relativity vs velocity addition
in [Relativity]
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From: Dirk Van de moortel on 27 Nov 2006 17:11 "Russell" <russell(a)mdli.com> wrote in message news:1164660301.883648.212360(a)l12g2000cwl.googlegroups.com... > Sue... wrote: >> Russell wrote: >> > Sue... wrote: >> > >> > ... >> > >> > > If you want to use a particle model then you need to learn QED. >> > > http://nobelprize.org/physics/laureates/1965/feynman-lecture.html >> > >> > This piece of advice is pretty hilarious coming from a >> > relativity disbeliever. >> >> What is a "relativity disbeliever" ? > > In your case, I meant someone who believes that the effect > known as the "twin paradox" does not exist, indeed is made > physically impossible by something or other having to do > with the moons of Jupiter. Indeed, and on this one "Sue" fully agrees with his friend Androcles: http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/StupidQuestion.html > > At least, I *think* that is your position. It is like pulling teeth > to get you to say what your position actually is. He is - just like Androcles - a retired engineer of the electrical kind, professionally. Both have exactly the same opinions. Dirk Vdm
From: Dirk Van de moortel on 27 Nov 2006 17:22 "Sue..." <suzysewnshow(a)yahoo.com.au> wrote in message news:1164659123.605309.175070(a)j44g2000cwa.googlegroups.com... > > Russell wrote: >> Sue... wrote: >> >> ... >> >> > If you want to use a particle model then you need to learn QED. >> > http://nobelprize.org/physics/laureates/1965/feynman-lecture.html >> >> This piece of advice is pretty hilarious coming from a >> relativity disbeliever. > > What is a "relativity disbeliever" ? > > Sue... A religious fanatic like yourself who thinks that everything he fails to understand must belong in a belief system, Dennis. Dirk Vdm
From: Sorcerer on 29 Nov 2006 16:33 "Sue..." <suzysewnshow(a)yahoo.com.au> wrote in message news:1164661735.894667.20970(a)14g2000cws.googlegroups.com... | Do you believe there is a real conflict with Maxwell's | equations and the principle of relativity ? Which one, Gauss, Faraday or Ampere?
From: lkoluk2003 on 1 Dec 2006 05:53 lkoluk2003(a)yahoo.com yazdi: > 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 Hi, I think I have managed to find a relativistic speed addition formula which gives the correct result for the symmetric twin problem. The formula is v+w-v.w/c where v and w are relative speeds in the same direction. By relative speed(is this a correct name for this?), I mean the following. Let x be the distance between two objects at a moment. After a time interval t, let the distance be x'. Then (x'-x)/t is the avarage relative speed of these two objects. A velocity addition formula based on a coordinate system should easily be derived from this formula. Now I will explain how I got it. As I have said before, the direction independent time dilation gives inconsistent result in the twin problem. So either there should not be a time dilation or it must be dependent on the direction of the speed. Let t1 and t2 are the total times spend by the twin A in outbound and inbound movement respectively. While twin A is in outbound movement, twin B is also in his/her outbound movement. The same thing is true for inbound movement also. The acceleration affects are ignored. Then let t1' and t2' are the total times spend by twin B as measured by twin A in outbound and inbound movements respectively. For the result to be consistent t1+t2=t1'+t2' must be true. The outbound relative speed of the twin need not be equal to the inbound relative speed. So we can write t1'=t1.B(v1), t2'=t2.B(-v2) where v1 and v2 are the outbound and inbound relative speeds and B(v) is the dilation factor. Then we get t1+t2=t1'+t2'=t1.B(v1)+t2.B(-v2) (x/v1) + (x/v2) = (x/v1).B(v1) + (x/v2).B(-v2) where x is the longest distance between the twin A&B. It seems that the only formula which satisfies this equation is B(v)=1+b.v where b is unknown. Now back to the experiment testing speed addition formulas. With this experiment, it is shown that the light speed must be direction independent relative to the source. Note that this result does not exlude the time& length dilation. The only difference is that the dilation factor must be applied to all coordinates now not just x and t. Let k(v) is the speed of light relative to the source. Of course k(0)=c and if k(v)=c then the correct transformations would be that of the Galilean type. For an observer in the train B the time required by a light beam to travel a distance x is x/c. From the point of view of the observer Oa, the time required is t=x'/k(v) for the same event where x'=x.B(v). Since x'/t'=c , the formula becomes t=t'.c/k(v) where t' is the time measured by the observer in the train B. From this and t'=t.B(v), we get c/k(v)=1/B(v) and then k(v)=c.(1+b.v). So the speed of light with respect to the observer Oa would be as v+k(v)=v+c.(1+b.v)=v+c+b.v.c. To obtain speed formula, do the same experiment but replace light sources with identical guns which gives a speed w' to the bullets when fired. By using two identical guns directed to opposite directions and identical bullets, we avoid a change in the speed of the train B due to a momentum change. However, we only need one bullet for the calculations. By similar logic, we find w'=w(1+b.v) where w' is the speed of the bullet relative to the source with respect to the observer Oa. Thus the speed of the bullet relative to the observer is found as v+w.(1+b.v) Now what is the value of b? The phsicists say that there are many experimental evidences showing c as an upper limit for the speed. So the formula would be v+w-v.w/c It also has the associative property. So if the calculations and the logic I have used are correct, this is the relativistic speed formula. However, if E=m.c^2 could not be derived from it, it has no value. One way of doing is to repeat the Einstein's study in his 1905 paper. However, to do this one needs the energy formula of light. As now the light speed is varying with respect to the observer, I wonder whether the correct formula is known. Anyway, I neither have sufficient experience to go beyond nor desire to go. Also I don't deal with a career in Physics. This is a relativistic solution to the twin paradox and as a logician and relativist it seems sufficient to me. However, in any case, I will form a full text consisting of what I write about this subject here and put somewhere. After the formula is verified by someone(s), I may send it to a journal. Lokman Kolukisa
