Masanov Posted June 13, 2005 Posted June 13, 2005 http://www.rainbow-calendar.hotmail.ru/Relativity_Refutation.htm To refute A.Einstein and do not break brains is the best solution.
Tom Mattson Posted June 13, 2005 Posted June 13, 2005 All the images on your site are broken. It is impossible to read it.
Ophiolite Posted June 13, 2005 Posted June 13, 2005 Tom, you may have a browser problem, or incompatibility: I had no problem seeing them.
losfomot Posted June 13, 2005 Posted June 13, 2005 I cannot see them either. I am on a mac. I think that as something to do with it.
swansont Posted June 13, 2005 Posted June 13, 2005 I cannot see them either. I am on a mac. I think that as something to do with it. I think that it's really because maybe he's on Windows? (i.e. a non-compliant browser system) Anyway, I'm on a mac, using Safari. No links. And yet, I can say with a high degree of certainty that the website is wrong. I just can't point out where the mistake(s) is(are).
Tom Mattson Posted June 13, 2005 Posted June 13, 2005 OK, switching from Mozilla to IE did the trick. I can now see the images. Sadly, the website doesn't make any more sense than before. Some questions: 1. From the website, "Relative speed of bunch along the perpendiculat line is [imath]c^{\beta}=c\frac{sin\alpha}{sin\beta}[/imath]." "Relative speed"? "Relative" to what? And where does the formula come from? To get a relative speed you would have to use a rule for adding velocities. What did you use? 2. "In the theory of relativity this formula [imath]c^{\beta}=c\frac{sin\alpha}{sin\beta}[/imath] was represented only in case [imath]\beta = 90^o[/imath] when sin alpha has the following presentation [imath]\sqrt{1-(\frac{v_{90^o}}{c})^2}[/imath] and [imath]sin{\beta}=1: c=c\sqrt{1-(\frac{v_{90^o}}{c})^2}[/imath]." This is wrong. Relativity is not derived by a thought experiment that even remotely resembles the above. It is derived from the requirements that Maxwell's equations take on the same form in every inertial frame and that the speed of any light pulse is determined by any inertial observer is c. And the results do apply to a beam that is directed at any angle relative to the line of motion of the observer. The last equation (the one after the colon) does not follow from relativity at all. Since I do not think you've started the analysis correctly, I do not see any point going further until these two points have been cleared up.
Masanov Posted June 15, 2005 Author Posted June 15, 2005 How would Einstein giggle if imagined 3 light pulses in his papallel mirrors, providing the pulses were shot at different angles to the direction you move. With you speed C*cos alpha you will see only one pulse perpendicular to your movement, other pulses will have another times. The more pulses you experiment with, the more times you receive.
Tom Mattson Posted June 15, 2005 Posted June 15, 2005 You're still talking as though you have proven something. I'm afraid that your website is unclear at best, as I have explained. Could you please answer my 2 questions?
Masanov Posted June 15, 2005 Author Posted June 15, 2005 In times's formula we always talk about relative speed of light pulse (to moving objects), [simply Einstein always changes relative speed to absolute]. T90/T = C90/C=[imath] {\sqrt{1-(\frac{v_{90^o}}{c})^2} [/imath], was twice mentioned. If we have 10 light pulses with their different relative speeds to a certain spaceship, he would change these speeds to one absolute and say, Hur, Now we have 10 different times in one spaceship!!!
Tom Mattson Posted June 15, 2005 Posted June 15, 2005 You have not answered my questions, and you are still not making any sense.
Masanov Posted June 16, 2005 Author Posted June 16, 2005 Answering your questions WHAT IS RELATIVE SPEED and ORIGIN OF THE FORMULA sin alpha/sin beta: 3 light pulses could be shot outside moving spaceship at angles alpha1 alpha2 and alpha3 (at a direction the spaceship moves). In the spaceship these three pulses will be viewed at angles beta1 beta2 and beta3 by one person. Relatively to this person the pulses will go relative distances, namely c sin alpha1/sin beta1 t , c sin alpha2/sin beta2 t , c sin alpha3/sin beta3 t , For further explanation I use bracket. We can say, that pulses move relatively to the person with relative speeds (shown in brackets): [c sin alpha1/sin beta1] t , [c sin alpha2/sin beta2] t , [c sin alpha3/sin beta3] t , or we can say that these relative distances, these pulses fly at relative times (shown in brackets): c [sin alpha1/sin beta1 t] , c [sin alpha2/sin beta2 t], c [sin alpha3/sin beta3 t]. Only for one pulse that move at an angle beta=90, Einstein thought a theory. So Einstein formula of time is a simple equation [when sin beta=1]: [imath] (\frac{[c sin{\alpha}] t}{[c] t })=(\frac{c [sin{\alpha} t]}{c [t]}) [/imath] From this formula relative time is: [imath] [t sin{\alpha}] =(\frac{[c sin{\alpha}] t c [t]}{[c] t c}) [/imath] Sin alpha when beta=90 is a well known phrase [imath] \sqrt{1-(\frac{v_{90^o}}{c})^2} [/imath], or [imath] \sqrt{(\frac{(c^2-v^2)}{c^2})} [/imath], or [imath] \sqrt{(\frac{([c*sin{\alpha}]^2)}{c^2})} [/imath] , or simply [c sin alpha]/c. So: [imath] [t sin {\alpha}] = (\frac{sin{\alpha} [t]}{...