VandD

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To keep speed of light constant in all intertial frames. Check f.ex. light clock experiment. Relativity of simultaneity, time dilation, length contraction are all a result of constant c. I think you have to face the fact that you miss some fundamental elementary knowledge about SR to discuss this kind of exercise and diagrams. Why don't you take a brake, surf a bit on the internet, learn about relativity of simultaneity etc, and how to read and draw all that in some preliminary spacetime diagrams?

Einstein's SR is block universe. The presentism alternative is linked to an ether and/or a preferred frame. There is no such thing. Think about this one: Screenshots teken here. I added notes to the diagram for clarity.

I apologize. I was thinking 'that's rubbish', but I should simply have written 'No'.

What is this for rubbish?

Don't you know? I thought you knew everything about length contraction?

My mind too. Briefly: In 4D spacetime there are only events. Ticking events of a clocks. No ticking rate. A clock has only a 'ticking rate' when a reference frame is introduced. Then we can compare which event of a clock happens simultaneously with a tick event of another clock, and thus conclude that a clock has or has not a slower 'ticking rate' relative to the other clock.

You cannot measure a projection of a length. What an observer mesures is the distance between two simultaneous events. If there is relative movement, the observer measures NOT the distance between the same simultaneous events as an observer 'at rest' would do. Do you understand this, or did you never read about relativity of simultaneity? No wonder you don't understand different 3D sections through 4D spacetime...

I don't think you will ever understand SR, but let's see whether this helps to trigger your mind: The 4D train (or car) indeed doesn't change. When Mr Magoo moves relative to the train he feels with his hands a different 3D section through that 4D train. And that section through the 4D train is shorter than a 3D section through the train if Mr Magoo is at rest relative to the train. You pretend there is only a 3D real train and Mr Magoo moving relative to the train 'observes' a distortion of that 3D train. That's wrong. You completely miss the fact that the different section throught the 4D train is a equaly valid different set of simultaneous events. And if you call one set of simultaneous events 'real', then all the other 3D sections are real as well. You simply have no valid physical reason for believing the contrary.

Of course, so what? If you don't accept he feels with his hands in front of him the contracted car, rear and front simultaneously, then you still haven't understood an inch about SR. But don't be desperate. You are not the only one I encounter during the last 25 years thinking that because the length contraction and time dilation is a reciprocal thing between the car passenger and Mr Magoo, it can only be possible if its a visual optical illusion effect. Do you think Einstein would have spent his time with such a stupid optical illusion effect?

Well, actually the 4D train indeed doesn't change. When Mr Magoo moves relative to the train he feels with his hands a different 3D section through that 4D train. You get it? But you pretend there is only a 3D real train and Mr Magoo moving relative to the train 'observes' a distortion of that 3D train. That's wrong. You completely miss the fact that the different section throught the 4D train is a equaly valid different set of simultaneous events. And if you call one set of simultaneous events 'real', then all the other 3D sections are real as well. You simply have no valid physical reason for believing the contrary.

No, it's correct.

Are you still pretending that the contraction of a train is nothing real? Let a small car stay at rest on road. An 'observer' (let's call him Mr Magoo) stands at rest relative to the car. He spreads his arms so that he can feel the front and rear of the car simultaneously (and his nose feels the middle of the car). No doubt about it: that car is physically present as a real object in his 3D real world. Now Mr Magoo will move at different speeds. For each speed Mr Magoo physically feels the car at a different length: for each different speed he has to spread his arms at different length in order to touch the front and rear of the car simultaneously (and his nose feels the middle of the car). No doubt about it: at each different speed a real short car of different length is physically present in Mr Magoo's 3D real world of simultaneous events.

Correct, because for the green guy to see the green 2013.4 he has to wait until the light from that green 2013.4 clock gets to him, and that's long after 2015. But red can SEE the green time dilation in a different set-up. Let's hope I don't push it too far for you at this stage: For example. Let the clock start at a distance left from red. Then when the green clock reaches him, there will be less time on that green clock than the time on his red clock. He sees it at an inch from his nose and can even feel the hands of the clock if he wants to. But... the tricky thing is that red first has to make sure his red clock and the still spatially removed green clock start ticking simultaneously. So he has first to go through the synchonization procedure... But that's another story. @Michel: I would like to highlight the following in that paper: For the traveler the earth HAS TO tick faster when traveler is turning, because when back home the earth observer is older than traveler. That's what's happening when moving around in 4D spacetime .

