studiot Posted July 23, 2018 Share Posted July 23, 2018 (edited) On 7/8/2018 at 5:29 PM, ravell said: The Special Theory of Relativity claims that an observer in a spacecraft moving with the speed v relative to e.g. Earth, will always see that the clocks on Earth run more slowly than the clock in the spacecraft, according to the formula for time dilation T = To / (1- (v / c) ^ 2)) ^ 0.5. The direct self- contradiction is shown in bold. On 7/21/2018 at 11:39 AM, ravell said: F1 = Fo *(1- (v /c)^2)^0,5 which also results directly from classical physics, where Fo is the frequency of the clock at rest (v = 0). Then you will be able to post the derivation - it's time to put up or shut up. The frequency that any clock ticks at in its own or any other frame is irrelevant to what is read on that clock by any observer. Every observer will make an identical sequence of readings of the clock say 1,2,3,4, because that is the nature of your digital clock and it can only show this sequence. They will, however, differ in the readings on their own clocks that coincide with each number on the broadcasting clock. This correspondence will depend upon two factors viz their own velocity relative to the broadcasting clock and also the time lag between when the broadcasting clock face shows say 4 and when the observer reads the clock face as showing 4. This time lag will increase as the outward journey progresses and decrease during the return. And yes, during this sequence of observations, the colour of the numbers on the clock face will change for the observer due to the Doppler shift. But these colours are irrelevant to the number displayed. As I said the only way to observe or see the broadcasting clock 'running slow' is to compare the readings broadcast in whatever colour, with the readings made on an identical clock carried along on the journey. This would occur naturally anyway as the separation, and thus the time lag, increases. Edited July 23, 2018 by studiot 1 Link to comment Share on other sites More sharing options...
Strange Posted July 23, 2018 Share Posted July 23, 2018 On 21/07/2018 at 12:39 PM, ravell said: Thus, the whole matter is clearly and unambiguously explained by classical physics. Hang on a minute ... I obviously skimmed this post too quickly before. Classical, you say? Non-relativistic, you say? But then you say this: Quote where the added factor Fo /F1 takes into account the length of the second of the light clock on board, which is longer on the moving spacecraft than the second on Earth In a non-relativistic world, the length of the second would not change. So you are not using classical physics at all You then go on to use the wrong formula for the time dilation and the wrong formula for the Doppler effect ... Link to comment Share on other sites More sharing options...
ravell Posted July 28, 2018 Author Share Posted July 28, 2018 On 23.07.2018 at 11:42 PM, studiot said: On 21.07.2018 at 12:39 PM, ravell said: F1 = Fo *(1- (v /c)^2)^0,5 which also results directly from classical physics, where Fo is the frequency of the clock at rest (v = 0). Then you will be able to post the derivation - it's time to put up or shut up. Well, I will try to show it as simple as possible. Here is a dense forest, forest, forest, forest, forest, forest, forest, forest . . . . . . L . . . L . . . . D . . . . . . . ---- Here is our train >> ------------- v --------------- train >> --------------------- A ------------------ S = v*T----------------------B D - Distance between the forest line and the railway line v – Speed of the train u - Speed of the sound pulse in open air. T - The tick time that elapses from the moment of sending sound pulse from the train (point A) to the moment of returning its echo (point B). The tick time of the sound pulse for train at rest (v=0) is To = 2D /u. The tick time of the sound pulse in open air for the train in motion at the speed v is T = 2L /u, where 2L is the path length of the sound pulse, and L is: L = ( D^2 +(S/2)^2)^0,5 = (D^2+(v*T/2)^2)^0,5 Thus, the tick time of the sound pulse in open air for the train in motion will be: T = 2L /u = 2(D^2+(v*T/2)^2)^0,5 /u Solving this equation with respect to T, we get: T = 2D/(u^2-v^2)^0,5 = (2D/u)/ (1-v^2/u^2)^0,5 2D/u = To then T = To/(1-v^2/u^2)^0,5 The above formula expressed in the form of the frequency F = 1 /T will take the form: F = Fo*(1-v^2/u^2)^0,5 If we now replace the word FOREST with the word MIRROR and the speed of sound u with the speed of light c, then our example with the sound changes into the light clock with the same formula as for sound: F = Fo*(1-v^2/c^2)^0,5 Thus, the classical (Newtonian) physics clearly shows that in case of a spacecraft or train in motion, the formula for the tick time of the light clock in a spacecraft is exatly the same as for the train and sound in the open air, and as shown above, this has nothing to do with relativity. Link to comment Share on other sites More sharing options...
