From: Dono. on 26 Sep 2009 13:41 On Sep 26, 10:16 am, "Dirk Van de moortel" <dirkvandemoor...(a)nospAm.hotmail.com> wrote: > > But note, Marcel, as always you fucked up where we take > the meanings of the variables into account. > > In special relativity, the equation > tA = t sqrt( 1-V^2 ) > says something about two events occuring at the same > place in A's frame, whereas the equation > tB = t sqrt (1-vB^2) > says something about two events occuring at the same > place in B's frame. > So, still in special relativity the combination > tB = tA sqrt( 1-v^2 ) / (1+ v V) > is only meaningful if it says something about two events > that are simultaneous in A's frame AND in B's frame. > That's only possible if v = 0, in other words words if > tB = tA sqrt( 1-0 ) / (1+0) > i.o.w. if Nooooo, you should have left the idiot in the dark.I gave him this exact part as an exercise to solve, he never answered. I was planning to keep him there forever (which may not be too long, given his advanced Alzheimer)
From: Dirk Van de moortel on 27 Sep 2009 05:23 Dirk Van de moortel <dirkvandemoortel(a)nospAm.hotmail.com> wrote in message j0svm.174383$4f4.105151(a)newsfe11.ams2 > mluttgens <mluttgens(a)orange.fr> wrote in message > 0e036cc7-d090-4958-8c55-6d872544f3d3(a)o35g2000vbi.googlegroups.com > > [snip] > >>>>>>>>>>>>>>> According to the CMBR observer (using c=1), >>>>>>>>>>>>>>> vA = (v-vB)/(1-vB*v) >>>>>>>>>>>>>>> tB = tCMBR * sqrt(1-vB^2) >>>>>>>>>>>>>>> tA = tCMBR * sqrt(1-((v-vB)/(1-vB*v))^2) >>>>>>>>>>>>>>> = tCMBR * sqrt(1-vB^2)*sqrt(1-v^2)/(1-vB*v) >>>>>>>>>>>>>>> Hence, tA = tB * sqrt(1-v^2) / (1-vB*v). > > [snip] > >> How muddled up are the analyses of the twin paradox in terms. >> of relativity theory! , see for instance >> http://www.physicsforums.com/showthread.php?s=12dddfc2f78c05c783324da355c4780a&t=209625&page=2 >> >> Anyhow, a simple analysis uses SR + acceleration. >> >> But the simplest one would use SR + cmbr, which >> straightforwardly shows the absence of the paradox. >> Note that the applicability of SR(cmbr) is wholly general: >> When a clock A moves at V relative to the CMBR, and >> a clock B moves at v relative to clock A, it can be shown >> that tB = tA * sqrt(1-v^2)/(1+Vv) > > You obviously derived this formula as follows: > > You took the velocy of B w.r.t. CMBR as > vB = (v+V) / (1+v V) > so, using t for the CMBR-time you used > tB = t sqrt (1-vB^2) > = t sqrt( 1 - (v+V)^2 / (1+v V)^2 ) > = t sqrt( (1+v V)^2 - (v+V)^2 ) / (1+v V) > = t sqrt( 1-v^2 ) sqrt ( 1-V^2 ) / (1+v V ) > and of course you also used > tA = t sqrt( 1-V^2 ) > so you took these together to find > tB = tA sqrt( 1-v^2 ) / (1+ v V). > Good, the algebra is correct. > > But note, Marcel, as always you fucked up where we take > the meanings of the variables into account. > > In special relativity, the equation > tA = t sqrt( 1-V^2 ) > says something about two events occuring at the same > place in A's frame, whereas the equation > tB = t sqrt (1-vB^2) > says something about two events occuring at the same > place in B's frame. > So, still in special relativity the combination > tB = tA sqrt( 1-v^2 ) / (1+ v V) > is only meaningful if it says something about two events > that are simultaneous in A's frame AND in B's frame. Typo. That should be: ... about two events occuring at the same place in both frames. > That's only possible if v = 0, in other words words if > tB = tA sqrt( 1-0 ) / (1+0) > i.o.w. if > tB = tA, > but we already knew that, didn't