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Homework Help: Gamma factor in legth and mass demostration

  1. Jan 8, 2007 #1
    1. The problem statement, all variables and given/known data

    I'm studing relativity(on extraschool) and known about gamma factor influences time, after some tries, I demostrate that if A is at Va, in relation to B: "light distance"=sqrt("light distance for B"^2-"distance traveled by A"^2)

    where we say that "light distance"=c*"ta"; "light distance for B"=c*"tb"; and "distance traveled by A"=Vatb resolving and we demonstrate it...i gess you know what i mean....

    now my problem is who to demonstrate that leght and mass are too influenced by gamma factor.

    2. Relevant equations

    wish i know then...

    3. The attempt at a solution

    don't have a tought...:X
  2. jcsd
  3. Jan 8, 2007 #2


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    You mean you wish to prove length contraction?
  4. Jan 8, 2007 #3
    indeed, and mass, but by maths method, not by "concept" method, what is the begining? what is the permiss that is missing me to acept that Y can too interfer in mass and leght(by math method)
  5. Jan 8, 2007 #4


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    Length contraction follows directly from time dilation. Time dilation can be 'proved' using a simple thought experiment. Length contraction follows directly from the lerentz transformations, so there is little in the way of a 'mathematical' proof, as far as I know.
  6. Jan 8, 2007 #5
    yes yes, till there i agree with it, but i wnat to demonstrate it mathematicaly...

    ok, my point is:

    for have time dilatation we have that:

    2 obeservers, A and B, A is moving at speed Va and B is still.
    "A" have a light clock(2 perfect mirrors, 1 photosensor, 1 laser making a 90º to mirrors).

    as light travels at same speed at diferent referencials, we have that B will see the light moving not at 90º but at a diferent angle.
    A will jut see light moving up and down like nothing was hapening...

    now, from trignometry, [tex] c_1=sqrt(H^2-c_2^2) [/tex]

    as [tex]V=d/t[/tex],

    we have that:
    distance that light takes, measured by B is H,
    distance travel by light, measured by A is [tex]c_1[/tex];
    distance travel A, measured by B, is [tex]c_2[/tex];

    so as [tex]d=V \cdot t[/tex];

    [tex]t_A \cdot c= \sqrt{(t_B \cdot c)^2-(t_b \cdot V_a)^2}<=>t_B= \frac{t_A}{ \sqrt{1 - \frac{v^2}{c^2}}}[/tex]

    now, i demonstrated with maths the time dilatation....how i demonstrate leght contraction and mass dilatation???
    Last edited: Jan 8, 2007
  7. Jan 9, 2007 #6
    still looking for some clue, i've search in wiki, but everything is already demonstrated, they only put the formulas there....the rest is explication of then...:(
  8. Jan 11, 2007 #7


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    consider a light clock [whose separation of mirrors is] oriented in the direction of motion, rather than perpendicular...
    ..and don't forget to use the principle of relativity.
  9. Jan 14, 2007 #8
    ok, after some tries, i think i reach a point it's correct, however, i would like a confirmation of what i've done.

    so, to demonstrate a leght contration, I used this logic:

    [tex]t_B= \frac{t_A}{ \sqrt{1 - \frac{v_a^2}{c^2}}}, V=\frac{d}{t}<=>t=\frac{d}{v}[/tex]
    with d as the leght travel by "a".

    so, as v_a is equal to both observers,

    [tex]\frac{d_b}{v_a}=\frac{d_a/v_a}{ \sqrt{1 - \frac{v_a^2}{c^2}}}[/tex]

    what is equal to:

    [tex]\frac{d_b}{v_a}=\frac{d_a}{v_a \cdot \sqrt{1 - \frac{v_a^2}{c^2}}}[/tex]

    what is equal to:

    [tex]d_b=\frac{d_a}{\sqrt{1 - \frac{v_a^2}{c^2}}}[/tex]

    what is equal to:

    [tex]d_a=d_b \cdot \sqrt{1 - \frac{v_a^2}{c^2}}}[/tex]

    so, to observer "a", who is moving, leght is everytime lower than to "b", and convertor factor is "Y"

    am I right??

    to demonstrate mass dilatation, I dout it's correct by my though, however, i putted it here:biggrin: :

    [tex]d_a=d_b \cdot \sqrt{1 - \frac{v_a^2}{c^2}}}[/tex]
    [tex]D= \frac{m}{V}[/tex]
    where m is mass, D is density, and V is Volume

    so, considering a rectangular object,

    [tex]m= D \cdot (H \cdot W \cdot d)[/tex]
    where H is High, w is Width, and d is Leght

    considering m_a the mass of "a" for the observer "a" and m_b the mass of "a" for the observer "b"

    for moving observer "a"
    [tex]m_a= D \cdot H \cdot W \cdot d_a[/tex]

    for rest observer "b":
    [tex]m_b= D \cdot H \cdot W \cdot \frac{d_a}{\sqrt{1 - \frac{v_a^2}{c^2}}}}[/tex]

    so, taking D off:

    [tex]\frac{m_a}{H \cdot W \cdot d_a}=\frac{m_b}{H \cdot W \cdot d_a} \cdot \sqrt{1 - \frac{v_a^2}{c^2}}}[/tex]

    as H, W, D and d_a are equal to both observers,

    [tex]m_a=m_b \cdot \sqrt{1 - \frac{v_a^2}{c^2}}}[/tex]


    [tex]m_b=\frac{m_a}{\sqrt{1 - \frac{v_a^2}{c^2}}}}[/tex]

    can it be done with this logic? or we need to use another way??

    thanks in advance, and hope wasn't too annoying
    Last edited: Jan 14, 2007
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