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

[Eise]

3 minutes ago, michel123456 said:

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

You are mixing up two things;

- in Earth's and X's FOR the distance between them simply does not change because some spaceship happens to travel from Earth to X.

- However, T is moving in this FOR, and so its spaceship and everything on board is length contracted.

Apply md65536's criteria: is it moving?

• no --> no length contraction (the distance between Earth and X is not moving!)
• yes --> that what is moving is length contracted (and time dilated)

9 minutes ago, michel123456 said:

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

I would say: for the traveler the distance between Earth and X is length contracted.

Just as the shadow of an object can be shorter as the object itself: it really is shorter. (a shadow is a projection of the object on the horizontal plane). Maybe (maybe...) it helps to see it that way. Two different inertial frames have different projections of events (and distances between events) on their respective time and space axes. But the observers agree on 'the real length' between two events , i.e. the distance as defined in Minkowskian spacetime.

[michel123456]

44 minutes ago, Eise said:

You are mixing up two things;

- in Earth's and X's FOR the distance between them simply does not change because some spaceship happens to travel from Earth to X.

- However, T is moving in this FOR, and so its spaceship and everything on board is length contracted.

Apply md65536's criteria: is it moving?

• no --> no length contraction (the distance between Earth and X is not moving!)
• yes --> that what is moving is length contracted (and time dilated)

I would say: for the traveler the distance between Earth and X is length contracted.

Just as the shadow of an object can be shorter as the object itself: it really is shorter. (a shadow is a projection of the object on the horizontal plane). Maybe (maybe...) it helps to see it that way. Two different inertial frames have different projections of events (and distances between events) on their respective time and space axes. But the observers agree on 'the real length' between two events , i.e. the distance as defined in Minkowskian spacetime.

I have no much problem with all of that.

You wrote:
 

Quote

 

Apply md65536's criteria: is it moving? (...)

yes --> that what is moving is length contracted (and time dilated)

 

So what about what I wrote:

 

2 hours ago, michel123456 said:

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.

 

In the twins paradox, that is exactly what happens: it is argued that, as seen from Earth, the twin traveled 1LH in 45 minutes. So the traveling twin is younger than his brother. There is a mistake somewhere.

[joigus]

1 hour ago, Eise said:

Just as the shadow of an object can be shorter as the object itself: it really is shorter. (a shadow is a projection of the object on the horizontal plane).

This is precisely the point I was trying to stress when I talked about introducing a very long object through a narrow opening. One foreshortens the object by rotating it with respect to the opening's diameter. If one doesn't, one can spoil both pieces. How real is that?

Another example from phenomenology: When high-energy particles approach a nucleus, you can picture the nucleus as a cluster of more or less static objects from the POV of the approaching particle --time dilation in action.

Also, particles with extremely high momentum appear to see the world to be two-dimensional (only 2 degrees of freedom), the so-called infinite-momentum approximation. Something that is perfectly explained by one of their dimensions to have become so hiper-dilated that efectively it has disappeared from the dynamics.

Edited October 2, 2020 by joigus

[Eise]

1 hour ago, michel123456 said:

In the twins paradox, that is exactly what happens: it is argued that, as seen from Earth, the twin traveled 1LH in 45 minutes. So the traveling twin is younger than his brother. There is a mistake somewhere.

Sorry, you are still mixing FORs:

• According to Earth's clock, the trip of T to X takes 1Lh/0.8c = 75 Minutes.
• From earth one sees the moment of arriving of T at X and looks at its clock, it only shows 45 minutes. (Of course one sees this only 1 hour after arrival of T at X, but that is just signal delay.)

 

[michel123456]

59 minutes ago, Eise said:

From earth one sees the moment of arriving of T at X and looks at its clock, it only shows 45 minutes.

That is exactly what I say: from Earth we are taking count of time dilation (looking at the clock) without taking count of length contraction (that shows nowhere). It is a mistake.

 

As seen from the Earth, we should look at both length contraction & time dilation, in such a way that when the clock ticks 45 minutes, the distance traveled is 0,6 LH.

Edited October 2, 2020 by michel123456

[Eise]

48 minutes ago, michel123456 said:

That is exactly what I say: from Earth we are taking count of time dilation (looking at the clock) without taking count of length contraction (that shows nowhere). It is a mistake.

 

As seen from the Earth, we should look at both length contraction & time dilation, in such a way that when the clock ticks 45 minutes, the distance traveled is 0,6 LH.

No, no and no again.

