Masanov Posted July 16, 2005 Author Posted July 16, 2005 Ct1 and Ct2 and Y belong to inside observer, outside observer cannot know ALPHA angle, only BETA angle. ALPHA times are the same. And the speed gamma*Y*cos Alpha proves that the distance is not simply Y, it should be [gamma*Y]. And in Y*sin ALPHA it should be then [gamma*Y]sin ALPHA. Then Ct1 and Ct2 are not relative distances for outside observer, between parallel mirrors we have relative distances to which postulates were applied. Here we don't have relative distances. E.G. relative distance [gamma Y sin Alpha] is shown without gamma.
Janus Posted July 16, 2005 Posted July 16, 2005 Ct1 and Ct2 and Y belong to inside observer' date=' [/quote']ct1 and ct2 are measured in S which is the frame of the "outside" observer. ct'1 and ct'2 are measured in S' the spaceship frame observer. Or weren't you paying attention outside observer cannot know ALPHA angle, only BETA angle. Why? What prevents him from knowing it? Unless you are claiming that the sphere is opaque and that the outside observer cannot see what happens in the spaceship, in which case this whole exercise is pointless. ALPHA times are the same. And the speed gamma*Y*cos Alpha proves that the distance is not simply Y, it should be gamma*Y. What speed [math]\gamma Y \cos \alpha[/math]? The only speeds in this example are v which can have any value <c and c, the speed of light in a vacuum.
Masanov Posted July 17, 2005 Author Posted July 17, 2005 Yes, I was inattentive. Sorry. But it is also all your fault and inattentiveness. I many times asked people to discuss the picture http://www.rainbow-calendar.hotmail.ru/7 I was thinking you have started discussing it. In this my picture light (3 pulses) was shot NOT in the moving sphere S', but vice versa in S. Sphere S' is catching up with these three light pulses. Relative distances seen [by the moving observer] BETA1, BETA2, BETA3 should be divided by C to receive times, different from each other. CONCLUSION of the picture is: we have three times in one spaceship. Your discussion is different, but nevertheless, sorry: your distances ct1 and ct2 are different, your distances ct1' and ct2' are the same. From Newton views: you cheat with C in ct1, because there should be relative speed instead of absolute, meaning vector C + vector V... Plus you cheat with gamma-factor... Nevertheless, what I have scooped, that you apply to relative distances ct1 and ct2 absolute speed C. So, pls understand my "7.png", divide relative BETA-distances by C and receive 3 times. Following my 7.png, I misunderstood you. As I followed your and Swansont thoughts, you are trying to say that ALPHA distances can have gamma-factor and beta-distances not. This means that you do not "allow" me to divide beta distances by C and receive their times, so we should use only alpha times + only gamma-factor. As for your experiment, pls be specific and slow. Try to imagine in the sphere S' three light pulses, do not return them back. Compare distances ct1, ct2 and ct3, compare distances ct1', ct2' and ct3'. In sphere S' times t1', t2', t3 are equal, in the sphere S times are different. So in one sphere three different times. This is the problem, not that one time, you have derived, having inserted gamma-factor.
Janus Posted July 17, 2005 Posted July 17, 2005 I did not ask you what the speed [math]\gamma Y \cos \alpha[/math] is? You said:And the speed gamma*Y*cos Alpha proves that the distance is not simply Y' date=' it should be gamma*Y.[/quote']To which I asked what speed[math] \gamma Y \cos \alpha[/math] you were talking about, since there is no such speed in the example. I told that distances ct1 and ct2 are [gamma * Y]... in your views as ALPHA distances for the moving outsider. ct1 and ct2 are distances the light travels in S before it hits the sphere. They contain gamma factor, which you do not want to assign to insider's [math]Y[/math]. because that is not what relativity predicts. If you wish to discuss Relativity you must first learn it, something you've obviously have neglected to do. Let's go further. As you poorly showed BETA distance and angle BETA at which outsider sees the light, I stand by my images, as they are accurate. If you cannot see that, then that is your problem. let's use the picture http://www.rainbow-calendar.hotmail.ru/7 Distances to 3 light pulses viewed by the moving otsider, that is BETA distances, are different and if divided by C, they will produce times BETA. So why you assign time ALPHA of the distance [math]\gamma Y[/math] which the moving outsider does not see to these three cases of times BETA. I start out with the assumption that the light is emitted at angle Alpha as seen by the inside observer. I then can calculate the path the light will take with respect to the outside obserever, based on that assumption and the velocity v and knowing the speed of light. Alpha distance cannot be seen by the moving ousider, so you have no rights to assign gamma factor to the alpha distance this outsider does not see. one does not have to see something to know that it is true. I didn't have to be around 2,000,000 yrs ago to see the light leaving the Andromeda galaxy to know that the light from it reaching the earth now left it 