The real twin paradox.
in [Relativity]
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From: colp on 26 Nov 2007 18:13 On Nov 27, 9:43 am, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > colp says... > > > > > > >On Nov 27, 7:49 am, stevendaryl3...(a)yahoo.com (Daryl McCullough) > >wrote: > >> colp says... > > >> >SR describes time dilation. SR does not describe time compression. > > >> That's incorrect. > > >According to your gamma equation. > > >> The Lorentz transform for time > >> has two factors: > > >> t' = gamma (t - vx/c^2) > > >> t' can be greater than t or less than t, > >> depending on the value of x. > > >According to Wikipedia the equation is: gamma = 1 / (sqrt (1 - v^2/ > >c^2)) > >http://en.wikipedia.org/wiki/Time_dilation > > >This means that time dilation occurs regardless of x and regardless of > >whether v is positive or negative. > > Did you notice that gamma is always greater than or equal to 1? > So the combination > > gamma (t - vx/c^2) > > is sometimes greater than t, and is sometimes less than t. > Let's take some specific examples. > > Let t = 1 second. > Let x = 0. > Let v = 4/5 c. > > Then gamma = 1/square-root(1-(4/5)^2) = 1/square-root(9/25) = 5/3. > Plugging everything into the equation gives: > > t' = 5/3 (1 second - 0) > = 5/3 second > > So in this case, t' > t. > > Try different numbers. Instead of x = 0, let's try x = 1 light-second. > > t' = 5/3 (1 second - 4/5 second) > = 5/3 (1/5 second) > = 1/3 second > > So, in this case, t' < t. > > So whether t' is greater than or less than t depends on > the choice of x, as I said. So how would that work using gamma = 1 / (sqrt (1 - v^2/c^2)) ?
From: Sue... on 26 Nov 2007 18:33 On Nov 26, 4:27 pm, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > Sue... says... > > > > > > > > >On Nov 26, 1:47 pm, stevendaryl3...(a)yahoo.com (Daryl McCullough) > >wrote: > > >> What SR says is that, from the point of view of any > >> *inertial* coordinate system, > > >No... the inertial modifier does not appear. > >That issue is considered with relation to > >mass energy equivalence > >http://www.bartleby.com/173/15.html > > >> 1. Light has speed c in all directions. > >> 2. Moving clocks run slow, by a factor of square-root(1-(v/c)^2). > > >SR says no such thing. It says moving clocks are *judged* to run > >slow from a different frame. > > I'm not sure what distinction you are making, but the > precise prediction is this: > > If a moving clock travels at constant velocity from event e1 > with coordinates (x1,t1) to event e2 with coordinates (x2, t2), > then the elapsed time on the clock will be given by > square-root((t2 - t1)^2 - (x2 - x1)^2/c^2), > which will be less than t2 - t1. That is clear enough. If I use my view of Big-Ben to cook an egg on the way to New York it will be overdone. Thank you. Sue... > > >"The Behaviour of Measuring-Rods and Clocks in Motion " > >http://www.bartleby.com/173/12.html > > >You are making a fine argument that the principle of > >relativity doesn't hold true regardless of the speed of light. > > You don't make any sense, Sue. I wish you would stop Googling > and start actually trying to read and understand some of the web > pages you cite. > > -- > Daryl McCullough > Ithaca, NY- Hide quoted text - > > - Show quoted text -
From: bz on 26 Nov 2007 18:40 colp <colp(a)solder.ath.cx> wrote in news:839b939c-528d-4b10-9507- 78495beb3de9(a)e10g2000prf.googlegroups.com: > > So how would that work using gamma = 1 / (sqrt (1 - v^2/c^2)) ? > > That is what Daryl did. He used (4/5 c)^2/c^2 ..8 = 4/5 right? ..8 c = (4/5) c = 4/5 c^2 right? and 4/5 c^2 / c^2 = 4/5 right? He said [quote] Then gamma = 1/square-root(1-(4/5)^2) = 1/square-root(9/25) = 5/3. [unquote] So, he is using exactly the formula you asked about. now if you take velocity away from ... as positive and toward ... as negative, your other great mystery will be cleared up. As he said: [quote] t' = gamma (t - vx/c^2) [unquote] So using the convention I just mentioned, you will see that when traveling away from the other ship (or they are traveling away from us) the effect will 'slow' the time we see on the other ships clock. When traveling toward the other ship, v changes sign, the effect changes to exactly the effect that you said relativity did not have. So, Dirk was right, Daryl was right, you were wrong. -- bz please pardon my infinite ignorance, the set-of-things-I-do-not-know is an infinite set. bz+nanae(a)ch100-5.chem.lsu.edu -- bz please pardon my infinite ignorance, the set-of-things-I-do-not-know is an infinite set. bz+spr(a)ch100-5.chem.lsu.edu remove ch100-5 to avoid spam trap