• From FOR of the Earth there is no length contraction, but only time dilation on the clock of T
• From FOR of T, there is only length contraction of the distance Earth -> X, but no time dilation on his own clock

The end result is the same: Earth and T agree that at the moment of arrival of T at X , the clock of T shows  45 minutes triptime.

Trip time calculation of T: distance x gamma /speed = 60Lm x 0.6 / 0.8 = 45 minutes

Trip time seen by Earth according Earth's clock: 60 Lm / 0.8 = 75 minutes.

But during the trip, Earth sees the clock running slow according the same factor of 0.6:

So, trip time as  Earth can calculate for T: distance (60LM / 0.8) x 0.6 = 45 Minutes. And this is also what Earth sees on T's clock when it arrives at X.

 

[michel123456]

1 hour ago, Eise said:

• From FOR of the Earth there is no length contraction, but only time dilation on the clock of T
• From FOR of T, there is only length contraction of the distance Earth -> X, but no time dilation on his own clock

Why such a difference? Earth & T should be reversible. Why is there length contraction on the other & no time dilation? Who made the choice?

Edited October 2, 2020 by michel123456

[Eise]

13 minutes ago, michel123456 said:

Why such a difference? Earth & T should be reversible. Why is there length contraction on the other & no time dilation? Who made the choice?

Because it is T that is traveling with 0.8c in Earth's and X's FOR. It is traveling the contracted distance between Earth and X.

There is also length contraction of T seen from Earth, but only for what is moving, i.e. T's rocket. It is shorter from the FOR of Earth and X. But Earth and X are not traveling from the exhaust and the frontside of the rocket in your scenario.

[michel123456]

44 minutes ago, Eise said:

Because it is T that is traveling with 0.8c in Earth's and X's FOR. It is traveling the contracted distance between Earth and X.

There is also length contraction of T seen from Earth, but only for what is moving, i.e. T's rocket. It is shorter from the FOR of Earth and X. But Earth and X are not traveling from the exhaust and the frontside of the rocket in your scenario.

In T' s FOR, the Earth is traveling. And Planet X is traveling.

[swansont]

8 hours 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

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

Yes, because you have numbers from two different frames of reference in that last bit.

IOW, there is no frame of reference where the distance is 1 LH and the described trip takes 45 min.

Mixing frames of reference leads to invalid and incorrect conclusions 

7 hours ago, michel123456 said:

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

Yes, they will. But in the example being discussed this is moot. The clock itself is contracted, but this has no impact on anything we’re calculating or measuring, which are the length of the trip and how long it takes.

Quote

Or to say it otherwise:

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

But that’s not saying it otherwise. It’s a different situation.

Quote

You wrote:

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

The traveling clock observed the trip’s length to be contracted.

Your habit of not making explicit references is a problem.

[michel123456]

34 minutes ago, swansont said:

IOW, there is no frame of reference where the distance is 1 LH and the described trip takes 45 min.

Yes exactly.

[swansont]

1 minute ago, michel123456 said:

Yes exactly.

And yet you’re the only on pairing those two values.

[michel123456]

44 minutes ago, swansont said:

The traveling clock observed the trip’s length to be contracted.

The clock is observing itself at rest. The "length of the trip" is the distance between 2 moving objects (the Earth & planet X)

 

7 minutes ago, swansont said:

And yet you’re the only on pairing those two values.

????????????? I am the one? who says that the clock traveled 1LH in 45 minutes?

I said that in Earths frame, the clock traveled 45 minutes until 0,6 LH, that it has to travel another 30 minutes to reach planet X , total trip 75 minutes.

I say that you cannot apply time dilation alone, or length contraction alone, you must use both, otherwise you don't get the 0,8c.

Edited October 2, 2020 by michel123456

[swansont]

5 hours ago, michel123456 said:

 ????????????? I am the one? who says that the clock traveled 1LH in 45 minutes?

Nobody is claiming the clock traveled 1LH in 45 minutes. You said “When you combine 1LH of distance and 45 min of travel, it is wrong.” and I agree: a wrong claim that nobody has made, is wrong. The only person who brought this up is you, with the implication that this argument exists somewhere. 

 

5 hours ago, michel123456 said:

I say that you cannot apply time dilation alone, or length contraction alone, you must use both, otherwise you don't get the 0,8c.

It’s fine that you say this. It’s true. It’s what everybody has explained. What’s not fine is the implication that anyone has suggested that it’s not the case. What’s the point of deny something that nobody is claiming?

 

 

 

[md65536]

1 hour ago, swansont said:

It’s what everybody has explained. What’s not fine is the implication that anyone has suggested that it’s not the case. What’s the point of deny something that nobody is claiming?