we? There's another way to have the events occur at the same place in both frames, even without v = 0, namely if they happen "on top of each other in time", i.o.w. when tB = tA = 0, in which case the equation says 0 = 0, but we already knew that as well, didn't we, Marcel? > >> (see http://pagesperso-orange.fr/mluttgens/twinpdx1.htm >> for more insight). > > Right. More insight in your failure to understand what > the variables represent. > >> Let's note that V and v can have identical or opposite signs. >> Let's also note that claiming that the CMBR can be moving >> wrt an object is physically nonsensical. >> So, in SR(cmbr), one is left with >> tB = tA * sqrt(1-v^2)/(1+Vv) >> tA = tB * (1+Vv)/sqrt(1-v^2) >> Clearly, the time dilation effect is not reciprocal. > > Well, with the necessary condition that this can only happen when > v = 0 , or when tA = tB = 0 , > one is left with > tB = tA > tA = tB > which looks quite reciprocal if you ask me. specially when we turn it into 0 = 0 0 = 0. > > Quod Erat Debunkandum. aka Quod Erat Deluttgandem > > Dirk Vdm
From: Dirk Van de moortel on 27 Sep 2009 05:32 Dono. <sa_ge(a)comcast.net> wrote in message cd418bd1-58a6-4a55-a1e4-ccaf5f7d8cf0(a)k13g2000prh.googlegroups.com > On Sep 26, 10:16 am, "Dirk Van de moortel" > <dirkvandemoor...(a)nospAm.hotmail.com> wrote: >> >> But note, Marcel, as always you fucked up where we take >> the meanings of the variables into account. >> >> In special relativity, the equation >> tA = t sqrt( 1-V^2 ) >> says something about two events occuring at the same >> place in A's frame, whereas the equation >> tB = t sqrt (1-vB^2) >> says something about two events occuring at the same >> place in B's frame. >> So, still in special relativity the combination >> tB = tA sqrt( 1-v^2 ) / (1+ v V) >> is only meaningful if it says something about two events >> that are simultaneous in A's frame AND in B's frame. >> That's only possible if v = 0, in other words words if >> tB = tA sqrt( 1-0 ) / (1+0) >> i.o.w. if > > > Nooooo, you should have left the idiot in the dark.I gave him this > exact part as an exercise to solve, he never answered. I was planning > to keep him there forever (which may not be too long, given his > advanced Alzheimer) Actually, there's still an exercise left for him. Let's see if he finds it. Dirk Vdm
From: Dono. on 27 Sep 2009 10:21 On Sep 27, 2:32 am, "Dirk Van de moortel" <dirkvandemoor...(a)nospAm.hotmail.com> wrote: > Dono. <sa...(a)comcast.net> wrote in message > > cd418bd1-58a6-4a55-a1e4-ccaf5f7d8...(a)k13g2000prh.googlegroups.com > > > > > On Sep 26, 10:16 am, "Dirk Van de moortel" > > <dirkvandemoor...(a)nospAm.hotmail.com> wrote: > > >> But note, Marcel, as always you fucked up where we take > >> the meanings of the variables into account. > > >> In special relativity, the equation > >> tA = t sqrt( 1-V^2 ) > >> says something about two events occuring at the same > >> place in A's frame, whereas the equation > >> tB = t sqrt (1-vB^2) > >> says something about two events occuring at the same > >> place in B's frame. > >> So, still in special relativity the combination > >> tB = tA sqrt( 1-v^2 ) / (1+ v V) > >> is only meaningful if it says something about two events > >> that are simultaneous in A's frame AND in B's frame. > >> That's only possible if v = 0, in other words words if > >> tB = tA sqrt( 1-0 ) / (1+0) > >> i.o.w. if > > > Nooooo, you should have left the idiot in the dark.I gave him this > > exact part as an exercise to solve, he never answered. I was planning > > to keep him there forever (which may not be too long, given his > > advanced Alzheimer) > > Actually, there's still an exercise left for him. > Let's see if he finds it. > > Dirk Vdm :-)
