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

Recommended Posts

2 minutes ago, michel123456 said:

You could imagine a 3rd "twin", bypassing X in the opposite direction.

But that is not necessary, you can consider the outbound trip only, it is sufficient. If there is a temporal difference between the traveling clock & the synchronized clock on planet X, then the paradox will arise.

I can only imagine your intent is to be banned>

• Replies 363
• Created

Posted Images

19 minutes ago, michel123456 said:

But that is not necessary, you can consider the outbound trip only, it is sufficient. If there is a temporal difference between the traveling clock & the synchronized clock on planet X, then the paradox will arise.

In previous examples on this thread, the concept of "bypassing" instead of "starting" is clearly used to avoid the effects of acceleration.

In such a way that all the components, Earth, planet X and traveling clock B are inertial.

OK, it was you who started to talk about the twin paradox again.

In the 'one-way, B passing by E and X' there is not even a 'seeming paradox'. Given that E and X's clocks are synchronised in their FOR, and B passes very close by E and X, all observers agree that B's clock showed 0:00 when it passed E, and B's clock shows 0:45 when passing X. Also, all observers agree that E's clock stands at 0:00 when B passes close by, and that X's clock stands at 01:15 when B passes by.

E-X and B only disagree about how much B traveled through time and how much through space. But they fully agree on the spacetime-distance of the events.

• Distances in space         do not change if you turn your coordinate system in space.
• Distances in spacetime do not change if you turn your coordinate system in spacetime.

Lorentz transformations are rotations in spacetime.

28 minutes ago, dimreepr said:

I can only imagine your intent is to be banned>

That remark is not necessary. Just because Michel has some blind spot, he should not be banned. If the mods would like to stop the discussion, they could close the thread. But in my opinion, one should be more rude than Michel is to get banned. Just my opinion.

Share on other sites

37 minutes ago, Eise said:

OK, it was you who started to talk about the twin paradox again.

In the 'one-way, B passing by E and X' there is not even a 'seeming paradox'. Given that E and X's clocks are synchronised in their FOR, and B passes very close by E and X, all observers agree that B's clock showed 0:00 when it passed E, and B's clock shows 0:45 when passing X. Also, all observers agree that E's clock stands at 0:00 when B passes close by, and that X's clock stands at 01:15 when B passes by.

E-X and B only disagree about how much B traveled through time and how much through space. But they fully agree on the spacetime-distance of the events.

• Distances in space         do not change if you turn your coordinate system in space.
• Distances in spacetime do not change if you turn your coordinate system in spacetime.

Lorentz transformations are rotations in spacetime.

But do you agree that they are all inertials?

Share on other sites

16 minutes ago, michel123456 said:

But do you agree that they are all inertials?

In the 'B-close-passing-E-and-X' example, yes. However, the FOR of E-X and the FOR of B have different time and space axis, so their respective projections on their space and time axis of the events 'B passes E' and 'B passes X' will differ. But they fully agree on the spacetime distance of these events.

Share on other sites

1 hour ago, michel123456 said:

You could imagine a 3rd "twin", bypassing X in the opposite direction.

Then it wouldn't be the twins paradox.

Quote

But that is not necessary, you can consider the outbound trip only, it is sufficient. If there is a temporal difference between the traveling clock & the synchronized clock on planet X, then the paradox will arise.

Then it's not the twins paradox.

Quote

In previous examples on this thread, the concept of "bypassing" instead of "starting" is clearly used to avoid the effects of acceleration.

In such a way that all the components, Earth, planet X and traveling clock B are inertial.

There are myriad examples one can come up with to discuss relativistic effects, but the twins paradox is a particular example — it has the element of the twin returning home to compare their clock with the one which stayed on earth, and thus has acceleration.

Share on other sites

17 minutes ago, Eise said:

In the 'B-close-passing-E-and-X' example, yes. (...)

In this case, why is it that the situation is not mirrored?

I mean,

_from the FOR of E, you said that B's clock ticks 45 min while reaching the goal (at X)

and

_from the FOR of B, you said that X's clock will show 1h:15 while ticking only 23 min.

Why such a difference?

If all are inertials, the situation should be exactly reversible, isn't it?

Share on other sites

2 hours ago, michel123456 said:

But that is not necessary, you can consider the outbound trip only, it is sufficient. If there is a temporal difference between the traveling clock & the synchronized clock on planet X, then the paradox will arise.

There is no paradox here since on the outbound leg the traveling twin must accelerate to another reference frame.  The traveling twins clock will indicate less time has passed than the clock on planet X, as he flies by.

2 hours ago, michel123456 said:

In previous examples on this thread, the concept of "bypassing" instead of "starting" is clearly used to avoid the effects of acceleration.

In such a way that all the components, Earth, planet X and traveling clock B are inertial.

If the ship and the planet's weren't ever in the same reference frame and just flew by each other, then both would see the other time dilated.  There would be no useful comparing of elapsed time since they are never in the same reference frame.

