michel123456

I ment deleted on the graph. With Tipp-Ex. You seem annoyed when I put a doubt on the premises. All observers will make correct measurements in their FOR. But at that speed, they cannot be in the same FOR, especially due to the tremendous acceleration in order to reach that speed. Giovanni & Raphael are decalcomania on their spaceship seats (I can't figure they go by foot). You must know that you cannot extract safe conclusions from a thought experiment based on unphysical situation in the first place. I always wanted to say that. But if you insist, Iggy may proceed.

If you are talking about Olber's paradox I have an inclination to accept the fractal star distribution explanation.

I like Leonardo at the center. You have talent. You are making someone run at 80% of the speed of light from Earth to Jupiter. You should have drawn a flame going out of his bottom. At this speed, t=1hour on the left & right should be deleted. And you must describe the method Leonardo used to measure 540 million miles in the first place, because to an alien it is not 540 million miles*. And it is not 1 hour. It is only for Leonardo. *Unless you accept the statement that Leonardo, who is in the same FOR of its rod, who holds the rod in his hands (wow) measures right and all the others around are wrong.

This is not my invention, You can find the yellow lines as bisectors in the wiki article under constancy of the speed of light.

Sorry, I have to apologize. After a second thought, I cannot imagine that either. I completely forgot that the spaceship is under uniform linear motion. IOW it is at rest in his own FOR. It cannot "track behind its own image". The only moment it can gain against its own image is during acceleration, that point we have negligate on diagram but explains the difference in the slope of the spaceship. Am I wrong there too? But even then, this "gain" will be lost when reducing speed at arrival. I suppose intuitively, due to symmetry.

Thank you for the reply. I can't recall exactly what was the next step. I always have a next step... What I can say now that I look again at the wiki entry, is that it is drawn in a way that people could believe point A actually looks at point B, and point B looks to C. That is not the case.

Funny conversation. Reflection? Lets take an even more reduced spacetime: only x,t I suppose the negative is induced by time. What does that mean from a geometric point of vue? Do euclidian laws of geometry change in such a spacetime?

enantiomorphic Like your left & right hand.

Does that mean it is mirrored?

Splitted from a tooo long thread: That is correct. What happens to Minkowski spacetime when one dimension is taken away? From the 4 dimensions x,y,z & t, eliminate z for example. The resulting reduced spacetime x,y,t looks quite a lot with Euclidian space, or do I miss something?

The speed of sound is not invariant. The "doppler effect" with light have nothing to do with waves going faster than others and getting closer to each other eventually joigning together. Waves of light go at the same speed no matter the speed of the source, that's what I have been told.

I ment all observations done from FOR of Earth. To show I am not completed stubborn. If at the U-turn point, a flash light is emitted, I can imagine a spaceship tracking behind the flash, and arriving at Earth a few instant after the flash arrival. But that is not exactly what happens. In fact the spaceship flashes constantly. The spaceship constantly follows constant flashes. I must have wood in my head.

The time you need to cross a specific distance is the expression of a speed. One hour to cross 1km, or one second to cross 1km, is speed. It means 1 km/hour and 1km/sec respectively. In the letter example, the speed of the letter is 1 [distancetraveledbybrother]/day, or 1 day. Bolded where it hurts. What I cannot swallow is that I observed the ship going slowly, and coming back in a blink of the eye, evoluting at the same speed.

In mundane observations, airplane or letters examples, you are making addition of speeds, something not allowed here. For example: the letters. Lets say your brother goes away, and every 2 days write a letter that makes half the time to come to you. That is because the time for the signal (the letter) is less than the time for travel. The first day, you get no letter. the 2nd day, your brother writes the letter. the 3rd day, you get the first letter. (2days of travel for your brother + one day for the letter). The next letter you will get 6 days from day zero (4 days brother+2days letter) The following at day 9 (6 days brother+3 days letter) The formula is, with [math]D[/math] number of days to wait, [math]N[/math] number of days to travel of brother, [math]C[/math] number of days for letter. [math]D=2N+C[/math] In fact, C is the speed of the signal: in this example it is the speed of the letter, in not "mundane" example, it is the speed of light. IOW you have speed summation, not allowed in Relativity. The example of letters must be dismissed, as any other "mundane'" example. In the OP, there are spacetime diagrams, these are not mundane examples. I may be wrong though. I guess so. There is something that bothers me very much in standard diagram: It is stated from the beginning that speed to go is the same as speed to come back. That is not negotiable. So we are looking in our telescopes, and after 50 days, we observe the spaceship making the U-turn. Till that moment, there is no problem. we can calculate the distance that the spaceship traveled, it is [math]X[/math] The speed, as calculated from Earth, is [math]X/50[/math]. Of course, we don't know what the spaceship measured, neither what another observer on another planet measured, we have only our measurement from our FOR. Maybe even the value of [math]X[/math] is wrong, and will be corrected afterwards, but what matters is that distance [math]X[/math] to go is the same with distance [math]X[/math] to come back, because it is the U-turn point. Then it is assumed that the spaceship will come back in only 25 days! It means that its apparent speed, as measured from our FOR, is [math]X/25[/math], which is different from the measured speed at departure. But that was not negotiable. Where am I wrong?