2,000,000 yrs ago. Thus just because the Outside observer doesn't see the light traveling at angle ALPHA does not mean he cannot know that the light travels at angle alpha as seen by the inside observer. You said contraction is possible, when it is seen, but distance ct1 or ct2 or gamma*Y cannot be seen by the moving outsider; apply your reason to beta distances. He sees light at angles beta, which are different. OF course the outside observer in S can see ct1 and ct2 As these are the distances the light travels as seen by him by definition!Sure, I could have determined angle BETA and worked from that, but to determine angle BETA, I have to start from angle ALPHA and determine it from that. Or conversely, I could start with Angle BETA and derive angle ALPHA from it. Either way you go,the two angle are co-dependent. Another question, why beta distance [math]Y\sin\alpha[/math] is mentioned without contractions? If you do not apply contractions to beta distances, then do not apply time induced by these contractions. [math]Y \sin \alpha[/math] is measured along a line perpendicular to the line of relative motion, Length contraction only applies to distances that are parallel to the line of relative motion. Again, if you had actually tried to learn Relativity, you would know this. -If you see light pulse at BETA distance at 90 degrees to the spaceship's movement, at which angle was it shot outside the spaceship, uh? [math]\arccos \left( \frac{v}{c} \right)[/math] Can you see this angle? Sure, just imagine that the spaceship is open and the light is passing through a series of glass plates. The light will slightly illuminate each plate as it passes though. If the outside observer draws a line through these spot of illumination they will follow this angle.You applied gamma-factor to the distance, you cannot follow, apply gamma-factor to BETA distances. As seen [math]Y\sin\alpha[/math] does not have gamma-factor. If you can't follow the example just say so, but your inability to understand it speaks more about you than it does the example.
Masanov Posted July 17, 2005 Author Posted July 17, 2005 Your message 9:25 PM is misleading and confusing, maybe because of my inattentiveness, pls delete it.
Janus Posted July 17, 2005 Posted July 17, 2005 Yes' date=' I was inattentive. Sorry.But it is also all your fault and inattentiveness. I many times asked people to discuss the picture http://www.rainbow-calendar.hotmail.ru/7 I was thinking you have started discussing it. In this my picture light (3 pulses) was shot NOT in the moving sphere S', but vice versa in S. Sphere S' is catching up with these three light pulses. Relative distances seen [by the moving observer'] BETA1, BETA2, BETA3 should be divided by C to receive times, different from each other. CONCLUSION of the picture is: we have three times in one spaceship. It does not matter whether you consider the light as being shot from the S or S', you get the same results either way. That is the whole point of Relativity. Your discussion is different, but nevertheless, sorry: your distances ct1 and ct2 are different, your distances ct1' and ct2' are the same. As they would be. From Newton views: you cheat with C in ct1, because there should be relative speed instead of absolute, meaning vector C + vector V... Plus you cheat with gamma-factor... Nevertheless, what I have scooped, that you apply to relative distances ct1 and ct2 absolute speed C. So, pls understand my "7.png", divide relative BETA-distances by C and receive 3 times. We don't live in a Newtonian Universe, we live in a Relativistic Universe. And in a Relativistic Universe the speed of light is invariant; Meaning that all observers will measure light in a vacuum as moving at c with respect to themselves regardless of any relative motion thye may have to the source. This is the second postulate of Relativity. Following my 7.png, I misunderstood you. As I followed your and Swansont thoughts, you are trying to say that ALPHA distances can have gamma-factor and beta-distances not. This means that you do not "allow" me to divide beta distances by C and receive their times, so we should use only alpha times + only gamma-factor. As for your experiment, pls be specific and slow. Try to imagine in the sphere S' three light pulses, do not return them back. Compare distances ct1, ct2 and ct3, compare distances ct1', ct2' and ct3'. In sphere S' times t1', t2', t3 are equal, in the sphere S times are different. So in one sphere three different times. This is the problem, not that one time, you have derived, having inserted gamma-factor. there is no problem here, it is just a different aspect of the fact that time is relative. In this case it is the idea of simultaneity that is relative. In S', the three pulses strike the walls at the same time In S they do not. It means that "at the same time" is different for S' than S. this is a result of living in a Relativistic Universe rather than a Newtonian one. PLS DELETE YOUR MESSAGE SEE NEXT PAGE, COMPLETE MISUNDERSTANDING,
Masanov Posted July 17, 2005 Author Posted July 17, 2005 You agree that three pulses in my 7.png have three different times?Really? T90 = T sin alpha1, tbeta2= T sin alpha2/sinbeta2, tbeta3=T sinalpha3/sinbeta3???
