relativity vs velocity addition
in [Relativity]
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From: Sorcerer on 12 Dec 2006 06:48 "Sue..." <suzysewnshow(a)yahoo.com.au> wrote in message news:1165921285.661554.32680(a)73g2000cwn.googlegroups.com... | lkoluk2003(a)yahoo.com wrote: | [...] | | > | > Now let the platform carrying the observer B is moving with a constant | > speed v with respect to the observer A. The clock is placed in such a | > way that the light pulse movement is in the same direction with the | > platform's speed. Assume there is a time dilation B. I.e. t'=t.B where | > t' is the time measured by observer B and t is the time measured by the | > observer A. Since according to the observer B, there is nothing | > changed, so (s)he will observe the tick time as t'=2x'/c or 2.x'=c.t'. | | Does the "B" have anything to do with magnetic field? [...] Does "magnetic field" have anything to do with relativity vs velocity addition? No! So shut up, ignoramus.
From: lkoluk2003 on 14 Dec 2006 04:31 > > > It seems that the assumption that the maximum distances between the > twins during inbound and outbound part are equal is not generally true. > I.e. the most general formula is t1=x1/v1 and t2=x2/v2 where x1 is not > equal to x2. In this case, the only explanation is that the clock rates > of both twins are the same even from the point of view of the twins. > > On the other hand, the relativity principle is fully compatible with > this. I copied the following from my text in another threat. > "Each tick in a clock is an event and an event's observed time can be > different from time dilation. For example one can set a clock by using > a light pulse > and two mirrors. The pulse is reflected between the mirrors and the > time interval between the reflection times of mirror 1 can be > considered as one tick of this clock. If the light speed is source > dependent then the duration of each tick is the same regardless of the > speed of the clock and the time delation." > > Assume there is a platform with the clock mentioned above and two > observers A&B. The tick time of this clock would be t=2.x/c where x is > the distance between the mirrors. > > Now let the platform carrying the observer B is moving with a constant > speed v with respect to the observer A. The clock is placed in such a > way that the light pulse movement is in the same direction with the > platform's speed. Assume there is a time dilation B. I.e. t'=t.B where > t' is the time measured by observer B and t is the time measured by the > observer A. Since according to the observer B, there is nothing > changed, so (s)he will observe the tick time as t'=2x'/c or 2.x'=c.t'. > > The relativity principle requires that the light speed is source > dependent. Let this relative speed is k(v). Then the tick time for > observer A would be > t=2.x'/k(v) = c.t'/k(v ) = c.t.B/k(v) > > >From here we deduce k(v)=B.c. On the other hand x'/t'=x/t must be true. Yep! There is something wrong here. This relation must be true to make tick times identical. I.e. it comes from the twin paradox experiment. The correct derivation should be as follows. Let the observer A has an identical clock. The tick time of this clock is found as ta=2.x/c. On the other hand, the observer A finds the tick time of B's clock as tb=2.x'/k. So we find k=c.B as shown above. From the twin paradox experiment we know ta=tb. From here, we obtain 2.x'/k=2.x/c. So there must be a length dilation such that x'=x.B. Also actually, the clock direction need not to be the same with the direction of the relative speed. This means that the same factor must be applied to all dimensions. Lokman Kolukisa > I.e. x'=x.B. So from here > > t=2.x'/k(v) = 2.x.B/(c.B) = 2.x/c > > same with if the speed was zero. As seen the observed tick time is > independent from the speed and from the dilation factor. The same thing > is true for any event including the movement of someting or at least > any event whose time is measured by distance/speed. This is a perfect > result because the twin paradox is fully resolved now(assuming the time > measure always involves something which has a movement) and the > dilation factor can be choosen without considering it. > > Best regards, > Lokman Kolukisa