From: Bryan Olson on 26 Nov 2007 22:54 colp wrote: > Bryan Olson wrote: >> colp wrote: >>> That is a link to this thread. You haven't show a solution. >> It's the message ID of the post. You can also find it at: >> >> http://groups.google.com/group/sci.physics.relativity/msg/1694288bbb6... > > Your post is not relevant. You stand refuted regardless of what you think relevant. > Nice try, but I was talking about the outgoing leg by itself. No, that is *not* what you were talking about: http://groups.google.com/group/sci.physics/msg/9a0b231a6cb3f53d > Outbound, twin A is in the frame of reference, so clock A runs > normally. Twin B's is time-dilated to run slow, as you said. When the > twins reach the original turnaround point twin A will be older than > twin B from twin A's frame of reference. If the twins had been sending > clock ticks by radio to each other, then SR predicts that twin B will > have sent fewer ticks than twin A sent. Since the twins are > symmetrical cases of each other, the number of clock ticks sent must > equal the number of clock ticks received. Thus SR predicts a false > result. Quite a change in your story, and still wrong. In twin A's outgoing frame, at A's turnaround, A has sent more ticks than B. A and B's paths are not symmetric in at that frame. In twin B's outgoing frame, at turnaround, B has sent more ticks than A. The turns-around are not simultaneous in either of those frames. >> For each twin, one leg is in the frame of reference, the other >> is time dilated. They arrive back at Earth showing equal time >> having passed. > > That was made clear in the original thought experiment. > >> Try sticking to any other single inertial frame of reference, >> and you'll get the same answer. > > False. You had to use 2 legs to get a sensible result. A single leg > demostrates the paradox. You've changed your story but you're wrong both ways. >> So where is the paradox that you >> just said this should demonstrate? > > The paradox is that SR predicts time dilation of a twin B is observed > from the frame of reference of twin A when the twins are symmetric > instances of each other. How do you get 'symmetric' when one is stationary in the frame you chose and the other is moving? They're symmetric with respect to the Earth and in that frame... hey, just like SR says! -- --Bryan
From: colp on 27 Nov 2007 01:05
On Nov 27, 12:40 pm, bz <bz+...(a)ch100-5.chem.lsu.edu> wrote: > colp <c...(a)solder.ath.cx> wrote in news:839b939c-528d-4b10-9507- > 78495beb3...(a)e10g2000prf.googlegroups.com: > > > > > So how would that work using gamma = 1 / (sqrt (1 - v^2/c^2)) ? > > That is what Daryl did. > He used (4/5 c)^2/c^2 O.K. He is using v = 4/5 c. > > .8 = 4/5 right? Yes. > .8 c = (4/5) c = 4/5 c^2 right? No. (4/5) c does not equal 4/5 c^2. > > and > > 4/5 c^2 / c^2 = 4/5 right? Yes. > > He said [quote] > Then gamma = 1/square-root(1-(4/5)^2) = 1/square-root(9/25) = 5/3. > [unquote] > So, he is using exactly the formula you asked about. Yes, gamma is 5/3 for v = 4/5 c using my formula. > > now if you take velocity away from ... as positive and toward ... as > negative, your other great mystery will be cleared up. Oh really? > As he said: > > [quote] > t' = gamma (t - vx/c^2) > [unquote] Lets try that again for v = -4/5 c with gamma = 1 / (sqrt (1 - v^2/ c^2)) -0.8 = -4/5 right? -0.8 c = (-4/5) c right? -4/5 c^2 / c^2 = -4/5 right? Then gamma = 1/square-root(1-(-4/5)^2) = 1/square-root(9/25) = 5/3. So we get the same gamma value regardless of the sign of the velocity. The same gammma value means that time dilation is observed regardless of whether the other frame of reference is approaching or retreating. > So using the convention I just mentioned, you will see that when traveling > away from the other ship (or they are traveling away from us) the effect > will 'slow' the time we see on the other ships clock. I'll take your word for it. > When traveling toward the other ship, v changes sign, the effect changes > to exactly the effect that you said relativity did not have. LOL v does change sign, but gamma remains the same using my formula. As in the case of your worked example which ignored relativistic effects, you have again sidestepped the real issue. Instead of using the same formula for both positive and negative velocities, you only did one half of the exercise and then quoted Daryl's formula for the second half. Using gamma = 1 / (sqrt (1 - v^2/c^2)) you arrive at the following conclusion: When traveling away from the other ship (or they are traveling away from us) the effect will 'slow' the time we see on the other ships clock. When traveling toward the other ship, v changes sign, but gamma remains the same. As gamma remains the same the time dilation effect is the same regardless of direction. > So, Dirk was right, Daryl was right, you were wrong. If you use my formula Dirk was wrong. Daryl was right as far as he went. And you haven't shown how I was wrong. |