Janus Posted July 17, 2005 Posted July 17, 2005 Here is an animation that shows what happens according to an observer in S for which the ship is moving. The white dots are the light pulses, the yellow lines are the path that the pulses follow with respect to S' the green lines show the paths the pulse's take with respect to S'(the ship}
Masanov Posted July 19, 2005 Author Posted July 19, 2005 Now in your experiment we have 3 pulses with three different times in the sphere S, while in the sphere S' they have common time. You inserted gamma-factor and proved common time T [see your calculations dated 07-12-2005, 12:40 AM] for all these 3 pulses. But! Time T is combination of times to the reflector and from the reflector. We cannot use T instead of t3, t2, t1, which are different and are parts of T. So, returning back to our discussions, Tom said that there cannot be different times t1,t2,t3 and we have to use time T for all directions ct1,ct2 and ct3. How can we use this time T instead of t1,t2,t3, when T is a combination of times to and from the reflector? Times t1,t2,t3 being only the part of T! Newtonian appologists can point these times t1,t2,t3 and ask, how on Earth these different times can coexist in one spaceship for one observer in the spaceship, uh? They will say: "Einstein simply changed relative speeds with times. Relative speeds can coexist in one spaceship, but not relative times." Newtonian apologists can also following your calculations derive common speed for all the pulses. The speed is the distance-factor, not time-factor, so gamma-factor [having speeds] is to be applied rather with Newton's speed, than with Einstein's time, and light speed gamma-contraction is more likely, than time gamma-contraction. But what they will do with this common speed, [the same is true with common T]. Difference in times t1,t2,t3 is the meaning of my thread and is not a special theory of mine, as was mentioned by someone. Your derivation of the common time T can be explained by the newtonians as the try to escape from refutation of the theory of relativity. Gamma-contraction cannot help: times t1,t2,t3 will remain defferent. Special remarks: in your experiment vectors of the speed C in spheres S and S' are directed differently; it cannot be so, one of the vectors should be relative, should be a combination of two vectors V and C, the newtonians could also point to that fact as a mistake: when you see the light pulse in the sphere S, shot in the sphere S', you see the different side of this pulse ["different side of the moon"]. There is this side movement of the light pulse along distance ct1... Plus in your animation this side movement is not shown + and it is plus to you, because you showed relative speeds of light pulses for some absolute time.
swansont Posted July 19, 2005 Posted July 19, 2005 Difference in times t1' date='t2,t3 is the meaning of my thread and is not a special theory of mine, as was mentioned by someone.[/quote'] If these are the one-way times, there is nothing in SR that requires these to be the same for the stationary observer, they will be different because the path lengths are different. It is the round-trip times that are equal.
Masanov Posted July 19, 2005 Author Posted July 19, 2005 Now it's Ok, you said times t1,t2,t3 difference is misapplication of postulates [06-27-2005, 09:14 PM]. Now, pls develop, how we could simultenously have three different times in a spaceship? . Decipher pls: t1 - we can see it turning the head to the direction ct1; t2 - we can have it turning the head to the direction ct2; t3 - we can have it turning the head to the direction ct3. So, time is where in our heads? On our watches? Then, if there would not be light return, what time T will you have? And ridiculous exp.: how twins should behave to dilate time t3, to hold the head to direction ct3 20-100 years?
swansont Posted July 19, 2005 Posted July 19, 2005 Now it's Ok' date=' you said times t1,t2,t3 difference is misapplication of postulates [06-27-2005, 09:14 PM']. Now, pls develop, how we could simultenously have three different times in a spaceship? . Decipher pls:t1 - we can see it turning the head to the direction ct1; t2 - we can have it turning the head to the direction ct2; t3 - we can have it turning the head to the direction ct3. So, time is where in our heads? On our watches? Then, if there would not be light return, what time T will you have? And ridiculous exp.: how twins should behave to dilate time t3, to hold the head to direction ct3 20-100 years? (You should use post numbers; times depend on the time zones.) I had previously said (post 48) that the pulses don't hit the wall at the same time, i.e. they aren't simultaneous. There's nothing mysterious about three separate events happening at three different times, so there's nothing to decipher. To demand that they hit the wall simultaneously in the stationary reference frame (the frame not moving with the ship, depicted by Janus in post 83) is a misapplication of the theory. To the observer in the center of the spaceship, in the spaceship frame, the events are simultaneous. But not to an observer in a different frame of reference. Simultaneity is not an absolute.
