.

19 hours ago, swansont said:

No, absent relativity, the clocks would/could tick at the same rate. The delay is simply the light travel time between the clocks. 

The rate slowing comes from c being invariant, not just being finite. That’s the basis of relativity.

 

And how would the delay come in when a clock travels? Wouldn't it change rate?

6 minutes ago, michel123456 said:

The delay is simply the light travel time between the clocks. 

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.

On 9/3/2020 at 12:40 AM, Markus Hanke said:

Because any conceivable clock - even an ideal one - must be massive, and therefore it cannot be comoving with a photon.
Mathematically speaking, you can't parametrise the length of a photon's world line using proper time (because ds=0); however, that doesn't mean that their world lines don't have a well defined length in spacetime. They do, you just need to use a different affine parameter.

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.

5 hours ago, swansont said:

No, in the observer’s frame the photon moves at c. There is no frame of the photon, so there is no time dilation for the photon. You can only observe time dilation with respect to some other frame of reference.

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.

22 hours ago, Ghideon said:

Can a photon do that? In what frame of reference does that happen? 

(Answer: 'No' and 'None'; as far as I know. In relativity the photon is not a valid frame of reference)

 

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.

On 8/26/2020 at 12:31 AM, swansont said:

You can be at rest with respect to space, but not time. 

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