dimreepr Posted July 28, 2018 Share Posted July 28, 2018 (edited) 39 minutes ago, ravell said: Well, I will try to show it as simple as possible. Thanks, but it just shows that even a simple layman such as myself with only a basic math understanding, can see you're talking bollox. Edited July 28, 2018 by dimreepr Link to comment Share on other sites More sharing options...
Strange Posted July 28, 2018 Share Posted July 28, 2018 43 minutes ago, ravell said: Well, I will try to show it as simple as possible. I think you will agree that both the person on the train and the person "stationary" on the ground will agree that the time, T, for the sound to be sent and received increases with the increasing speed of the train? If so, then that trivially disproves your claim. But thanks for playing. Link to comment Share on other sites More sharing options...
Janus Posted July 28, 2018 Share Posted July 28, 2018 4 hours ago, ravell said: Well, I will try to show it as simple as possible. Here is a dense forest, forest, forest, forest, forest, forest, forest, forest . . . . . . L . . . L . . . . D . . . . . . . ---- Here is our train >> ------------- v --------------- train >> --------------------- A ------------------ S = v*T----------------------B D - Distance between the forest line and the railway line v – Speed of the train u - Speed of the sound pulse in open air. T - The tick time that elapses from the moment of sending sound pulse from the train (point A) to the moment of returning its echo (point B). The tick time of the sound pulse for train at rest (v=0) is To = 2D /u. The tick time of the sound pulse in open air for the train in motion at the speed v is T = 2L /u, where 2L is the path length of the sound pulse, and L is: L = ( D^2 +(S/2)^2)^0,5 = (D^2+(v*T/2)^2)^0,5 Thus, the tick time of the sound pulse in open air for the train in motion will be: T = 2L /u = 2(D^2+(v*T/2)^2)^0,5 /u Solving this equation with respect to T, we get: T = 2D/(u^2-v^2)^0,5 = (2D/u)/ (1-v^2/u^2)^0,5 2D/u = To then T = To/(1-v^2/u^2)^0,5 The above formula expressed in the form of the frequency F = 1 /T will take the form: F = Fo*(1-v^2/u^2)^0,5 If we now replace the word FOREST with the word MIRROR and the speed of sound u with the speed of light c, then our example with the sound changes into the light clock with the same formula as for sound: F = Fo*(1-v^2/c^2)^0,5 Thus, the classical (Newtonian) physics clearly shows that in case of a spacecraft or train in motion, the formula for the tick time of the light clock in a spacecraft is exatly the same as for the train and sound in the open air, and as shown above, this has nothing to do with relativity. The problem is that in your original post you said "clock". Which can be taken to mean any general time keeping device and not one based on an y particular mechanism. In fact, in such discussions, the term "clock" is assumed to mean an "ideal clock" (one that does not lose or gain time due to local conditions.) It isn't until much later that you mention a light clock. And even then you restrict yourself to just a single type of light clock. So lets take your latest example. You have two light (or sound) clocks, one is moving, and one is at "rest" : The red numbers keep track of the ticks for each clock. The white dots are the light/sound pulses and the expanding circles show how the light travels at a set speed relative to our chosen frame. In this example, it does appear that Relativity and Newtonian physics will give the same answer as to how fast the moving clock ticks with respect to the stationary clock. However, what if we add another set of mirrors/sound reflectors, laid out left to right? Then if we use sound or a Newtonian fixed speed for light we get this: With the stationary clock, the two pulses keep time with each other. But with the moving clock, the up and down pulse makes one complete round trip before the left-right pulse has even made one leg of the round trip. In Relativity there is length contraction so you get this: The distance between the left and right mirrors is length contracted as measured from our rest frame, and this causes the two pulses to stay in sync in terms of round trip time. So right here we have a fundamental difference between Newtonian and Relativistic predictions. Or let's go further. Let's enclose the sound clock so that it carries it air with it, and design or light clock to use fiber optic cables to carry the pulses rather than free space. In both these cases, Newtonian physics says that the speed of sound for the sound clock or the speed of light in the light clock is a constant relative