Janus Posted July 19, 2005 Posted July 19, 2005 Now it's Ok' date=' you said times t1,t2,t3 difference is misapplication of postulates [06-27-2005, 09:14 PM']. Now, pls develop, how we could simultenously have three different times in a spaceship? . Decipher pls:t1 - we can see it turning the head to the direction ct1; t2 - we can have it turning the head to the direction ct2; t3 - we can have it turning the head to the direction ct3. So, time is where in our heads? On our watches? Then, if there would not be light return, what time T will you have? And ridiculous exp.: how twins should behave to dilate time t3, to hold the head to direction ct3 20-100 years? To any observer traveling with the ship, this is what happens: No matter what direction he is facing.
Masanov Posted July 22, 2005 Author Posted July 22, 2005 Animation of the light pulse between the parallel mirrors: Along the perpendicular line, the light pulse flies with the relative speed, along skewed line it flies with the absolute speed. Along perpendicular line the light pulse flies askew, by its side. http://www.rainbow-calendar.hotmail.ru/Relativity_Experiment(2).htm In Janus's calculations we saw two vectors C and V added as Y*sin alpha and Y*cosalpha, and finally Janus divided the resulting distance by C, receiving time t1. And the same was the case with all three paths in our new experiment with three light pulses ct1, ct2, ct3. So, to check times in directions ct1, ct2, ct3 we have to divide these distances by [C and V], not just by C, and the resulting times should be equal. You missed factor-V! You can argue and say, that only one observer can see relative speed [C + V], another sees C. Then I would ask, which observer saw the relative speeds [C + V], observer S or S'? The observer S could also have the same experiment as the observer S' and could also have common time t for all his pulses: in this case S has time t and S' has times t1,t2,t3. So the sphere S can have times t and t1,t2,t3, and the sphere S' can too. In your calculations distance ct1 was presented as (vector V + vector C) time t. The time t1 was presented as [(vector V + vector C) time t / C], so the formula to derive is: (vector V + vector C) time t = C time t1, which is relativity formula. For me this formula only means, that the observer S shooting the pulse in direction ct1 will wait time t1 until ct1 is passed by his pulse, while he would wait only time t, watching the other pulse [shot in S'] moving with speed (vector V + vector C) along the same path ct1. To prove that the pulse S' is much quicker than the pulse S along the same distance ct1 see the animation: http://www.rainbow-calendar.hotmail.ru/Relativity_Experiment(3).htm. This experiment being done officially, it will prove the relativity of light speed.
Masanov Posted July 22, 2005 Author Posted July 22, 2005 Janus Post: 88 Isn't your sphere S a spaceship? Nevertheless in the sphere S you can also have time t, equal to time t in sphere S':imagine they both have this experiment, post 88 But can't you be more attentive? Or should I follow each my message with pictures or animations? To be more specific, OK, instead of the word SPACESHIP in my post 86 use the phrase "immovable sphere S in the experiment with three light pulses, proposed by Janus, see posts 83, 84 + 68(with one light pulse)+Animation 89. In this sphere, shown in the post 83 by Janus (the sphere immovable in the animation, not this shown in logo), we could see three times t1, t2, t3, which is impossibility, that is, refutation of the theory of relativity: Defferent distances divided by C give different times, which means nonsence.
swansont Posted July 22, 2005 Posted July 22, 2005 So' date=' to check times in directions ct1, ct2, ct3 we have to divide these distances by [C and V'], not just by C, and the resulting times should be equal. You missed factor-V! Not according to SR. The speed of light is c for all inertial observers. Janus was giving the SR result.
swansont Posted July 22, 2005 Posted July 22, 2005 In this sphere' date=' shown in the post 83 by Janus, we could see three times t1, t2, t3, which is impossibility, that is, refutation of the theory of relativity:[b']Defferent distances divided be C give different times, which means nonsence.[/b] No, it means that the events are not simultaneous in that frame. Simultaneity is not absolute.