That it is not physically possible to observe behind an object, except made of glass, of course.

from another thread: ------------------- That was sarcasm. ------------------- Without sarcasm now. Your words reflect what you know and what you have learned. Not what you are thinking. I hope.

Here is how to describe the situation: At day 75, we are looking "through" the spaceship, observing its position far away although it is closer to us. The spaceship is supposed to run behind its own image. We have an object 25 days from us, that we cannot see, but we observe its image behind it. Is that what you are assuming?

You seem to accept that the velocity of the spaceship (as observed from earth) is different when coming back.

Forget the 100 days. Say you know nothing. You don't know the speed of the spaceship, you don't know the distance, you know nothing about its time dilation or length contraction. You are only observing in your telescope. After 50 days you observe it in your telescope making a U-turn. How many days will you have to wait for arrival when you know its speed will be exactly the same?

Think of it twice. Theory is about measurements made from Earth's FOR. -------------------------- Answer this question: The spaceship goes away. You observe it constantly in your telescope. After 50 days you observe it making a U-turn. How many days will you have to wait for arrival when you know its speed will be exactly the same? --------------------------- It is speculated that the "twins paradox" happens at point C of the following diagram:

My graph gives the correct result. Your post #156 don't give the correct result. I appreciate that you didn't edit it. You should have known that [math]S[/math] cannot be bigger than the radius. That was my point. The radius, which represents distance, don't change, [math]S[/math] does. You may notice that for [math]t=x[/math], [math]S=0[/math], which is the expected result.

[math]S^2=25-16=9[/math] [math]S=3[/math] ------------------- edited [math]S=\pm3[/math]

NOW I am interested in numerical value.

Here below a graphic representing the twin paradox in Minkowski space-time. The diagram is supposed to be valid for the Frame Of Reference of the non moving observer, spending time doing nothing from point O to point A, while the spaceship travels from O to B, then makes a U-turn and comes back to Earth at point A. Here is the problem I want to discuss: When the spaceship arrives at point B, at half the travel, and makes the U-turn, the observer on earth at point H cannot see it simultaneously. He will observe the U-turn after a while, from point M. At point H, 50 days have passed, and the spaceship is not observed at point B. At point M, 75 days have passed, the spaceship is observed at point B making the U-turn manoeuvre. At point A, 100 days have passed, the spaceship is back to Earth. What is surprising is that the paceship has been observed to travel the distance d in 75 days when going, and 25 days when coming back. The spaceship did not have the same apparent velocity on both parts of the trip. That's my problem, because I always made the interpretation of the diagram stating that the angle on the graph represents speed. Since the angle of the first blue segment is exactly symmetric with the angle of the second segment, it should mean that both speeds were the same. How is it possible then that the spaceship has been observed making the return trip in only 25 days? I hope it is clear till now. There is an inconsistency. Now, another graph here below: Here I was deeply thinking the opening statement: "The diagram is supposed to be valid for the Frame Of Reference of the non moving observer" We don't care about what is the observation of the traveler, we care only to represent what is observed from Earth's FOR. What we observe is this: The spaceship goes for a trip. After 50 days (as observed from Earth's FOR) he makes the U-turn. He comes back after another 50 days. We observe that on the going trip, the spaceship traveled the measured distance d at speed d/50. During the return trip, its speed was d/50: the same speed. What is important is that triangle O-H-B has the same basis and same height with triangle H-A-B. They have the same surface. The angle of the blue segments with the vertical do not represent speed. The ratio between height & basis of the triangle does. And there is no inconsistency. Michel.

I was asking because i struggled with it, entering weird situations. I thought maybe someone already did that. Now I have trouble with a simple Minkowski one. I'll start another thread in speculations.