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What happens in the restframe with lightsource? 
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#19
Mar512, 11:58 AM

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I read somewhere that special relativity and time dilation has been proven to really occur. Isn't this a contradiction according to what has been said so far?
Consider the following experiment: There is a rest frame of reference, with available clocks anywhere. There is another person with a clock traveling very close to light speed such that the traveling clock should almost stop its ticks relative to the restframe. After traveling in a straight line for some seconds, the person stops traveling to compare the differences of the clocks. According to experiments the traveling clock has been ticking slower. I am wondering why? Because according to the traveling clock, it was the other clock which slowed down? Is it because the moving frame had to stop and deaccellerate? Doesn't this mean that time dilation does not really occur because of traveling faster, but that accelleration is what causes time dilation? There is some contradiction here about that special relativity slows time, since it is simply a contradiction that both clocks are slow relative to eachother. I would appreciate if somebody could help me out again to understand this better. 


#20
Mar512, 02:15 PM

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In the scenario that you describe now, where B goes from Earth to some distant location and stays there, and A makes the same journey some time later, then they will certainly be the same age when they meet. 


#21
Mar512, 03:22 PM

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Why is nobody replying to my questions anymore? :/ If there's anything I need to explain better please let me know and I'll try..



#22
Mar512, 03:45 PM

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The accelerated observer attributes the difference to a combination of factors, but mainly failure of simultaneity. Suppose your rest observer, with clocks stationed everywhere, in particular has a clock next to himself where the race begins, and a clock at the "finish line" where the traveling observer slows down and stops, and that these clocks are synchronized in his frame. Then they will not be synchronized in the traveling frame. The traveling observer observes both clocks running at the same slow rate, but the finish line clock will read a value that is vastly ahead of the starting line clock. So it is no surprise to him that when he gets there, it is reading many years ahead of his own clock, even though it runs more slowly. 


