October 2002 - rwth-netware-admin - lists.rwth-aachen.de

uncertainty principle is untenable !!! (new)
by guest 28 Oct '02

28 Oct '02

please reply to hdgbyi(a)public.guangzhou.gd.cn or bdpgong(a)hotmail.com, thank you. UNCERTAINTY PRINCIPLE IS UNTENABLE By reanalysing the experiment of Heisenberg Gamma-Ray Microscope and one of ideal experiment from which uncertainty principle is derived , it is found that actually uncertainty principle can not be obtained from these two ideal experiments . And it is found that uncertainty principle is untenable. Key words : uncertainty principle; experiment of Heisenberg Gamma-Ray Microscope; ideal experiment Ideal Experiment 1 Experiment of Heisenberg Gamma-Ray Microscope A free electron sits directly beneath the center of the microscope's lens (see the picture below or AIP page: http://www.aip.org/history/heisenberg/p08b.htm). The circular lens forms a cone of angle 2A from the electron. The electron is then illuminated from the left by gamma rays--high energy light which has the shortest wavelength. These yield the highest resolution, for according to a principle of wave optics, the microscope can resolve (that is, "see" or distinguish) objects to a size of dx, which is related to and to the wavelength L of the gamma ray, by the expression: dx = L/(2sinA) (1) However, in quantum mechanics, where a light wave can act like a particle, a gamma ray striking an electron gives it a kick. At the moment the light is diffracted by the electron into the microscope lens, the electron is thrust to the right. To be observed by the microscope, the gamma ray must be scattered into any angle within the cone of angle 2A. In quantum mechanics, the gamma ray carries momentum, as if it were a particle. The total momentum p is related to the wavelength by the formula p = h / L, where h is Planck's constant. (2) In the extreme case of diffraction of the gamma ray to the right edge of the lens, the total momentum in the x direction would be the sum of the electron's momentum P'x in the x direction and the gamma ray's momentum in the x direction: P'x + (h sinA) / L', where L' is the wavelength of the deflected gamma ray. In the other extreme, the observed gamma ray recoils backward, just hitting the left edge of the lens. In this case, the total momentum in the x direction is: P''x - (h sinA) / L''. The final x momentum in each case must equal the initial x momentum, since momentum is never lost (it is conserved). Therefore, the final x momenta are equal to each other: P'x + (h sinA) / L' = P''x - (h sinA) / L'' (3) If A is small, then the wavelengths are approximately the same, L' ~ L" ~ L. So we have P''x - P'x = dPx ~ 2h sinA / L (4) Since dx = L/(2 sinA), we obtain a reciprocal relationship between the minimum uncertainty in the measured position,dx, of the electron along the x axis and the uncertainty in its momentum, dPx, in the x direction: dPx ~ h / dx or dPx dx ~ h. (5) For more than minimum uncertainty, the "greater than" sign may added. Except for the factor of 4pi and an equal sign, this is Heisenberg's uncertainty relation for the simultaneous measurement of the position and momentum of an object . Reanalysis To be seen by the microscope, the gamma ray must be scattered into any angle within the cone of angle 2A. The microscope can resolve (that is, "see" or distinguish) objects to a size of dx, which is related to and to the wavelength L of the gamma ray, by the expression: dx = L/(2sinA) (1) It is the resolving limit of the microscope, and it is the uncertain quantity of the object's position. Microscope can not see the object which the size is smaller than its resolving limit dx. Therefore, to be seen by the microscope, the size of the electron must be larger than the resolving limit dx or equal to the resolving limit dx. But if the size of the electron is larger than or equal to the resolving limit dx, electron will not be in the range dx. dx can not be deemed to be the uncertain quantity of the electron's position which can be seen by microscope, dx can be deemed to be the uncertain quantity of the electron's position which can not be seen by microscope only. dx is the position's uncertain quantity of the electron which can not be seen by microscope To be seen by the microscope, the gamma ray must be scattered into any angle within the cone of angle 2A, so we can measure the momentum of the electron. dPx is the momentum's uncertain quantity of the electron which can be seen by microscope. What relates to dx is the electron which the size is smaller than the resolving limit .The electron is in the range dx, it can not be seen by the microscope, so its position is uncertain. What relates to dPx is the electron which the size is larger than or equal to the resolving limit .The electron is not in the range dx, it can be seen by the microscope, so its position is certain. Therefore, the electron which relate to dx and dPx respectively is not the same. What we can see is the electron which the size is larger than or equal to the resolving limit dx and has certain position, dx = 0.. Quantum mechanics does not relate to the size of the object. but on the Experiment Of Heisenberg Gamma-Ray Microscope, the using of the microscope must relate to the size of the object, the size of the object which can be seen by the microscope must be larger than or equal to the resolving limit dx of the microscope, thus it does not exist alleged the uncertain quantity of the electron's position dx. To be seen by the microscope, none but the size of the electron is larger than or equal to the resolving limit dx, the gamma ray which diffracted by the electron can be scattered into any angle within the cone of angle 2A, we can measure the momentum of the electron. What we can see is the electron which has certain position, dx = 0, so that none but dx = 0��we can measure the momentum of the electron. In Quantum mechanics, the momentum of the electron can be measured accurately when we measure the momentum of the electron only, therefore, we can gained dPx = 0. Therefore , dPx dx =0. (6) Ideal experiment 2 Experiment of single slit diffraction Supposing a particle moves in Y direction originally and then passes a slit with width dx . So the uncertain quantity of the particle position in X direction is dx (see the picture below) , and interference occurs at the back slit . According to Wave Optics , the angle where No.1 min of interference pattern is , can be calculated by following formula : sinA=L/2dx (1) and L=h/p where h is Planck��s constant. (2) So uncertainty principle can be obtained dPx dx ~ h (5) Reanalysis According to Newton first law , if the external force at the X direction does not affect particle ,the particle will keep the uniform straight line Motion State or Static State , and the motion at the Y direction unchangeable .Therefore , we can lead its position in the slit form its starting point . The particle can have the certain position in the slit, and the uncertain quantity of the position dx =0 . According to Newton first law , if the external force at the X direction does not affect particle,and the original motion at the Y direction is unchangeable , the momentum of the particle at the X direction will be Px=0 , and the uncertain quantity of the momentum will be dPx =0. Get: dPx dx =0. (6) It has not any experiment to negate NEWTON FIRST LAW, in spite of quantum mechanics or classical mechanics, NEWTON FIRST LAW can be the same with the microcosmic world. Under the above ideal experiment , it considered that slit��s width is the uncertain quantity of the particle��s position. But there is no reason for us to consider that the particle in the above experiment have position��s uncertain quantity certainly, and no reason for us to consider that the slit��s width is the uncertain quantity of the particle��s position. Therefore, uncertainty principle dPx dx ~ h (5) which is derived from the above experiment is unreasonable . Concluson

