ralfcis Posted June 11, 2019 Report Share Posted June 11, 2019 That's not an answer to the question I asked, it's just another one of your smoke bombs. Why are both used concurrently when it comes to the constancy of c. My train example showed how the MMX can be explained as either due to td or useless lc. Lorentz used lc to explain the MMX but along comes Einstein and he needs both lc and td concurrently to explain it. The speed of light is constant because the ratio of lc to td is always constant. You either explain why or you throw out the need for lc in any example if everything is already taken care of by td. Otherwise you get guys like Sluggo who start saying lc is real on its own. So what's your choice. Einstein was wrong about how he explained the MMX or Einstein was wrong copying Lorentz's belief that length contraction was real. No more smoke bombs. Quote Link to comment Share on other sites More sharing options...
Vmedvil2 Posted June 11, 2019 Report Share Posted June 11, 2019 (edited) One: There is no round trip. You need to restate your explanation without using a round trip. As I said you don't know the difference between time dilation and age difference. The round trip gives one answer and the light signal gives another. In the round trip example, the planes agree on the same time but their clock is one yr less than the earth clock. In the pure time dilation example with the starting light signal, all 3 clocks agree on the same time upon co-location with earth. Me saying the same thing over and over will not make it sink into your head. You'll just ignore the answer and keep repeating the question much like when my brains got scrambled in a car accident, I sounded very much like you. Two: the ship's occupants are not interested in measuring anything about the other ship. They are only comparing their time with the others time when they are all located at point E In both the light signal and round trip examples, the planes have the same time. Three: Why would the occupant want to use the SR convention? At this stage, you are still explaining how that convention could possibly be true. I'm explaining the answer SR gets, you're saying you get another answer not using SR. So what. Are you asking why a wrong answer is different from a right answer if you use a right and wrong method to arrive at that answer. You have no way of telling which method is right or wrong therefore you have no way of telling which answer is right or wrong. That's a brilliant insight but your conclusion that relativity must be wrong because its answer is biased to that method is not so brilliant. Four: Why would the "delay" of any time comparison have any effect of the physics of this experiment? The two ships are first, not interested in comparing their clocks until they are at the position E. They knew that that started off on equal footings. So any "delay" is going to be identical for each ship, the time comparison is going to be exactly the same from the beginning of the experiment to the end, i.e, there will be no observed discrepancy for any observer. As the ships come closer and closer their lines of simultaneity get shorter and shorter and there is less time difference between them at the ends of those lines until when they meet there is no time difference between them. The relative velocity is unchanged and so is the value of the reciprocal time dilation but their co-location prevents the time dilation from causing a reciprocal time difference between them. But you'd have to understand what a line of simultaneity is and how to draw it on an STD so there's no hope of you understanding my answer. Five: even though we are not interested in the outbound and inbound times, why do you say that "the outbound and inbound times process, times are not equal." Einstein's postulate says that the time for a two way trip A to B and return must have equal times, and its also claimed in Galilean relativity. So now for some reason, you are claiming that this is NOT true? I have no idea what you're saying here. Two round trips in opposite directions relative to each other will create no time difference between them but it will create a permanent time difference between them and the earth at co-location. The ships will have aged 1 yr less than the earth in that example. Six: So you seem to be saying that somehow, during the trip, the clocks on the ships differ, but then by magic, when that pass each other, there is now no difference? This is also counter to SR theory, which says that there is a real, lasting time dilation, remember the two ships meet at point E, BUT they are still both moving at the same velocity as they were throughout the entire trip. Point E is NOT THE DESTINATION, its a position they both pass at the same time. In the light signal example, from each ship's perspective of his own clock in causal proper time, their clocks do not differ. But when they send out light signals to each other of what their time is and the delay of the light between them is taken into account, they will each calculate from their perspective that less time has passed on the other guy's clock. This is an illusion of perspective that Einstein renamed as perspective reality and the rest of his theory falls apart from there. Time dilation was invented to explain the results of the MMX. There is no real lasting time dilation, there is only real lasting age difference and that has a completely different cause than the cause for time dilation as I've explained over and over. To answer your question when they sync clocks there will be no difference in the time unless you count the movement of earth which has a velocity as well, but if they proceed to earth at the same velocity there will be no difference in time clocks though their clocks will differ from earth by the dilation effect, if they proceed to earth with a different velocity there will be 3 different rates of time that of Earth, Ship 1, and Ship 2 and all three will disagree as the dilation effect was different for ship 1 and ship 2 due to difference in velocity to answer your question all you have to ask yourself is if the velocity of object A,B, and C are different if the velocities are different then the clocks will be too. Let's say VEarth , Vship1 , and VShip2 are equal then there will be no dilation effect noticed and all clocks reading the samething, if VEarth , Vship1 , and VShip2 are different then it will read time dilation and the clocks will be different, even if they sync at some point during the journey then the clocks will still change but by a lesser amount as the dilation effect will happen for a different amount of proper time or proper distance depending on the measurement of movement you are using. Edited June 11, 2019 by VictorMedvil Quote Link to comment Share on other sites More sharing options...
