Jump to content

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


michel123456

Recommended Posts

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.

Link to comment
Share on other sites

8 hours ago, Eise said:

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.

Adding to that: A ruler doesn't measure "accumulated" distance, but an odometer does. While a twin makes a round-trip at constant speed, if its clock records half the time Earth's does, it will measure that it traveled half the distance Earth measured it traveling. Its odometer retains the accumulated effects of length contraction.

(The trip's clock and odometer dicrepancy ratios would generally differ is the speed wasn't constant.)

Link to comment
Share on other sites

17 hours ago, michel123456 said:

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

Sigh... First, I expect from you that you really explain to us how it is possible that muons reach the earth's surface. The (didactic) importance of that is either you see that you can't, in which case you know that your intuitions are wrong; or you think you can explain it, we see how you do it, and pinpoint precisely to the point where you make an error. 

Secondly, already said many times, you should start with the explanation why the twins did not age equally, and they agree on this difference. What observers see during the traveler's trip unnecessarily complicates the situation. But I know that for many people this is some kind of show stopper, so, look here in Wikipedia. Ask us questions if you do not understand it. And 'not understanding' is not the same as 'it is wrong'.

Thirdly, just another try in the hope you will see the light:

Say the traveler flies with 0.8c to the planet Solaris, 10 light years away, and to keep it simple, earth and Solaris are in the same frame of reference. Now for the traveler the distance, due to length contraction, is only 6 light years. Because the home stayer and the traveler agree on the speed of the traveler, that means that the traveler does the trip in 0.6 the time compared to the view from the earth. Moving back home of course the same. So the traveler grew 15 years older, the home stayer grew 25 years older. The asymmetry lies in the fact that the twins do not agree on the distance traveled. But because the twins agree on their different ages now, it means also that they do not agree on their time measurements. Taking that into account, the twins can be sure they live in the same reality: they agree on their age difference.

And now, PLEASE, instead of throwing another conflicting intuition of yours at my post, work as carefully as you can through these points. If you get stuck with the muons, tell us where you are stuck. If you get stuck on the Wikipedia article, tell us exactly where, and why (no, not simply a conflict with your intuition, but where you cannot follow the logic of the explanations given). And tell us why you do not understand that the situation of the twins is not symmetric. 

Or simply refuse to understand relativity, but then say so honestly. In that case I would suggest to close the thread. So what it wanna be boy, trying to understand, or refusing to understand, yes or no?

 

Link to comment
Share on other sites

5 hours ago, Eise said:

Or simply refuse to understand relativity, but then say so honestly. In that case I would suggest to close the thread. So what it wanna be boy, trying to understand, or refusing to understand, yes or no?

Maybe he can figure it out by the dashboard lights?

Link to comment
Share on other sites

21 hours ago, michel123456 said:

Thus the traveling twin has seen his brother time compressed.

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.

Link to comment
Share on other sites

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.

Edited by michel123456
Link to comment
Share on other sites

39 minutes ago, michel123456 said:

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.

At last... 

Though I doubt that's the last we hear about it...

Link to comment
Share on other sites

2 hours ago, Eise said:
2 hours ago, Bufofrog said:

Maybe he can figure it out by the dashboard lights?

😄 Yes, maybe. You recognised what was in the back of my head when I wrote that...

Let him sleep on it … He'll give you an answer in the morning.

Link to comment
Share on other sites

1 hour ago, michel123456 said:

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.

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? 

Link to comment
Share on other sites

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.

Edited by michel123456
Link to comment
Share on other sites

2 hours ago, michel123456 said:

They are observing THE SAME THING.

This is literally your thread now, I'm curious how this adds up in your examples, without relativity.

You said that when traveling twin A stops, it doesn't immediately see Earth as stopped because of the delay of light. Relativity says Earth doesn't see A stop immediately, and you're saying A sees Earth appearing the same.

For example, suppose Earth is at rest at the 0 mark on a ruler, and A travels to a 1 LY mark, and then stops. The whole time while traveling, A sees that Earth appearing to stay at the 0 mark, agreed?

When A stops at the 1 LY mark, how far away does it see the Earth? I think it should be 1 LY. It sees Earth at 0 and itself at 1 LY and the distance is "normal", not contracted, and it is 1 LY. If A sees Earth continuing to move away, how is it appearing to move? Does it appear to move farther than 1 LY away? How do those numbers add up, in your view?

Link to comment
Share on other sites

Since michel123456 just asks the same questions over and over I think I will look through his threads to copy and paste the answers to his questions to one post.  I can then number the answers so that when he asks one of his group of repeating questions you can just write down the number.  It will take some time up front but in the long run it will save time for everybody.