From: lkoluk2003 on 14 Dec 2006 04:47 Sorcerer yazdi: > <lkoluk2003(a)yahoo.com> wrote in message news:1165919683.448586.288430(a)16g2000cwy.googlegroups.com... > | It seems that the assumption that the maximum distances between the > | twins during inbound and outbound part are equal is not generally true. > | I.e. the most general formula is t1=x1/v1 and t2=x2/v2 where x1 is not > | equal to x2. In this case, the only explanation is that the clock rates > | of both twins are the same even from the point of view of the twins. > | > | On the other hand, the relativity principle is fully compatible with > | this. I copied the following from my text in another threat. > | "Each tick in a clock is an event and an event's observed time can be > | different from time dilation. For example one can set a clock by using > | a light pulse > | and two mirrors. > > The second is the duration of 9 192 631 770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the cesium 133 atom. > > http://physics.nist.gov/cuu/Units/second.html > > See anything about setting a counter with two mirrors? > > Send a caesium atom to Proxima Centauri and back, COUNTING transitions. > > The count will match an identical caesium atom that remains here. > > During the journey it will *appear* not to match due to > > transitions being "in flight", aka Doppler shift. > > There are no missing or additional counts, hence no count dilation, Ok. But what is the mechanism behind this light emmitting? Does the frequency relate to distance/speed relation? For example is the frequency proportional to (d1-d2)/(v1-v2) where d1&v1 is the radius of an orbital and the speed of an electron in this orbital respectively and d2&v2 is the radius&electron speed at the orbital where the electron drops to after it emits a photon? > > hence no time dilation. > > Einstein was an idiot. > I don't think so. His mistake was to assume the light speed is the same for all inertial frames and the others vary respectively in the relation light speed=distance/duration. However, the relativity principle requires that the duration is the same for all inertial frames and the others vary respectively. Lokman Kolukisa > > > > > > > > > > > > > > > The pulse is reflected between the mirrors and the > | time interval between the reflection times of mirror 1 can be > | considered as one tick of this clock. If the light speed is source > | dependent then the duration of each tick is the same regardless of the > | speed of the clock and the time delation." > | > | Assume there is a platform with the clock mentioned above and two > | observers A&B. The tick time of this clock would be t=2.x/c where x is > | the distance between the mirrors. > | > | Now let the platform carrying the observer B is moving with a constant > | speed v with respect to the observer A. The clock is placed in such a > | way that the light pulse movement is in the same direction with the > | platform's speed. Assume there is a time dilation B. I.e. t'=t.B where > | t' is the time measured by observer B and t is the time measured by the > | observer A. Since according to the observer B, there is nothing > | changed, so (s)he will observe the tick time as t'=2x'/c or 2.x'=c.t'. > | > | The relativity principle requires that the light speed is source > | dependent. Let this relative speed is k(v). Then the tick time for > | observer A would be > | t=2.x'/k(v) = c.t'/k(v ) = c.t.B/k(v) > | > | >From here we deduce k(v)=B.c. On the other hand x'/t'=x/t must be true. > | I.e. x'=x.B. So from here > | > | t=2.x'/k(v) = 2.x.B/(c.B) = 2.x/c > | > | same with if the speed was zero. As seen the observed tick time is > | independent from the speed and from the dilation factor. The same thing > | is true for any event including the movement of someting or at least > | any event whose time is measured by distance/speed. This is a perfect > | result because the twin paradox is fully resolved now(assuming the time > | measure always involves something which has a movement) and the > | dilation factor can be choosen without considering it. > | > | Best regards, > | Lokman Kolukisa > |
From: Sorcerer on 14 Dec 2006 05:34 <lkoluk2003(a)yahoo.com> wrote in message news:1166089679.248549.246580(a)f1g2000cwa.googlegroups.com... | | Sorcerer yazdi: | > <lkoluk2003(a)yahoo.com> wrote in message news:1165919683.448586.288430(a)16g2000cwy.googlegroups.com... | > | It seems that the assumption that the maximum distances between the | > | twins during inbound and outbound part are equal is not generally true. | > | I.e. the most general formula is t1=x1/v1 and t2=x2/v2 where x1 is not | > | equal to x2. In this case, the only explanation is that the clock rates | > | of both twins are the same even from the point of view of the twins. | > | | > | On the other hand, the relativity principle is fully compatible with | > | this. I copied the following from my text in another threat. | > | "Each tick in a clock is an event and an event's observed time can be | > | different from time dilation. For example one can set a clock by using | > | a light