The real twin paradox.
in [Relativity]
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From: colp on 24 Nov 2007 10:37 On Nov 25, 3:57 am, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > In article <4191946b-0701-4b77-986c-77373f0b4...(a)s36g2000prg.googlegroups.com>, > colp says... > > >> >> The contradictory outcome was colp's own theory. > > >> > Wrong. The contradictory outcome is a result of the theory of > >> > relativity predicting that an observation will disagree with a > >> > logically expected observation. > > >> That's just more of the same wrong theory. > > >Yes, the theory of relativity is wrong. > > If the theory of relativity is wrong, then > there are two simple ways to demonstrate that > it is wrong: (1) Show that it makes predictions > that are contradictory, or (2) Show that it makes > predictions that are proved false by experiment. In this thread I have pursued the first option. > > But in both cases, you have to actually use > the *actual* theory, and use the *actual* > predictions it makes. You can't just make > things up and call it relativity, and prove > that that's contradictory. Of course. > > What are the *actual* predictions of Einstein's > Special Relativity? For most of the thought experiments > discussed here, the predictions consist of the > following: > > 1. There is a special collection of coordinate systems, > the inertial coordinate systems. If x,y,z,t are coordinates > for an inertial coordinate system, then so are x', y', > z', and t' obtained from (x,y,z,t) through the following > transformations: > > A. Translations: x' = A + x > y' = B + y > z' = C + z > t' = D + t > > B. Rotations: x' = x cos(A) + y sin(A) > y' = y cos(A) - x sin(A) > z' = z > t' = t > (and similiarly for rotations about yz and xz planes) > > C. Lorentz transformations: x' = gamma (x - vt), t' = gamma(t - vx/c^2) > y' = y, z' = z. (And similarly for boosts in the y and z directions) The paradox that I have described incorporates the Lorentz transformation for time. The equation that I have referred to has a different form to the one that you cite: http://en.wikipedia.org/wiki/Time_dilation
From: Sue... on 24 Nov 2007 10:39 On Nov 24, 9:57 am, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: [...] > > 3. If an ideal clock is at rest in an inertial coordinate > system, then the time tau shown on the clock will advance > at the rate (change in tau)/(change in t) = 1, where t is > the time as measured in that coordinate system. It should be noted that a light-clock moving in dielectric media will give this response. It is not an *ideal* clock in the normal sense. It is specifically designed to respond to motion but it inherently encodes the retardation to allow for inertial transactions with its frame of reference. Sue... [...] > Daryl McCullough > Ithaca, NY
From: colp on 24 Nov 2007 10:52 On Nov 24, 12:24 pm, Bryan Olson <fakeaddr...(a)nowhere.org> wrote: > colp wrote: > > Coming from someone what can't event get SR time dilation right for > > decreasing relative distance in an inertial frame, your opinion isn't > > very credible. > > I think you misunderstood Dirk on that. He wrote: > > After clock A has made its turnaround, it has shifted to > another inertial frame, in which according to clock A, > each tick on clock A is simultaneous with some tick on > clock B with a larger time value. > > He's saying that in the returning frame of A, clock B is now > ahead of clock A; running slower due to time dilation, but ahead. I understand that that interpretation is necessary in order to maintain appearances. However it remains that Dirk's statements imply that time dilation occurs on the the outbound leg but not on the inbound leg, when according to SR time dilation occurs on both legs. Dirk says: While clock B is coasting, according to clock B, each tick on clock B is simultaneous with some tick on clock A with a smaller time value. After clock A has made its turnaround, it has shifted to another inertial frame, in which according to clock A, each tick on clock A is simultaneous with some tick on clock B with a larger time value. After clock B has made its turnaround, it has shifted to another inertial frame, in which according to clock B, each tick on clock B is simultaneous with some tick on clock A with a larger time value.
