michel123456's relativity thread (from Time dilation dependence on direction)

[joigus]

20 hours ago, md65536 said:

The funny thing is, he's already accepted that a Doppler effect is acceptable in his definition of reality, and it's an easy modification of a twin paradox setup to make neither twin inertial and make it truly symmetric. Then just say "that's what they both see."

It's also funny because for me, seeing the asymmetry in the Doppler analysis of the twin paradox is probably what fully sold me on the predictions of SR, and I never doubted the resolution of the paradox after that, even though I still would have struggled with "the Earth's clock jumps forward with the traveler's change in inertial frame."

Then if you cherry pick some predictions of SR, you can get something that fits Michel's reality and doesn't add up (which is not a problem for Michel). For example, if you let the traveling twin have a lightyear-long ruler attached behind it, and you make it so the entire ruler stops simultaneously in Earth's frame, then you can see something like this: Say v=.6c, from Earth the receding ruler appears compressed by the Doppler factor of 1/2. Then when the 1 LY mark on the ruler reaches Earth, that part of the ruler stops, but the traveler appears to keep moving until it reaches 1 LY rest distance. All along the ruler, a "wave" of successive lengths of the ruler being seen coming to stop and returning to normal length spreads down the ruler, the wave moving at an "apparent rate" of c, so that it takes 1 year to see the traveler and the end of the ruler coming to a stop 1 LY away. That's something SR predicts and sounds similar to what Michel has described the traveler seeing (instead of SR's prediction of the traveler seeing Earth's entire ruler appearing normal instantly, when the traveler---not Earth's ruler---stops and comes to rest in Earth's frame). I'm not positive I got the details right.

Course, you'd have to sell your soul to argue that the predictions of SR describe reality and show that SR is wrong.

I think your argument is quite correct. If you pressed the theory too much, you would need a working approximation of special relativity for the case when instantaneous forces act on bodies, either to stop them suddenly or to set them in motion instantly. In the case that a rigid ruler stops, by an instantaneous impulsive force applied when its tip reaches Earth, applied at that point, a wave would have to propagate carrying both the momentum of the impulsive "braking" and a Lorentz contraction factor, if I'm thinking correctly. In fact, you have pointed to a well-known logical consequence of special relativity: Namely: That no perfectly-rigid bodies can exist, for the simple reason that they would contradict relativistic causality. The reason being that one tip of the ruler would have to stop instantly and it would take some time for the local perturbation to reach the other end. This is in close analogy to the Doppler effect, in which a kinematic factor (length contraction/time dilation) travels along with a propagation effect by the receding speed that would be present even if Galilean relativity were exact.

That's why in the last analysis you need waves (fields) to implement special relativity in an absolutely watertight manner. Rigid bodies can't play the part of physical objects. Point-like particles can do the job, but at the price of introducing densities, which are far removed from direct intuition and have nontrivial transformation properties.

This is an extremely clever point. But I don't think @michel123456's arguments have to do with it.

Edited September 27, 2020 by joigus

[md65536]

3 hours ago, joigus said:

Namely: That no perfectly-rigid bodies can exist, for the simple reason that they would contradict relativistic causality. The reason being that one tip of the ruler would have to stop instantly and it would take some time for the local perturbation to reach the other end.

True... if Earth alone stopped the ruler and the ruler approached the limit of perfect rigidity, it would take at least 2 years for Earth to see part of the ruler come to rest 1 LY away, twice as long as if the ruler stopped all at once. But that's why I specified that the entire ruler was "made to stop" simultaneously in a given frame, rather than that it just stopped. I avoid worrying about practical details because SR isn't limited to what's practical. But it would mean this is a poor example to use as a thought experiment to explain what can be expected in real experiments. Not that anyone's argued that exactly!

Practically, you could stop a very long ruler all at once (in a given frame), eg. with a very long synchronized brake.

