michel123456

Now what I am wondering is this: Calculation shows that E.T.'s velocity from Earth's perspective is 365243.09 days. Maybe, I say "maybe" it is wrong to "correct" this number by substracting the time for light (365242 days) and obtaining the "1 day" figure. IOW maybe the astronomer does not "observe the ET coming in 1 day" but in 365243.09 days as the math suggest.

Yes, I acknowledge that the astronomer can calculate that, the same way he calculated the distance 1000LY in the first place. In his FOR. But what he sees, what he lives, what he feels, the motion that crashes upon the Earth has nothing to do with a velocity "less than c". As you said it takes 1000 years + 1 day ( or 854,4 days I have lost my mind on this), it does not take 1 day as the astronomer have seen. I know that the 1 day figure is wrong, it is a mistake, it is false. But that is what have been observed, all the way long, from the boarding till the crash. Like the piano of George Clooney in the coffee commercial. You see it coming very (very very) fast. Then comes the astronomer and tells you that no, the piano was gently floating in the air and landed smoothly because that is what his calculations say. That baffles my mind, forget it, I am afraid I will die with that misconception.

Ha. back to the beginning. He didn't travel in his own FOR, he was at rest. The Earth traveled to him in 854.4 days, and the distance was not 1000LY in his FOR (If I am not messing things again).

Fair enough. Yes I committed errors but truly, genuinely, I don't understand, and I am sure now that you don't understand my question. Just to put an end at this scenario so that i can learn something from it: and that other people reading this long thread get a final correct idea. O.K. the astronomer observes the motion of the E.T. coming from a planet 1000LY away in 1 day. He observes this hilariously fast motion from the beginning till the end until the E.T. crashes on Earth. The E.T. made the travel in 854.4 (of his) days. He appeared (falsely) as if he made the travel in one of our days. The astronomer observed the E.T. live in fast motion. Is that it? Where is Delta?

O.K. the astronomer observes the motion of the E.T. coming from a galaxy 1000LY away in 1 day. He observes this hilarously fast motion until the E.T. crashes on Earth. The E.T. made the travel in one (of his)day. He appeared (falsely) as if he made the travel in one of our days. Is that it?

If the diagonal is c squared then units are m^2/sec^2 You are correct that the pythagorean method uses addition. That's an issue, yes. Usually you cannot add apples and oranges. But many diagrams use different units on X and Y axes and work well. Of course in this case what the diagonal represents is not always so evident. I understand your idea about the horizontal axis being the same as the diagonal, in which case the vertical value is zero. I understand your idea about representing mass zero (a photon). That gives you units m^2/sec^2 on the X axis (the units of c^2) And units of Mass on the Y axis. Having Mass multiplied by c^2 then the surface of the diagram is Energy (e=mc^2) I remember having seen that somewhere. That makes sense. (edit) But I doubt the diagonal has a value of c^2. I don't know what would represent the length of the diagonal. The other problem is that c^2 is a constant. Only mass is a variable. The next problem is that in this diagram, for m=0, Energy=0 which is not what happen for a photon (a photon has energy)

My bad. So if i understand clearly, it is physically possible to travel in one day to the other side of the universe at 0.9999999999999999999c And at c there is no horizon problem.

(emphasis mine) No? One does not "observe" speed? You surely don't mean that the E.T. aged only 20h during the travel. For him the travel had a duration of many years. So the astronomer observed the E.T. living in fast motion. That's what I ment.

Read again what you wrote: "The diagonal is c." If the diagonal is c then why do you calculate the diagonal again stating : This becomes c squared for the diagonal. = 89875517873681764 m/s this is wrong. you should have stated otherwise, do you know trigonometry? What is the angle needed in order to make correspond c for the diagonal and c/2 on the X axis? ------------ (edit) You' re not trying to add apples and oranges, you are multiplying apples by oranges, I think physicists do that all the time.

(emphasis mine) the bold part is not correct. You can look at yourself entirely, from head to feet, in a mirror half your width and about 3/4 your height.(25 cm width, 140 cm height) Make this little experience: stand in front of a mirror at about 60 cm of distance and mark upon the mirror the height and width of your reflected face (using a permanent marker). My face on the mirror enters a rectangle 7cm width over 12 cm height, about half the real dimensions, at 60 cm distance. The rest of the post looks correct to me.

I feel really alone. I need Spielberg to direct the scene. An astronomer looks in his telescope and observe an E.T. coming to Earth. In his telescope, the E.T. comes from a galaxy one thousand LY away and made the trip in one day, apparently , since the astronomer during 24h was really observing the travel of the E.T., each second of it, from the beginning when he embarked till the end, traveling 1000 LY in one day. (that is apparently - mistakenly 365000 times c) and aging about 400000 times faster than us. Then, after 24h of observing, the astronomer sees the E.T. crashing on Earth at velocity less than c. At which moment did the false impression disappear from the telescope and changed into the correct "less than c" velocity? -------------------- Michel's answer: at no moment the astronomer observed the correct speed. The astronomer observed the E.T. crashing upon Earth at velocity 365000 times c. Only after calculating can the astronomer obtain the correct result.

There are 2 solutions. Either Swansont did not understand the question either Michel does not understand the answer. Actually Michel understands that the answer does not answer the question. @Delta Did you understand Swansont's answer? You seem to have a good grasp at understanding other people problematics.

