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Relativity, force, acceleration, momentum

  1. Jan 12, 2007 #1
    let q=relativistic factor
    then F=q*m0*a (where m0 is rest mass and a=acceleration)

    if E=m*c^2 =>dE=dm*c^2 (i'm not sure about this)
    and dL=F*dx (is the infinitesimal work completed by an external strenght F on the system)

    F*dx=dm*c^2 =>

    q*m0*a*dx=dm*c^2 => dm/m0=(q*a/c^2)*dx

    m'(x)=q*m0*a(x)/c^2 ==> F=m'(x)*c^2 .


    is right?
  2. jcsd
  3. Jan 13, 2007 #2
    No,I suggest you to study "Introduction to special relativity"
    Robert Resnick.
    I can provide you the detailed computations,if you want.
    Last edited by a moderator: Jan 15, 2007
  4. Jan 15, 2007 #3
    someone can explain easy, where's the mistake?
  5. Jan 15, 2007 #4


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    Well, this is wrong for a start
    I suggest you read http://math.ucr.edu/home/baez/physics/Relativity/SR/mass.html" [Broken], specifically the addendum at the end, to begin with.
    Last edited by a moderator: May 2, 2017
  6. Jan 15, 2007 #5
    F is defined as dp/dt and p = qm0v. But v is contained in the factor q so when you derive p you actually obtain:

    F = qm0a + q^3(m0/c^2)(v°a)v

    where ° means scalar product.
    Last edited: Jan 15, 2007
  7. Jan 15, 2007 #6


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    Just a light correction on lightarrow's post (and to make things a little easier to read);

    [tex]\vec{F} = \gamma m_{0}\vec{a} + \gamma^{3}m_{0}\frac{\vec{v}\cdot\vec{a}}{c^2}\vec{v}[/tex]

    Edit: Ahh lightarrow's on the ball today :wink:
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