michel123456

OR, because in all the above there is no mention of Length contraction, things may be more simple and go like this:

When length contraction is applied, Earth see the length traveled by the clock as contracted by a factor of 0.6.

IOW when the traveling clock (that looks contracted) ticks 45 minutes, it looks, from Earth, having traveled 0.6 of the 1LH distance. And not that it has reached the 1 LH in 45 minutes.

In 45 minutes the traveling clock has reached point w. (as seen from earth)

But that have not moved point y. (as seen from Earth)

Thus, if I am correct, the contracted traveling clock needs another 30 minutes to reach Planet X at coordinates Y. (as seen from Earth)

And everything comes in place nicely. No gap needed.

 

This below a long post. It is a genuine attempt to clarify the situation.

I am using diagrams with only Doppler effect first, then Relativity. With calculations.

Those who do not have patience can go straightforward to diag.4 , although IMHO a careful look at the whole thing is necessary to spot the eventual errors.

I have put into diagrams taking the same values inserted by  @Janus from the "Time dilation dependence on direction"thread:
 

Quote

 

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

(...)

 

So we have departure time 12:00, distance 1Light Hour to planet X where the second clock is synchronized with the clock on Earth.

Let's begin with something simple: A flash of light leaves the Earth in all directions at 12:00. It will arrive at Planet X 1 hour later, bounce on a mirror and come back to Earth 1 more hor later, total 2 hours to make the all journey. The flash will arrive back at Earth at 14.00.

.

On this diagram, Time goes from down to up on the Y axis, distance is represented on the X axis.

Now, the situation presented by @Janus but ONLY with DOPPLER SHIFT, No Relativity involved (yet).

On diag.2 above:

At velocity 0.8c, the traveling clock is a little slower than the flash of light from diag.1.

The clock will take 1h & 15 minutes to get to Planet X or 75 min=60 min/0,8. The clock on Planet X will read 13.15. The same reading will count for all observers since no Relativity (yet), thus no time dilation (yet).

From Planet X, the departure will be seen not at 12.00, but 1 hour later (because of the delay), at 13.00.

IOW from Planet X, the reading on the clock will run from 12.00 to 13.15 (arrival time) in only 15 minutes. The clock will look like running faster than the clock on Planet X. How much faster: 75min/15min= 5 times faster. Although no Relativity involved (yet).

As seen from the Earth, the same clock going out will be seen arriving at Planet X at local time (Earth) 14:30. It means that the Earth will see the traveling clock ticking 75 minutes during an interval of 135 min. The traveling clock will be seen as going late by a ratio of : 75min/135min=0.55 late. Although no Relativity involved (yet).

Which means that the Doppler effect on its own gives a similar effect with Relativity: the outgoing clock is seen ticking late, the incoming is seen ticking fast. The values of course are not the same with Relativity.

The 13:07 in the middle of the diagram is the time when the traveling clock will be hit by the flash of light reflected by the mirror. This value will be used afterwards.

Take note that so far all observers agree on all clocks reading: the simultaneity lines are horizontal & the traveling clock is ticking as usual. The "fast" and "late" values are observer dependent. Nothing happen to the clock, time for it does not dilate, the clock does not experience any contraction. It is all in the eye of the observer because of the delay.

Now let's insert the values given by Relativity: quoting @Janus again:
 

Quote

 

The math for what you will see is

 

fofs1−vc1+vc−−−−√

Where f₀ is the observed tick rate.

f_s is the source tick rate

v is the velocity of the clock relative to you(positive if receding and negative if approaching)

c is the speed of light.

 

For the outgoing v/c=0.8, for  the incoming v/c=-0.8

Wich give the following values

0.333 for the outgoing (the traveling clock will look like ticking 0,333 slower than the clock on Earth)

and

3 for the incoming ( the same traveling clock will look like ticking faster as seen planet X on the first leg)

The Earth sees the traveling clock arriving at time stamp 14:15, that is to say after 135min.

135 min at a ticking rate of 1/3=45 min.time of travel as seen by the Earth.

So lets introduce those numbers in the previous diagram

Note that the insertion is somehow problematic: the numbers do not come out from the diagram itself as in diag.2. They have been introduced by force: the simultaneity lines of the background (horizontal between Earth & Planet X do not correspond to the values of the traveler.