to the medium it is traveling through and is completely independent of any motion these clocks may have. So two self-enclosed sound clocks would measure no difference in tick rate between each other no matter which one was moving relative to the outside air, and two light fiber clocks would tick at the same rate regardless of their relative motion. This gets back to what strange was saying with your sound clock example. Someone riding along with the moving sound clock would agree with the stationary observer that his sound clock was ticking slow. Which is in contrast to what Relativity says about moving clocks, in that each will say that it is the other clock ticking slow and their clock is ticking at normal rate. You took one particular set up, measured by one particular frame, and when you got an answer you liked, you stopped and didn't think it all the way through. You failed to consider whether or not the relationship held for more general cases or not. This also means that with the sound clock example the person riding along with the moving light clock would measure a stationary light clock as ticking at the "normal" rate, and the only difference he would visually perceive in its tick rate would be due to Doppler shift. Basically what this mean is that in Newtonian physics, the Doppler shift measured between two receding sources will be different when measured by the Moving source and when measured by the stationary source, While in Relativity all that matter is the relative velocity and they both measure the same Doppler shift. This has practical consequences. when we send space probes to other planets, their velocity has to be planned out very carefully, if they are moving too fast or too slow by just a small amount, not only will they arrive at the target planet's orbit too soon or too late, their actual trajectory will be changed so that they will arrive at the wrong point of the orbit. In other words, it wouldn't take much to make the probe miss the target planet by a wide margin. Once the craft is in space, the only real way NASA has to measure its speed is by Doppler shift. They know the base frequency of the craft and they can check for a Doppler shift to get the relative speed with respect to the Earth. If you use a Newtonian Doppler shift model, both the Earth and Probe are moving objects (the whole solar system for that matter, unless you are proposing a heliocentric universe) And thus all these various motions would have an effect on the Doppler shift measured from the probe, even what time of the year the measurement was made would make a difference. with the Relativistic model, all that counts is the relative velocity between probe and Earth. The two models would lead to different conclusions as to the probes velocity with respect to the Sun when measuring the same Doppler shift. Thus If NASA used the relativistic model to work out what the craft's velocity was, and the Newtonian model was what the universe worked by, then NASA could measure the Doppler shift, conclude that the craft had the right velocity, and have it go on to miss the target planet. ( or conversely, they would conclude that the craft somehow missed its target velocity and that they had screwed up somewhere, only to have it make a perfect meet-up) 3 Link to comment Share on other sites More sharing options...
studiot Posted July 28, 2018 Share Posted July 28, 2018 On 08/07/2018 at 5:29 PM, ravell said: and has an external, very large visible from space, digital display of the date measured by this clock Ravell this is the clock you have specified in the opening post so this is the clock I have been working with. Other clocks are a red herring as is the train in you last post. Since you specified this, it's pretty obvious that I would work with this and consequently all my questions would pertain to the situation whereby the traveller reads that digital display and indeed I took great pains to describe this several times. The machine frequency that generates the reading on this display is irrelevant. The display is certainly not read by any 1kHz EM wave, which is not light or even a radio wave. Each time I have asked you and I ask you once again. What numbers will the traveller see on the clock when his clock reads say 1 month, 6 months, 1 year by your predictions? Your system must have some predictions for this situation or it is totally worthless. Link to comment Share on other sites More sharing options...
Markus Hanke Posted July 29, 2018 Share Posted July 29, 2018 Another excellent post by Janus...+1. Link to comment Share on other sites More sharing options...