 

But, yikes... what happens if the traveler's 1 LY-long ruler is stopped by the traveler, and the rest of it stopped like a wave propagating across it at a speed of c? Before the traveler moving at .6c stops, it sees the ruler behind it at rest, 1 LY long, and Earth is .8 LY away (1/gamma) just before stopping, but appears .5 LY away (Doppler, receding). Then just after the traveler stops, but before the moving ruler stops, the ruler appears 2 LY long (Doppler, approaching). But it's length-contracted to .8 LY which means the far end has already passed the Earth (not yet seen) and really is still traveling at .6c... The closing speed of the far end of the ruler and the "stop" wave is 1.6c which means half a year for it to stop, it stops .2+.3 LY = halfway between Earth and the stopped traveler, as measured in the Earth's frame. Which means that stopping an object at its front would physically compress it by the Doppler factor (or more)! This agrees with what Earth would see: If the moving ruler stopped as if by a light signal traveling from its far end, it would appear to stop all at once, and it would appear .5 LY long (Doppler, receding) before that happened.

This seems like an even worse "impractical example", sorry! Is it pointless to even consider such a thing? (an object in which the speed of sound equals c)

Back to this thread........... the above example would be what a traveler might see if a ruler accelerated along with the traveler. However, using long rulers as representing lengths in different inertial frames, the rulers would remain inertial; they'd remain in their respective inertial frames, and no part of any ruler would be seen moving at a different speed than the rest of the ruler. A moving ruler would have different parts appear distorted differently (compressed moving away from you, stretched moving toward you). Using these inertial rulers, it is obviously (with hand waving) paradoxical to measure an inertial Earth appearing to move closer to you after you've come to a rest relative to it.

Edited September 28, 2020 by md65536

[Halc]

9 minutes ago, md65536 said:

Practically, you could stop a very long ruler all at once (in a given frame), eg. with a very long synchronized brake.

It cannot be done 'all at once' since different parts of the ruler need different levels of proper acceleration.

I did a topic once positing the minimum time it would take a 100 light year Born-rigid object to move one light hour in frame X, beginning and ending at rest in that frame.  We had to carefully define independent acceleration all along its length to keep it always stationary in its own frame.  It can be done, but it is necessarily not all moving at the same speed in any other frame, which doesn't contradict born rigidity. The only rule was that at no point could there be either stress or strain.  Keeping it stationary in its own frame wasn't a requirement, but we couldn't find a quicker solution.

It takes over 55 days to move the object as measured in any frame since the speed never got very high. The math is pretty trivial until you investigate alternate methods, but none of them yielded a shorter duration.

[md65536]

38 minutes ago, Halc said:

It cannot be done 'all at once' since different parts of the ruler need different levels of proper acceleration.

[...] The only rule was that at no point could there be either stress or strain.

You're right, though it could but not really but actually it could, but not really. "All parts accelerating equally and simultaneously" in a single frame basically describes Bell's paradox, but kind of the inverse because you're slowing it instead of speeding it. Definitely you can't do it as a rigid object. If you stop a ruler all at once, you change its rest length. Its former length-contracted length becomes its new proper length, it must be compressed. To do this would require an opposite rule, that strain or stress doesn't affect the ruler's behavior, so it could be arbitrarily compressed.

I guess a normal ruler would disintegrate if it was forced to stop quickly from relativistic speeds all at once, or if slowed gradually enough, it would have internal forces pushing it to resist being compressed, which would cause different parts of it to have different speeds. A realistic material ruler could survive not being perfectly Born rigid.

Edited September 28, 2020 by md65536

[michel123456]

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.

 

 

 

Edited September 29, 2020 by michel123456

[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.

 

[md65536]

Credit where due, you seem to be making progress. Using conventions like time on the y-axis and light signals shown at 45 degree angles really helps in communicating ideas.

In cases where you start with bad assumptions and then show what happens, the conclusions aren't going to be useful. If you put "garbage in", you get "garbage out." Eg. the Doppler factor of 5 from assuming the clocks tick at the same rate, is not useful. You *can* assume the clocks tick at the same rate, but then you'll find that the speed of light isn't constant etc., which doesn't match reality.

Where you lined up "lines of simultaneity" by connecting moving clocks reading the same time, is not correct. That's not what simultaneity means.