})=[t] [c sin{\alpha}]/c. [/imath] To know relative time we first should know [c sin alpha]/c, not necessarily the speed of the spaceship! This means if we experiment with more then one apha angle, use more pulses, we come to more times in one spaceship. [More brainstraining]. When you say relative speed of light it looks naiive, but when you say relative time, it looks fashionable and gives brainstraining. Though relative speed and relative time deal with the same notion - relative distance (their common notion). 2. You asked the second question: where I got sin alpha/sin beta. If we follow pulse of light along perpendicular line to the movement of the spaceship, this pulse goes the distance [ct sin alpha]. If the angle beta will not be 90 and less than that, we will see the pulse at the angle beta <>90 degrees and the distance along the beta line is [ct sin alpha]/sin beta. Relative speed in this formula is (shown in brackets): [c sin alpha/sin beta] t Relative time in this formula is (shown in brakets): c [sin alpha/sin beta t] I ANSWERED ALL YOUR QUESTIONS about relative speed of light, and about my formula origin. In my site-page I wanted to point that brackets in the theory of relativity move: Original position is simple [t sin alpha c]. Time equation is: [t sin alpha] = [t] [c sin alpha]/c. As you can see here on the left side of this Einstein's time equation the brackets seesaw to the notion "relative time", On the right side brackets seesaw to the notion "relative speed of light pulse", because in the formula the meaning of [imath] \sqrt{1-(\frac{v_{90^o}}{c})^2} [/imath]is [c sin alpha]/c.
mezarashi Posted June 16, 2005 Posted June 16, 2005 You are confusing the propagation of light with that of sound in air. Moving towards a light source will not make the light appear any faster. You may at most experience a doppler shift if you are moving fast enough. If you align two sensors collinearly to any of the three pulses, you will find that the light waves will pass the two sensors at an interval that is equal to (distance-between-sensors/c). Your assumption of "relative speed" of light is already wrong. That is not what Einstein assumed. Tom Mattson had already quoted what the postulates were. I would say that to disprove a theory, you either 1. show an observation in which disagrees with the prediction of the theory or 2. show somehow that the theory's original postulates are incorrect. What you are doing is attempting to derive relativity using different assumptions (I don't even know what your postulates are), and then claiming that your results are different from Einstein's.
Masanov Posted June 17, 2005 Author Posted June 17, 2005 ABOUT RELATIVE SPEED OF LIGHT. If you chase the light pulse and catch it up with, its relative speed becomes 0. If you say, that the speed of light is always absolute and the light pulse cannot be reached, or even be close to, then you deny any movement to the light pulse, and even more: you deny paradox of twins, which needs the movement close to light speed, and deny relativity of time. So, in the theory itself there's no contradiction between relative time and relative speed. 1)somebody outside the spaceship can shoot the light pulse close to the direction where the spaceship moves at an angle alpha [close to 0] and this spaceship can move with the speed c*cosalpha to catch this pulse up with. This pulse being caught up with, its relative speed is 0. [time cannot be close to 0!] 2)reduce the size of parallel mirrors to microscopic dimensions, leave them with the caught up light pulse in your apartment and ride away from it at a velocity 10 km per hour. The relative speed of the pulse (left in the appartment) would be 10 km per hour. If you would not leave the appartment, then the relative speed is 0, not time? So [imath] t90/t = [c90/c=\sqrt{1-(\frac{v_{90^o}}{c})^2}] [/imath] is not a contradiction in the theory itself: relative speed and relative time together. Personally I see many contraditions in the theory of relativity. I explained Tom Mattson the origin of the formula and he in return has to agree or disagree with the experiment proposed. The Postulates are not only for 90 degrees relative distance to the light pulse, and if you explain relativity on 3 light impulses between parallel mirrors, you should use three different relative distances and different formulas. By the way, another contradiction is time 0, when the light pulse is caught up with in case no.1 (here above).
mezarashi Posted June 17, 2005 Posted June 17, 2005 Personally I see many contraditions in the theory of relativity. Contradictions with YOUR theory of relativity? The theory assumes that light will be c to any observer in an inertial frame. YOUR theory does not. Light cannot take on all these velocities. You can say travel at 99% the speed of light from A to B and get there shortly after light traveling from A to B does. However on your journey it will appear to you that that light is traveling at the speed of light due to time dilation. Time slows down, and although you are travelling a great distance, velocity = distance/time you will find the same velocity, c. Again I restate my point that you are finding contradictions because you do not completely understand the workings of the original theory or atleast fail to accept its postulates, and if so why are saying anything about it? The theory is consistent given that the postulates hold. You should instead go about proving that the postulates are wrong.