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Einladung zum Computertreff im Oktober 2002
by "Pätzold, Thomas" 22 Oct '02

22 Oct '02

Thema: Viren & Co. Die Scannengine der Firma Sophos Datum: 30.10.2002 ab 14:00 Uhr Ort: Hoersaal der Rechen- und Kommunikationszentrums der RWTH Aachen Ab 01. August 2002 hat die RWTH Aachen und FH Aachen eine Lizenzvereinbarung mit der Firma Sophos geschlossen, die es allen Studenten, Mitarbeitern und Einrichtungen der RWTH Aachen und FH Aachen ermöglicht, die Virenscanengine der Firma Sophos auf jedem Rechner kostenlos zu installieren. Informationen sowie das Anmeldeformualr zu dieser Veranstaltung finden Sie unter: http://www.rz.rwth-aachen.de/computertreff/ Alle Interessenten sind hierzu herzlich eingeladen. Mit freundlichen Gruessen Thomas Paetzold

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Datenübertragung per GSM, die 2.
by Norbert Keilen 11 Oct '02

11 Oct '02

Moin! Ich war wohl nicht präzise genug... Deshalb noch Ergänzungen: Ich habe hier im Institut einen PC mit ISDN-Karte und möchte von diesem aus auf ein Gerät zugreifen, welches die Möglichkeit besitzt, ein Modem anzuschließen. Eine Telefonleitung ist nicht verfügbar. Ich bin also auf Mobilfunk angewiesen. Ich brauche eine Lösung, die es ermöglicht, OHNE Handy ein GSM-Modem an dieses Gerät anzuschließen. Dieses Modem muß also im Prinzip die gleichen Eigenschaften haben, wie ein normales Modem (serieller Anschluß, AT- Befehlssatz...), nur eben über GSM. Das Gerät und das Modem sollen in einem Schaltschrank installiert werden. Danke und nochmal ein schönes Wochenende Norbert **************************************************************** - EDV-Beratung und -Support - Planung, Installation und Administration von PC-Netzwerken Norbert Keilen Achterstr. 29-31 52062 Aachen Fon (privat) : 0241 / 54 11 35 Fon (Buero) : 0241 / 80 239 70 e-Mail: EDV-Service-Keilen(a)Web.de ****************************************************************

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Datenübertragung per GSM
by Norbert Keilen 11 Oct '02

11 Oct '02

Moin! hat jemand von euch Erfahrungen mit dem Übertragen von Daten über GSM-Mobilfunk (D1, E+ o.ä.). Konkret geht es darum per PCAnywhere o.ä. auf einen externen PC zuzugreifen, diesen zu steuern und Daten zu übertragen. Mich interessieren folgende Punkte: Welche Modems können für diesen Zweck verwendet werden? Welche maximalen Übertragungsraten sind machbar? Muß für die Datenübertragung ein spezieller Vertrag (T-D1-Data?!) geschlossen werden, oder funktioniert das auch mit einer normalen Prepaid-Karte im GSM- Modem? Muß ein GSM-Modem mit Richtantenne ausgestattet sein und muß das Modem für den gewünschten Standort eingemessen werden, oder läuft das Ganze per Plug-and- Play? Fragen über Fragen... Besten Dank für eure Antworten! Schönes Wochenende Norbert **************************************************************** Norbert Keilen Institut fuer Siedlungswasserwirtschaft der RWTH Aachen (ISA) EDV-Abteilung Mies-van-der-Rohe-Strasse 1 D-52074 Aachen Germany Phone : (++49) (0)241 / 80 239 70 FAX : (++49) (0)241 / 80 22 285 E-Mail: keilen(a)isa.rwth-aachen.de WWW : http://www.isa.rwth-aachen.de ****************************************************************

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