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michel123456 last won the day on July 12 2019

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About michel123456

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  1. So you seem to agree that A & B would see different things although getting the same light. If that does not hurt your feelings at the departure, why is it so mind blowing at the arrival at planet X?
  2. Let's take it from the start. At time zero A & B are at rest looking at planet X that is 1LH away. Out of magic, B steps instantly into a FOR that travels at 0.8c. What does B sees? Doesn't he sees planet X length contracted? And closer to him? Instantly? While observer A sees it normal as usual? Or am I wrong there too?
  3. I meant the Earth will look larger, because closer. And length contracted. 1 hour ago, the traveler was closer to the Earth, and the Earth was closer to the traveler.
  4. And 1 hour behind, the Earth was closer to him. So he is observing the earth larger (because of regular perspective). And because of length contraction, he is observing the Earth flattened.
  5. But if the traveler looks behind him just before stopping, he will see the earth as it was in the past 1 hour ago. At this time (1hour ago) the Earth was closer to him. Because he is in a state of motion he sees things differently than the observer at rest at the bar.* You know that you are right, and I know it. If you read my previous posts more carefully you may understand than I don't want to counter the mathematics of Relativity. I am against some interpretations of relativity. Like the "multiple reality" argument, or like the present discussion. * And yes I was wrong. I forgot that when the traveler stops he jumps into another FOR and he must see what other people are seeing from this FOR.
  6. There must be a delay. If I am wrong, then the delay is somewhere else. Go and find it. As observed from the Earth, the image of the traveler going away is delayed. So logically speaking, the image of the Earth that as seen by the traveler is delayed too. There is no reason why one observer would see a delay and the other not.
  7. As I wrote above: if he stopped, it is 60 minutes aka1 hour (because he is 1 HL away). If he makes the U-turn immediately it gets complicated because you have to count for the velocity after the U-turn. It is this instant that Janus describes when the traveler is going back but observer on the Earth haven't seen the U-turn yet. But we have not reached an agreement on what is happening. Basically I am the bad guy disagreeing with everybody.
  8. When the traveler makes the U-turn, he sees the Earth still getting away from him. He will see the Earth stop going away and begin the rush at him some minutes after he made the U-turn, because there is a delay. The image of the Earth takes some time to go to the traveler. See it otherwise: if the traveler stopped for a drink at destination, he would see the image of the Earth stop getting away from him after 1 hour. (approx 4 beers in Belgian units)
  9. Because what the traveler sees is the same (the mirror) of what the observer on earth sees. When the traveler goes out, as much the distance to the Earth increases, as much the delay increases too. And on the return trip, as much the distance reduces, so reduces the delay. The turning point will not be reached at the middle point (in time) of the travel. Although it will be the middle point in distance. If the clock makes the U-turn at the middle point in time, at the end of the travel the clock will miss the Earth by a distance corresponding to the delay (if I am correct, this is more a guess than an accurate calculation). Check with Janus example.
  10. Galilean relativity yes., but not Relativity (because of SOL). But even with Galilean relativity, if the Earth is also moving in the direction of traveler B, but at a lower velocity (which is the equivalent of the delay caused by SOL), the outbound will not be the same with the inbound. Because of the delay, the outbound will look longer than the inbound, it appears in all examples posted by Janus.on page 1 of this thread.
  11. Specifically Half the rate for one hour, so the outbound traveler sees Earths clock clicking only 30 minutes. Double the rate for returning: Earths click 60 minutes for a return travel of 30 minutes. The traveler sees at his clock that he traveled 1h30 minutes, the same as the observer on Earth. That is what I call "symmetry". The trip lasted the same time for the traveler & for the guy at rest on Earth. You have been fooled.
  12. . (previous answer erased, obvious mistake) The half and double rate are observed from the Earth. From this FOR, the outbound & inbound travel will not be observed the same. I suspect that the same goes for the traveling twin, he will not measure the same time for the outbound & for the inbound, because of the Doppler shift taking place only in 1 direction (toward the Earth). So it is not "Half the rate for one hour, and double the rate for one hour" it is "Half the rate for one hour, and double the rate for less than an hour.
  13. Fine. Why the relativistic scenario does not end to be symmetric? While presenting a totally symmetric equation for frequencies? see my post above. A sees B traveling 2 and 1/4 hrs while going away. And you say that A sees B traveling back in 15 minutes? Or 2 and 1/4hrs divided by 3 = 45 min? Why do you divide 45 minutes (as observed by the traveling clock)? instead of the time observed by A?
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