Masanov Posted June 17, 2005 Author Posted June 17, 2005 Do you agree that the time is [t*sin alpha/sin beta] in the experiment proposed for discussion, do you agree that expression\sqrt{1-(\frac{v_{90^o}}{c})^2} is sin alpha. If not, pls express you thoughts using proofs. See my e-mail from yesterday 8.02 PM.
mezarashi Posted June 18, 2005 Posted June 18, 2005 I'll say it one last time. Your neat manipulations work given that you assume that light is not constant and you are finding your own contradictions based on this. There is no reason for me to even go through your workings given that your initial assumptions are not "correct".
Masanov Posted June 18, 2005 Author Posted June 18, 2005 I think that you knew Einstein theory's postulates from yesterday. I expect advanced scientists to visit my thread, I will notice administator about your abuse. Or else, let us agree, that you say no more than 3 sentences and do not waste my time. By the way, your original place is "Earth" and your emblem is a kid, are you kidding? I expect only advanced scientists to enter my thread. Answer only the 2 question!!!In short!!! the first one is not for you, pls understand me. 1 (Only for advanced scientists). There's no need for absolute speed in the formula [imath]\frac{T90^o}{Tabsolute}=sin{\alpha}[/imath]. If you know the angle ALPHA, upon which the light impulse was shot, you don't even have to know the observer's speed [imath]V90^o[/imath], which is [cos alpha*c] when beta=90 degrees, and even know the meaning of c. All you have to know is the time, observed outside the spaceship, in which the observer moves. 2 (For beginners).You did not answer about APPROACH to the light impulse. Is it possible to approach light impulse, which moves at a speed [c]? You say No, or Yes, pls answer in short. If not possible, then there's no relative speed in your thinking, and V90 in the formula [imath]\frac{T90^o}{Tabsolute}=sin{\alpha}=\sqrt {1-v90^2/c^2}[/imath] must be 0. You do not allow the spaceship to move in the direction of the light pulse. 3. (For advanced scientists). To say relative speed is not possible is the same as to say the speed is always equal to absolute c. With such rules, you cannot approach the light pulse if your relative speed is impossible, so V90=0, and the expression [imath]\sqrt {1-v90^2/c^2}[/imath] is 1, so t90/t=1. No twins paradox. To have twins paradox, one has to admit relative speed of light. This proves that in Einstein formula [imath]\sqrt {1-v90^2/c^2}[/imath] expression is [sinalpha] or [relative speed/absolute speed], and this is a very advanced discovery.
[Tycho?] Posted June 19, 2005 Posted June 19, 2005 edit at mods request: This is a public forum and anyone can post here.
YT2095 Posted June 19, 2005 Posted June 19, 2005 OKAY! enough`s enough now, before this degenrates into a slanging match about further trivia such as avatars and the like, let me make something clear, No ones doing themsleves any favors getting off topic, least of all Masanov. Masanov, you`ve been asked a few questions by Tom Mattson, please answer them if you can and read the other posts of relevance also, THEN reply if your idea can be taken a single step at a time, I`m confident that we can find out what or where the problem may lie in your assertion. Swansont will also be able to get a handle on this and explain also. `till then, NO NAME CALLING!
Masanov Posted June 19, 2005 Author Posted June 19, 2005 I 'm afraid Tom quitted and went to other pages, expressing no interests. I explaned what he had pointed, though it was difficult to understand his demands. Maybe he once again after careful consideration expresses his points. I will copy them and show my steps between his lines. Tomorrow I will try copy his old opinion, but opinions change, you know.
Masanov Posted June 19, 2005 Author Posted June 19, 2005 I worked the whole day, now here's the reference to the article. Anyway, it looks as if it is published. http://www.rainbow-calendar.hotmail.ru/Einstein Relativity Refutation (2).htm shortly: If you move after light impulse with the speed c/2, you approach it with the speed c/2, and relative speed is c/2 also. In the same way, if someone moves after the light impulse at an angle alpha with the speed [c*cos alpha], you receive the relative speed [c*sin alpha]. If someone denies relative speed, than v should be zero. By the way I expect a good friendly guy to join soon my thread, I expect to appear between his reviews, not between attacks.
Janus Posted June 19, 2005 Posted June 19, 2005 I worked the whole day' date=' now here's the reference to the article.Anyway, it looks as if it is published. http://www.rainbow-calendar.hotmail.ru/Einstein Relativity Refutation (2).htm shortly: If you move after light impulse with the speed c/2, you approach it with the speed c/2, and relative speed is c/2 also. In the same way, if someone moves after the light impulse at an angle alpha with the speed [c*cos alpha'], you receive the relative speed [c*sin alpha]. If someone denies relative speed, than v should be zero. The above is only true as measured by an observer to which you are moving at c/2. As measured by you, the light will have a relative speed of c with respect to yourself, per the second postulate of Relativity. By the way I expect a good friendly guy to join soon my thread, I expect to appear between his reviews, not between attacks.
Recommended Posts