Link to comment
Share on other sites

4 hours ago, michel123456 said:

But A is also at rest.

At rest with respect to what? Relativity duscussion makes no sense without this detail.

Quote

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.

It might be better if you didn’t keep switching examples. Before, A was at rest wrt the earth. Now A is moving. And now you’re looking at a different part if the trip. (like someone has already noted, like you’re hoping to get a different answer)

Yes, A’s distance to earth is length contracted. Before anyone changes frames, A and B will think the other’s clock is running slow.

 

Link to comment
Share on other sites

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.

Edited by michel123456
Link to comment
Share on other sites

3 hours ago, michel123456 said:

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.

“When he arrives” is ambiguous 

Is he moving, or has he come to a stop wrt earth?

If moving, he measures the distance to X as contracted. If stopped, he measures what the earth observer measures

The answers won’t change from asking the same thing over and over

Link to comment
Share on other sites

On 9/18/2020 at 5:26 AM, michel123456 said:

The same questions arise all the time, the same answers, no satisfaction.

What is wrong with you?  Seriously, what is wrong with you??

You ask the same questions and get the same answers.  Wierd, isn't it?  Why do you think the answers to your questions are consistent?  Your not satisfied with the answers, but for some reason nobody will change their answers to satisfy you!  I guess you will just have to keep asking the same questions; the answers are bound to change eventually and satisfy you, right?

Link to comment
Share on other sites

On 9/25/2020 at 2:20 PM, michel123456 said:

I am asking about the traveling twin when he passes by planet X.

I didn’t really follow the entirety of this thread too closely, so if there was a ‘Planet X’ mentioned, then I missed that.
So then the answer depends on exactly what form this world line of the travelling twin takes. It is of course possible that the travelling twin is almost fully inertial, meaning that the vast majority of his world line is inertial - if your Planet X coincides with those inertial portions, then you are right, the two frames will be symmetric. However, there must be at least one portion of the journey - however short and small - that is not inertial; during that portion the symmetry is broken, which leads to the difference in total proper time recorded on the clocks.

This is all closely related to the difference between coordinate quantities, and proper quantities - understanding the difference is crucial to understanding relativity.
 

Link to comment
Share on other sites

3 hours ago, swansont said:

The answers won’t change from asking the same thing over and over

@michel123456 please also note that if there would be any changes or inconsistencies among the many answers to the same questions, that would just mean that a member made a typo or similar. It would definitely not point at inconsistencies in special relativity.

Link to comment
Share on other sites

I'm not 100 % sure this is at the root of @michel123456's problem, but I have a feeling that it may be. Please, pay utmost attention to this:

On 9/17/2020 at 12:43 AM, Janus said:

Again, you are conflating "what someone visually sees", with what they say is actually happening.

(Emphasis mine.) I think this observation by Janus is absolutely essential. Here's something you said a while ago, that I think is very significant and, to me at least, betrays that you haven't understood what Janus was trying to say:

19 hours ago, michel123456 said:

But A is also at rest. He sees (observes, measures) the Earth length contracted & the distance to the Earth is also contracted.

The length and time that undergo dilation and contraction in relativity are not the naive "as seen" lengths and times that one intuitively gets from common experience, the projections on your retina, so to speak, and on your inner clock. Your use of "see" and "measure" as interchangeable has set off my alarms.

In relativity you must be careful about what you mean by position and time (and thereby by length and duration) from the get go. So much so that I think you're not there yet.

Coordinatizing events is the groundwork for everything else you do in special relativity. When objects are distant from you, you must establish some kind of measurement standards to guarantee that you know what you're talking about when you say "it is at x at time t." This is at the core of what Janus meant by "see" and "say" in my previous quotation of him. Distinction, this one, that becomes absolutely crucial when discussing the relativistic Doppler effect, because there is a mixture there of both aspects. But it is very important in general.

The standard observer in SR is not one that simply believes what she sees. She sets up measuring devices so as to be able to say where and when something distant happened, knowing full well that the more distant something is, the later it gets to her. Einstein pictured a frame of reference as an infinite set of observers relatively at rest with respect to each other, and with clocks that have been sync'd at t=0. That's not necessary. You can set, as @md65536 said before, one reliable clock and get any timing from light signals back and forth from it. The standard observer in SR already knows objects are expected to be "seen" different from what they "are."