From: bz on 24 Nov 2007 10:57 colp <colp(a)solder.ath.cx> wrote in news:dcc9435e-926e-414e-9af9-2a636e701816(a)e23g2000prf.googlegroups.com: > On Nov 24, 10:59 pm, bz <bz+...(a)ch100-5.chem.lsu.edu> wrote: >> colp <c...(a)solder.ath.cx> wrote >> innews:2145a8d3-287f-4ac0-a232-46a77fb32680(a)s19g2000prg.googlegroups.com >> : >> >> >> >> > On Nov 24, 5:01 pm, bz <bz+...(a)ch100-5.chem.lsu.edu> wrote: >> >> colp <c...(a)solder.ath.cx> wrote >> >> innews:93dca4c9-c5c5-4708-9d0c-258e521aeb9b(a)s36g2000prg.googlegroups. >> >> com >> >> : >> >> >> > O.K. I see what you are saying now. But saying that they see a 1 >> >> > GHz signal for "a while" is a bit misleading because they only see >> >> > that signal for one instant in the entire journey. They start out >> >> > seeing a redshifted signal and end up seeing a blue shifted signal >> >> > so of course they've got to see the signal at it's natural >> >> > frequency at some point. >> >> >> No! It is much longer than a single instant. >> >> > The reason that it is only an instant is that redshift/blueshift >> > would be expected to occur for the receiver as well as the sender, >> > based on symmetry. Typically earth-bound receivers don't have to deal >> > with this issue, so it's no suprise that it didn't occur to you. >> >> > Your example said that the maximum signal transit time was longer >> > than the "a while" that it took for turnaround to occur. So by the >> > time the signal sent during turnaround got to the other twin, that >> > twin would be accelerating or cruising, so the 1GHz signal would be >> > blueshifted during observation and thus observed to be more than >> > 1GHz. >> >> There is NO doppler shift when both ships are moving in the same >> direction at the same velocity. > > That condition never occurs in the experiment. True as seen by an omnicient observer, but NOT true as seen by the occupants of the ships. > The point is that a paradox exists due to the time dilation expected > by SR. The length of time that a 1 GHz signal will be observed in your > example is incidental. [I quote myself as I said this in another article] There is NO Doppler shift when both ships are moving in the same direction at the same velocity. [or appear to be doing so from the viewpoint of either ship]. Ship A can not know that B has turned around for a period of time. Let us see how long that will take. At .5 c it will take our ship 1 year to return to earth. By our clock on the ship, that will be 316 days. 365 days if relativity had no effect on our clock. Signals we transmitted at turn around took 182.5 days to reach earth. That was 158 days by our clock. We will be receiving signals from ship B at 1 GHz until we 'see' his ship turn around. During our mutual, 1 year trips away from earth, before turn around at exactly 1 ly from earth, we will both have transmitted 2.731x10^7 one second ticks. we will both have only received 9.104x10^6 'one second ticks' from the other ship up to the time we turn around. We assume instant turn around so no ticks at that instant. We need to pick up 1.821x10^7 more ticks at 1 tick per second before we see B turn around. That is 210.733 days by our clock. So we will see 1 GHz signals from B for 210.733 days. But it will take us 316 days to reach earth, so we will be left with 105.366 days during which we will see signals transmitted by B after turn around. That means we will receive 2.731x10^7 1 second ticks (take the number of seconds in 105.366 days multiplied by 3[remember the Doppler shift]) And guess what? That turns out to be the exact number of ticks that each of us transmitted during our trip out and that will be the same number of ticks transmitted during our trip back. No ticks are missing at all. Amazing, isn't it. No paradox. Not even a single doc. [unquote] It all worked out exactly correct using relativistic formula. I can send you the mathcad program I used for the computations if you don't believe me. -- 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: Daryl McCullough on 24 Nov 2007 11:50
colp says... > >On Nov 25, 3:57 am, stevendaryl3...(a)yahoo.com (Daryl McCullough) >wrote: >> If the theory of relativity is wrong, then >> there are two simple ways to demonstrate that >> it is wrong: (1) Show that it makes predictions >> that are contradictory, or (2) Show that it makes >> predictions that are proved false by experiment. > >In this thread I have pursued the first option. No, you haven't. As I said, you have to look at what relativity *actually* predicts, not your own distorted version of relativity. >> 1. There is a special collection of coordinate systems, >> the inertial coordinate systems. If x,y,z,t are coordinates >> for an inertial coordinate system, then so are x', y', >> z', and t' obtained from (x,y,z,t) through the following >> transformations: >> >> A. Translations: x' = A + x >> y' = B + y >> z' = C + z >> t' = D + t >> >> B. Rotations: x' = x cos(A) + y sin(A) >> y' = y cos(A) - x sin(A) >> z' = z >> t' = t >> (and similiarly for rotations about yz and xz planes) >> >> C. Lorentz transformations: x' = gamma (x - vt), t' = gamma(t - vx/c^2) >> y' = y, z' = z. (And similarly for boosts in the y and z directions) > >The paradox that I have described incorporates the Lorentz >transformation for time. No, they don't. The Lorentz transformations describe the relation between two *inertial* coordinate systems. They don't say anything about what an accelerated rocket sees or measures. >The equation that I have referred to has a different form to the one >that you cite: > >http://en.wikipedia.org/wiki/Time_dilation Everything on that page is simply mathematical consequences of the claims that I gave. What you seem to have not understood is that time dilation is *not* a relationship between two clocks. It is about the relationship between *one* clock and an inertial coordinate system. If an ideal clock is moving, then the relationship between the time tau shown on the clock and the coordinate time t is given by tau = tau_0 + Integral from t_0 to t of square-root(1-(v/c)^2) dt where v and t are measured in any *inertial* coordinate system, and tau_0 is the time shown on the clock when t = t0. According to this formula, if two clocks accelerate identically (as measured in an inertial coordinate system) then they will always have the same time shown on the clock as measured in that coordinate system. There is nothing contradictory about the application of this time dilation formula. -- Daryl McCullough Ithaca, NY |