rhertz Posted June 11, 2019 Report Share Posted June 11, 2019 Time dilation is the same thing as length contraction just upon another dimension of space-time, they are both different figures of the same thing, thus both happen simultaneously. May I say that this is not true? You can use only one "mind perceived" sensorial distortion per frame of reference. You can't use them simultaneously, as Iproved above, in my long post using Lorentz mathematics and the problem of cosmic muons (for which I discussed very muchhere and at another forum. In that forum, full of narcisists twats with a closed mind, once I got convinced that it was a lostbattle to talk there, I wrote a post telling them so and then I DELETED my account. That forum is from England too, and hasa little more traffic than this, but the "moderators" from England and Australia are arrogant beings, making a stark differencewith moderators here. I always remember this classic explanation about an spaceship approaching a "black hole" (I don't believe in them). This is very interesting, because it's an example of the distortive perception of "REALITY" that relativity forces you to have. You have three things: A black hole, an spacecraft with a real person within it approaching the BH at a moderate speed, andan EXTERNAL OBSERVER. This is what I remember (and this memory is from 40 years ago, minimum): 1) The observer OUTSIDE the spacecraft reference frame watch HOW the ship became slower and slower while approximating the black hole until it completely FREEZES when reaching the edge. So, for this observer, TIME DILATION approached INFINITY. 2) The observer INSIDE the spacecraft reference frame watch HOW the distance between HIM and the edge of the Black Hole becames larger and larger. This is LENGTH EXPANSION (similar to length contraction for an external observer). So, for the astronaut, it is IMPOSSIBLE to reach the edge of the Black Hole, as it is farther and farther. This was a classic explanation of relativity and black holes in the '70s. I don't know if it's used anymore. But, as you can read, nobody can apply both "time dilation" and "length contraction" effects simultaneously. Quote Link to comment Share on other sites More sharing options...
ralfcis Posted June 11, 2019 Report Share Posted June 11, 2019 That again has nothing to do with the question asked. Quote Link to comment Share on other sites More sharing options...
ralfcis Posted June 11, 2019 Report Share Posted June 11, 2019 (edited) Was that the forum with the mad duck? The mods here are great. I haven't been thrown off once for stuff 10 times more offensive than I've been throw off on other forums for. "But, as you can read, nobody can apply both "time dilation" and "length contraction" effects simultaneously." Nobody except Einstein (wrt c constancy) and my question is why. So far no answers. Edited June 11, 2019 by ralfcis Quote Link to comment Share on other sites More sharing options...
Vmedvil2 Posted June 11, 2019 Report Share Posted June 11, 2019 (edited) May I say that this is not true? You can use only one "mind perceived" sensorial distortion per frame of reference. You can't use them simultaneously, as Iproved above, in my long post using Lorentz mathematics and the problem of cosmic muons (for which I discussed very muchhere and at another forum. In that forum, full of narcisists twats with a closed mind, once I got convinced that it was a lostbattle to talk there, I wrote a post telling them so and then I DELETED my account. That forum is from England too, and hasa little more traffic than this, but the "moderators" from England and Australia are arrogant beings, making a stark differencewith moderators here. I always remember this classic explanation about an spaceship approaching a "black hole" (I don't believe in them). This is very interesting, because it's an example of the distortive perception of "REALITY" that relativity forces you to have. You have three things: A black hole, an spacecraft with a real person within it approaching the BH at a moderate speed, andan EXTERNAL OBSERVER. This is what I remember (and this memory is from 40 years ago, minimum): 1) The observer OUTSIDE the spacecraft reference frame watch HOW the ship became slower and slower while approximating the black hole until it completely FREEZES when reaching the edge. So, for this observer, TIME DILATION approached INFINITY. 2) The observer INSIDE the spacecraft reference frame watch HOW the distance between HIM and the edge of the Black Hole becames larger and larger. This is LENGTH EXPANSION (similar to length contraction for an external observer). So, for the astronaut, it is IMPOSSIBLE to reach the edge of the Black Hole, as it is farther and farther. This was a classic explanation of relativity and black holes in the '70s. I don't know if it's used anymore. But, as you can read, nobody can apply both "time dilation" and "length contraction" effects simultaneously. Sure you can, solve for gamma and that will tell you the time dilation and length contraction at the sametime. The Gamma Factor solves for both including mass increase. Now with the gamma factor Edited June 11, 2019 by VictorMedvil Quote Link to comment Share on other sites More sharing options...
