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Lorentz factor semplification

  1. Aug 10, 2009 #1
    Hello, my name is Ivan and I come from Italy, so sorry for my English.
    I propose a simple argument that leads to the Lorentz factor. It seems to me far more orthodox (of course, I'm talking about the simplifications) of the theorem of Pythagoras to the known motion of the mirror that reflects the beam. Indeed, in that case, it claims to get the time (t) of the system in motion (which we call X), with reference to the vertical cathetus (whose length is equal to tc); the cathetus as is known, would correspond (in said about the methodological approach) to the square root of the hypotenuse (t'c) square minus the horizontal cathetus (t'v) square: this is the triangle that would be observed by those who are in a fixed system (or of linear motion uniform speedless than the motion of the mirror), which we call Y.
    In this regard, it seems clear that it expects to apply Pythagora to a triangle which has the vertical cathetus in a space-time (stationary system) different than that forming the hypotenuse and the horizontal cathetus (moving system); and the second is that this calculation gives, in fact, the measure of the perpendicular dimension (theoretical) to the direction of motion always guessed by anyone present in the system (Y), and nothing more. So, to get the time (t) of the system, we appeal to two allegations of support: 1) it assumes (without proofs) that the vertical dimension is invariant, and 2) it assumes( without proofs) that the time of this invariant corresponds to which that the other subject measures in his system (even if when this person doesn’t do the measurements to the vertical motion, the same for everyone, but the other dimensions).
    To me, however, it seems more logical and correct the simplification obtained by the following example.

    A traveller who is in the stellar ship X shoots (with an electric gun) the classical photon to a target which is located on the bow (along the direction of motion); he measures in 100 nanoseconds to the time needed for the photon to reach target: he deducts that the target is 30 meters (tc) from the point of departure of the photon (according to him, who judges himself stationary, that point always coincides with the current position of the mouth of the electric gun). Then he directs another photon in the opposite direction (exactly inverse to that of motion: the stern), where is, again at 30 meters, another target: this time too, the time needed by the projectile to reach the target is, according to the traveller's clock, of 100 nanoseconds.
    An observer, located in the system Y (stationary or in uniform linear motion with speed constant smaller), finds, with logical-mathematical procedure (I’m not talking about visual-geometric findings, because the question would be more complex by the fact that observer sees the X system shorter, given the known contraction in the direction of motion), that, with the constant of light, the photon of stern reaches the target before that the photon of bow does center.
    This observer also knows that the target of bow can never travel (by boat), in a given time fraction, a journey longer than that of the photon in the same time (because you can not win the speed of light). For ease of calculation, therefore, he assumes a "infinite space" to bow, corresponding almost to the twice of the travelling over distance by the photon in a given unit of time: in practice, equal to the sum obtained by adding to the path travelled by the photon the path travelled ( in the same unit of time) by the bow of the stellar ship (when this has reached nearly the speed of light). He assess the phenomenon of the stern, instead, in a diametrically opposed way, assuming a spatial zero, which coincides with the difference obtained by subtracting from the path travelled by the photon the path travelled by the target of the stern (which meets the bullet) in the same unit of time (spatial zero that, even here, there will be when the ship reaches almost the speed of light). Furthermore, and importantly, the observer can see that with the changing speed of the ship, the value of the sum (the one concerning the succession of bow) is always inversely proportional to the difference (the one that concerning the succession of stern) .
    Now, knowing that for the traveller, the measurements are identical and independent from the speed of the ship (as we know, he, who is subject to relativistic effects, measures the execution on the ship as if he were stationary), the observer makes the following equation: in regard, it is obvious (and this applies also to the proposal later) that the time t' and t are intended as number of nanoseconds; it is equally obvious that the value of t'c + t'v always has to be considered in relation to the almost infinite, while the value of t'c - t'v always has to be considered in relation to almost zero. The observer thinks: "The area calculated by me (with reference to the successions of the bow: t'c + t'v) stays to the space measured by traveller (with reference to the executions of bow: tc) as the space measured by traveler (with respect to the events of the stern: for he is always tc) stays to the the area measured by me (with respect to stern: t'c - t'v).
    so he defines in the next proportion:
    (t'c + t'v): tc = tc: (t'c - t'v)
    Developing the easy equation, we find that
    tc square = (t'c)square – (t'v)square, and so
    t'= t divided by square root of 1 – v square / c square: Lorentz factor.
    Thus, the number of nanoseconds counted by the traveler (t') will hold, as v increases towards infinity (compared to those counted by the traveller); or, and is the same, the nanoseconds t' (of the observer) gradually will scroll faster than nanoseconds t (of the traveller): in short, the duration of the nanosecond (or of nanoseconds) of the traveller will hold (compared to the duration of nanoseconds of the observer) to infinity.
    This example seems to me more plausible for the practical understanding of the invariability of the transversal dimension, because it is none other than the abstract (in practice, impossible to invoice) vertical ring (with respect to the direction of motion) along which the photon travels without tilt (even minimally), neither to the side of the bow neither of the stern: the observer so understands that, in the ideal circle of that ring, t'c + t'v is equal to t'c - t'v, since v is zero (and therefore, there is no motion).
    And yet, since the phenomenon was assessed in its extreme dynamics (perfect direction of motion and perfect opposite), we agree that the inconsistency between the two times (t and t') concerns the phenomena that occur at all points of the two systems: in our case, at any point of the ship (in one part) and at all points of the system where is the observer (in the other).
     
  2. jcsd
  3. Aug 10, 2009 #2
    Hi Rell-
    The use of the Pythagorean theorem is well known in relativistic calculations. As you may know, the total energy E of a relativistic particle is related to its rest mass mc2 and momentum pc (in energy units) by the relation
    E2 = (mc2)2 + (pc)2
    Before the days of pocket calculators (late 1960's), people used to use the sine and cosine trig tables to "calculate" the various variables in the above equation. It was a lot easier than using a slide rule.
    Bob S
     
  4. Aug 12, 2009 #3
    Hello Bob,
    thanks for your answer,
    but what do you think about my equation?
     
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