Here below the simultaneity lines for the traveling clock:

The simultaneity lines are not horizontal anymore. They go in diagonals up & down accordingly to the direction of travel. They are parallel to each other for each part of the trip.

The diagram is symmetric.

There is a gap appearing on the left side (the blue vertical line from 12:45 to 13:45)

There is a discontinuity at point Y at time 13:15 on planet X where there are 2 different readings on the same clock (12:45 & 13:45) during the U-turn.

Note that in the regular explanation of the travel, there is no discontinuity at point Y, it is assumed that the traveling clock continues ticking as usual during the U-turn, which cause the time gap to take place at arrival (the wins do not have the same age). But when the gap is introduced in the center of diagram, the discontinuity is inevitable & the twins meet with the same age. This gap is not my invention.

The same gap appears here: https://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_vase.html#gap

How to make an interpretation of this gap?

It may be understood as this:

from https://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_gap.html

Quote

The apparent "gap" is just an accounting error, caused by switching from one frame to another.

 

or as this:

from https://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_gr.html

Quote

As an added bonus, the Equivalence Principle analysis makes short work of Time Gap and Distance Dependence Objections.  The Time Gap Objection invites us to consider the limit of an instantaneous turnaround.  But in that limit, the pseudo gravitational field becomes infinitely strong, and so does the time dilation.  So Terence ages years in an instant—physically unrealistic, but so is instantaneous turnaround.

Note: Terence is the guy on Earth.

I needed more, so I made a diagram of what the traveller experiences. The traveller is in a rest frame & the Earth is moving. Planet X moves together synchronized with the Earth.

The traveller is at rest on the vertical line. No Relativity involved (yet). Now, if you go back to diag.2, you will remeber the 13.07 value. We find it back here, with the ray of light bouncing on the traveling Planet X & hitting the traveler path at 13:07. That looks nice. However what is not nice at all is the path of light from Earth, the path 12:00, 13.00, 14.00 that shows on diag.1.

On this diag. 5, the path does not make much sense, as if light had different velocities.

And with Relativity

 

The values extracted from Relativity are shown. The gap appears again from 12.45 to 13.45 (as seen by the traveller). And the discontinuity appears when the Earth makes the U-turn (!).

Which I am thinking shows that the gap & the discontinuity are observer dependent. Nothing really happen on Earth, and nothing really happen on the clock.

M.

 

 

 

Thank you Empress of Everything @hypervalent_iodine. Better than expected!

Dear Staff

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On 9/25/2020 at 8:11 PM, Bufofrog said:

(...)

 

15 hours ago, joigus said:

(...)

 

14 hours ago, The victorious truther said:

(...)

 

5 hours ago, md65536 said:

(...)

 

 

17 hours ago, Ghideon said:

(...)

I have put the (..) in order to make this post shorter. I have read all your comments. I will proceed in 2 steps with diagrams as suggested by @joigus & @The victorious truther.

Let's go back to something simpler: a ray of light from the Earth to planet X & back, a mirror is placed on planet X

 

Oops the import image is disabled??? 

Ihave used all of the attachment space  allowed, I will continue this post after the issue will be resolved.

11 hours ago, swansont said:

A and B will think the other’s clock is running slow.

Thank you.

I'll keep this answer in the drawer.

------------------------------------

Now take the return trip:

A goes back to Earth.(the switch between A & B was made by another member, and i followed)

When he arrives and look back at planet X, what does he observe?

Hint, Planet X & A are both inertial.

What kind of pipe? Metallic? What diameter (inside & outside)? Where does this pipe comes from (the roof?) & where does it go?

52 minutes ago, Eise said:

No. You should know if you had read my posting: the distance to X is not the same for A and B. For A the distance to X is contracted. 

And now stop with this 'observing'. The changing signal delay only makes the example more complicated. Just assume that A and B are intelligent enough to account for the signal delay, OK? 

But A is also at rest. He sees (observes, measures) the Earth length contracted & the distance to the Earth is also contracted. A & B must be considered on the same ground. The situation is symmetric, reversible, how to say. Motion is relative: there is no Earth at rest & A travelling.

For A, the Earth is travelling.

26 minutes ago, Markus Hanke said:

The travelling twin, upon his return, will find that this brother who stayed behind is older than himself. This is as expected, since the travelling twin is not fully inertial, so these frames are not globally related by a Lorentz transformation - there is no symmetry between them. Note that both of them agree on who is older and who is younger.