pulse | > | and two mirrors. | > | > The second is the duration of 9 192 631 770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the cesium 133 atom. | > | > http://physics.nist.gov/cuu/Units/second.html | > | > See anything about setting a counter with two mirrors? | > | > Send a caesium atom to Proxima Centauri and back, COUNTING transitions. | > | > The count will match an identical caesium atom that remains here. | > | > During the journey it will *appear* not to match due to | > | > transitions being "in flight", aka Doppler shift. | > | > There are no missing or additional counts, hence no count dilation, | | Ok. But what is the mechanism behind this light emmitting? Strike a match. That emits light. Does the | frequency relate to distance/speed relation? No, frequency relates to the inverse time law, f = 1/t. Blind Poe can explain inverse laws to you. | For example is the | frequency proportional to (d1-d2)/(v1-v2) where d1&v1 is the radius of | an orbital and the speed of an electron in this orbital respectively | and d2&v2 is the radius&electron speed at the orbital where the | electron drops to after it emits a photon? No. | | > | > hence no time dilation. | > | > Einstein was an idiot. | > | | I don't think Of course you don't. Nobody ever said you did. That's why I have to tell you Einstein was an idiot, you can't work it out for yourself. You are an idiot too, you CAN'T think. | so. Exactly. | His mistake If he made a mistake he was an idiot. | was to assume the light speed is the same | for all inertial frames Einstein never said it was. That's your mistake. | and the others vary respectively in the | relation light speed=distance/duration. However, the relativity | principle requires that the duration is the same for all inertial | frames and the others vary respectively. What relativity principle? Read this, published in the British Journal of Theoretical Physics http://www.androcles01.pwp.blueyonder.co.uk/PoR/PoR.htm | | Lokman Kolukisa | > | > | > | > | > | > | > | > | > | > | > | > | > | > | > The pulse is reflected between the mirrors and the | > | time interval between the reflection times of mirror 1 can be | > | considered as one tick of this clock. If the light speed is source | > | dependent then the duration of each tick is the same regardless of the | > | speed of the clock and the time delation." | > | | > | Assume there is a platform with the clock mentioned above and two | > | observers A&B. The tick time of this clock would be t=2.x/c where x is | > | the distance between the mirrors. | > | | > | Now let the platform carrying the observer B is moving with a constant | > | speed v with respect to the observer A. The clock is placed in such a | > | way that the light pulse movement is in the same direction with the | > | platform's speed. Assume there is a time dilation B. I.e. t'=t.B where | > | t' is the time measured by observer B and t is the time measured by the | > | observer A. Since according to the observer B, there is nothing | > | changed, so (s)he will observe the tick time as t'=2x'/c or 2.x'=c.t'. | > | | > | The relativity principle requires that the light speed is source | > | dependent. Let this relative speed is k(v). Then the tick time for | > | observer A would be | > | t=2.x'/k(v) = c.t'/k(v ) = c.t.B/k(v) | > | | > | >From here we deduce k(v)=B.c. On the other hand x'/t'=x/t must be true. | > | I.e. x'=x.B. So from here | > | | > | t=2.x'/k(v) = 2.x.B/(c.B) = 2.x/c | > | | > | same with if the speed was zero. As seen the observed tick time is | > | independent from the speed and from the dilation factor. The same thing | > | is true for any event including the movement of someting or at least | > | any event whose time is measured by distance/speed. This is a perfect | > | result because the twin paradox is fully resolved now(assuming the time | > | measure always involves something which has a movement) and the | > | dilation factor can be choosen without considering it. | > | | > | Best regards, | > | Lokman Kolukisa | > | |
From: Sylvia Else on 14 Dec 2006 06:06