It's a strawman argument, and his entire argument.

"These numbers don't make sense; there is something wrong with relativity."

> Those aren't the numbers that special relativity predicts.

"But they have to be! It's the only way that relativity can be right!"

[michel123456]

20 hours ago, swansont said:

Nobody is claiming the clock traveled 1LH in 45 minutes. You said “When you combine 1LH of distance and 45 min of travel, it is wrong.” and I agree: a wrong claim that nobody has made, is wrong. The only person who brought this up is you, with the implication that this argument exists somewhere. 

 

It’s fine that you say this. It’s true. It’s what everybody has explained. What’s not fine is the implication that anyone has suggested that it’s not the case. What’s the point of deny something that nobody is claiming?

 

 

 

If you read the example from Janus, you will see that time dilation is included while length contraction is forgotten: here below in bold where it is argued that the clock reads 45 min. when arriving.
 

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

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

 

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

 

Let's see the question from the FOR of the traveling clock: the clock is constantly at rest, the Earth & planet X are moving.

At 12:00, the clock T is at rest. For some mysterious reason the Earth goes away at velocity 0,8c. At the same instant, at 12:00 on planet X (that is synchronized with Earth), planet X starts toward T, also at velocity 0,8c.

Regarding the Earth: clock T sees Earth going away immediately, there is motion, and thus there is time dilation & length contraction.

BUT regarding planet X, there is a delay: clock T continues to see planet X at rest, at distance 1LH. There is no motion yet, there is no length contraction yet, there is no time dilation yet. Motion of planet X will appear after the delay has passed, that is to say after 1 hour.

IOW after 1 hour Earth has left, clock T will see planet X start its travel.

If you agree with the above then you may also re-estimate the 45 minutes mentioned by Janus: 45 minutes is less than 1 hour, so that would mean that planet X would have arrived at T before the signal transmitted by light. This is impossible: the 45 min are wrong.

In fact, after waiting 1 hour, clock T will see planet X rush and crash on clock T after 15 minutes. The dial on clock T will show 1h & 15 minutes for the trip, or 75 minutes.

[Janus]

On 10/2/2020 at 9:09 AM, michel123456 said:

The clock is observing itself at rest. The "length of the trip" is the distance between 2 moving objects (the Earth & planet X)

 

????????????? I am the one? who says that the clock traveled 1LH in 45 minutes?

I said that in Earths frame, the clock traveled 45 minutes until 0,6 LH, that it has to travel another 30 minutes to reach planet X , total trip 75 minutes.

I say that you cannot apply time dilation alone, or length contraction alone, you must use both, otherwise you don't get the 0,8c.

You have to apply them properly, and you need to allow for the relativity of simultaneity while also taking into account what frame you are measuring from.

This is how events unfold according to Earth and Planet x.

B and it clock are length contracted and B's clock is time dilated. The length contraction doesn't really play a role in the out come here.

B leaves the Earth and take's 1 hr 15 min as measured by Earth's and planet X's clock to cross 1 Lh at 0.8c. B's clock is time dilated and only ticks of 45 min during the trip.

Note that the animation pausing at the end is not meant to mean B stopping at X, only that the animation "freezes" at moment so that we can compare clock readings for that moment.

If we now consider things from B's inertial frame of reference, you get this.

B is at rest while the Earth and planet X move to the left at 0.8c.

the Earth, planet X, their clocks, and the distance between them is length contracted. B is not.

It take 45 min by B's clock for planet X to travel from being 0.6 lh away to B at 0.8c

Earth and planet X's clocks are time dilated and only accumulate 27 min during this time.

However, due to Relativity of Simultaneity, the Planet X clock already reads 48 min later than Earth's clock when B and the Earth are next to each other. Thus the 27 minutes it advances brings it to 1 hr 15 min, as it passes B.  The times on Earth's clock and B's clock when they are next to each other agree with the first animation, and the Times shown on B's clock and Planet X's clock when they are next to each other also agree with the first animation.

[swansont]

45 minutes ago, michel123456 said:

If you read the example from Janus, you will see that time dilation is included while length contraction is forgotten: here below in bold where it is argued that the clock reads 45 min. when arriving.
 

The moving clock does read 45 min when arriving - that’s what the rocket’s frame reads. At 0.8c, the trip was 0.6 LY, meaning both length contraction and time dilation were accounted for.

In the earth frame, its clock ticked off 1:15 for the 1 LY trip. No dilation or contraction.

Just as Janus said.

[Lan Todak]

11 hours ago, Janus said:

the distance between them is length contracted

What are you referring to, for " the distance"?. I mean like if we want to point " the distance" of a runner running on the track, usually we will refer to the track he is running on.

