Time dilation is propagation delay (split from What is time? (Again))

[michel123456]

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.

[swansont]

28 minutes ago, michel123456 said:

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.

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.

 

[michel123456]

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.

Edited September 9, 2020 by michel123456

[Eise]

18 hours ago, Markus Hanke said:

I don’t think it is counterintuitive - on the contrary, I would struggle to imagine what a universe would look like where this isn’t true. Saying that c is invariant between inertial frames is saying that all inertial observers experience the same laws of physics - whether you turn on your laptop in your living room, or while travelling in a very fast rocket, it will function the same in both cases. Intuitively, this is exactly what I would expect to happen. But maybe that is just me again  

Hmm, you are already too long in relativity. People like you and Janus know relativity so well, that it has become intuitive for you. 'Intuition' is not a given, it is something to develop, it comes with experience. So to answer the question if relativity is intuitive, I think you should go back to the first time you heard about it. I remember my first 2 encounters: me reading a children-level book about astronomy, in which was an example of time dilation, a rocket flying to the Andromeda galaxy and back again. I think an important factor in accepting this was that when I told my father, he confirmed that he knew that (and that it was not understandable how that is possible).

The second encounter I found really astonishing: that of the invariance of the speed of light.

Now I am also pretty used to it, and I understand the absolute basics of special relativity. But still... You, @Markus Hanke take one of the postulates that nearly everybody would agree upon. Everybody who has been travelling in a train or even an airplane, knows that everything works exactly the same as on the surface of the earth. It is only in combination with the fact that the speed of light is invariant, and not infinite, that we get at the non-intuitive results of special relativity, like time dilation and the equivalence of mass and energy.

Personally, I feel that a dynamic solution, based on the æther, is much more intuitive. It compares with the experience of air. I you walk slowly, you do not notice it, but as you go faster, you feel the effects, and see them. Airplanes are just not fast enough to feel the æther pressure...

2 hours ago, michel123456 said:

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.

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.

[michel123456]

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.

[Eise]

17 minutes ago, michel123456 said:

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.

You cannot explain the time dilation of the muon with just delay. Time dilation and delay are not the same. So you cannot deduce time dilation by a delay. If you think you can, please show me with my example of the muon.

[swansont]

3 hours ago, michel123456 said:

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

It changes rate because c is invariant and time is relative. That's relativity. In your scenario there is no reason for the clock to change rate.

 

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.

If you think otherwise, you need to do an actual analysis to show it. 

(note that there would be other implications of c not being invariant, because breaking the laws of physics has implications everywhere, but these are being ignored as they are not relevant)

 

Quote

If this statement is correct, then all the rest is simple logical deduction. There is no need for C being invariant.

c being an invariant is a constraint put on us by nature. The "need" is that the model be consistent with experiment.

And "logical deduction" (especially when you skip the steps showing this deduction) is less powerful than actual mathematical analysis.

 

Quote

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.

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

[michel123456]

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).

Edited September 9, 2020 by michel123456

[swansont]

11 minutes ago, michel123456 said:

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.

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.

 

11 minutes ago, michel123456 said:

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).

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.

 

[michel123456]

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.

Edited September 10, 2020 by michel123456

[swansont]

1 hour ago, michel123456 said:

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?

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. A knows that it was 9:00 one hour ago, i.e. that it is now 10:00. IOW, the clocks agree. It is 10:00 everywhere in that frame.

There was never a change in the rate of clock B. If there had been, it would have accumulated more or less phase (time), and the clocks would not agree. (this is what happens in relativity, and is the reason for time dilation, but in our example we are assuming this is not happening)

This is exactly the same as the example that Janus gave about the chiming of the bell at noon. If it takes 10 seconds for the sound to get to you, you know it was noon 10 seconds before you hear it. I have seen a map of London where this is done for Big Ben, showing the delay so people could properly set their clocks.

 

1 hour ago, michel123456 said:

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.

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.

[Eise]

@michel123456:

Now turn your example the other way round:

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? And then, B travels back to A. Now, at ten o'clock according to A's own clock, B's clock shows 11:00h. So time expanded? And instead of a delay we have an 'advancement'?

[michel123456]

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).

Edited September 10, 2020 by michel123456

[swansont]

23 minutes ago, michel123456 said:

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, 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.

It does not look like it is ticking slower as long as you realize that there is a delay in the signal. But if you don't, and do the analysis improperly, you will the wrong answer.

 

Quote

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.

 

It only reads 9:00 because you are sending the signal at 9:00 to have it arrive at 10:00. If you waited 24 hours to send the signal, you would send it when the clock reads 10:00. A will get that signal at 11:00

Your scenario is flawed. 

"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."

Clock be reads 9:00 because you waited 23 hours, not 24. And there should be nothing mystifying about a clock reading 9:00 23 hours after reading 10:00

Quote

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).