No. What you show is not what is meant by time dilation (time slowing). The time dilation is here: For RED simultaneity: When red clock indicates 2015, the green clock indicates 2013.4 I.ow. when the red clock ticked from 2007 to 2015 (8 years), the green clock ticked from 2007 to 2013.4 (=6,4 years). This is also what gamma tells us for v=.6c: 8 / 1.25 = 6.4 For GREEN simultaneity: When green clock indicates 2015, the red clock indicates 2013.4 I.o.w. when the green clock ticked from 2007 to 2015 (8 years), the red clock ticked from 2007 to 2013.4 (=6,4 years). This is also what gamma tells us for v=.6c: 8 / 1.25 = 6.4 A. s. o. For example: (not made bold in the diagram) For GREEN simultaneity: When green clock indicates 2013.4, the red clock indicates ... 2012.12. Can you find which green simultaneity line I'm talking about? When the green clock ticked from 2007 to 2013.4 (6.4 years), the red clock ticked from 2007 to 2012.12 (=4.12 years). This is also what gamma tells us for v=.6c: 6.4 / 1.25 = 4.12

Here it is. The time dilation is not what a black light path connect. For time dilay read green and red lines of simultaneity. Also check with gamma factor for .6c. For red the green clock ticks 1.25 slower. For green the red clock ticks 1.25 slower. Does your diagram show this?

It's not. But I forgive you because it's a classic mistake a lot of people people make. I will show you loedel diagram: same unit lengths and showing time dilation. Watch it.

Nonsense. A light path is not a simultaneity line. In a Minkowski diagram a light path is always at 45°. Not the simultaneity lines. Please return back to your SR books. Learn what events are, what simultaneity means, and how constant speed of light is measured. Etc etc. Until then it's hopeless to continue this discussion. Please use proper quotation method. Unless moderator allows different colors when answering? What you draw is what is seen by the observer. That's not time dilation. Time dilation is what's happening with time indication on clocks as per reference frame. In your case you don't even draw the reference frame of the traveler. Where is it? You probably make the mistake that because a Minkowski diagram shows greater time units on the rocket worldline THIS meaning time dilation. It's not. To avoid different time unit lengths one can draw a Loedel diagram. I prefer loedel diagrams for the simple reason people misread the greater time unit lengths lenths as being time dilation. The problem is that the loedel diagram is not that good in showing the full twin story in one diagram. You have to cut it in two, one for outbound, one inbound. Are you interested in a Loedel diagram (same unit lengths) showing time dilation for the right diagram? The relative speed between frames.

The diagrams with light paths:

Glad you accept you missed something. You said: "Because as you have all agreed when the ships are put side by side, they have the same length". This is when the ships are AT REST relative to each other. They have NOT the same length when they move relative to each other, as shown correcly in bvr's drawings (see his 2. sketches) When both astronauts cross they are not in the same set of simultaneous events. In fact you don't understand relativity of simultaneity, correct? Because the traveler is going left, flying toward the earth. In that case the simultaneity line is reversed. Lorents transformations and Minkowski diagrams work that way Wait a second. Read what I wrote. I told you that the green and red lines are the frames of SIMULTANEITY. These are not the paths of light (I left these out but I can add them if you want. Give me some time I'll do it so that you can see the difference.) No, while turning, the rocket catches up on what's happening on earth in a fast way The total earth worldline of 10 years lays out there in 4D spacetime, and by travelling away and then getting back he ends up at 2017 earth clock.

Pure nonsense. To help you I add a diagram for the rocket not returning back home. In both diagrams I show where and when -per rocker frame i.o.w. the red 3D spaces of simultaneous events- the earth clock ticks slower (see blue). But... In the first diagram the added earth clock ticking of 3.6 years (= 10 – 6.4 ) is a result of the rocket changing direction in 4D spacetime. As long as you refuse to understand 4D spacetime you'll never understand SR. But no offence it seems you are not interested in learning how SR works. And it won't help you by repeatring us a hundred times that proper length is all there is. Do you actually see and understand what the diagrams show?

No, they are both correct. The problem is that you refuse to see why. They have not the same length when they cross in 2. (see below). You probably missed that. They have both the same length when they get back together and are at rest relative to each other. Wrong. See 2.

Yes. They are both true. For ship A astronaut the two A-shot events occur simultaneously with event E. But none of the B-shot events occurs simultaneously with event E. For ship B astronaut the two B-shots occur simultaneously with event E, but none of the A-shots. Relativity of simultaneity at work. Welcome to SR At event E both astronauts consider a different 3D space of simultaneous events existing 'now' at event E. The different 3D spaces of simultaneous events are only 3D reality sections of the bigger 4D spacetime reality (see also Einstein's quotes here.) One more thing: movement is a relative thing: for A, B moves. For B, A moves. Period.

Yes, for a split second he feels me between his two fingertips. And you think it's madness because you don't accept that for the train passenger I'm the one moving? Is that it?

Do you perhaps think a moving train in front of your eyes is NOT contracted, but you only see it contracted (kind of an optical illusion)? Stretch your arm and show two fingers. A very fasty moving train in front of your arm will fit exactly between your two fingers. You physically feel the rear of the train touching one finger simultaneously with the front of the train touching the other finger. Now tell me whether you consider that contracted train physically real or not? And yes, the passenger in that train will have a physically very small you standing on a physically very small embankment: a split second you fit perfectly between two fingers of the train passenger. Seems you don't understand how relativity of simultaneity works. You cannot consider one set of simultaneous events as real, but not another set of simultaneous events.