JohnMnemonic Posted August 5, 2018 Share Posted August 5, 2018 Quote Or let's go further. Let's enclose the sound clock so that it carries it air with it, and design or light clock to use fiber optic cables to carry the pulses rather than free space. In both these cases, Newtonian physics says that the speed of sound for the sound clock or the speed of light in the light clock is a constant relative to the medium it is traveling through and is completely independent of any motion these clocks may have. So two self-enclosed sound clocks would measure no difference in tick rate between each other no matter which one was moving relative to the outside air, and two light fiber clocks would tick at the same rate regardless of their relative motion. This gets back to what strange was saying with your sound clock example. Someone riding along with the moving sound clock would agree with the stationary observer that his sound clock was ticking slow. Which is in contrast to what Relativity says about moving clocks, in that each will say that it is the other clock ticking slow and their clock is ticking at normal rate. That's exactly, what I'm wondering about lately... Let's leave the light for now and focus on the sound waves. We can as well use waves on a water surface - they will work just as fine. If we would enclose the air or water within the moving clock, time in which the wave would travel from one end of clock to another, should be exactly the same, as in the stationary frame. If the velocity of a medium is constant, it's motion doesn't matter for the wave propagation. If it would, sound couldn't propagate in the direction of frame's motion, if it's velocity would be higher, than the speed of sound - so a fighter jet pilot, sitting in front of me, wouldn't be able to hear me at all. If I take a bowl with water and start to move it at a constant velocity, waves on the water surface would propagate in all directions with the same speed, as they would do in a stationary frame - wave would move from one end of the bowl to the other, in the same time, as inside a stationary bowl. In such case, both animation doesn't show the right result Let's now consider the second scenario, in which medium remains stationary in relation to a moving sound (water) clock. In this case, animations are valid only if the direction of wave propagation is parallel to the motion of a clock - what can be nicely explained with the Doppler's effect. However, if the direction of propagation would be perpendicular to the clock's motion, effects would be different, than on those animations - observer in motion would observe, that a wave emitted perpendiculary to it's motion would "stay behind" and it's motion wouldn't be perpendiculary for the clock in motion. But the wave emitted from a source in motion, would be perpendicular for a stationary observer. It means also, that after the emission, clock would move further away, while the wave wouldn't follow it and it would completely miss the moving receiver - so the perpendicular clock wouldn't work at all... This leaves me with the conclusion, that those animations are valid only for the light - at least I think so... Link to comment Share on other sites More sharing options...
ravell Posted August 5, 2018 Author Share Posted August 5, 2018 On 28.07.2018 at 9:32 PM, Janus said: The problem is that in your original post you said "clock". Which can be taken to mean any general time keeping device and not one based on an y particular mechanism. In fact, in such discussions, the term "clock" is assumed to mean an "ideal clock" (one that does not lose or gain time due to local conditions.) Thank you for your post with the beautiful animation. In the opening post, I pointed out that my questions concern light clocks in motion. Quote Someone riding along with the moving sound clock would agree with the stationary observer that his sound clock was ticking slow. Which is in contrast to what Relativity says about moving clocks, in that each will say that it is the other clock ticking slow and their clock is ticking at normal rate. Two questions: 1. What in the theory of relativity means the ticking of the clock at a normal rate, and based on what data, the person on the spacecraft in motion finds that his light clock is ticking at a normal rate? 2. As we have stated already, in our example the observer on the spacecraft will receive the frequency 500 Hz of the clock on Earth when is receding, and 2000 Hz when is approaching. Thus on the basis of what data or in what way, the person on the spacecraft will determine or experience the illusion, that the clock on Earth, from which he took off, ticks slower than the light clock on board, regardless of the spacecraft approaching or receding from the Earth? Quote Basically what this mean is that in Newtonian physics, the Doppler shift measured between two receding sources will be different when measured by the Moving source and when measured by the stationary source, While in Relativity all that matter is the relative velocity and they both measure the same Doppler shift. This is unfortunately not true. The light clocks in motion tick slower and it is a real phenomenon. This is not a relativistic illusion experienced by an observer on Earth. This is the immanent technical feature of these clocks. Doppler measurements according to Newtonian physics will thus show for observers on both sides the same frequency of the clocks but related to the local second, which on the spacecraft is longer than on Earth. On 29.07.2018 at 1:10 AM, studiot said: Each time I have asked you and I ask you once again. What numbers will the traveller see on the clock when his clock reads say 1 month, 6 months, 1 year by your predictions? I have already answered this question in my post dated July 15: 1. What date shown on the clock display on Earth, will be seen and photographed by the observer on the spacecraft the day before reaching Earth? Traveling at a speed of v = 180 000 km /s back and forth to an object 1,5 light years away, will take 5 years according to the clocks on Earth. Thus, the day before the spacecraft returns, the clock on Earth will show on its screen the date 31.12.2024. And this date will be seen and photographed by the observer on the spacecraft. Link to comment Share on other sites More sharing options...
