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

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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?

 

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12 minutes ago, michel123456 said:

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

Yes.

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5 hours ago, michel123456 said:

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?

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.

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.

In both cases, the missile can see nothing behind it because it is superluminal, similar to how a supersonic aircraft cannot hear any noise from behind.

Both theories predict different results than what is actually observed, but the actual theory has no different answer for the problem since it forbids the missile from having a space-like worldline. As you say, relativity forbids that you go there, but the other two theories do not.

Edited by Halc

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.

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?

 

Edited by michel123456

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7 hours ago, michel123456 said:

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?

 

Galileo would have assumed that light followed the rules of Galilean Relativity and that the speed of the light emitted by B relative to A would depend on the velocity of B relative to A.

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.  If B then stops relative to A at a 1 light hr distance*,  A would eventually see this two hrs after B left**,  If both A and B read 10:00 while near each other, B will read 11:00 when it reaches its destination, and this light will reach A when its reads 12:00. 

Thus A would see B tick normally up until it starts moving away at c, and then A would see nothing until 12:00, when it would suddenly see B one light hr away and reading 11:00. Knowing that the light took an hour to get to him, that clock B at that moment also reads 12:00.  Galileo would assume that clock B just ticked normally the whole time, and there were just portions of the trip that A simply could not see.

If B were moving away at greater than c, A would also not be able to see it as it receded, because, once again, the light emitted by B would never get to A (in fact, it would have a velocity away from A.)

If B were then were to suddenly reverse direction(for simplicity we'll assume an instantaneous change of velocity)  upon reaching 1 light hr from A and returns at c, Galileo would expect A to the see the following.

For 1 1/2 hours after B leaves, A sees nothing, Then at 11:30 it will see B at 1 light hr away, reading 11:00 and coming towards A at c.  During the next half hr,  It will see B close the distance, while B's clock appears to tick twice as fast, accumulating 1 hr of time until both clocks meet with both reading 12:00.

If B make the same type of trip at 2c,  A sees nothing for 50 min, then at 10:50 B suddenly appears, at a distance of 1 light hr reading 10:30. For the next ten minutes, A sees B close the distance while B's clock ticks 3 times faster, accumulating 30 min, to read 11:00 once it returns to A.

In short, Galileo would not assume that anything out of the ordinary happens to Clock B in terms of its tick rate, just that the combination of the SOL and the relative velocity of B to A will alter how the information about B reaches A.  And as long as A takes this into account, A will say that clock B always keeps perfect time with clock A

Note that this is completely different for Relativity, where A, after taking into account the SOL effects on how the info arrived would conclude that Clock B did not keep time with clock A.

Thus in my earlier example, When A sees a 1 hr and 75 sec difference between itself and clock B, 1hr of that is due to light propagation delay, and the remaining 75 sec is due to time dilation that occurred during the outbound trip.

 

 

*Galileo didn't have a good idea of how fast light traveled, only that it was much faster than sound, so here, we'll just assume some finite value for c.

** This light will pass the stationary(relative to A) light left behind by B on its outbound trip.

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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.

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1 hour ago, michel123456 said:

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.

No photons = no image

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@michel123456: I am still waiting how you see the situation in the opposite direction: A seeing the clock of B at a lighthour distance, and the B traveling to A. Can you explain the example of muons reaching the earth's surface, even if they would not be able to reach it, according their velocity (near light speed) and their half life. So take A to be an observer on the earth's surface, and B the location high in the atmosphere where the muons are created. For simplicity of argument, you can assume that the muons on their path send regular light signals to earth (of course they don't, but it would not change anything in your example).

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3 hours ago, michel123456 said:

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.

Janus is describing the ballistic theory of light, which is actually a much better fit to Galilean relativity which asserts no preferred frame.

One cannot 'continue to see B as "frozen"' '.  Light from that moment can at best be measured once, not continuously.

Oddly enough, if the missile moves at 2c, with either theory, if they look at a 'stationary' Earth from the missile,  Earth will appear to run backwards.  At 11:00 missile time, it will see Earth at 9:00 given a 10:00 departure.  The 9:00 light from Earth gets to the same location as where the missile is at that time. Funny thing is, the image of Earth will appear in front of the missile, not behind it like intuition suggests.  Sound works the same way.

The post of mine discussed both, but the part to which you responded involved more of a Newtonian preferred-frame ether theory which denies Galilean relativity.

22 hours ago, michel123456 said:

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.

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'.

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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?

 

 

 

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27 minutes ago, michel123456 said:

Is that correct so far?

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

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1 hour ago, michel123456 said:

 

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?

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?

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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.

Edited by michel123456

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1 hour ago, michel123456 said:

Yes.(1)

Because you are thinking that A knows the answer right from the beginning.

I think that A knows the speed if light us finite, yes. That seems like a reasonable assumption, since it can be tested independently of this scenario, and would have been noticed already. 

I’m going with what was known about light prior to Einstein postulating c being invariant could be applied to kinematics. People already knew about the effects of the finite time it takes for information to travel.

 

Quote

But in reality A makes an observation: a clock is observed ticking at unusual rate. Then he must figure the reason why.

If A is ignorant of the signal delay, and does not account for it.

 

Quote

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.

Yes it is. Janus has made dozens of post on the topic, explaining the difference between what an observer “sees“ and what they measure the clock doing

such as

“It is important here not to confuse what someone visually sees happening to a clock approaching or receding with how fast that clock is actually ticking according to that person”

https://www.scienceforums.net/topic/105185-time-dilation-dependence-on-direction/?tab=comments#comment-986159

Put another way, it is taken into consideration because the people involved understand and account for propagation delay. It’s not explicitly pointed out because it’s assumed that the reader already knows about the phenomenon 

 

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2 hours ago, michel123456 said:
3 hours 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) IIRC this effect is not taken into consideration in the twins paradox.

Of course it is! I think if you google a little, you will find descriptions what A and B observe. That what they see is a combination of time dilation and delay because of the increasing distance. You are just saying something without being informed. However, only the time dilation is really responsible for the different age of the twins.

57 minutes ago, swansont said:

And when the object is receding faster than SOL (I wrote: receding) what is the clock showing I don't know.

No, of course not. You argument is based on non-physical assumptions. Assuming SOL is not invariant, without giving a premise what is the case then, you cannot know. 

I see 2 possibilities:

  • light behaves as material objects, i.e. its velocity depends on the movement of the source only. If B moves away faster than SOL, you see nothing anymore because the light is also receding from you
  • or there is a fixed medium, like sound in the air, in which case you continuously see B's clock ticking slower than A's clock. Now the frequency of the ticks depends on the velocities of A and B in relation to the medium.

However, nothing of this gives you the time dilation of SR, which you should see now: if B moves to A, in 'your universe', you get a time contraction.

Edited by Eise

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