From: mluttgens on 29 Sep 2009 18:11
On 26 sep, 19:16, "Dirk Van de moortel" <dirkvandemoor...(a)nospAm.hotmail.com> wrote: > mluttgens <mluttg...(a)orange.fr> wrote in message > > 0e036cc7-d090-4958-8c55-6d872544f...(a)o35g2000vbi.googlegroups.com > > [snip] > > >>>>>>>>>>>>>> According to the CMBR observer (using c=1), > >>>>>>>>>>>>>> vA = (v-vB)/(1-vB*v) > >>>>>>>>>>>>>> tB = tCMBR * sqrt(1-vB^2) > >>>>>>>>>>>>>> tA = tCMBR * sqrt(1-((v-vB)/(1-vB*v))^2) > >>>>>>>>>>>>>> = tCMBR * sqrt(1-vB^2)*sqrt(1-v^2)/(1-vB*v) > >>>>>>>>>>>>>> Hence, tA = tB * sqrt(1-v^2) / (1-vB*v). > > [snip] > > > How muddled up are the analyses of the twin paradox in terms. > > of relativity theory! , see for instance > >http://www.physicsforums.com/showthread.php?s=12dddfc2f78c05c783324da.... > > > Anyhow, a simple analysis uses SR + acceleration. > > > But the simplest one would use SR + cmbr, which > > straightforwardly shows the absence of the paradox. > > Note that the applicability of SR(cmbr) is wholly general: > > When a clock A moves at V relative to the CMBR, and > > a clock B moves at v relative to clock A, it can be shown > > that tB = tA * sqrt(1-v^2)/(1+Vv) > > You obviously derived this formula as follows: > > You took the velocy of B w.r.t. CMBR as > vB = (v+V) / (1+v V) > so, using t for the CMBR-time you used > tB = t sqrt (1-vB^2) > = t sqrt( 1 - (v+V)^2 / (1+v V)^2 ) > = t sqrt( (1+v V)^2 - (v+V)^2 ) / (1+v V) > = t sqrt( 1-v^2 ) sqrt ( 1-V^2 ) / (1+v V ) > and of course you also used > tA = t sqrt( 1-V^2 ) > so you took these together to find > tB = tA sqrt( 1-v^2 ) / (1+ v V). > Good, the algebra is correct. > > But note, Marcel, as always you fucked up where we take > the meanings of the variables into account. > > In special relativity, the equation > tA = t sqrt( 1-V^2 ) > says something about two events occuring at the same > place in A's frame, whereas the equation > tB = t sqrt (1-vB^2) > says something about two events occuring at the same > place in B's frame. > So, still in special relativity the combination > tB = tA sqrt( 1-v^2 ) / (1+ v V) > is only meaningful if it says something about two events > that are simultaneous in A's frame AND in B's frame. > That's only possible if v = 0, in other words words if > tB = tA sqrt( 1-0 ) / (1+0) > i.o.w. if > tB = tA, > but we already knew that, didn't we? > > > (seehttp://pagesperso-orange.fr/mluttgens/twinpdx1.htm > > for more insight). > > Right. More insight in your failure to understand what > the variables represent. > > > Let's note that V and v can have identical or opposite signs. > > Let's also note that claiming that the CMBR can be moving > > wrt an object is physically nonsensical. > > So, in SR(cmbr), one is left with > > tB = tA * sqrt(1-v^2)/(1+Vv) > > tA = tB * (1+Vv)/sqrt(1-v^2) > > Clearly, the time dilation effect is not reciprocal. > > Well, with the necessary condition that this can only happen when > v = 0 , > one is left with > tB = tA > tA = tB > which looks quite reciprocal if you ask me. Dirk, clock B moves at v relative to clock A in the scenario. Whn v = 0, both clocks are at rest relativ to each other. Then, of course, both clocks will show the same time. Marcel Luttgens > > Quod Erat Debunkandum. > > Dirk Vdm |