 

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

 

 

 

 

[Janus]

4 minutes ago, michel123456 said:

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.

But that is just light propagation delay, which generally is factored out when we deal with SR.

Suing your example, with B traveling to a distance of 1 light hr in 1 day:

To do this it has to travel at 1/24c.

At that speed, the relativistic Doppler shift ( which determines the rate at which A would visually see B tick at as it recedes), is ~0.95917

Since it takes an additional hour for light to travel from a 1 light hr distance,  A doesn't see B arrive at the 1 light hr distance until 25 hrs after B leaves.

Thus for 25 hrs, A will see B tick at a rate of 0.95917 and accumulate ~86,325 of ~sec, while accumulating 90,000 sec itself.  this is a difference of ~ 3675 sec.

3600 of those sec is accounted for by the 1 hr light propagation delay, leaving a 75 sec real time difference.

Thus, A, at 10:00, will see B reading 08:58:45, but after accounting for the 1 hr propagation delay, will determine that B actually reads 09:58:15.

[swansont]

8 minutes ago, Janus said:

But that is just light propagation delay, which generally is factored out when we deal with SR.

Suing your example, with B traveling to a distance of 1 light hr in 1 day:

To do this it has to travel at 1/24c.

At that speed, the relativistic Doppler shift ( which determines the rate at which A would visually see B tick at as it recedes), is ~0.95917

Since it takes an additional hour for light to travel from a 1 light hr distance,  A doesn't see B arrive at the 1 light hr distance until 25 hrs after B leaves.

Thus for 25 hrs, A will see B tick at a rate of 0.95917 and accumulate ~86,325 of ~sec, while accumulating 90,000 sec itself.  this is a difference of ~ 3675 sec.

3600 of those sec is accounted for by the 1 hr light propagation delay, leaving a 75 sec real time difference.

Thus, A, at 10:00, will see B reading 08:58:45, but after accounting for the 1 hr propagation delay, will determine that B actually reads 09:58:15.

We are assuming there is no time dilation in this scenario. There is no relativistic anything. The premise is that light travels at a finite speed, but not that it is invariant.

[michel123456]

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).

[Janus]

35 minutes ago, michel123456 said:

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).

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?

[swansont]

It’s as Janus says. There are no effects other that signal delay, which is rather mundane, since it’s part of everyday life.

[joigus]

Because the main arguments have been established more than satisfactorily enough by Hanke, Eise, Janus, and Swansont, IMO, I would like to concentrate on the fact that the world michel123456 is trying to picture would be awfully incongruous and we wouldn't have an invariant picture of phenomena. Relationships between events would be distorted depending on how fast you're moving with respect to particles/fields. For some collisions, we would see one particle bouncing off before the other colliding particle reached there.

The underlying mathematical reason is that the Lorentz group with the choice c=infinity is perfectly reasonable and gives homogeneous transformations of intervals that make timings and placings of phenomena mutually consistent. Relationships between events would be congruences.

c finite and invariant would be OK (Einstein's relativity)

c infinite (thereby invariant) would be OK (Galilean relativity)

But taking c finite and observer-dependent is not. It does not produce anything in the way of congruences of events.

Edited September 10, 2020 by joigus

[Markus Hanke]

12 hours ago, swansont said:

The premise is that light travels at a finite speed, but not that it is invariant.

Just to add to all the excellent points already made - the invariance of the speed of light, i.e. Lorentz invariance, is fundamental to the Standard Model of Particle Physics. If you take away Lorentz invariance, you 'break' all the fundamental interactions. You wouldn't even end up with the same elementary particles! In other words, we wouldn't be here to discuss this (let alone type on anything resembling a computer) if this symmetry didn't hold. And signal delays due to finite speed of light isn't the same as Lorentz invariance.

In terms of the bigger picture, special relativistic effects (invariance of spacetime interval, time dilation, length contraction, relativity of simultaneity, Thomas rotation) arise from the geometry of spacetime, which now is no longer Euclidean. This goes far beyond signal delays (which isn't a relativistic effect btw).

Edited September 11, 2020 by Markus Hanke

[michel123456]

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?

 

Edited September 11, 2020 by michel123456

[swansont]

2 hours ago, michel123456 said:

Where am I trying to go: where Relativity does not allow me to go.

What's the point? Relativity is how nature behaves.  

 

2 hours ago, michel123456 said:

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.

Why is it 10:00? The missiles did not send the signal at 10:00. At 10:00, the clocks were co-located. There is no other time on the trip the clock would read 10:00. 

At B, the clock said 11:00

2 hours ago, michel123456 said:

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"

Why would it stop ticking?

Do you have a model which predicts this, or is this just more flawed logic?

 

[michel123456]

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

Edited September 11, 2020 by michel123456

[swansont]

56 minutes ago, michel123456 said:

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

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.