Something like that.

Is there incomprehension here? I was not asking how the traveler accelerates (decelerates) into my frame. -------------- never mind. I hope that someone around here understands my question.

That is not acceleration. We are talking about the transit between an "apparent velocity" (that dubious unphysical effect you have seen with your own eyes looking into your telescope) and "real physical velocity" when the writer knocks your door. When did that change happen?

O.K. Good example (good to Delta too) Very understandable. Now what I cannot understand: In the letter example, one can imagine the letter traveling on the back of a donkey and the writer traveling with an airplane. In such a a way that the traveler can knock at your door and enter gently even before the letter. With light and massive objects, there are some conditions: 1. the "letter" takes the faster existing airplane and the writer can take only a slightly slower airplane and 2. the straight path of the letter is the same as the straight path of the writer. When the letter arrives, it crashes upon your house because its velocity is c. I don't understand how the hell the traveler will knock your door gently* (traveling at less than c) when you have seen him with your eyes coming to you at "apparent velocity" thousands of times that of the airplane. *exaggerating a bit for the sake of the argument.

-------------------- Here you are (hoping not being toooo wrong) I needed to introduce a "help observer" at point R. The velocity that observes the observer at o is the same as observed at R with a delay caused by c. (words in italic edited) It is clear like a crystal. So the statement: Is correct. and this part Is correct too.

It is clear that you want to know the 'apparent velocity' of the blue line from the perspective of an observer 1.333 light-years from earth. This diagram shows what earthlings observe. It does not show what an observer 1.333 light-years from earth observes. It does not show what Tom observes either. Nor any other observer in any other FOR. _You want me to draw a diagram that shows an observer 1.333 light-years from Earth, in the same FOR with the Earth, observing Tom that leaves the Earth at 0,5c (as measured in Earth's FOR), then turning back to Earth at 0,5c (Earths FOR). I suppose that you want Tom to approach the observer. That's a bit tuff. Working on it. Which of my statements was wrong? This one?

(emphasis mine) i didn't even read the question. from the moment you propose to put the observer out of his world line, it stops to make sense. You can't make a diagram for observer Michel in Greece describing what Iggy observes where he is. either you will make a diagram for Michel and the Y axis will be his world line, or make another diagram for Iggy and then Iggy will be on the Y axis. Putting the observer out of the Y axis is wrong.

No. If you are the observer, you are on your world line. You cannot take the diagram and state that the observer is anywhere on it.

"Light takes 1 year to make the trip. And you have seen in one year a spaceship making the trip." That was with 0,5c What about going at 0,99c? In this case I observe in my telescope Tom waving his hand, embarking, and landing the next day! If I follow this logic, I may observe in my telescope an E.T. coming from the other side of the Universe in just one day because he traveled at 0,99999999999 c. And I would have observed something that we deny light can do. (emphasis mine) You can't do that. The observer is always upon his own world line, the observer is always on the Y axis.

If I use the same method I should use a part of time twice. Explaining The green triangle shows me what the earth observes till Tom reaches point R. I hope everyone agrees on that. From this moment and after, Tom leaves point R. I cannot use the time before he left (the orthogonal projection of R on the time axis) because this part of time has been already used in the green triangle. (emphasis mine) Not your bookkeeping, your eyes. Light takes 1 year to make the trip. And you have seen in one year a spaceship making the trip.

What i present is that the projection onto the time axis is through the 45 degrees diagonal. It is not possible to observe velocity in "real time" (with a horizontal projection). At time 2, the Earth observes Tom reaching point R. The velocity is 0,5 c (the slope of the line as you said). So earthlings calculate that after 2 years earths time, Tom has reached a point 2 LY away (Earths distance) minus the time (earths time) the light had to reach the Earth. That is point R. So i believe until point R we should not have any disagreement, except the fact that I call "apparent velocity" the ratio D/T. I will stop doing that. The question is: _is that the point R where Tom makes the turn? or _at what point should Tom make the turn so that his velocity (as measured from Earths FOR) is the same when coming back? I mean: in the standard explanation it blows my mind to think about observing Tom waving hand on a planet 1 LY away and one year after shaking my hand after landing at "regular" velocity. The astronomer looking in his telescope would be incapable of seeing Tom traveling back. Tom would be apparently moving as a photon! In other words the astronomer would see Tom landing on Earth popping out of nowhere. On the other hand the velocity of Tom landing must be the same as the velocity when he left. In my diagram, the time lapse between the waving and shaking is 2 years, because Tom comes from a planet 1LY away and travels at 0,5c. Which is the regular definition of velocity V=D/T Imagine an E.T. that you observe in your telescope. The E.T. is on a planet 1 LY away. He waves his hands and embark a spaceship that travels at 0,5c. In how much time do you expect to see him landing? (edit) If you answer "next year" (counting 2 years of travel minus 1 year delay of the waving hand image), that means from earth the E.T. is seen as a photon.

What is the terminology of velocity if it is not V=D/T?

That's what blocked my mind so many times. Then I thought that we interpret the velocity as the slope of a line. The diagram does not speak. We interpret the diagram as we wish. So in this diagram, the slope of a line is velocity only for c. But it is a special case. In this diagram, as a generalization the slope is not velocity. But it is always what the equation tells us: it is the ratio height/basis.