JohnMnemonic Posted August 5, 2018 Share Posted August 5, 2018 Quote This is unfortunately not true. The light clocks in motion tick slower and it is a real phenomenon. This is not a relativistic illusion experienced by an observer on Earth. This is the immanent technical feature of these clocks. Well, problem is, that EVERYTHING in the Universe is in motion and everything except photons move at velocities, which are anything but definitive. For me something can move at 99,9% of c, while for you the same thing will move at 0,01% of c - the only difference, is the direction, in which you and I are moving. Which one of us has right? Link to comment Share on other sites More sharing options...
Janus Posted August 5, 2018 Share Posted August 5, 2018 26 minutes ago, ravell said: 1. What in the theory of relativity means the ticking of the clock at a normal rate, and based on what data, the person on the spacecraft in motion finds that his light clock is ticking at a normal rate? The fact that the person on the spacecraft measures the speed of light in a vacuum relative to himself to be c. In other words, if you take a bunch of observers each with different velocities with respect to each other, they will all measure light as moving at c relative to themselves. This is completely different than the case with sound, where everyone would agree that the speed of sound is measured relative to the medium, to which each observer has his own velocity with respect to. Thus for light in a vacuum, if a flash of light is emitted at the same time as two observers meet moving relative to each other and from the spot where they meet, then each of these observers would measure the front edge of this light flash as expanding outward in a sphere from them at c and the other observer as moving away from the center of the expanding sphere. For sound, what each observer would measure would depend on what his own velocity with respect to the medium carrying the sound wave was. If observer A was stationary with respect to the medium, then he measure the sound expanding outward from him at S(the speed of sound for the medium) while observer B moves with respect to the center of the expanding sphere. Observer B will agree. He will also say that observer A stays at the center of the expanding sound wave and that he(B) is moving with respect to it. 1 hour ago, ravell said: 2. As we have stated already, in our example the observer on the spacecraft will receive the frequency 500 Hz of the clock on Earth when is receding, and 2000 Hz when is approaching. Thus on the basis of what data or in what way, the person on the spacecraft will determine or experience the illusion, that the clock on Earth, from which he took off, ticks slower than the light clock on board, regardless of the spacecraft approaching or receding from the Earth? So we are assuming a 1000 hz signal emitted at Earth as measured by the Earth and a relative velocity of 0.6c. If we account for the effect caused by the changing distance between the spacecraft and Earth ( A factor of 0.4 when receding and 1.6 when approaching), Then we get 0.4/0.5 = 0.8 and 1.6/2 = 0.8 In other words, A time dilation of 0.8 for the Earth clock regardless of whether the ship is receding or approaching. And this is not an "illusion". 1 hour ago, ravell said: This is unfortunately not true. The light clocks in motion tick slower and it is a real phenomenon. This is not a relativistic illusion experienced by an observer on Earth. This is the immanent technical feature of these clocks. Doppler measurements according to Newtonian physics will thus show for observers on both sides the same frequency of the clocks but related to the local second, which on the spacecraft is longer than on Earth. No. Continually repeating the same falsehood over and over again does not make it true. Two light clocks maintaining a constant velocity with respect to each other, will both measure the other as running slow. As long as they maintain this constant velocity with respect to each other, there is no objective way to say which clock is really ticking slow. There is no absolute reference by which "motion" can be objectively be measured. The only objective way to compare how much total time two light clock tick off is to separate them and then bring them back together again. But that means that one or the other or both did not remain in a single inertial reference frame the whole time. And changing inertial frames has an effect on what a clock will measure. 