Not the traveling twin upon his return, no no.

I am asking about the traveling twin when he passes by planet X. This traveling twin (A) is observing twin at rest (B) as aging less than him. Because he uses the laws of Relativity exactly in the same way. So B observes A aging less, and A observes B aging less. They are observing THE SAME THING.

27 minutes ago, Markus Hanke said:

Because both twins agree that the world line of the travelling twin is shorter than that of the stationary twin (one reality!). A purely inertial frame always represents the longest possible world line between two given events - since the travelling twin is not purely inertial, his world line will be shorter, so he ages less in comparison.

Thus the traveling twin has seen his brother time compressed.

4 hours ago, Eise said:

As others also notice: you are consistently avoiding to explain from your view why muons make it to the surface of the earth. It is time you take the challenge.

Another remark (which might help you with the 'muon-challenge'): you assert that in the so called 'twin paradox', there is an asymmetry, because the effect of the time dilation stays (the traveler has not grown older so much as the home-stayer), but the length contraction has gone (the twins are still equally sized). Truth is that you comparison is wrong. After arriving back home the twin's clocks tick at the same rate, so the time dilation itself is gone, just as the length contraction. However this is not true during the travelling: the length contraction of the length of the trip for the traveler is real, and its 'mirror' for the home-stayer is the real time dilation. As space and time are relative, but spacetime is not, this is no problem at all. Important is that the twins agree on their observations when they are in the same FOR again: they agree that the traveling twin has not aged so much as the home-stayer twin. So they live in the same reality.

But What the trveler twin is observing? he is observing that the guy at rest is older. How is that possible?

1 hour ago, joigus said:

You don't strike me as someone who agrees too quickly on anything.

You got me.

1 hour ago, joigus said:

Michel spending another 20 years doubting relativity.

1.I hope I have that much.

2. Again, Relativity is OK to me as long it is considered as a Theory that describes (between others) a phenomena comparable to perspective: a kind of effect caused by the different states of motion of different observers, a Theory that explains how one observer can relate his observation to another observer. Taking the example of length contraction, it seems evident (to all of us I hope) that once the traveler stops, the resting observer does not observe length contraction anymore,  the phenomena has vanished. In fact, in his own FOR, the traveler was never contracted at all: length contraction is an observational effect that appears from the FOR of reference (at rest). Exactly as the forearm presented in perspective: it does not change length. Is this doubting of Relativity?

33 minutes ago, md65536 said:

It is, see image (b). Image (c) is what the dice look like when you receive (at the same time) light that left the dice at different times and traveled different distances.

That is highly confusing, I assumed the dices were equidistant. Thanks for the explanation.

17 hours ago, joigus said:

Time for the free falling trajectory is actually a maximum, not a minimum.

Unintuitive? Perhaps, but that's the way it is.

I was still thinking about this statement, when I agree too quickly it means I haven't think enough.☺️ The frequency equation gives results clocks ticking slower but also faster than normal. In Janus example, with 0.8c, the clock that goes away ticks at 0.3 rate, and the returning clock ticks at rate 3. As seen from the observer at rest, this clock ages faster. So, why do you say that time for the free falling trajectory is a maximum?

As I wrote, for each answer I get, more questions raise:

( my error corrected)

Quest1. Why is the distance d2 larger than the distance d1? Shouldn't it be contracted?

Quest2. Why is the red distance smaller than the orange one? I am counting 4 intervals for the red one (4 squares) and 8 or more for the orange one. If the cube is moving at constant velocity, the 2 lines should be equal.

And I have to admit that I had in mind picture B. I was wrong again as it seems. But I still don't know why: in this picture B everything looks fine.

 

 

8 hours ago, joigus said:

Time for the free falling trajectory is actually a maximum, not a minimum.

Unintuitive? Perhaps, but that's the way it is.

Good point. That's a good input for the "what is time" thread.

 

3 hours ago, Janus said:

So let's go back to my last example:

A and B both see light coming from planet X as they pass each other.  They see the same light.  The difference is that due to aberration, A sees the light come from a smaller angle, and visually "sees" a smaller image.  Now this is not inconsistent with what I said about what he concludes, which is that the light he is now seeing left planet X when it was 3 light hrs from him, thus he would expect to "see" a smaller image.

