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Thinking bout the Special Theory of Relativity

  1. Sep 7, 2009 #1
    I understand that Einstein derived his Special Theory from two assumptions:

    1. That classical physics holds in all inertial frames
    2. That the speed of light will always be measured to be the same.

    So, what I was trying to do was to derive the equations starting out from these two assumptions. I got a bit messed up though. Consider this:

    Three particles are lined up:
    .
    .
    .

    The top one (a photon, if you will) travels at the speed of light. The middle one travels at some velocity v<c. And the bottom one stays still. Now what if the middle and bottom particles are holding a ruler? After some time t, the picture will look like this:
    -------------.
    -------._____________
    ._____________

    (I put the hyphens in so it would come out right; the underscore is the ruler)
    The distance along its own ruler the bottom particle measures the photon to have traveled will be d=c*t, since it observes the photon's velocity to be c. Now what the bottom particle observes to be the distance the photon has traveled along the middle particle's ruler is c*t - v*t, or (c-v)t. So, since the middle particle has to observe the photon's speed to be c, we can argue that its flow of time is altered:

    c*t = (c-v)t'

    t'= tc/(c-v) where t' is from the bottom particle's perspective how much time appears to pass for the middle particle during a unit time interval (in its own frame).

    Not only is this an incorrect formula for time dilation, we could go the other way and say that the ruler contracts for the moving particle by a factor k, instead of saying time is altered:

    c*t=k(c-v)t

    k=c/(c-v)
    And this is obviously an incorrect formula for Lorentz contraction.

    The last thing is that instead of assuming either time is slowed or the ruler contracts (from the still reference frame), couldn't one say both happened to some extent?

    Obviously I'm doing something really wrong here, since there usually is no confusion about time dilation/length contraction, and my formulas are messed up. But I'm trying to get it from first principles instead of following along a text, so here I am.
     
  2. jcsd
  3. Sep 7, 2009 #2

    Janus

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    There are three effects to account for:

    Time runs slow for the middle particle according to the bottom particle.
    The middle particle's ruler is length contracted according to the bottom particle.
    Clock's placed along a line parallel to the motion of the particles that synchronized according the middle particle will not be according to the bottom particle.

    Your example requires you to account for all three at once.

    It would be better to isolate the effects.

    For time dilation, have the photon bouncing back and forth between two mirrors traveling with the middle particle (so that it bounces up and down in your example while middle particle moves to the right.)

    Now compare the path that the particle takes according to both particles, and the time it takes at c to travel the paths.
     
  4. Sep 7, 2009 #3

    Integral

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    Have you read http://www.fourmilab.ch/etexts/einstein/specrel/www/" [Broken] Note that AE did NOT assume things like time dilation or length contraction. They are results. He used the coordinates of a object at certain locations to derive a Differential Equation. The Lorentz transforms are a solution to his differential equation. You will not arrive at the correct solution with purely algebraic operations.
     
    Last edited by a moderator: May 4, 2017
  5. Sep 7, 2009 #4

    JesseM

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    I think it is possible to derive the Lorentz transformation in an algebraic way--I tried to do this in post #14 of this thread, though I don't know how rigorous it is. And in post #4 of this thread I tried to follow along with Einstein's derivation from the 1905 paper you linked to, the way I explained his steps didn't really require any calculus equations, although in one step I did rely on the conceptual calculus argument that if you zoom in on an infinitesimal region of a continuous function of two variables it will look like a flat plane...as explained by DrGreg in post #6 of this thread, that step would normally be done with the chain rule from calculus.
     
    Last edited by a moderator: May 4, 2017
  6. Sep 8, 2009 #5
    Sorry about posting it in the wrong forum! I'll be more careful next time.
    Okay, I will read the links first.
     
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