Share on other sites

26 minutes ago, Bufofrog said:

There is no paradox here since on the outbound leg the traveling twin must accelerate to another reference frame.  The traveling twins clock will indicate less time has passed than the clock on planet X, as he flies by.

Indeed.

Share on other sites

1 hour ago, Bufofrog said:

(...)

If the ship and the planet's weren't ever in the same reference frame and just flew by each other, then both would see the other time dilated.

Show me.

Share on other sites

1 hour ago, michel123456 said:

Show me.

Deleted

Edited by Bufofrog
Clarity
Share on other sites

4 hours ago, michel123456 said:

Show me.

$\Delta t' = \frac{\Delta t} {\sqrt{1 - \frac {v^2}{c^2}}}$

Where $\Delta t$ is the clock rate of the observer (stationary frame).

Share on other sites

4 hours ago, michel123456 said:

Show me.

What are you trying to achieve here? I'd say the question of whether or not you will learn relativity is answered, your refusal to do so is just too strong. I think you've convinced others that you're interested in relativity, even though you've stated that you're not. I don't see what's in it for you, to waste time on this. Do you hope to have your mind changed? Do you hope to change anyone's mind? I'm fairly certain, no one's changing their minds here. This will go on to page 140+. Would you persist, knowing you'll never change the mind of someone who understands relativity?

For others, how long is it worth persisting if the result is what we currently have? (For myself, it's only worth it to write about relativity in this thread if I'm doing it for myself, not to try to inform michel123456.)

When you say "hyperbolic rotation", michel123456 reads "pirouette". You can't force-feed understanding to someone willing to put in the effort to avoid it.

Share on other sites

10 hours ago, Bufofrog said:

Δt=Δt1v2c2

Where Δt  is the clock rate of the observer (stationary frame).

But both are in a stationary frame.

It has been stated here that for the observer "at rest" the time to reach planet X is 1h:15min

For the "moving" observer it has been stated that it is 45min.

Now, if you keep in mind that the situation must be so that you cannot know who is the "moving" and who is the "at rest", you get the confusion I struggle with.

Share on other sites

33 minutes ago, michel123456 said:

Now, if you keep in mind that the situation must be so that you cannot know who is the "moving" and who is the "at rest", you get the confusion I struggle with.

You don't know how to tell if something is moving?

Share on other sites

3 hours ago, michel123456 said:

But both are in a stationary frame.

That is a nearly empty statement. Every inertial observer is in its own stationary frame, per definition. What is important what velocity the 2 observers, and therefore automatically their FOR's have relative to each other.

3 hours ago, michel123456 said:

It has been stated here that for the observer "at rest" the time to reach planet X is 1h:15min

For the "moving" observer it has been stated that it is 45min.

Really? This is what I said:

On 10/19/2020 at 10:59 AM, Eise said:

So once again, this is the way to see it, all the rest is irrelevant complication from your side:

• According the FOR E-X, B's clock ticks slower. So E-X observes that B's clock only shows 45 minutes travel time when it arrives at X.
• According to B, he travels only a distance of 0.6 Lh, which costs him 45 minutes according his own clock.

So E-X and B agree: B's clock shows 45 minutes when it arrives at X (or X arrives at B...).

Where do you see a statement about which FOR is at rest and which FOR is moving?

The situation is symmetrical:

• E-X sees the clock of  B having a slower ticking rate
• B sees the clocks of E-X having a slower ticking rate
• E-X sees everything in B's FOR length contracted
• B sees everything in the FOR of E-X length contracted, which includes the distance E-X

What is not symmetrical is:

• The distance traveled between E and X, which are both in a common the FOR E-X; length contractions in the FOR of B from E-X's view play no role
• The clocks of E and X are synchronous in the FOR E-X, so they are not in the FOR of B.

We are interested in the events "E's location coincides with B's" and "X's location coincides with B's", whereby E and X are in the same FOR. So we have to look at what this space distances are in the respective FORs: 1Lh in E-X, and 0.6 Lh in B. If we look at what the respective clocks show at these 2 events, we must take time dilation and the relativity of simultaneity in account: for E-X, B travels simply from E to X, its time is dilated, so it takes only 45 minutes;  for B, E's clock is at 0:00, and X's clock is already 48 minutes ahead (not synchronised in his frame!) when E passes B, and because of time dilation, he observes B's clock ticking away only 27 minutes, which makes that when X passes by, he sees X's clock at 48 + 27 = 75 minutes.

3 hours ago, michel123456 said:

Now, if you keep in mind that the situation must be so that you cannot know who is the "moving" and who is the "at rest", you get the confusion I struggle with.

I hope I succeeded to show that only the relative velocity plays a role, not who is in rest or is moving.

Can you please let this sink in, and ask about (or criticise) the points I made above. Don't, really don't introduce some other of your confusion's at this moment.

Share on other sites

According to your scenario one could argue that X is traveling and measures 45 min of travel.

Share on other sites

6 minutes ago, michel123456 said:

According to your scenario one could argue that X is traveling and measures 45 min of travel.

Share on other sites

X is traveling toward B.

Exactly as B is traveling toward X

The setting is one, these are not two different situations. Simply the observer changes, once in the FOR of B, once in the FOR of X.