That is the point, why is it not inconsistent? As you say, the image of planet X  at 3 LH that the traveler A observes when bypassing B is different than the image of planet X at 1 LH as seen by resting observer B. It is not a small difference. 3 times farther means 9 times smaller.

6 hours ago, joigus said:

About 'twins' trips:

Exactly the same as you can go from one corner of a square 9 m² room (of 3 m-wide walls) to the contiguous one along the wall, measuring a distance of 3 meters, you may decide go to the contiguous corner the long way, by sequentially covering the other two corners, and cover 9 m (3+3+3) around the other direction. That's exactly the same that happens to the non-inertial twin that went around to the same place on a non-inertial trip. The first twin has gone along a flat wall and has noticed nothing significant. The non-inertial twin has had to cross two corners and has noticed curvature.

Nicely presented. Except that intuition would say that the guy who covered the long path through the corners would be older than the one going straight away, while the result of the twin paradox is the contrary.

@Janus could you please continue your example & describe what traveler A observes when reaching planet X? Thank you.

6 hours ago, The victorious truther said:

 

On 9/18/2020 at 10:44 AM, michel123456 said:

Not to say that the laws of optics should be derived from Relativity, since it is a theory that deals with what is being observed.

This is an issue I see all the time in people trying to grasp relativity. In the theory (with minimal ontological assumptions of Minkowski spacetime) while you have this effect of length contraction that is rather mathematically explicit even by theory or ontology what you would actually observe is something like. . . 

In fact I'm pretty sure there is even a further different classical perspective of the cube that you would expect which does differ from the visual image seen above that special relativity would, being approximately correct, in the end expect. 

 

That is worth a separate thread. Some Moderator will send me a warning & do the job (I hope) with a link here for those interested.

A comment from Swansont about the same topic some posts above says:

21 hours ago, swansont said:

This is a geometry/perspective issue, not one of relativity. They are distinct effects and have to be treated as such. The perspective issue should be easy to incorporate and separate from analysis, because it's an everyday effect. But it doesn't go away simply because of relativity. 

 

The laws of perspective (that try to represent our visual  everyday experience) say that an object that gets away from you is seen as if its size was diminishing, and an object that gets close to you is seen as getting larger. Of course the object does not change size in "reality", it is simply an effect of optics, but as I read here, it is completely ignored in Relativity. The moving examples posted above by V.T. do not care about this perspective effect.

 

From Wiki

Quote

Linear perspective is an approximate representation, generally on a flat surface, of an image as it is seen by the eye. The most characteristic features of linear perspective are that objects appear smaller as their distance from the observer increases, and that they are subject to foreshortening, meaning that an object's dimensions along the line of sight appear shorter than its dimensions across the line of sight. All objects will recede to points in the distance, usually along the horizon line,

What is foreshortening:

https://drawpaintacademy.com/foreshortening/

 

The analogy with length contraction is pure coincidence, I guess.

Or not. Are the laws of physics a single thing that enties everything ( the way we see things on a daily basis), or are the laws of perspective totally independent of Relativity? (as it seems to be the case at first sight).

And respectively, can Relativity ignore the fact that objects appear smaller as their distance from the observer increase? And not taking count of this effect when representing the distortion of objects that move at near to c velocity?

1 hour ago, swansont said:

B would see X as length-contracted (not that he can actually see through the planet to notice this), and also see planet X (that is 1 LH away according to A) as being only 0.6 LH away. According to B's clock, the trip will take 45 minutes. According to A's clock it will take an hour and fifteen minutes. 

All of this has been explained, multiple times, starting with Janus in the fourth post of the thread. It's not going to change.

 

 

So you seem to agree that A & B would see different things although getting the same light. If that does not hurt your feelings at the departure, why is it so mind blowing at the arrival at planet X?

3 hours ago, Eise said:

Oh my... Our traveler sees the same light arriving from earth as the man in the bar. So they see the same.

Let's take it from the start.

At time zero A & B are at rest looking at planet X that is 1LH away.

Out of magic, B steps instantly into a FOR that travels at 0.8c.

What does B sees? Doesn't he sees planet X length contracted? And closer to him? Instantly? While observer A sees it normal as usual?

Or am I wrong there too?