If you want to make further progress, try this: In diag 2, where it shows a light signal from Planet X to Earth when the traveler turns around, draw a light signal from Earth to Planet X at that same time, and see where (and when) it reaches the traveler. You'll see that Earth and the traveler can't see the same thing, and there must be asymmetry. That diagram would show what each observer actually sees. Any other diagram would have to agree with it. Diag 5 does not show the same thing as diag 2. If diag 5 showed the same scenario but from a different perspective, it would still show that the traveler sees the same things that diag 2 shows the traveler seeing.

4 hours ago, michel123456 said:

Nothing really happen on Earth, and nothing really happen on the clock.

In my opinion, that interpretation is not incompatible with SR. But, however you interpret what is "really" happening, it still has to match the measurements, and the twins "really" do measure a difference in age.

2 hours ago, michel123456 said:

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

 

Earth doesn't see that distance length-contracted, the clock does.

To wrap your head around it, you can ask "is the ruler that I'm using to measure this distance, moving?" If yes, length contraction will apply, otherwise it won't. For example if you measure the distance between Earth and Planet X is 1 LY according to Earth, that ruler is not moving. If something travels from Earth to X, the ruler doesn't move. Earth measures distances along that ruler without length contraction.

Meanwhile if a traveler going from Earth to X would see that same ruler connecting Earth and X moving along with them; the distance between Earth and X is length-contracted.

[Eise]

15 hours ago, michel123456 said:

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

Of course not, as md65536 explained. The earth sees the travelling clock running slower by a factor of 0.6.

Further your diagrams 5 and 6 do not make sense. Spacetime diagrams should always be drawn from one single inertial frame. But the travelling clock changes its inertial frame.

And the gap is an artifact of the physically not realisable instantaneous change of inertial frame. In reality it would be something like this:

From here.

You do not need to invoke General Relativity (we do not have curved spacetime), SR can handle this.

[michel123456]

7 hours ago, Eise said:

The earth sees the travelling clock running slower by a factor of 0.6.

Yes, OK. The earth sees the travelling clock running slower by a factor of 0.6.

And what about length contraction?

The earth sees the traveling clock ....<insert your comment here>

[Eise]

1 hour ago, michel123456 said:

The earth sees the traveling clock ....<insert your comment here>

... length contracted. But not the length traveled.

[md65536]

2 hours ago, michel123456 said:

Yes, OK. The earth sees the travelling clock running slower by a factor of 0.6.

And what about length contraction?

The earth sees the traveling clock ....<insert your comment here>

All from the Earth's viewpoint:

Is the traveling clock moving?

Is a ruler attached to the traveling clock moving?

Is a ruler attached to the Earth moving?

Do you know which are length contracted?

[Bufofrog]

5 hours ago, Eise said:

... length contracted. But not the length traveled.

According to who?  An Earth bound observer and the traveling astronaut will not agree on the length traveled.  The astronaut will say the length traveled is shorter.

Edit:  I see that you clearly implied the Earths frame, my bad.

Edited September 30, 2020 by Bufofrog

[michel123456]

The lack of comments gives a mixed feeling...

15 hours ago, Eise said:

17 hours ago, michel123456 said:

The earth sees the traveling clock ....<insert your comment here>

... length contracted. But not the length traveled.

I hope that I am misreading your comment and that you don't suffer from the same syndrome with Swansont in this post:

On 9/23/2020 at 5:39 PM, swansont said:

Is the “length” moving? No. So it’s not contracted.

On 9/23/2020 at 6:33 PM, swansont said:

On 9/23/2020 at 5:48 PM, The victorious truther said:

In the description is declares that it's a row of dice moving. Not a single (die) pictured at different times while it was moving if I recall correctly. 

My mistake. The questions are still focusing on irrelevant details, given the fundamental misunderstanding of relativity.

 

The Earth sees the traveling clock length contracted. No matter if there is a rod between the clock & Earth, contraction is observed in all cases. It is a geometric effect, it does not act on material things only, it acts on everything.

14 hours ago, md65536 said:

All from the Earth's viewpoint:

Is the traveling clock moving?

Is a ruler attached to the traveling clock moving?

Is a ruler attached to the Earth moving?

Do you know which are length contracted?

My answers in bold below for clarity (not shouting).

All from the Earth's viewpoint:

Is the traveling clock moving? Yes the clock is moving

Is a ruler attached to the traveling clock moving? No, it does not matter.

