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What happens in the restframe with lightsource? 
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#37
Mar612, 10:13 AM

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 within the specific example he posed, there was no ambiguity in his wording  for the general case, you cannot use such wording (agreeing with me about this). At this point, there was perfect closure and mutual understanding. You chose to try and turn it into a debate. For some reason, you don't like the idea that there is anything wrong with the statement "the twin that spends more time at higher speed ages less"; I believe, most everyone else understands that for the general case, interpreted as words are normally used, it is wrong. For suitably constrained cases, it is accurate. 


#38
Mar612, 11:08 AM

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#39
Mar612, 11:23 AM

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#40
Mar612, 11:59 AM

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#41
Mar612, 12:21 PM

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#42
Mar612, 01:09 PM

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Here's your example in post #24 of Dumb twin paradox question:



#43
Mar612, 01:34 PM

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d tau^2 = d t^2  (dx^2 + dy^2 + d z^2)/ c^2 From this: d tau/ dt = sqrt (1  ((dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2)/c^2) d tau/dt = sqrt (1  v^2 /c ^2) = 1/gamma Then, for any segment of constant v, compute sqrt(1v^/c^2) times coordinate time of that segment, and add them up (no need to actually integrate if v is constant). So, in the given example we have, we have let's make total coordinate time be 10. Then for twin A we have: 10 * sqrt (1 .4^2) = 9.165 appx. For twin B we have: 8 * sqrt( 1.1^1) + 2 * sqrt(1  .99999^2) = 7.9688 appx Thus twin A ages more even though they were going 4 times faster 80% of the time, and only about 2.5 times slower 20% of the time. What does any of this have to do with the point under discussion except to establish the obvious fact that the line element contains all information needed to compute proper times? 


#44
Mar612, 01:48 PM

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The equation finds the relation between the times of each side, it doesn't consider how many times the photon goes to the top and bottom of a clock. So then the answer should give you a direct translation of how much time has occured in one frame and give you how much time has occured in another frame, since c is the same on two sides. I am sorry you have not concinced me yet, because I have gotten really good at algebra. They made me retake it twice from switching colleges and then I took it in high school and junior high 3 times. I got an A in every course, if they hadn't have done that we may not have had this problem... The theory here just doesn't seem to flow like older well done proofs. I apoligize for not being that forum savy. Also if you do short substitution for the time dialation equation you don't get the length contraction equation. If you substitute t' in the equation L'=vt', then say L=ct you get the wrong relation between the two equations to gamma, and v≠v'. If a different value for v is found then put back into the equation it doesn't give the same value's for L and t. But, both observers should agree on the relative velocity. 


#45
Mar612, 03:40 PM

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Here's how I analyzed your scenario: Remember, the whole point of this discussion is to counter the idea that acceleration alone is what results in an aging difference and my only point is although accleration causes a change in speed, the twin has to accumulate time at that speed in order to effect his aging, the more (coordinate) time and the more (coordinate) speed, the lower his aging during that interval. I will attempt to be more precise in my future descriptions so as to not require you to correct me, OK? I will either stipulate that my statement applies only for those Frames of Reference where two speeds exist, one for each twin, but when analyzed from a different Frame, there will be more than two speeds and you have to calculate the partial aging for each speed segment separately, or for twins that accelerate more than the minimal number of times, you also have to calculate the partial aging for each speed segment separately, and for twins that accelerate over a long period of time instead of instantaneously, you have to actually integrate the acceleration to get a speed profile and do really complicated computations. OK? 


#46
Mar612, 04:38 PM

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Why don't they just start out teaching relativity using (τ) if they are not going to derive the equation anyways? But, tau doesn't give the same answer as 1/t'. How would you then convert t' into the proper time?



#47
Mar612, 05:33 PM

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 Acceleration by at least one twin is needed so the twins can get back together. Thus, acceleration somewhere is a necessary (but not sufficient) condition.  The acceleration isn't the cause of differential aging, nor can the age difference be localized to the acceleration or any other part of the path, in any objective way.  This formulation of yours: "my only point is although acceleration causes a change in speed, the twin has to accumulate time at that speed in order to effect his aging, the more (coordinate) time and the more (coordinate) speed, the lower his aging during that interval." is fine, I would never dispute it. It is precise and accurate in noting you have to consider segment by segment in some (any) frame.  I would add that which segments of a journey are associated with 'slower aging' is frame dependent, and not objectively meaningful. But any frame will come up with the same total for a journey. 


#48
Mar612, 06:31 PM

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#49
Mar612, 10:40 PM

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OK, well in the picture of the stationary light clock, they say that the period of the clock is Δt and they calculate it based on the speed of light and the distance between the mirrors. So I agree that t is the amount of time it takes for light, traveling at c, to travel a certain distance. But I don't see why you bring up solving for cisn't c a known constant and you're solving for Δt, the time between two ticks? So I don't know why you think this leads to a wrong value for c. I have no idea what your concern is. In any case, the only way you would get a larger value for Δt (why are you dropping the delta?) is if the distance between the mirrors is greater. And if that is the case, the light clock is taking longer between ticks and therefore ticking at a slower rate. And then you talk about the distance the object traveled, but there is no object traveling, just the light is traveling between the two stationary mirrors. But you shouldn't read into this explanation any more than just a simple declaration of the fact that a moving light clock will take longer between ticks than a stationary one. They are not addressing any issue with regard to Lorentz Transformation and you shouldn't associate the Δt and Δt' nomenclature with similar nomenclature in the Lorentz Transform equations or even in any time dilation equation. 


