The real twin paradox.
in [Relativity]
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From: Daryl McCullough on 27 Nov 2007 10:24 colp says... >Because an observer on the ground sees the satellites in motion >relative to them, Special Relativity predicts that we should see their >clocks ticking more slowly (see the Special Relativity lecture). You don't know what you are talking about. Relativity is not about what this or that observer *sees*. I gave you the formula for proper time: In differential form, dTau = square-root(|sum over u,v of g_uv dx^u dx^v|) This gives the proper time tau in terms of the *path* taken by the clock through spacetime. Time dilation is *not* about comparing two clocks, it is about computing the proper time for a path through spacetime. Different paths correspond to different proper times. >Special Relativity predicts that the on-board atomic clocks on the >satellites should fall behind clocks on the ground by about 7 >microseconds per day because of the slower ticking rate due to the >time dilation effect of their relative motion. Work through the math! That's what my formula predicts. For near-Earth, slow-velocity motion, the formula for dTau is approximately given by: dTau = (1 - GM/(c^2 r) - 1/2 v^2/c^2) dt There are two effects at work: dTau depends on height: -GM/(c^2 r)dt is a negative contribution which gets less negative at larger r. So higher clocks "run faster". The second term -1/2 v^2/c^2 dt, is a negative term that gets more negative the larger v gets. So velocity makes the clocks "run slow". Both effects are included in the formula for proper time. It is *not* a comparison of two clocks, it is a formula giving the time on *one* as a function of its path through spacetime. -- Daryl McCullough Ithaca, NY
From: Daryl McCullough on 27 Nov 2007 10:32 colp says... > >On Nov 27, 9:51 am, stevendaryl3...(a)yahoo.com (Daryl McCullough) >wrote: >> SR doesn't say *anything* about the frame of reference of an >> accelerated twin. It only talks about how things work within >> a single inertial coordinate system. If the twins are accelerating, >> then they are not in an inertial coordinate system. > >The original thought experiment that I described in the OP (that Dirk >quoted) does talk about inertial coordinate systems. What happened when >the twins are accelerating and decellerating doesn't affect the >paradox. Well, that's completely incorrect. You only think that because you haven't actually done a calculation. Do the calculation! Don't go saying "the acceleration has no effect" without *proving* it. >> >> >Special relativity says that each twin must observe that the other's >> >> >clock is running slow >> >> >> No, it doesn't say that. >> >> >Yes it does. >> >> You don't know what you are talking about. > >The velocity term is squared in the Lorentz factor. But *LOOK* at the equations! t' = gamma (t - vx/c^2) Yes, gamma is always greater than 1. But t' is not just proportional to t. There are *TWO* terms in the expression for t': 1. gamma t 2. - gamma vx/c^2 t' is the *SUM* of those terms. The first term is always greater than t, but the second term can be positive, negative, or zero, depending on the values for x and v. >> >In special relativity, clocks that are moving with respect to an >> >inertial system of observation are measured to be running slower. >> >> Yes, but in the case you are talking about, *neither* twin >> is in an inertial coordinate system. > >Wrong. The experiment described in the OP is what I am talking about. Your experiment described two twins traveling away from each other and then *TURNING* around and traveling back in the opposite direction. Turning around means *changing* velocity, which means *NONINERTIAL* motion. >> >> SR does not say that each clock observes the other clock running >> >> slow. What it says is that any *inertial coordinate system* measures >> >> any moving clock to be running slow. Time dilation is *not* a >> >> relationship between two clocks, but is a relationship between *one* >> >> clock and one inertial coordinate system. >> >> >Time dilation is a relationship between two clocks in the case of GPS >> >sattelites. >> >> No, it is not. You don't know what you are talking about. > >GPS clocks have to be corrected for SR (& GR) time dilation. Yes, I gave you the formula for that. >> Relativity tells you how much proper time a clock experiences >> for a given path through spacetime. That's an invariant. It's >> not a comparison of two different clocks. > >In the real world different clocks do get compared. What gets compared are two *different* proper times, for two *different* paths through spacetime. What GR tells you is how to compute proper times for a path through spacetime. -- Daryl McCullough Ithaca, NY