Please, take a good look at Fig. 2.8 here (from https://www.academia.edu/39463786/R_dInverno_Introducing_Einsteins_relativity) by Ray D'Inverno:

27-bb225b977a.jpg

28-3d9a402dae.jpg

It is plain to see that, for any event, t2 is the time when you see it happen, but that's not the time when you infer (measure, in your wording) that it happened. The time when you infer (measure) it happened is,

\[t=\frac{1}{2}\left(t_{1}+t_{2}\right)\]

The difference between one and the other is source of much confusion for relativity students. When Janus says "see" he refers to t2 in Fig. 2.8 (or increments of it). When he says "say" he refers to coordinate time (or increments of it), which is different, and corresponds to t.

The timing of events in relativity, the coordinate timing, is the time that you say (guess, if you know SOL), not the time that you see (paraphrasing Janus.)

The times and lengths that you see are only too obviously deceiving you already, because the image of the head of the train, e.g., that's coming to you, obviously predates the image of the tail, as seen by you at the same "psychological" time.

I really hope this helps. If it only adds confusion, feel free to ignore it. But in that case, I don't know what it is you don't know.

Link to comment
Share on other sites

59 minutes ago, joigus said:

I'm not 100 % sure this is at the root of @michel123456's problem, but I have a feeling that it may be.

I think this is an example of hope overcoming reason.  Michel has to literally see a clock go in slow motion or see a meter stick shrink, anything other than that is fake news to him.  He is a lost ball in high weeds...

Link to comment
Share on other sites

1 hour ago, Bufofrog said:

I think this is an example of hope overcoming reason.  Michel has to literally see a clock go in slow motion or see a meter stick shrink, anything other than that is fake news to him.  He is a lost ball in high weeds...

You're very, very likely right, @Bufofrog. For me it's a last-ditch attempt. After this I think I'm done. But I'll still keep an eye on the thread, only to read alternative complementary explanations, for my toolkit.

Edited by joigus
Link to comment
Share on other sites

On 9/25/2020 at 8:20 AM, michel123456 said:

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.

If you make a spacetime diagram of this situation in which one twin remained inertial while the other had a device to measure that they flipped frames of reference (remember it's relativity of inertial frames of reference not relativity of NON-INERTIAL frames of reference) then with a instantaneous turn around the non-inertial twin would notice the other twin actually jump ahead in time in a way the other twin (inertial one) would not see. If you plot the diagram you'll notice the time loss as a missing triangular section if I recall.  

On 9/25/2020 at 10:27 AM, michel123456 said:

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.

No, both WORLD LINES cannot be treated entirely as the same as the moving twin CHANGES frames of reference half way through his journey (his proper time). So it isn't symmetric just as if a car drives away and comes back we know it was the car that accelerated (changed inertial frames of reference) as it was not the person given they both had accelerometers. If they both remained inertial then they would never meet back up again and if they were both non-inertial then their accelerometers would have noticed the inclusion of some force fictitiously into each of their respective frames of reference but the person on the ground never noticed such inertial forces arise so the only conclusion is that the car was moving, ergo it WASN'T symmetric. Why is this so hard to understand when you literally can go into a parking lot and forget about special relativity to then show this to be the case. Remember that special relativity has to simplify down to classical physics in the low speed limit. 

Also, acceleration is not relative in the sense that velocity is or inertial frames of reference are just as in CLASSICAL PHYSICS or in GALILEAN SPACETIME. In special relativity (as in classical physics) it isn't relative whether you are ACCELERATING or are ROTATING. We can disagree the magnitude of said quantities but whether you are or are not accelerating is not frame dependent but can rather frame independent. 

@michel123456 Do you understand that the postulates of special relativity are:

1. All inertial frames of reference are to be treated as equivalent in that the laws of physics behave the same in any inertial frame of reference.

2. The speed of light as measured from any arbitrary inertial frame of reference in free space is constant. 

This would mean that if you had a ANY non-inertial motion then by definition we could detect it and know that we are in a non-inertial reference frame because these postulates wouldn't apply. Further, this would also mean the situation of one non-inertial observer compared to an inertial one (the twin paradox) would mean by these postulates that their world lines COULD NOT be treated symmetrically or in other terms equivalently as if the non-inertial reference was inertial. AGAIN, the postulates are not, 

\( 1^{*} \). All frames of reference have the laws of physics behave the same within them. 

 \( 2^{*} \). The speed of light as measured from any frame of reference in free space is constant. 

You seem to be under the delusion that the starred ones are what define special relativity when in relativity it's the un-starred ones. 

Link to comment
Share on other sites

10 hours ago, Bufofrog said:

Michel has to literally see a clock go in slow motion or see a meter stick shrink, anything other than that is fake news to him.

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.

 

Edited by md65536
Link to comment
Share on other sites

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.

Edited by michel123456
na attachment space available
Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.