#50
Mar712, 12:30 AM

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But τ is equal 1/t' if you use the formula for Δt' in the wiki article on Time Dilation and you use the formula for Δτ in the wiki article on Time Dilation (and you equate Δt to ΔT). However, the main point is that the two articles are talking about two different times that are the reciprocal of each other. But your last question about converting t' into Proper Time is nebulous because as I pointed out in post #25, the term t' is defined in different ways in different contexts. I tried to clear this up for you in that long post but maybe I misunderstood what your concern was. You just have to recognize that each context can have a different meaning for the same nomenclature. 


#51
Mar712, 02:39 AM

P: 249

Also, Δt√(1v^2/c^2) ≠ √(1v^2/c^2)/Δt v ≠ ΔL√(1v^2/c^2)/(Δt/√(1v^2/c^2) v ≠ ΔL√(1v^2/c^2)/(√(1v^2/c^2)/Δt) v = ΔL/τ the gamma cancels, (v ≠ ΔL/Δt) So since this does not equal the velocity put into the equation, if you found a velocity with dialated spacetime using ΔL and Δt, the theory breaks down. You could then take that new velocity and then find new values for ΔL and Δt, and then find another new velocity and so on etc. This is why I think the time dialation and the length contraction equation should both be directly porportional to gamma. Δt'=Δt√(1v^2/c^2) solving for the time variables reversed ΔL'=cΔt' I assume the distance traveled by the photon is a different value since the speed of light is constant ΔL'=cΔt√(1v^2/c^2) substitute Δt' from the first equation ΔL'=ΔL√(1v^2/c^2) ΔL=cΔt, this also works for the direction of motion ΔL'=vΔt' if you assume v=v' Solving for length in the vertical dimension doesn't assume space contraction in that direction since both observers would agree that the photon reached the same position after being measured. Neither one says the photon traveled a greater vertical distance and is a requirement of this calculation to form the right triangle. I then found an equation that describes time dialation for an object under constant acceleration. It assumes that because of the MichelsonMorely experiment that an object under acceleration still measures the photon to travel in a straight line since the Earth was accelerating and the effects of gravity where balanced out. Δt'=Δt√(1(vi+vo)^2/4c^2) I still need to look into how this differs from constant motion 


#52
Mar712, 05:23 AM

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Are you looking at the current English version of the wikipedia.org article on Time Dilation? I cannot see much of what you say you are seeing.



#53
Mar712, 05:53 AM

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#54
Mar712, 05:37 PM

P: 249

My simple proof. An observer in motion observers a photon to travel straight out in a line a distance cΔt'. An observer at rest measures the photon to travel out at an angle along with the objects direction of motion cΔt. The observer at rest also measures the object to travel a distance vΔt. This gives three sides of a right triangle and then the relation to how they measure Δt can be found. (ct')^2+(vΔt)^2=(cΔt)^2 a^2+b^2=c^2 c^2Δt'^2=c^2Δt^2v^2Δt^2 distribute the square and subtract both sides by (v^2Δt^2) Δt'^2=(c^2Δt^2v^2Δt^2)/c^2 divide both sides by c^2 Δt'^2=c^2Δt^2(1v^2/c^2)/c^2 factor out a (c^2Δt^2) from (c^2Δt^2v^2t^2) Δt'=Δt√(1v^2/c^2) the c^2 cancels and take the square root of both sides It then follows that the MichealsonMorley experiment also measured the photon to travel out in a straight line even though the experiment was under constant acceleration due to the Earths movement. So then, the distance traveled by the object in constant acceleration can be replaced with Δt(vi+vo)/2. s=Δtvo+(aΔt^2)/2 s=Δtvo+(Δt^2(vivo)/t)/2 substitute from a=(vivo)/Δt s=Δtvo+(ΔtviΔtvo)/2 cancel a Δt from the substitution and distribute the Δt s=2Δtvo/2+Δtvi/2Δtvo/2 multiply Δtvo by 2/2 and seperate the factor s=Δt(vi+vo)/2 add like terms and factor out Δt (cΔt')^2+(Δt(vi+vo)/2)^2 = (cΔt)^2 Pythagorean Theorem c^2Δt'^2 = c^2Δt^2  Δt^2(vi+vo)^2/4 distribute the square c^2Δt'^2 = c^2Δt^2(1(vi+vo)^2/4c^2) factor out c^2Δt^2 from the right side Δt'^2 = Δt^2(1(vi+vo)^2/4c^2) divide both sides by c^2 Δt'=Δt√(1(vi+vo)^2/4c^2) take the square root Then you have an equation for the relation between time dialation and constant acceleration. I noticed that by adding the initial velocity and the final velocity that you can get double that amount of velocity than just the time dialation equation alone. Then when it is squared it comes to be counteracted by the 4 in the denominator. This is because 2 in the acceleration equation is squared to give 4. So you have 4c^2 instead of just c^2, that seems to work out. But my proof seems to imply that as a velocity becomes c then the amount of time that is dialated comes to zero instead of being undefined. 


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