lkoluk2003(a)yahoo.com wrote: > Hi, > Although the symmetric twin paradox can be explaied by ALT(Aether > theory with Lorentz Transformations) , I am a relativist. So after I > was sure SR(special relativity) is incorrect, I started to search > explanation(s) of the paradox in a relativist way. According to me the > starting point ought to be the velocity addition rule, because every > huge leap in physics is achieved by understanding the secrets of > velocity. Galileo set up a new phsics by the concepts of inertia and > independence of velocities in different axes(vector addition). SR and > GR(General Relativity) is also set up by claiming the velocity > additition rule is not a simple algebraic sum. I don't try it, but it > seems that the lorentz transformations can be derived from the velocity > addition rule which is (v+w)/(1+vw/c^2) if v and w have the same > direction. Now I will try to show that if relativity principle(i.e. if > there is no absolute inertial frame) is true, then the speed of light > must be a constant relative to the source. > > Let there are two platforms A and B and within each platform there are > two observers Oa and Ob respectively. Let the platforms are two trains > and Ob is in the middle of the train B with a detector D. On each of > the two far sides of the train there is a clock and a light source. > When the clock ticks a predefined times, the light source fires a light > beam such that it will hit the detector on the middle of the train. > I.e. the light source Sf fires light beam from left to right and Sb > fires in opposite direction as shown in the following. > > ------------------ > -------------------------------------------------------------- > | | | Sf --------> D > <--------- Sb | > | Oa | | Cf Ob > Cb | > ------------------ > --------------------------------------------------------------- > Train A Train B -----> > x axis > > The distance between each light source and detector D is the same. > Detector gives two results: the two light beams hit at the same time > or in different times. > > My postulates are the followings: > > 1. The experiments within a train does not affected by the outside > objects which have a constant speed relative to it. > 2. The speed of light is direction independent within a train. > > > Experiment1: > Synchronize the clocks and set up such that the light sources will be > fired after n ticks. So they will fire at the same time according to > observer Ob. The relative speed of trains A and B is zero. So the same > thing is true for observer Oa. Of course , from the Ob's reference > frame the two lights must hit the detector at the same time with the > given postulates. This is the same for Oa. > > Experiment2: > Synchronize the clocks and set up such that the light sources will be > fired after n ticks. Place the clocks and light sources on the two far > sides of the train B as mentioned. The relative speed of trains A and > B is zero. So the clocks are synchronized according to both Oa and Ob. > Now let train B accelerates and reach a constant speed v relative to > train A after a while along the x axis. Then wait for the experiment > to be completed. According to Ob the experiment gives the same result. > I.e. the lights hit at the same time. Now examine what Oa see with the > assumption that the speed of light is always the same according to the > observer. > >>From Ob's reference frame: The clocks are still synchronized since they > share the same movement and so get the same affects. So the two light > beams are fired at the same time. The speed of the light train fired > from Sf is c and from Sb is -c. Still the distance between Sf and D is > the same with the distance between Sb and D although they are shorter > now. Let this distance be x. So, the travel time of the light beam > fired from Sf would be x/(c-v) and the travel time of the light beam > fired from Sb would be x/(c+v). Since v is greater than zero these > times are not equal and Oa predicts a different result from that of Ob. > So relativity principle conflicts with the postulate that the speed of > light is always the same according to the observer. > > Actually what above experiments show that if the relativity principle > is true and the speed of light is direction independent, then the speed > of light is direction independent relative to the source. Since the > direction independence of light speed is a proven fact(Michael&Morley > experiment and others), any theory conflicts with this also conflicts > with relativity principle. This means that the Lorentzian velocity > addition law conflicts with relativity principle. > > Lokman Kolukisa > The Lorentz transformations just describe rotations in a four dimensional space. As such, they are known to be mathematically self consistent. It follows that regardless of whether SR is a correct discription of reality, it cannot be invalidated by thought experiments which seek to demonstrate internal contradictions. So why do people keep trying? Every such attempt that is posted here constitutes nothing more than a request for an explanation of where the poster has got his or her maths wrong. That's what schools are for. Who'd have thought that a group on relativity could be so tedius? Sylvia. |