Speed of Time
in [Relativity]
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From: Androcles on 26 Dec 2009 00:19 "Ste" <ste_rose0(a)hotmail.com> wrote in message news:c6d4ea67-9711-4b79-b35f-7bc54e086b76(a)n35g2000yqm.googlegroups.com... On 26 Dec, 00:30, "Androcles" <Headmas...(a)Hogwarts.physics_q> wrote: > > You should read Dork's twin paradox analysis. He can > pick up the origin of frame of reference and move it, > so if you go from London to New York you can do it > twice without ever going from New York to London. > > He says > quote/ > "We use 3 inertial reference frames. > S: The frame of the "stay at home" twin. > S': The frame of the "outbound part of the trip". > S": The frame of the "inbound part of the trip". > http://users.telenet.be/vdmoortel/dirk/Physics/TwinsEvents.html > > So if T = 5 years and v = 0.8c, then the stay at home twin will > have aged 10 years while his travelling twin sister will have aged > 6 years. > /unquote I'm afraid I don't understand how he arrives at that conclusion. =========================================== You don't understand how Dork is a dork? I doubt anybody does. =========================================== It rubbishes relativity. Relativity says that both twins will perceive the same effects, relative to their own frame of reference. If that is true, then the astronaut cannot return younger than the homebody - it cannot happen. Because if you change the analysis and have the astronaut in the fixed frame of reference, and have the universe accelerate around him, then by exactly the same logic the *homebody* will be the younger twin when the astronaut returns to earth. If the astronaut is younger when he returns, then either relativity is false, or there is another factor in play that applies differently to the astronaut than the homebody. Incidentally, what evidence is there exactly to suggest that such twins ages would differ once the astronaut returned to earth? ========================================= Evidence? You want EVIDENCE? Bwhahahahahahaha! "That is, we can reverse the directions of the frames which is the same as interchanging the frames, which - as I have told you a LOT of times, OBVIOUSLY will lead to the transform: t = (tau-xi*v/c^2)/sqrt(1-v^2/c^2) x = (xi - v*tau)/sqrt(1-v^2/c^2) or: tau = (t+xv/c^2)/sqrt(1-v^2/c^2) xi = (x + vt)/sqrt(1-v^2/c^2)" -- Paul Bigot Andersen Note that the Andersen Transformations tau = (t PLUS xv/c^2)/sqrt(1-v^2/c^2) xi = (x + vt)/sqrt(1-v^2/c^2) differ from the Lorentz Transformations tau = (t MINUS xv/c^2)/sqrt(1-v^2/c^2) xi = (x - vt)/sqrt(1-v^2/c^2) when the frames are reversed. Further note that Einstein was a fruitcake.
From: Inertial on 26 Dec 2009 00:53 "Androcles" <Headmaster(a)Hogwarts.physics_q> wrote in message news:0fhZm.188$K96.184(a)newsfe04.ams2... > > "Ste" <ste_rose0(a)hotmail.com> wrote in message > news:c6d4ea67-9711-4b79-b35f-7bc54e086b76(a)n35g2000yqm.googlegroups.com... > On 26 Dec, 00:30, "Androcles" <Headmas...(a)Hogwarts.physics_q> wrote: >> >> You should read Dork's twin paradox analysis. He can >> pick up the origin of frame of reference and move it, >> so if you go from London to New York you can do it >> twice without ever going from New York to London. >> >> He says >> quote/ >> "We use 3 inertial reference frames. >> S: The frame of the "stay at home" twin. >> S': The frame of the "outbound part of the trip". >> S": The frame of the "inbound part of the trip". >> http://users.telenet.be/vdmoortel/dirk/Physics/TwinsEvents.html >> >> So if T = 5 years and v = 0.8c, then the stay at home twin will >> have aged 10 years while his travelling twin sister will have aged >> 6 years. >> /unquote > > I'm afraid I don't understand how he arrives at that conclusion. > > =========================================== > You don't understand how Dork is a dork? I doubt anybody does. > =========================================== > > > It > rubbishes relativity. Relativity says that both twins