Is a ruler attached to the Earth moving? No, it does not matter

Do you know which are length contracted? Yes. From the Earth's viewpoint the traveling clock is contracted, the distance that it travels is contracted, its time is dilated.

Also from the Earth's viewpoint, the destination (planet X at coord y on the diagram) is not moving. So the distance from Earth to planet X is still 1 LY: this distance is not length contracted.

The question is: after 45 minutes (as read by the Earth on the traveling clock), did the traveler reach destination? (as observed by Earth)

My answer is No: after 45 minutes (as read by the Earth on the traveling clock), the traveler has moved 0,6 LY (as seen by the Earth). It misses destination by 30 minutes (as read by the Earth on the traveling clock), and 0,4 LY (as seen by the Earth).

Edited October 1, 2020 by michel123456

[Eise]

1 hour ago, michel123456 said:

I hope that I am misreading your comment and that you don't suffer from the same syndrome with Swansont in this post:

No, you are not, and the syndrome that Swanson, Markus, md65536, Janus, Bufofrog, and I  have, is that we have a more than superficial knowledge of relativity.

Let's assume, as you do, that Earth and planet X are in the same inertial frame. Two spaceships, T1 and T2 (Traveler) make up for planet X, 1 light hour away. T1 travels with 0.8c, and T2 with 0.6c.

What is the distance from Earth to Planet X?

When T1 arrives at Planet X how far did T1 travel from the view of Earth's and planet X's FOR?

When T1 arrives at Planet X how far did T2 travel from the view of Earth's and planet X's FOR?

So how far is planet X from Earth?

 

[michel123456]

5 minutes ago, Eise said:

No, you are not, and the syndrome that Swanson, Markus, md65536, Janus, Bufofrog, and I  have, is that we have a more than superficial knowledge of relativity.

Let's assume, as you do, that Earth and planet X are in the same inertial frame. Two spaceships, T1 and T2 (Traveler) make up for planet X, 1 light hour away. T1 travels with 0.8c, and T2 with 0.6c.

What is the distance from Earth to Planet X?

When T1 arrives at Planet X how far did T1 travel from the view of Earth's and planet X's FOR?

When T1 arrives at Planet X how far did T2 travel from the view of Earth's and planet X's FOR?

So how far is planet X from Earth?

 

In your opening statement you wrote that planet X is 1 light hour away. From Earth.

And from planet X, the Earth is 1 LH away.

So it is 1 LH in all cases.

[Eise]

 

1 hour ago, michel123456 said:

So it is 1 LH in all cases.

On 9/29/2020 at 4:50 PM, michel123456 said:

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

So how much does the clock travel? 1LH or 0.6 x 1LH?

 

[michel123456]

4 minutes ago, Eise said:

 

So how much does the clock travel? 1LH or 0.6 x 1LH?

 

That is the question.

IMHO if the clock reached destination, as seen by Earth it has traveled 1LH. If it traveled 0,6 LH (as seen by Earth) it has not reached destination. Because in Earth's FOR, the destination did not moved. As seen by Earth, at 0,6 LH away, there is no planet X.

[Eise]

1 hour ago, michel123456 said:

IMHO if the clock reached destination, as seen by Earth it has traveled 1LH. If it traveled 0,6 LH (as seen by Earth) it has not reached destination. Because in Earth's FOR, the destination did not moved. As seen by Earth, at 0,6 LH away, there is no planet X.

Right. So why do you say then something like this?

On 9/29/2020 at 4:50 PM, michel123456 said:

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

 

[md65536]

6 hours ago, michel123456 said:

That is the question.

IMHO if the clock reached destination, as seen by Earth it has traveled 1LH. If it traveled 0,6 LH (as seen by Earth) it has not reached destination. Because in Earth's FOR, the destination did not moved. As seen by Earth, at 0,6 LH away, there is no planet X.

