CMBR's motion wrt the Earth
in [Relativity]
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From: Dirk Van de moortel on 30 Sep 2009 03:57 mluttgens wrote: > 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. Yes of course. That's what *I* was telling *you*, imbecile. Good grief, you are so stupid. Dirk Vdm
From: mluttgens on 1 Oct 2009 07:39 On 30 sep, 09:57, "Dirk Van de moortel" <dirkvandemoor...(a)nospAm.hotmail.com> wrote: > mluttgens wrote: > > 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. > > Yes of course. That's what *I* was telling *you*, imbecile. > Good grief, you are so stupid. The formula tB = tA * sqrt(1-v^2)/(1+Vv) leads to the same results as SR in the analysis of the twin paradox, see http://pagesperso-orange.fr/mluttgens/twinpdx1.htm : Luttgens: "If you had used (v+v) / (1+v^2) instead of -tanh(2 * acrtanh(.99) ) as the velocity addition formula, and made the necessary simplifications, you would have obtained 2 / sqrt(1-v^2) for the round-trip time" Shuba: "Of course I do not disagree that 2 * 7 = 14. But 2 * \gamma is not a general result that can be used to solve this type of problem. It works in this case only because the FAQ problem was designed to be very simple." SR gets rid of the paradox by resorting to acceleration. The above formula is wholly general. Interestingly enough, tA = tB, not only when v = 0, but also when 1-v^2 = 1+Vv. Btw, I find your vulgar behaviour pityful. Bye! Marcel Luttgens > > Dirk Vdm
From: Dirk Van de moortel on 1 Oct 2009 11:26 mluttgens <mluttgens(a)orange.fr> wrote in message 2197966f-2954-4458-9017-1785526a6b2d(a)37g2000yqm.googlegroups.com > On 30 sep, 09:57, "Dirk Van de moortel" > <dirkvandemoor...(a)nospAm.hotmail.com> wrote: >> mluttgens wrote: >>> 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. >> >> Yes of course. That's what *I* was telling *you*, imbecile. >> Good grief, you are so stupid. > > The formula tB = tA * sqrt(1-v^2)/(1+Vv) leads to the same results > as SR in the analysis of the twin paradox, In special realtivity, with *your* set-up the formula tB = tA * sqrt(1-v^2)/(1+Vv) is only valid if v = 0 or if tB = tA = 0 and I'm not sorry if that ruins your life's work. Honestly not. Dirk Vdm
From: Androcles on 1 Oct 2009 18:15
"Paul B. Andersen" <paul.b.andersen(a)utopia.no> wrote in message news:4AC50ACD.6090301(a)utopia.no... http://www.androcles01.pwp.blueyonder.co.uk/SR.GIF > Paul, still laughing |