From: bz on 27 Nov 2007 10:56 "Sue..." <suzysewnshow(a)yahoo.com.au> wrote in news:1df927b1-290c-44b4-9353-613eb5b3291c(a)s12g2000prg.googlegroups.com: > On Nov 27, 7:41 am, bz <bz+...(a)ch100-5.chem.lsu.edu> wrote: >> "Sue..." <suzysewns...(a)yahoo.com.au> wrote in >> news:1666352e-81c8-4ec7-9a90- >> fc0dd2440...(a)s12g2000prg.googlegroups.com: >> >> > I am not aware of any calculations that that involve a >> > real geostationary clock but they could contribute a >> > lot of confidence in the 7us component if they exist. >> >> Gosh, Sue, it is very easy to do such calculations. I would think that >> an expert on SR and time, such as yourself, would have already done >> them. My calculations show that the gravity at GPS altitude (orbital >> radius 26562.463 km) is 3.746% g and at geosync altitude (orbital >> radius 42165.282 km) >> the g field is 1.725% g. >> >> That gives a 590.217 ppt difference in clock rate (as opposed to >> 528.428 ppt for the GPS clocks). >> Or 50.995 us for GeoSync (as contrasted with the 45.532 us for the >> effect of altitude on GPS clocks). > > What is the name of the SV? Does it matter? There are plenty of 'Real' SVs up there in geosync orbits. How about a satellite named Sue? > Where did you get the raw data on its clock stability? > How was it compared to GPS clocks ? I am just doing some calcs to see if it would be practical to do such a test. Answer, yes. >> So, the crystal oscillators (and frequencies of transmission) for >> Geosync satellites should be off by 590.217 ppt. >> That amounts to a 0.59 Hz shift in a 1 GHz transmitter's output >> frequency. >> >> Certainly measurable, but probably swamped by drift due to component >> aging and temperature variations as it will NOT be cumulative (unless >> someone decides to count cycle on the carrier frequency over a period >> of time, THEN it would be cumulative AND would allow a long term test >> WITHOUT orbiting a cesium clock). >> Telemetry of temperature conditions could be used to compensate >> somewhat, as well as comparison with a similar circuit kept on earth >> under similar pressure and temperature conditions. > > Calculations look great! What is the status of your oscillator? Which one? I have some 100 KHz clocks that are pretty stable. Used them when I was working on the river. One was the clock for my home built frequency counter. I checked for zero- beat with wwv 10 MHz many times and it was within +/- 1 Hz at 10 MHz. That put the 156 MHz ship board transmitters within FCC specs when I set them on channel with the counter. I am sure that there is much better xtals available now. I mentioned this because there are many satellites NOW in orbit that HAVE 'clocks' that can be tested to see if the current frequency is more consistent with "Sue's relativity" or Einsteins. > Proceedings of the 2003 IEEE International > << Summary: A science-quality space GPS receiver is being studied > for the primary atomic reference clock in space (PARCS) mission. > The PARCS flight experiment is an International space station (ISS) > payload that will conduct investigations into the laser cooling of > atoms, time interval measurement, and fundamental physics. > The receiver will make GPS carrier phase observations, to transfer > the frequency measurements made by other PARCS subsystems to the > ground and to determine the experiment's precise position and > velocity. > The receiver is based on the Jet Propulsion Laboratory's BlackJack > radiometric instrument. This is a dual frequency, codeless design > that > is a veteran of multiple spaceflights. The major challenges for its > use on PARCS derive from the ISS environment, for example, the > antenna field of view, multipath sources, and potential > electromagnetic interference. Simulations indicate that the > restricted field of view will be the main limitation, and that > the receiver antenna should be tilted away from the ISS structure > by /spl sim/30/spl deg/ for better results. The use of GPS > ground networks and data analysis techniques to provide a total > measurement system adequate to meet PARCS' requirements will > need to be examined further. >> > http://ieeexplore.ieee.org/iel5/8990/28531/01275085.pdf (subscription) That would certainly give interesting information but would not answer the question you raised: how do clocks in GeoSync orbits behave. Hence, it is another 'distraction' from the case in question, which you raised as a distraction from the other case in question. -- 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 27 Nov 2007 11:44 colp says... > >On Nov 27, 9:43 am, stevendaryl3...(a)yahoo.com (Daryl McCullough) >wrote: >> 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)) ? That's what used. v = 4/5 c, (1 - v^2/c^2) = 9/25. square-root(1 - v^2/c^2) = 3/5. gamma = 1/square-root(1 - v^2/c^2) = 5/3. -- Daryl McCullough Ithaca, NY
From: Daryl McCullough on 27 Nov 2007 11:51
Sue... says... > >On Nov 26, 4:22 pm, stevendaryl3...(a)yahoo.com (Daryl McCullough) >wrote: >> However, >> a precise enough clock (atomic clock, for instance) >> would notice a difference. If you left one atomic >> clock at Big Ben, and carried the other one with >> you to New York, then when you get back to Big Ben, >> there will be differences in the elapsed times on >> the two atomic clocks. > >There were no clocks in motion. The proper time intervals >were taken from views of the clocks that never left >their stations in London and New York. You are making no sense whatsoever. Proper time is time along a spacetime path. It has nothing to do with "views of clocks". -- Daryl McCullough Ithaca, NY |