From: Sue... on 1 Dec 2006 07:56
lkoluk2003(a)yahoo.com wrote: > lkoluk2003(a)yahoo.com yazdi: > > 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 > > Hi, > I think I have managed to find a relativistic speed addition formula > which gives the correct result for the symmetric twin problem. The > formula is v+w-v.w/c where v and w are relative speeds in the same > direction. By relative speed(is this a correct name for this?), I mean > the following. Let x be the distance between two objects at a moment. > After a time interval t, let the distance be x'. Then (x'-x)/t is the > avarage relative speed of these two objects. A velocity addition > formula based on a coordinate system should easily be derived from this > formula. Now I will explain how I got it. > > As I have said before, the direction independent time dilation gives > inconsistent result in the twin problem. So either there should not be > a time dilation or it must be dependent on the direction of the speed. > Let t1 and t2 are the total times spend by the twin A in outbound and > inbound movement respectively. While twin A is in outbound movement, > twin B is also in his/her outbound movement. The same thing is true for > inbound movement also. The acceleration affects are ignored. Then let > t1' and t2' are the total times spend by twin B as measured by twin A > in outbound and inbound movements respectively. For the result to be > consistent t1+t2=t1'+t2' must be true. The outbound relative speed of > the twin need not be equal to the inbound relative speed. So we can > write > > t1'=t1.B(v1), t2'=t2.B(-v2) > > where v1 and v2 are the outbound and inbound relative speeds and B(v) > is the dilation factor. Then we get > > t1+t2=t1'+t2'=t1.B(v1)+t2.B(-v2) > (x/v1) + (x/v2) = (x/v1).B(v1) + (x/v2).B(-v2) > > where x is the longest distance between the twin A&B. It seems that the > only formula which satisfies this equation is B(v)=1+b.v where b is > unknown. > > Now back to the experiment testing speed addition formulas. With this > experiment, it is shown that the light speed must be direction > independent relative to the source. > Note that this result does not exlude the time& length dilation. The > only difference is that the dilation factor must be applied to all > coordinates now not just x and t. Let k(v) is the speed of light > relative to the source. Of course k(0)=c and if k(v)=c then the correct > transformations would be that of the Galilean type. For an observer in > the train B the time required by a light beam to travel a distance x > is x/c. From the point of view of the observer Oa, the time required > is t=x'/k(v) for the same event where x'=x.B(v). Since x'/t'=c , the > formula becomes t=t'.c/k(v) where t' is the time measured by the > observer in the train B. From this and t'=t.B(v), we get > c/k(v)=1/B(v) and then k(v)=c.(1+b.v). So the speed of light with > respect to the observer Oa would be as v+k(v)=v+c.(1+b.v)=v+c+b.v.c. > > To obtain speed formula, do the same experiment but replace light > sources with identical guns which gives a speed w' to the bullets when > fired. By using two identical guns directed to opposite directions and > identical bullets, we avoid a change in the speed of the train B due to > a momentum change. However, we only need one bullet for the > calculations. By similar logic, we find w'=w(1+b.v) where w' is the > speed of the bullet relative to the source with respect to the observer > Oa. Thus the speed of the bullet relative to the observer is found as > v+w.(1+b.v) > > Now what is the value of b? The phsicists say that there are many > experimental evidences showing c as an upper limit for the speed. So > the formula would be > > v+w-v.w/c > > It also has the associative property. So if the calculations and the > logic I have used are correct, this is the relativistic speed formula. > However, if E=m.c^2 could not be derived from it, it has no value. One > way of doing is to repeat the Einstein's study in his 1905 paper. > However, to do this one needs the energy formula of light. As now the > light speed is varying with respect to the observer, I wonder whether > the correct formula is known. Anyway, I neither have sufficient > experience to go beyond nor desire to go. Also I don't deal with a > career in Physics. This is a relativistic solution to the twin > paradox and as a logician and relativist it seems sufficient to me. > However, in any case, I will form a full text consisting of what I > write about this subject here and put somewhere. After the formula is > verified by someone(s), I may send it to a journal. Why are you comparing bullets to light? If you have some success don't send it to a journal, send it to the Nobel committe. <<Now, does not the prize to Einstein imply that the Academy recognised the particle nature of light? The Nobel Committee says that Einstein had found that the energy exchange between matter and ether occurs by atoms emitting or absorbing a quantum of energy,hv . As a consequence of the new concept of light quanta (in modern terminology photons) Einstein proposed the law that an electron emitted from a substance by monochromatic light with the frequency has to have a maximum energy of E=hv-p, where p is the energy needed to remove the electron from the substance. Robert Andrews Millikan carried out a series of measurements over a period of 10 years, finally confirming the validity of this law in 1916 with great accuracy. Millikan had, however, found the idea of light quanta to be unfamiliar and strange. The Nobel Committee avoids committing itself to the particle concept. Light-quanta or with modern terminology, photons, were explicitly mentioned in the reports on which the prize decision rested only in connection with emission and absorption processes. The Committee says that the most important application of Einstein's photoelectric law and also its most convincing confirmation has come from the use Bohr made of it in his theory of atoms, which explains a vast amount of spectroscopic data. >> http://nobelprize.org/physics/articles/ekspong/index.html Sue... > > Lokman Kolukisa |