Masanov Posted July 22, 2005 Author Posted July 22, 2005 Any comments to animation? In the animation (3) the light pulse of S appeared too quick, but generally, I think, it's clear. Swansont, you repeated many times, that according to SR... You agreed with three times, for me this is important and enouph. I understood at once, that three times in a spaceship for you is not a nonsence: a cockpit can have 9:00 AM, a restroom can have 13:00 AM, a dining room 18:00 AM, as for you this is ok. Yesterday is afore, tomorrow is astern. I understand that you want to beat records by short comments and be Almighty 3000 posts, playing games and giving short comments. Don't butt in between real discussions. If you show in, give food for minds, prove. Or else I will not take you seriously. I erred in my post. Serves me right' date=' trying to watch the game and post at the same time.[/quote']
Janus Posted July 22, 2005 Posted July 22, 2005 Any comments to animation? In the animation (3) the light pulse of S appeared too quick' date=' but generally, I think, it's clear. Swansont, you repeated many times, that according to SR... You agreed with three times, for me this is important and enouph. I understood at once, that three times in a spaceship for you is not a nonsence: a cockpit can have 9:00 AM, a restroom can have 13:00 AM, a dining room 18:00 AM, as for you this is ok. [/quote'] As determined by someone to which the spaceship is moving this can[i/] be true. Provided that by this one means that if, As determined by someone in the ship, clocks at these three positions all show the same time at any given instant. Then, according to the person that the ship is moving with respect to, these three clocks at any given instant could read 9:00 , 13:00 and 18:00 (For the differences to be this large the ship would have to be very large and moving at a high velocity.) This is not nonsense, it is a consequence of living in a Relativistic universe, in which the concept of simultaneity is not absolute, but relative.
Janus Posted July 22, 2005 Posted July 22, 2005 Janus Post: 88 Isn't your sphere S a spaceship? Nevertheless in the sphere S you can also have time t' date=' equal to time t in sphere S':[/quote'] I have no sphere S or sphere S'. I have frames S and S'. I have one spaceship/sphere, to which I did not give any designation. The animations show this same sphere and the same light pulses but as seen from either S or S'. Post #83 show these events according to S, and Post #88 shows these events according to S'.
Masanov Posted July 22, 2005 Author Posted July 22, 2005 I see, what you mean. Did you notice, that the light speed of the pulse S' is relative to the sphere S?
swansont Posted July 22, 2005 Posted July 22, 2005 Any comments to animation? In the animation (3) the light pulse of S appeared too quick' date=' but generally, I think, it's clear. Swansont, you repeated many times, that according to SR... You agreed with three times, for me this is important and enouph. I understood at once, that three times in a spaceship for you is not a nonsence: a cockpit can have 9:00 AM, a restroom can have 13:00 AM, a dining room 18:00 AM, as for you this is ok. Yesterday is afore, tomorrow is astern. I understand that you want to beat records by short comments and be Almighty 3000 posts, [b']playing games and giving short comments[/b]. Don't butt in between real discussions. If you show in, give food for minds, prove. Or else I will not take you seriously. People are trying to help you understand relativity, which you obviously do not grasp. Showing a little respect wouldn't be out of line. Simply sticking to the physics at hand would be sufficient. Reading this might help you with your manners. But let me make perfectly clear that I do not give a rodent's posterior if you take me seriously. Is that supposed to be some kind threat? Nor do I give a damn about 3000 posts. My short comments are directly to the point of your errors, nothing more. Don't try to have a private conversation in a (semi-) public place. Originally Posted by swansont I erred in my post. Serves me right' date=' trying to watch the game and post at the same time.[/i'] Was there any point to quoting this, out of context? Anything other than to show that I am willing to acknowledge that I have made a mistake? Now on to your post: are you really saying that you can't have one event occur in the cockpit at 9:00, another event happen at the restroom at 13:00 and yet another happen at the dining room at 18:00? That is nonsense.
