michel123456

Marx Lysenkoism flourished under totalitarian stalinism. Other totalitarian governments of the time (the fascists) also wanted to influence everything in the society, including science. It is not a privilege of socialism. Lysenko was presented as a new kind of scientist coming from the proletarian mass, his ideas look successful, it may have been sufficient. The rest is most probably just a monumental error (like WW1 and WW2).

Is this an allusion to Lysenkoism?

Correct. The green triangle has height D and basis T (numbers are not important) The blue triangle has height D and basis T The formula for velocity is V=D/T So the 2 rectangles represent the same value V=D/T Meaning simply that the ratio of D/T for the 2 rectangles is the same. Of course the product DT/2 is the same too. It is the triangle area but it is irrelevant here.

You see, "day 360" is Earth's time. In fact Tom makes the turn at half his time. What happens is that Tom travels X days forth and X days back in his FOR, because he travels at the same velocity. So the symmetric diagram is from Tom's FOR and not from Earth's FOR. IOW distances & times on the conventional diagram are Tom's, and not Earth's. (edited) You see, at the end, your calculation show that the flashlight is observed flashing once a day (Earth's FOR) although at the beginning it was assumed that the flashlight was flashing once a month (Earth's FOR). There has been a switching somewhere. In fact the "once a month" is always in Toms-flash-light FOR and never in Earth's FOR. If that is the case that means that at day 59 the Earth does not observe the first flash. In Iggy's graph, time is Tom's, and distances are Tom's. Quoting myself: ----------------------------------------------------------------------------------- . . . 299 792 458 m / s It is the red triangles on my diagram (see below).

I cut out your post for the sake of clarity and emphasis mine: "29 light days" as seen from which FOR? Are this the words of Tom or the words of Earth? This is a distance you are talking about and distances are measured differently from one FOR to another. "the photon take 29 days to reach the Earth" as seen from which FOR? This is a time. Is this the time as seen from Tom? (as seen by a neutrino?) "Day 30": _that is clearly Earth's time. "Flashlight flickers 29 light days from Earth (velocity: 29/30c or 0.967c)" _that is clearly a distance in Earth's FOR. (or Am I wrong? Tom-flash-light didn't traveled 29 days of his time.) "Day 59": _that is clearly Earth's time. "Earth sees flashlight flicker 29 light days away (apparent velocity: 29/59c or 0.492c)" _Here you have supposed that Tom-flash-light traveled 29 light-days-earths-FOR in 29 days-earths-FOR, But for the flash to click, one has to wait a month in Tom's FOR. Or was that each month in Earth's FOR? It is a mess. I prefer working with diagrams. That is the basic of a spacetime diagram. A velocity is a ratio, and an angle is a ratio too. Oh? I thought velocity was frame dependent. There is no one "distance traveled" and "time traveled". There are an infinity of "distances observed" and "time observed", each one for each FOR which divided give a "velocity observed". That is not my invention. You are speaking like there was an absolute velocity. ---------------------------- (edit) I could understand the following: Tom leaves the Earth and for 2 years he travels to a point. Then he comes back, traveling 2 years and landing on Earth. In this case, Time and distances are both in Tom's FOR. Not in Earth's FOR. The diagram is the same as in a conventional diagram, the labeling is different.

That's about buoyancy, out of the atmosphere it does not work. I was thinking more at an early concept (1865). To make a projectile. And to incorporate inside the projectile a device that can play with momentum: deploy it for velocity or sustain it for stopping.

I am very glad that you raised this argument. The situation you describe is different, you have moved the goal post. The common understanding of the twin experiment consists in drawing a spacetime diagram as one would have drawn the path of an object bouncing on a wall, a kind of billiard game. My diagram consists simply in stretching the diagram and in such a way that the deformation equal to c. This stretching reflects the fact that earthlings observe all along the path the velocity of Tom being 0,5c. That includes also the instant where Tom is landing back on Earth. Note again that the 2 triangles have the same area, the same basis, the same height: they represent the same velocity. The stretching also reflects the fact that earthlings observe the turning point with a delay of c, although the turning appears to them at half the trip (as observed by them). In you argument, the earthlings did not send anything to make the supernova explode. There is no delay to include before the explosion. But if you whish, I engage you to make the diagram of the twin experiment with neutrinos sended from Earth and back and provide here the results of the standard solution. At what speed will the earthlings observe the (real) neutrinos?

Off-topic but funny. You know what scotus means in Greek: darkness. (an death in some circumstances). From the word σκότος came the verb σκοτώνω: to kill.

You must agree that velocity is relative. Each observer, from his own FOR, measures a different velocity. So each observer measures what I call an "apparent velocity" which is different for another observer in another FOR. The "apparent velocity" of light is the same for all observers. In this diagram velocity of light is a line at 45 degrees (or 135 degrees) oriented from the past to the future, always going from down to up in the diagram. Never going up to down. For all observers. In this diagram, from Earth's FOR, the apparent velocity is distance divided by time. It is not necessarily the angle of a line. It is the distance as read by earthlings divided by time as read by earthlings. ONLY for light the angle represents accurately velocity.