1 hour ago, ravell said: 1. What date shown on the clock display on Earth, will be seen and photographed by the observer on the spacecraft the day before reaching Earth? Traveling at a speed of v = 180 000 km /s back and forth to an object 1,5 light years away, will take 5 years according to the clocks on Earth. Thus, the day before the spacecraft returns, the clock on Earth will show on its screen the date 31.12.2024. And this date will be seen and photographed by the observer on the spacecraft. None of which changes the fact that during the outbound and inbound legs of the trip, the spaceship would say that the Earth clock ran slower than his own. Once again: You are conflating "time dilation" (the moment to moment comparison of tick rates) to "total accumulated time" Which is the end result of All the relativistic effects that occurred over the course of the trip. That's like saying that if two cars take two different routes to the same destination, one along a straight road and one along a curved path, that just because the car driving the straight route took less time, he drove faster. Depending on the curvature of the other route, the car on this route could have actually driven at a higher speed and still end up taking longer. Relativity can and does explain why in the twin paradox, the spaceship twin ends up younger upon his return without invoking or using the idea of absolute motion or claiming that his clock objectively ticked slower than the Earth clock at any given point of the trip. The fact that you will not, or can not, accept this does not constitute a refutation. At this point I see no reason to continue this discussion. You obviously have your own view on how the Universe "must" work that you doggedly adhere to no matter what the evidence to the contrary. The thing is that the universe has its rules by which it operates, and it doesn't care one wit as to whether they make sense to you personally. Just because doesn't appear to make sense to you, doesn't mean that it isn't true 1 Link to comment Share on other sites More sharing options...
studiot Posted August 5, 2018 Share Posted August 5, 2018 3 hours ago, ravell said: On 29/07/2018 at 12:10 AM, studiot said: Each time I have asked you and I ask you once again. What numbers will the traveller see on the clock when his clock reads say 1 month, 6 months, 1 year by your predictions? I have already answered this question in my post dated July 15: 1. What date shown on the clock display on Earth, will be seen and photographed by the observer on the spacecraft the day before reaching Earth? Traveling at a speed of v = 180 000 km /s back and forth to an object 1,5 light years away, will take 5 years according to the clocks on Earth. Thus, the day before the spacecraft returns, the clock on Earth will show on its screen the date 31.12.2024. And this date will be seen and photographed by the observer on the spacecraft. NO this was not an answer to my question. I asked what would be read on the Earth clock, by the traveller when his clock reads specific values. You told me what would be read when the Earth clock showed specific values. So I still await an answer. When the traveller reads his clock as 30 days , 60 days, 180 days, what will he see on the Earth clock display, by your reckoning? (I have converted months to days to avoid length of month confusion.) Link to comment Share on other sites More sharing options...
studiot Posted August 5, 2018 Share Posted August 5, 2018 (edited) As a matter of interest I am banging on about this for a very good reason. We don't have the technology to perform the experiment you describe and confirm with actual measurements. However we do have the technology to take two clocks, running as identically as possible, leave on eon Earth and take the other for a few trips round the Earth at speed. This experiment has been done many times now and always confirms the predictions of SR. Indeed, Swansont makes his living from it, and look how well he lives. If you ask him nicely he will probably know the record for the biggest and smallest differences ever measured. Edited August 5, 2018 by studiot Link to comment Share on other sites More sharing options...