rhertz Posted June 11, 2019 Report Share Posted June 11, 2019 Was that the forum with the mad duck? The mods here are great. I haven't been thrown off once for stuff 10 times more offensive than I've been throw off on other forums for. "But, as you can read, nobody can apply both "time dilation" and "length contraction" effects simultaneously." Nobody except Einstein (wrt c constancy) and my question is why. So far no answers. I don't know, but possibly. You made me laugh with your last comment about Einstein. I'm preparing a math reply to help. Quote Link to comment Share on other sites More sharing options...
ralfcis Posted June 11, 2019 Report Share Posted June 11, 2019 (edited) Poof. Smoke bomb. Poof. Poof. More smoke bombs. Quick get Amp in here to post more wiki articles as a diversion. Maybe a little tap dance across the screen from the green slime ball farquad. Nothing to see here folks. Edited June 11, 2019 by ralfcis Quote Link to comment Share on other sites More sharing options...
rhertz Posted June 11, 2019 Report Share Posted June 11, 2019 That's not an answer to the question I asked, it's just another one of your smoke bombs. Why are both used concurrently when it comes to the constancy of c. My train example showed how the MMX can be explained as either due to td or useless lc. Lorentz used lc to explain the MMX but along comes Einstein and he needs both lc and td concurrently to explain it. The speed of light is constant because the ratio of lc to td is always constant. You either explain why or you throw out the need for lc in any example if everything is already taken care of by td. Otherwise you get guys like Sluggo who start saying lc is real on its own. So what's your choice. Einstein was wrong about how he explained the MMX or Einstein was wrong copying Lorentz's belief that length contraction was real. No more smoke bombs. Victor, if you don't mind me trying to supply some Lorentz's math in order to help. I'll try to simplify it: I'm using d (for DELTA) to express differences in length or times, just by substracting two equationsat different points and times: dx' = Y (dx - v.dt)dt' = Y (dt - v.dx/c2) Now, I divide both equations side per side, giving a single equation: dx'/dt' = (dx - v.dt)/(dt - v.dx/c2) = (dx/dt - v) / (dt/dx - v/c2) As you can see, you have a single equation with two variables: dx and dt (remember they are DELTAs, not part of derivates). As you can see, you can't resolve dx'/dt' because the theory say that dt is measured at different times than dx, so this can'tbe solved simultaneously. Sorry, but I didn't the theory. If it was for me, I'd abolish it at everything around both relativities. Quote Link to comment Share on other sites More sharing options...
Vmedvil2 Posted June 11, 2019 Report Share Posted June 11, 2019 (edited) Victor, if you don't mind me trying to supply some Lorentz's math in order to help. I'll try to simplify it: I'm using d (for DELTA) to express differences in length or times, just by substracting two equationsat different points and times: dx' = Y (dx - v.dt)dt' = Y (dt - v.dx/c2) Now, I divide both equations side per side, giving a single equation: dx'/dt' = (dx - v.dt)/(dt - v.dx/c2) = (dx/dt - v) / (dt/dx - v/c2) As you can see, you have a single equation with two variables: dx and dt (remember they are DELTAs, not part of derivates). As you can see, you can't resolve dx'/dt' because the theory say that dt is measured at different times than dx, so this can'tbe solved simultaneously. Sorry, but I didn't the theory. If it was for me, I'd abolish it at everything around both relativities. That is not necessarily true you could solve for both, both would be solved for the same velocity when dx'1/dt'2= 1 , Different velocities when dx'1/dt'2 > or < 1 though i'll admit it is not something I would usually do. It becomes a ratio when dx'1/dt'2 > or < 1 the ratio between two frames. Edited June 11, 2019 by VictorMedvil Quote Link to comment Share on other sites More sharing options...
rhertz Posted June 11, 2019 Report Share Posted June 11, 2019 That is not necessarily true you could solve for both, both would be solved for the same velocity when dx'/dt' = 1 , Different velocities when dx'/dt' > or < 1 though i'll admit it is not something I would usually do. Victor, don't be confused with what LOOKS LIKE velocity: dx/dt and dx'/dt'. THEY ARE NOT! Those expressions are the outcome to divide two equations, side by side, which are used at different spaces and times. Better to go back to their original expressions, without dividing them: dx' = Y (dx - v.dt)dt' = Y (dt - v.dx/c2) Now you have FOUR PARAMETERS to mess with: dx and dt can't be simultaneously measured.dx' and dt' are RESULTS calculated at different xK positions in different tQ times. So, you have to manage how to explain you obtaining x1, x2, t1 and t2, which generate dx and dt, as: dx = x2 - x1dt = t2 -t1 and you have to prove that they are concurrent/simultaneous. I mean, you measured everything (time and distance) at the same time! But if you measure them at the same time, then t2 = t1 and dt = 0. Do you see the problem now? Simultaneity of time or space can't be concurrent for measurementsbecause any attempt gives you either dt = 0 or dx = 0. You have to do ONE thing, and after time passed, do the other thing. Sorry, it's not me. It's Lorentz. Quote Link to comment Share on other sites More sharing options...