What's the point of working through the details of relativity, like length contraction, but not accepting the predictions of SR, and instead replacing them with your own opinions? Does the length contraction as you're describing it, actually make sense to you? Or are you hoping to show that if you replace a few details with nonsense, then all of SR can be clearly seen as nonsense? Have you convinced yourself of that? Do you expect to convince others? Do you actually believe what you're saying, do you really believe that if length contraction as predicted by SR is applied, then Earth sees the traveler having gone only 0,6 LY in the example above? Or is the above example meant to show that length contraction must be nonsense?

If it's the latter... Can I convince you that addition is nonsense, by adding 3 plus 2 together and getting 9? Are you convinced? Do you think I should spend 20 years telling people that addition is nonsense because I think addition, in my opinion, gives a ridiculous answer of 9?

Edited October 1, 2020 by md65536

[michel123456]

5 hours ago, Eise said:

Right. So why do you say then something like this?

 

I should have said : When length contraction is applied, Earth see the length traveled by the clock <in 45 minutes> as contracted by a factor of 0.6.

[swansont]

12 hours ago, michel123456 said:

 

I hope that I am misreading your comment and that you don't suffer from the same syndrome with Swansont in this post:

My “syndrome” was misreading the circumstance of the example (the “length” was, indeed moving) 

The physics was not wrong

12 hours ago, michel123456 said:

 

All from the Earth's viewpoint:

...

Do you know which are length contracted? Yes. From the Earth's viewpoint the traveling clock is contracted, the distance that it travels is contracted, its time is dilated.

The distance to planet X is not contracted. There is no motion between earth and X; they exist independent of a rocket and its motion.

1 hour ago, michel123456 said:

I should have said : When length contraction is applied, Earth see the length traveled by the clock <in 45 minutes> as contracted by a factor of 0.6.

The earth sees 1 LY.  There is no motion between earth and X, thus there is no length contraction of this distance.

[Eise]

11 hours ago, michel123456 said:

I should have said : When length contraction is applied, Earth see the length traveled by the clock <in 45 minutes> as contracted by a factor of 0.6.

Which makes no sense at all. From the FOR of the earth, you just see T traveling with 0.8c. A ruler attached to T will be length contracted with a factor of 0.6, but the distance between Earth and planet X will still be 1 Lh because they are still in the same FOR. So after one hour, T has simply traveled 0.8Lh in the FOR of Earth and X.

[michel123456]

Let's say it differently

With 1LH of distance and 75min of travel you get 0,8c (which was the assumption)

With 0,6LH of distance and 45min of travel you get 0,8c (which was the assumption)

And everything is fine.

BUT

When you combine 1LH of distance and 45 min of travel, it is wrong.

[Eise]

13 minutes ago, michel123456 said:

Let's say it differently

With 1LH of distance and 75min of travel you get 0,8c (which was the assumption)

With 0,6LH of distance and 45min of travel you get 0,8c (which was the assumption)

And everything is fine.

But why don't you say it the simplest way? The speed of T is 0.8c in the FOR of earth and planet X. There is no length contraction of the distance in the FOR of earth and X.

15 minutes ago, michel123456 said:

When you combine 1LH of distance and 45 min of travel, it is wrong

Yes, in the FOR of Earth and X.

What an observer from Earth sees is time dilation: the clock of T seems to run slow by a factor of 0.6. So the Earth's observer, using T's clock, T needs only 0.6 x 60 minutes = 36.66 minutes.

According T however, the distance between Earth and X is length contracted 0.6 x 1Lh, and his own clock runs normally, so he does that distance in ... 0.6 x 1Lh x 60 minutes = 36.66 minutes. 

So both Earth's observer and T see the same: T's clock shows that the trip took 36.66 minutes^*. But according to Earth's own clock it took 1 hour/(0.8c) = 1 hour and 15 minutes.

* That it takes another hour before he actually sees T arrive a X plays no role. He can see T arrive at X, and that his clock shows 36.66 minutes 'trip time'.

[michel123456]

35 minutes ago, Eise said:

What an observer from Earth sees is time dilation:

And no length contraction? An observer never observes a particle flatten?

Or to say it otherwise:

As viewed by the traveling clock, what is the distance between Earth & Planet X? You wrote:

40 minutes ago, Eise said:

According T however, the distance between Earth and X is length contracted 0.6 x 1Lh

So, I understand that the traveling clock is observing length contraction.