beecee Posted August 6, 2018 Share Posted August 6, 2018 (edited) 1 hour ago, JohnMnemonic said: Let's wait a year or two and see, if SR will still be considered as the best solution. Funny, that after 100 years, it is still considered as a THEORY and not a functional model, like QM, EM, MHD and many other ones... 113 years actually, but yeah let's wait a year or two or three or three hundred. Oh, and you say it is a theory and not a model? Gee I was pretty sure they were synonymous. Or are you being purposely obtuse again? Quote It doesn't invalidate SR - but in many ways replaces it. Perhaps you would like to link me to some reputable company and/or scientific orginization that doesn't use SR but some other model? I'm in and out of here all day to day so I have plenty of time. Quote Look, how long ago those papers were published - most of them around 2 years ago. Give them another 2 to 5 years and we'll see... None will replace SR that is certain....GR? not replace either but possibly in time extended on with a validted QGT. Quote And yet you repeat some statements, without giving a single thought about them... "Time doesn't flow for light..." Why? "Because the theory says so...", "Because everybody accept the SR - so it has to be correct...". Because any logical extension of what we see happen at sub relativistic speeds, can extended to relativistic speeds and logically be also assumed at "c". And it is certainly far more logical for me as a lay person to accept 113 years of research, tests, experiments, discoveries etc that have supported and aligned with SR/GR, then to swallow the rhetoric of another lay person on a science forum open to any Tom, Dick or Harry...or even John. Quote You see, this is the difference between us - I won't accept any claim, without spending some time, thinking about it... You accept claims, just because everybody tell you, that you have to accept them... I accept them, only when I understand their meanings and only after they will make sense for me... If something doesn't make sense, or can be easily disproved by well known facts, then I reject it - and until someone won't prove me, that it is in fact correct, I will treat it as useless junk (no matter, if it was Einstein, who said so, or if everyone around tries to force my acceptance) I call that delusions of grandeur...no formal qualifications, no hard yards and or degrees etc, no access to particle accelerators or any other state of the art experimental facility, just your own intuition, and pride...sorry matey, that's not what science is based on. You have none, zilch, nada empirical evidence to show any part of SR/GR is invalid or wrong, and you lack any model that predicts, anything over and above the incumbent model/s. That's the state of the nation at this time. Quote Simple and well known fact, that each photon has it's own timeline/history proves clearly, that it has to experience flow of time - this is for me more than enough, to conclude, that Einstein made in this case a mistake. For the umpteenth time wrong...firstly the frame of a photon is unrealistic despite your unsupported denial of that fact, secondly as experiments have shown time dilation increases as "c" is approached until, you guessed it, time for want of a better term stands still, thirdly taking the time dilation into consideration and length contraction in the unrealsitic frame of reference of the photon, it could/would traverse the observable universe in an instant. Quote But it's most likely because I'm not a professional scientist... Great!! Bingo!!! So now swallow your pride, and study up on the incumbent theory you are trying to claim fault with, formulate the proper maths, put in the hard yards, the 5 to 10 years and more of blood sweat and tears, accept that your incredulity at what SR/GR has shown us and what it has successfully predicted, is no scientific reason to reject it. The world of science and the scientific methodology proceeds on regardless of the dime a dozen amateurish folk with no qualifications etc and their rhetorical nonsense that they intend to swamp science forums with. Edited August 6, 2018 by beecee Link to comment Share on other sites More sharing options...
beecee Posted August 7, 2018 Share Posted August 7, 2018 2 minutes ago, JohnMnemonic said: Tell this to people from nasa, who make pseudo-scientific models of magnetosphere. Gravity has no chance with electromagnetism... Again, this is off topic. If you want to discuss the long defunct Plasma or Electric universe, start another thread. Link to comment Share on other sites More sharing options...
phyti Posted August 11, 2018 Share Posted August 11, 2018 (edited) On 8/5/2018 at 9:34 AM, ravell said: 1. What date shown on the clock display on Earth, will be seen and photographed by the observer on the spacecraft the day before reaching Earth? Traveling at a speed of v = 180 000 km /s back and forth to an object 1,5 light years away, will take 5 years according to the clocks on Earth. Thus, the day before the spacecraft returns, the clock on Earth will show on its screen the date 31.12.2024. And this date will be seen and photographed by the observer on the spacecraft. The date seems incorrect. It would be 5yr-1.6 days. Edited August 11, 2018 by phyti Link to comment Share on other sites More sharing options...
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