In the FOR of B the distance to travel is contracted (following your arguments). So in the FOR of X the distance to B should be contracted too, because B is moving.

So, being symmetrical, B's time to travel is 45min (following your arguments) and X's time to travel should also be 45 min.

Which does not give the expected result.

Share on other sites

8 hours ago, michel123456 said:

But both are in a stationary frame.

It has been stated here that for the observer "at rest" the time to reach planet X is 1h:15min

For the "moving" observer it has been stated that it is 45min.

Now, if you keep in mind that the situation must be so that you cannot know who is the "moving" and who is the "at rest", you get the confusion I struggle with.

You changed the scenario to eliminate acceleration so the only thing we had to work with is 2 inertial frames, so my answer was based on that.  Maybe if you stop trying to complicate or change the scenario it would be less confusing for you.

You said:

Now, if you keep in mind that the situation must be so that you cannot know who is the "moving" and who is the "at rest", you get the confusion I struggle with.

This is wrong!  We know exactly who has accelerated to a new reference frame and who stayed in the same reference (who is the "moving" and who is the "at rest")

We know the rocket accelerated and the earth did not.  An astronaut on the rocket would almost lose consciousness from the acceleration and the earth would feel no acceleration.  The rocket is 'moving' and the earth is not*.

This is based on the origin of the coordinate system being starting frame of the rocket on the earth.

Do you understand this?  If you don't understand, then please don't say you do!

Edited by Bufofrog
Share on other sites

2 hours ago, michel123456 said:

X is traveling toward B.

Exactly as B is traveling toward X

The setting is one, these are not two different situations. Simply the observer changes, once in the FOR of B, once in the FOR of X.

Until here, yes.

2 hours ago, michel123456 said:

In the FOR of B the distance to travel is contracted

Right, because the distance to travel is 1 Lh in E-X, and because B moves with 0.8c, it sees this distance as 0.6Lh.

2 hours ago, michel123456 said:

So in the FOR of X the distance to B should be contracted too, because B is moving.

Wrong. Which distance? See above: we started from the fact that in the FOR of E-X, the distance between E and X is 1 Lh. So X must travel this distance to get to the place where E was, and B still is. And in X's FOR, this is still 1Lh.

This is exactly the point where the situation is not symmetrical, as I said in my first point above (made a few typos, I correct them here):

4 hours ago, Eise said:

What is not symmetrical is:

• The distance traveled between E and X, which are both in the common FOR of E-X; length contractions in the FOR of B from E-X's view play no role

There is no distance of 1Lh in B's FOR, so this is simply invalid:

2 hours ago, michel123456 said:

So, being symmetrical, B's time to travel is 45min (following your arguments) and X's time to travel should also be 45 min.

So no, it is not exactly as symmetrical as you think.

Edited by Eise
Share on other sites

1 hour ago, Eise said:

Wrong. Which distance? See above: we started from the fact that in the FOR of E-X, the distance between E and X is 1 Lh. So X must travel this distance to get to the place where E was, and B still is. And in X's FOR, this is still 1Lh.

This is exactly the point where the situation is not symmetrical, as I said in my first point above (made a few typos, I correct them here):

Applying MD's rule:

From the FOR of X, Is B moving?

If B is moving then the distance is contracted. Which, I agree with you, was not the opening statement. Thus we have an inconsistency.

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

Look, I know the distance is 1LH, I am trying to show you that the situation as you are presenting it is not symmetric. That is why @Bufofrog is insisting so much on acceleration. But IMHO it is wrong. There is no account for acceleration in any of the maths formulas that have been used so far. The setting can be imagined with bypassing clocks, not even jumping from one FOR to another, but simply reading instantly the clock in the other FOR. The change of FOR by acceleration is an excuse to avoid the inconsistency.

But I believe there is another way to explain what is happening.

Look at my diag1 and diag2 again.

If you spot an error there please tell me.

Share on other sites

1 hour ago, michel123456 said:

Applying MD's rule:

From the FOR of X, Is B moving?

If B is moving then the distance is contracted.

What rule is that? That's not my rule.

Share on other sites

2 hours ago, michel123456 said:

Applying MD's rule:

From the FOR of X, Is B moving?

If B is moving then the distance is contracted.

Provide a quote where anyone said that,

Share on other sites

11 hours ago, swansont said:

Provide a quote where anyone said that,

Ouaouch the syndrome hit again.

Share on other sites

14 hours ago, michel123456 said:

If B is moving then the distance is contracted. Which, I agree with you, was not the opening statement. Thus we have an inconsistency.

Uh?

'The distance'? Here you go again. Not mentioning which distance from which frame

In E-X's FOR, the distance between E and X is 1Lh.

So for B, moving with 0.8c, he sees the distance he must travel as 0.6Lh.

There are no other relevant distances for this situation. Lengths of poles connected to B just play no role. Why, B could be a rocket of 1 Lh long. The only thing that interests us is the passing of the observer in B with his clock, passing E and X.

Create an account

Register a new account