With all the risks, this below is what i think happens in the twins experiment: Tom is heading at 0,5c to point P. When Earth is at time M =2 years (half the waiting time), then earthlings observe Tom making the U-turn at point R. After that, earthlings wait for another 2 years before Tom comes back. Analysis: -notice that the blue and green triangles have the same area. Same basis & same height. _the velocity of Tom during the outbound is distance/time (the height of the green triangle divided by its basis) _ the velocity of Tom during the return travel is distance/time (the height of the blue triangle divided by its basis) The 2 velocities are equal to each other. Tom never reached point P.

You said "One year after the turnaround, earth would see the turnaround. I remember you knowing that" Yes I know that. But i cannot swallow that. I changed my mind. Let me explain. And tell me where I am wrong: a. On your graph, earthlings observe the turnaround when they are at time 3. b. After that, earthlings observe Tom landing back. That happens at time 4. c. the difference between the 2 events is 4-3=1 year . d. Earthlings have observed Tom coming from a planet 1 LY away in 1 year . e. Thus, the apparent velocity of Tom as observed by earthlings is c. ----------------- (edit) take some time and make a similar diagram with 0.66c (1 by 1.5) and you will be surprised of what the apparent velocity of Tom becomes. Here i did it for you: Tom is observed from Earth at point 2.5 Tom lands on earth at point 3 Earthlings have observed in their telescopes Tom coming from a planet 1 LY away in half a year. That is twice c! How can one observe an object traveling faster than light? It is wrong. (edit) I mean the diagram is wrong. Relativity is correct.

Re-read your own post and you will conclude that the earth actually observes Tom making the trip from 1LY away in 1 year. After that, replace 0,5c with 0.6c and tell everybody what the earthlings observe in their telescopes.

that makes no sense. The Earth does not welcome the image of Tom, but Tom himself.

Let's say that observers from Earth look in their telescope and 1 year ago look at Tom making the turnaround. in this case observers from Earth will be astonished to see in their telescope Tom coming in 1 year from a point 1 LY away. That is Tom is observed traveling at c. I am pretty sure that changing slightly the numbers in this scenario observers from Earth could observe Tom coming back faster than c! and that is not allowed.

---------------------- IOW the problem states: Tom is leaving the Earth at T=0 at v=0,5c At t=2 (Earth time) observers from Earth look in their telescope and observe Tom making a turn. At t=4 (Earth time) observers on Earth welcome back Tom landing at velocity 0,5 c.

Explaining my post #54 what does it mean "a velocity of 0,5c"? _it means that, as observed from Earth, the object travels an observed distance of 0,5 LY in an observed time of 1 Year. The common explanation of the twin experiment says otherwise. The common explanation says that the velocity as put on the diagram is 0,5 c and as observed from the Earth is different. That;s the nail of the question.

This is wrong. 1_What is the velocity of Tom at the begin of the travel, when leaving the Earth: it is 0,5 c as observed by the Earth 2_What is the velocity of Tom at the end of the travel, when coming back to Earth: it is 0,5 c as observed by the Earth The travel is totally symmetric as observed by the Earth. The Earth observes after 2 years that Tom has reached a point 1 LY away because Tom travels at 0,5 c as observed from the Earth. And not after 3 years. The symmetric diagram is made from the FOR of the Earth. (edited) _In your concept, when leaving, what is the velocity of an object that, as observed from the Earth, travels 1LY in 3 years? It is 0,33 C and not 0,5c It is contrary to what was stated in the beginning. _In your concept,when returning, what is the velocity of an object that as observed from the Earth travels 1LY in 1 year? that is c, that is not 0,5c and it is not allowed. At no moment the traveler is "chasing its own light signal", that is BS.

Yes I had the same question. It has been answered by (valuable) members in another thread that when the traveler remains inertial there is no paradox, everything is fine and everybody has the same age. I still wonder (and I am not alone) how one can figure out a twin experiment entirely based on inertial observers. See the interminable other thread. When someone "forget' what happens at the turning point the result is that would be observed for a twin that would not have turned back and finally reached a point very far away.

I think it is caused by the way human understanging after Galileo works. We are making measurements. A balance measures one kilogram, a ruler measures one meter, a clock measures one second, etcetera. Then people like Newton try to combine the measurements.

(...)This is obviously false, as we're moving in an orbit.(...) You may have noticed that the paths of many orbits alltogether give the shape of a spiral. Like in the solar system. See here and you can check that out everywhere you want.

I don't think it would be a good idea to make the scale going through living organisms.

Oops, old thread.

So you are not surprised that after so many posts full of equations the argument is still not resolved. You don't need math to solve this argument. Relativity is based on the principle that there is no preferential inertial F.O.R. When you describe a situation where all observers are inertial ones there should exist no paradox. It is a simple way to CHECK the maths. If you obtain a discrepancy, your maths are wrong. Md's argument is that one can construct an experiment where all observers are inertial ones and at the end obtain a discrepancy. He should look for the error. It is disappointing that he refuses to search for it. Worse, when showed where the "trick" is, he proceeds and invents another one. He behaves exactly like all those who present a free energy device. And you are supporting him.

I guess she must be christian for wearing clothes upon her sexual organs and not upon her face or her hair. Oh she's got no hair.

Event BC is special because there is a frame jumping. A similar frame jumping happens at AB and at AC. I'd suggest to open a new thread about frame jumping. There is no need to consider the particular case of a turning point: any frame jumping implies weird results that can be solved only by considering frame jumping as an extreme case of acceleration. IMHO of course. And IMHO xyzt is right against everybody. If my opinion counts...