5 minutes ago, swansont said:

The earth doesn't "know" where an observer is. The lights leaving the earth doesn't care about any potential observers. It will spread out, decreasing in intensity with approximately a 1/r^2 dependence. The earth will look smaller to the observer as they move away. The thing that the delay will affect is if there is an event. If there is an explosion, the observer will not notice until the light gets to them. Events happen at particular times. But the earth just sitting there sends out the same signal continuously. Only time-tagged events will be affected by the delay.

 

I meant the Earth will look larger, because closer. And length contracted.

1 hour ago, the traveler was closer to the Earth, and the Earth was closer to the traveler.

1 minute ago, Eise said:

As long he is moving away from earth, yes, he sees earth clocks go slow, but as soon as he stops at the bar, his clock and earth clocks go in the same pace. But, yes, the earth clocks seem to be one hour behind. That is the delay due to the distance.

And 1 hour behind, the Earth was closer to him. So he is observing the earth larger (because of regular perspective). And because of length contraction, he is observing the Earth flattened.

3 minutes ago, swansont said:

There is a delay, which they both see. The light from the earth takes an hour to get to the traveler. But the light is continuously sent. It's not sent every hour, like the clock signals everyone has discussed. But you identified two delays - the light travel time, and the delay before the twin sees the earth getting closer, after turnaround. The first exists. The second does not.

If you go 1 LH away and stop, you will get light that was sent from earth an hour earlier. If you turn around and go back, when you've moved 1 LS, you will get light that left the earth 59 minutes and 59 seconds ago.The earth will take up an incrementally larger solid angle on your screen (assuming sufficient resolution) because you are closer to the source. There is no delay in noticing this; that light was already en route.

Yes, but (quoting myself)

44 minutes ago, michel123456 said:

But if the traveler looks behind him just before stopping, he will see the earth as it was in the past 1 hour ago. At this time (1hour ago) the Earth was closer to him. Because he is in a state of motion he sees things differently than the observer at rest at the bar.*

 

15 minutes ago, Ghideon said:

Other members are doing a good job describing what actually happens in nature. So I try a different path, describing consequences of your description and see where that may take the discussion. I hope the questions may trigger som thinking regarding your current understanding in relation to the models available in mainstream physics.

I have some trouble with all the paradoxes that follows from the description and why you would believe in such scenarios. Maybe I have not understood your ideas. Let's assume that your description correctly describes what happens.

Questions:

1: If the traveler* and the bar tender** takes simultaneous photos of the earth from the bar, will the images look different according to your understanding? If they switch cameras with each other will that change anything?

2: If the traveler had followed a curved or "s"-formed path earlier, how does that affect what he is supposed to see? While the earth is still getting away from the traveler 1 hour after the traveler has stopped, does the earth also move sideways to reflect the earlier curves os "s"-bends?

 

*) That is, the person that just traveled
**) an individual that was at the destination already, and has been there for an extended amount of time

I was wrong. See above.

1 hour ago, Eise said:

The delay is in what any observer in the bar sees happening on earth. So the traveler and the person living there see exactly the same: what happens on earth one hour ago. That is your delay.

But if the traveler looks behind him just before stopping, he will see the earth as it was in the past 1 hour ago. At this time (1hour ago) the Earth was closer to him. Because he is in a state of motion he sees things differently than the observer at rest at the bar.*

1 hour ago, Eise said:

And when do I get your explanation that we can see muons reaching the surface of the earth? What does it look like for an observer on earth, and what does it look like from the FOR of the muon? 

I assume you cannot answer it, if I don't get an answer.

You know that you are right, and I know it. If you read my previous posts more carefully you may understand than I don't want to counter the mathematics of Relativity. I am against some interpretations of relativity. Like the "multiple reality" argument, or like the present discussion.

* And yes I was wrong. I forgot that when the traveler stops he jumps into another FOR and he must see what other people are seeing from this FOR.

9 hours ago, swansont said:

No. A signal will take an hour to get there, but light is continuously sent. The traveler will immediately see the earth get closer, from photons emitted an hour (or slightly less) earlier

 

Disagreeing without scientific justification. If you think there is a delay, show the math.

There must be a delay. If I am wrong, then the delay is somewhere else. Go and find it.

As observed from the Earth, the image of the traveler going away is delayed. So logically speaking, the image of the Earth that as seen by the traveler is delayed too. There is no reason why one observer would see a delay and the other not.