Vmedvil2 Posted June 11, 2019 Report Share Posted June 11, 2019 (edited) Victor, don't be confused with what LOOKS LIKE velocity: dx/dt and dx'/dt'. THEY ARE NOT! Those expressions are the outcome to divide two equations, side by side, which are used at different spaces and times. Better to go back to their original expressions, without dividing them: dx' = Y (dx - v.dt)dt' = Y (dt - v.dx/c2) Now you have FOUR PARAMETERS to mess with: dx and dt can't be simultaneously measured.dx' and dt' are RESULTS calculated at different xK positions in different tQ times. So, you have to manage how to explain you obtaining x1, x2, t1 and t2, which generate dx and dt, as: dx = x2 - x1dt = t2 -t1 and you have to prove that they are concurrent/simultaneous. I mean, you measured everything (time and distance) at the same time! But if you measure them at the same time, then t2 = t1 and dt = 0. Do you see the problem now? Simultaneity of time or space can't be concurrent for measurementsbecause any attempt gives you either dt = 0 or dx = 0. You have to do ONE thing, and after time passed, do the other thing. Sorry, it's not me. It's Lorentz. I wasn't saying that dx'/dt' was a velocity but rather a ratio between two frames at this point, I was talking about the velocity of the object in the original equation, if the ratio dx'/dt' = 1 then you have solved for time dilation and length contraction at the same time with a equal velocity of the object being V not dx'/dt'. This isn't quantum mechanics where you can't solve for position and momentum simultaneously there is no uncertainty principal here, so yes that is solving for both, you just did it and you know i don't mean time as temporal dimension but rather in the same equation. Edited June 11, 2019 by VictorMedvil Quote Link to comment Share on other sites More sharing options...
sluggo Posted June 11, 2019 Report Share Posted June 11, 2019 (edited) It's a useless effort to present evidence to someone with preconceived notions, and who misinterprets text, reading in ideas that aren't there. Edited June 11, 2019 by sluggo Quote Link to comment Share on other sites More sharing options...
ralfcis Posted June 11, 2019 Report Share Posted June 11, 2019 The irony. Especially those who start a dialog then slink away when they can't answer subsequent questions. I guess self-awareness is not your strong suit. Quote Link to comment Share on other sites More sharing options...
sluggo Posted June 11, 2019 Report Share Posted June 11, 2019 The irony. Especially those who start a dialog then slink away when they can't answer subsequent questions. I guess self-awareness is not your strong suit.My time is more important than to waste it on forums. My initial purpose is complete. Quote Link to comment Share on other sites More sharing options...
marcospolo Posted June 12, 2019 Author Report Share Posted June 12, 2019 I don't mean that you take the effect twice, I am saying that they are the same effect upon two connected dimensions, they mathematically cannot be used twice or you get a wrong answer but they both effect time and space simultaneously being the time dilation is equal to length contraction upon a space dimension and vice versa, meaning if you contract space it is equivalent to dilating time as the length of time is shortening its cycle.Neither you are ralfcis are being intellectually honest. SR has three effects that happen concurrently, TD, LC and Mass increase. You can't separate the three, that are the result of Special Relativity. Nowhere in the theory of SR does it say that the three effects can occur independently. So you can't just pretend suddenly that you are NOT going to consider length contraction when it suits you. The ONLY reason why you say that you are not considering length contraction in the case of muons, is because you get an answer that does NOT FIT YOUR DESIRED OUTCOME. Strangely, in other scenarios, you are happy to have TD and LC function concurrently. This my friends is being intellectually dishonest.The fact is that the concocted example of Muons, the prime experimental example of SR tie dilation, FAILS because you are not faithfully applying the rules of your own devising. There are many reasons apart from this nonapplication of ALL the effects of SR, that destroy using Muons as any evidence in support of SR. It's supposed to be the best example? Yet its failed on several levels. The ONLY possible conclusion is that Special Relativity is unsupported by rational analysis, by mathematics, by geometry, and by empirical evidence. Quote Link to comment Share on other sites More sharing options...
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