#23
Mar512, 04:10 PM

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#24
Mar512, 04:23 PM

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#25
Mar512, 04:43 PM

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Part of the confusion comes from a misunderstanding of the nomenclature. In the usual presentation of the Lorentz Transform the equations are stated with t' and x' on the left side of the equation and t and x on the right side of the equations. You shouldn't think of the primed symbols as being associated exclusively with just a primed frame like S' and the unprimed symbols associated exclusively with an unprimed frame like S. Rather, you should think of the LT as providing a way to convert the coordinates of any frame into the coordinates of any other frame and back again if you want. I think it is better to refer to each frame with unprimed symbols to avoid this confusion. So, for example, let's say that you have an event in frame A at t=18 and x=10 and you want to find the coordinates in frame B moving at 0.6c in the x direction with respect to B. (I'm going to use compatible units where c=1.) First we need to calculate gamma, γ and note this is not the English letter y, it is the Greek letter γ. You can get it by going to the advanced editing mode and clicking on the fourth Quick Symbol in the second row. Unfortunately, it looks very much like the letter y but you can tell the difference with the tail going straight down instead of angled. OK, since β=0.6, we can calculate γ: γ = 1/√(1β^{2}) = 1/√(10.6^{2}) = 1/√(10.36) = 1/√(0.64) = 1/0.8 = 1.25 Now we are ready to apply the LT formulas: t' = γ(txβ) = 1.25(1810*0.6) = 1.25(186) = 1.25(12) = 15 x' = γ(xtβ) = 1.25(1018*0.6) = 1.25(1010.8) = 1.25(0.8) = 1 So if we represent the coordinates as [t,x], then [18,10] in frame A is [15,1] in frame B. Now if we want to go back the other way, we do the same thing except that now frame A is moving at 0.6c along the x axis with respect to frame B. Gamma is of course the same value. t' = γ(txβ) = 1.25(15(1)*(0.6)) = 1.25(150.6) = 1.25(14.4) = 18 x' = γ(xtβ) = 1.25(115*(0.6)) = 1.25(1+9) = 1.25(8) = 10 We have confirmed that [15,1] in frame B transforms to [18,10] in frame A, the original coordinates that we started out with. So you see, the primed symbols can apply temporarily to either frame, they are just intended to mean the new values in whatever frame you are converting the coordinates into. It's also important to recognize that the values that t and x refer to are coordinate values, that is, coordinate time and coordinate distance. Now I want to get to the confusion over the Time Dilation factor. Since we use t' and t in the Lorentz Transform for two different frames, we don't want to use both of these symbols in the Time Dilation formula because they are both coordinate times (t' being used just temporarily). The Time Dilation formula applies in just a single frame and allows us to determine the time on a moving clock in that frame so we use Einstein's symbol for Proper Time, the Greek letter tau, τ. Einstein presented this formula as τ=t√(1v^{2}/c^{2}) but it is more common to see it τ=t/γ. When presented these ways, there is no confusion because τ is always the Proper Time for the moving clock and t is always the Coordinate Time for the stationary clocks. So if we want to find out how much time progresses on a clock moving at 0.6c in any frame compared to the coordinate clocks, it is simply 0.8 times the progress of time on the coordinate clocks (since 1/1.25=0.8). Now let's consider a clock that starts at the origin in our Frame A from above and moves in a straight line at a constant speed to the event [18,10]. What speed will it move at? That's pretty easy to calculate because it has traveled a distance of 10 in a time of 18 so the speed is 0.555556c. If the time on the clock was 0 at the start of the trip, what time will be on the clock when it reaches 10? First we have to calculate gamma, like we did before: γ = 1/√(1β^{2}) = 1/√(10.555556^{2}) = 1/√(10.308642) = 1/√(0.691358) = 1/0.831479 = 1.202676 Then we divide the coordinate time of 18 by 1.20267 and we get 14.96663. We could have used the Lorentz Transform calculate the coordinates in Frame C moving at 0.555556c to arrive at this same value as follows: t' = γ(txβ) = 1.202676(1810*0.555556) = 1.202676(185.55556) = 1.202676(12.44444) = 14.96663 x' = γ(xtβ) = 1.202676(1018*0.555556) = 1.202676(1010) = 1.202676(0) = 0 Well, look at this, the distance coordinate came out zero and the reason why is because we have gone from Frame A in which the clock was moving at 0.555556c to Frame C in which it is stationary so the xcoordinate better come out zero because that is where it started and it isn't moving in Frame C. Now the final point of confusion has to do with length contraction and why the distance coordinate isn't also divided by gamma in the Lorentz Transform. The answer to that question has to do with that zero coordinate for distance. In order to properly arrive at the correct time dilation factor, we had to use an event in Frame A that would result in a coordinate of zero (because the clock started at location zero in Frame A) for distance in Frame C. This happens automatically when we calculate where the clock is, based on its speed in Frame A. But in order to see the correct length contraction, we have to use two events in Frame A that result in the time coordinates being the same at both ends of the desired length in Frame C. How do we do that? There are several ways but the easiest is to simply multiply the distance coordinate by the speed and use the same xcoordinate as before. Remember, what we are trying to do is calculate the distance that the clock thinks it has traveled in Frame C. So starting from Frame A the clock moved from [0,0] to [18,10]. In Frame A, the distance is 10 and we expect it to be divided by gamma resulting in a value of 8.314794. All we have to do is multiply the distance coordinate by the speed so that is 10 times 0.555556 which is 5.55556 and we create a new event of [5.55556,10] in Frame A which we transform to Frame C: t' = γ(txβ) = 1.202676(5.5555610*0.555556) = 1.202676(5.555565.55556) = 1.202676(0) = 0 x' = γ(xtβ) = 1.202676(105.55556*0.555556) = 1.202676(103.08642) = 1.202676(6.91358) = 8.314794 Here we see the requirement that the time coordinate is the same as it was when the clock started moving and this gives us the same length contraction for the distance the clock traveled as when we simply divide the original length by gamma. Keep in mind that this is an easy calculation because one end was at the origin. If you are calculating the length of an object or a distance in which neither end is at the origin, you need to multiply the speed by the delta between the two ends and then add that number to the time coordinate of the first event. I hope if you study this, all your confusions will be cleared up and you won't have to publish your own theory. 


#26
Mar512, 05:04 PM

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Here the twins are doing exactly the same acceleration, but spend different amounts of time at the higher speed relative to Earth. The one who spent the most time at the higher speed ends up being younger when they're both back on Earth. 


#27
Mar512, 06:00 PM

P: 249

http://en.wikipedia.org/wiki/Time_dilation I think this calculation should give the proper time, but it doesn't because it misassigned the time variables. It says the time dialation forumla is t'=yt, but I think it should be t'=t/y. If the wikki is correct there then it would in effect be a different theory, unless you are saying that it is also wrong.... The observer in motion doesn't see the light ray to travel a longer distance, it is the opposite. The observer in motion sees the light ray travel straight up and down. They then use their own time to measure this, that would have to be smaller. If you used the the time dialation equation given there, you would get a larger value for time not a smaller one. That is because it assumes that the observer in motion see's the light to travel a longer distance, so then they would have to experience more time to allow that to happen. 


#28
Mar512, 08:40 PM

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But regarding the rest of my post, does it help you to understand the other points of confusion so that you don't feel the need to publish your own theory of Special Relativity? 


#29
Mar512, 11:22 PM

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http://physicsforums.com/showpost.ph...9&postcount=33 


#30
Mar612, 02:18 AM

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In general, there's no substitute for the statement that the number displayed by a clock is the proper time of the curve that represents the motion the clock has done so far. 


#31
Mar612, 03:30 AM

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#32
Mar612, 07:48 AM

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Mar612, 08:42 AM

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#34
Mar612, 08:45 AM

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#35
Mar612, 09:04 AM

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#36
Mar612, 09:59 AM

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