Masanov Posted July 27, 2005 Author Posted July 27, 2005 People are trying to help you understand relativity, which you obviously do not grasp. ... are you really saying that you can't have one event occur in the cockpit at 9:00, another event happen at the restroom at 13:00 and yet another happen at the dining room at 18:00? That[/b'] is nonsense. Repeat pls what do you mean by one event occur in different times? And what is nonsence. If you try to "help", then be specific and do not mix my arguments. We discuss Janus calculations: paths ct1. ct2. ct3, seen from the frames S. Janus said that we can have simultaneously in one spaceship different times: in a cockpit t1, in a restroom t2, in the dining room t3. You also agreed that paths ct1, ct2, ct3 are different and have different times. So, you have to admit that nonsence t1, t2, t3 is the requirements of the theory of relativity. +when you say everybody is trying to help me is the same... "think". Now I am trying to help you to understand all the nonsence you try to help me understand. If in the dining room you observe million directions, you observe million times. You ask me, whether it is a nonsence? Yes, it is the nonsense, but we live in a relativity world: this is a result of living in a Relativistic Universe rather than a Newtonian one.
Masanov Posted July 27, 2005 Author Posted July 27, 2005 I have no sphere S or sphere S'. I have frames S and S'. I have one spaceship/sphere' date=' to which I did not give any designation. The animations show [i']this same sphere[/i] and the same light pulses but as seen from either S or S'. Animation shows your experiment vice versa in S and S', innovation. The light pulse S was shot in frames S and along the alpha distance it is slower than the light pulse S' shot in frames S'. The main thing in the animation is to notice that the speed of the light pulse S' is bigger and is a combination of vectors C and V. In the sphere S' the light pulse S' has its time t, in the sphere S the light pulse S has its time t. So, we have the common time t. So the light pulse S' along alpha-distance has different from C speed, and the difference is famous V1-v2/c2.[/b] Einstein refers this difference to time t, not speed C so that in direction alpha we have time [time t * V1-v2/c2]. Any other light pulse of the sphere S' has different direction and different relative speed and accordingly different time from the point od view of S in its direction. These directional times [we discussed them as t1,t2,t3] cannot influence a simple cooking process in the sphere S, cannot slowing it down or speed it up. The spheres S and S' as we said have common time t, which cannot be substituted by the directional times. So, from the point of view of the observer S he can use his time t to know the speed of the light pulse S' (see animation) along the alpha distance, which is [C* time t * V1-v2/c2]. Expression according to the theory of relativity should go to time t, but since the observer S measures everything by his time t, the expression V1-v2/c2 goes to C and as animation shows the light pulse is really much faster than C but strictly to this expression V1-v2/c2. Animation (3): http://www.rainbow-calendar.hotmail.ru/Relativity_Experiment(3).htm. 3)Animation(4) http://www.rainbow-calendar.hotmail.ru/Relativity_Experiment(4).htm. shows different angles of light direction in the frames S'. These angles change relative alpha-distances, from which relative times [directional times] can be dirived. These angles cannot cause a certain time change in the sphere S? Relative times are non-obligatory times! They can be millions and millions and one cannot use them to measure processes, for example in cooking, because they are different from the time t in S and S'! Relative speeds as vector C + vector V do not contradict the theory of relativity and its postulates, as an example see the relative speed in "V2/C2" of the famous V1-v2/c2 expression. If relative speeds contradicted postulates, they would not be widely used in calculations? Relative time formula is "time t * relativespeed/absolutespeed" More over, the alpha-distance passed by the light pulse S' (as was said) is C * alpha-factor V1-v2/c2 * time t. So, the observer S having time t is not obliged to combine alpha-factor with time t: he combines alpha-factor with C, seeing the light pulse S' faster than C. ***alpha-factor: relative speed/absolute speed, or [vector C+ vector V/C]. If you change the angle of the light beam in the sphere S', do processes slow down or speed up in the sphere S? (See animation) Pls think carefully before you give any answer!!!! NEW HERE IS POSSIBILITY TO HAVE TIME T IN S AND S'. So, if the observer S sees the light pulse S' along alpha-distance he uses his time T and alpha-factor goes to C.
swansont Posted July 27, 2005 Posted July 27, 2005 Repeat pls what do you mean by one event occur in different times? And what is nonsence. If you try to "help"' date=' then be specific and do not mix my arguments[/i']. We discuss Janus calculations: paths ct1. ct2. ct3, seen from the frames S. Janus said that we can have simultaneously in one spaceship different times: in a cockpit t1, in a restroom t2, in the dining room t3. You also agreed that paths ct1, ct2, ct3 are different and have different times. So, you have to admit that nonsence t1, t2, t3 is the requirements of the theory of relativity. It's not one event. Each light pulse striking the wall is an independent event, and while they occur simultaneously in the spaceship frame, they are not simultaneous in the stationary observer's frame - they happen at different times.
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