Edited October 4, 2020 by Lan Todak

[michel123456]

13 hours ago, Janus said:

However, due to Relativity of Simultaneity, the Planet X clock already reads 48 min later than Earth's clock when B and the Earth are next to each other. Thus the 27 minutes it advances brings it to 1 hr 15 min, as it passes B.  The times on Earth's clock and B's clock when they are next to each other agree with the first animation, and the Times shown on B's clock and Planet X's clock when they are next to each other also agree with the first animation.

Thank you for the animations, very enlightening.

So let me see if I understand correctly:

If all three are at rest (clock T standing on Earth & planet X at rest) and a missile M is sent from planet X to Earth at 0.8c, how long will it take (as seen from T & Earth) to reach the Earth (and T)?

[swansont]

3 hours ago, michel123456 said:

Thank you for the animations, very enlightening.

So let me see if I understand correctly:

If all three are at rest (clock T standing on Earth & planet X at rest) and a missile M is sent from planet X to Earth at 0.8c, how long will it take (as seen from T & Earth) to reach the Earth (and T)?

1:15

It’s no different than the trip in the other direction.

6 hours ago, Lan Todak said:

What are you referring to, for " the distance"?. I mean like if we want to point " the distance" of a runner running on the track, usually we will refer to the track he is running on.

It’s space. There is no track, and it’s not a mechanical effect. This helps avoid the confusion of people who think that there is a force (i.e. compression) involved

 

[Eise]

16 hours ago, michel123456 said:

Regarding the Earth: clock T sees Earth going away immediately, there is motion, and thus there is time dilation & length contraction.

Without saying what is time dilated and what is length contracted, and for whom this is a very imprecise remark. Nothing follows from it.

16 hours ago, michel123456 said:

BUT regarding planet X, there is a delay: clock T continues to see planet X at rest, at distance 1LH. There is no motion yet, there is no length contraction yet, there is no time dilation yet. Motion of planet X will appear after the delay has passed, that is to say after 1 hour.

No, of course not. T travels in the direction of X, so he will see it coming closer immediately. Again you are mixing up what actually occurs, and what any observer sees. Signal delay has nothing to do with relativity. You only need to take signal delay in account when you want to calculate what the observers actually see.

Let's change the scenario a little: Earth, T and X register their time readings and measured distances in a logbook. After T returns at earth, they can compare their logbooks. Or T can stay at X and send the data from his logbook to Earth, and the other way round. In this way we can leave out effects of time delay. 

3 hours ago, michel123456 said:

If all three are at rest (clock T standing on Earth & planet X at rest) and a missile M is sent from planet X to Earth at 0.8c, how long will it take (as seen from T & Earth) to reach the Earth (and T)?

Sigh... Why don't you let T travel back from X to Earth? And your 'as seen' is again ambiguous. 

Let first try my 'logbook approach'. So the missile has its logbook too. And it shows that the trip took 45 minutes for the missile. The missile also registered the time of X's clock, and Earth's clock when it arrives, and sees that according to these (synchronised) clocks, his trip took 1 hour and 15 minutes, exactly 1Lh/0.8c, what is expected from the Earth's and X's FOR.

When you actually mean 'observes' with 'seen': Earth will only see that the missile started from X after one hour, the time light takes to reach Earth. But then, oh shock, the missile arrives only 15 minutes later. So it looks as if the missile took only 15 minutes. But the earth observer is not so stupid. He knows that he got the signal of the missile's start after 1 hour. So he concludes that the trip took 1 hour and 15 minutes. So it all fits.

17 hours ago, Janus said:

However, due to Relativity of Simultaneity, the Planet X clock already reads 48 min later than Earth's clock when B and the Earth are next to each other. Thus the 27 minutes it advances brings it to 1 hr 15 min, as it passes B. 

This I would have missed. One can see that you have relativity in your blood. With me it is only in the top of my head, and it sometimes costs me an awful lot of thinking. 

[Lan Todak]

55 minutes ago, swansont said:

It’s space. There is no track, and it’s not a mechanical effect. This helps avoid the confusion of people who think that there is a force (i.e. compression) involved

So, do you mean that space can have its length contracted, dont you?

[swansont]

30 minutes ago, Lan Todak said:

So, do you mean that space can have its length contracted, dont you?

Yes. Spacetime is affected by relativistic effects. Space (length contraction) and time (dilation)

[dimreepr]

5 hours ago, michel123456 said:

So let me see if I understand correctly:

No you don't!!!