will perceive > the same effects, relative to their own frame of reference. If that is > true, then the astronaut cannot return younger than the homebody - it > cannot happen. Because if you change the analysis and have the > astronaut in the fixed frame of reference, and have the universe > accelerate around him, then by exactly the same logic the *homebody* > will be the younger twin when the astronaut returns to earth. > > If the astronaut is younger when he returns, then either relativity is > false, or there is another factor in play that applies differently to > the astronaut than the homebody. > > Incidentally, what evidence is there exactly to suggest that such > twins ages would differ once the astronaut returned to earth? > ========================================= > Evidence? You want EVIDENCE? Bwhahahahahahaha! > "That is, we can reverse the directions of the frames > which is the same as interchanging the frames, > which - as I have told you a LOT of times, > OBVIOUSLY will lead to the transform: > t = (tau-xi*v/c^2)/sqrt(1-v^2/c^2) > x = (xi - v*tau)/sqrt(1-v^2/c^2) > or: > tau = (t+xv/c^2)/sqrt(1-v^2/c^2) > xi = (x + vt)/sqrt(1-v^2/c^2)" -- Paul Bigot Andersen > > Note that the Andersen Transformations > tau = (t PLUS xv/c^2)/sqrt(1-v^2/c^2) > xi = (x + vt)/sqrt(1-v^2/c^2) > > differ from the Lorentz Transformations > tau = (t MINUS xv/c^2)/sqrt(1-v^2/c^2) > xi = (x - vt)/sqrt(1-v^2/c^2) Derr .. because v is in the opposite direction when you change frames, so the opposite sign. You're such an idiot, Andy .. nice of you to confirm it every so often.
From: Inertial on 26 Dec 2009 04:29 "Henry Wilson DSc" <..@..> wrote in message news:pkkbj5da4m916llnnejiae7027uuj4qicl(a)4ax.com... > On Sat, 26 Dec 2009 14:28:03 +1100, "Inertial" <relatively(a)rest.com> > wrote: > >>"Ste" <ste_rose0(a)hotmail.com> wrote in message >>news:ee5c4ca0-faf6-46a8-8565-c830f685d3b9(a)m16g2000yqc.googlegroups.com... >>> I was just wondering, can anyone tell me at what rate time advances on >>> earth? >> >>There is no such thing as a rate of time, as 'rate' implies change over >>time. That is a problem with the English language (and I suspect most if >>not all human languages). >> >>But one can compare the rates of ticking of clocks (which measures time) >>and >>compare the rates in different locations or at different relative >>velocities. >> >>There is no absolute measure of 'rate' of time (whatever that means) > > Time flows at 1 second (t1) per second (t2) What a load of bullshit
From: Sue... on 26 Dec 2009 10:15 On Dec 24, 3:51 pm, Ste <ste_ro...(a)hotmail.com> wrote: > I was just wondering, can anyone tell me at what rate time advances on > earth? <<invariance with respect to time translation gives the well-known law of conservation of energy>> http://en.wikipedia.org/wiki/Noether%27s_theorem#Applications See also: E. Noether's Discovery of the Deep Connection Between Symmetries and Conservation Laws Authors: Nina Byers http://arxiv.org/abs/physics/9807044 Sue...
From: Ste on 26 Dec 2009 16:59
On 26 Dec, 03:28, "Inertial" <relativ...(a)rest.com> wrote: > "Ste" <ste_ro...(a)hotmail.com> wrote in message > > news:ee5c4ca0-faf6-46a8-8565-c830f685d3b9(a)m16g2000yqc.googlegroups.com... > > > I was just wondering, can anyone tell me at what rate time advances on > > earth? > > There is no such thing as a rate of time, as 'rate' implies change over > time. That is a problem with the English language (and I suspect most if > not all human languages). > > But one can compare the rates of ticking of clocks (which measures time) and > compare the rates in different locations or at different relative > velocities. > > There is no absolute measure of 'rate' of time (whatever that means) Yes, precisely, so *how* do we know it is flowing forwards? It's a pretty simple question to ask of physics. And anyway I think I've just solved the twins paradox to explain why the astronaut will indeed come back younger. More on that later. |