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Relativistic mechanics again

  1. May 5, 2013 #1
    I am still trying to derive all the relations in relativistic mechanics, and have narrowed one of my problems down to the following. If E = m (c=1), then shouldn't dE/dt = dm/dt ? The left side of that equation is the work done on the particle per unit time, or the velocity vector -dot- force vector. I have been unable to get the right side of that equation to also equal that simple expression for power. I get (m*(u -dot- du/dt)) / (1-u^2) instead of the time derivative of the momentum vector (dp/dt), etc. Any ideas on how to show that? Thanks.

    PS. m=m_o/sqrt(1-u^2), p=mu, vectors, etc.
     
    Last edited: May 5, 2013
  2. jcsd
  3. May 5, 2013 #2

    mfb

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    Please do not use a "relativistic mass". No one in physics does this, you just find it in ancient text books and bad TV documentations. "Mass" is used as "rest mass".
    ##E = \gamma m##.

    The acceleration (du/dt) is not the same as it is in classical mechanics, this could cause the difference.
     
  4. May 6, 2013 #3
    That's very helpful. Thanks. Could you elaborate a little more?
     
  5. May 6, 2013 #4
    Try to write the equations. (c=1)

    If: [itex]E = m_{o}/\sqrt{1-u^{2}}[/itex]
    Then: [itex]\dot{E}=\dot{m}[/itex]
    And:[itex] \bar{u}\cdot\bar{F}=m\frac{\bar{u}\cdot\dot{\bar{u}}}{1-u^{2}}[/itex]
    So: [itex]\bar{F}=\frac{m}{1-u^{2}}\dot{\bar{u}}[/itex]
    But isn't the force supposed to be the derivative of the momentum?
    i.e., [itex]\bar{F}=m\dot{\bar{u}}+\dot{m}\bar{u}[/itex]

    These two expressions don't seem to match.
     
  6. May 6, 2013 #5

    mfb

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    Relativistic momentum is ##\gamma m u## and not m*u.
    Your derivative of the inverse square root looks wrong.
     
  7. May 6, 2013 #6

    pervect

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    One approach is the Hamiltonian approach. We define the Hamiltonian (which for simple systems of the sort we are going to talk about can be thought of as the total energy) in terms of momentum and position.

    This is typically found in advanced college textbooks, but the math is really easy.
    So for a free particle, we'd write:

    [tex]
    H(p,q) = \sqrt{p^2 + m^2}
    [/tex]

    where p is the (generalized) momentum and q is the position coordinate. v, velocity would be dq/dt.

    Note that this is just the well-known relationship E^2 -p^2 = m^2.

    If we wanted to consider a particle in a potential well, we'd add a potential term V(q) that was a function of position to the hamiltonian. But we don't really need it for what we're going to do.

    Then we get the equation of motion from Hamilton's equations:
    http://en.wikipedia.org/wiki/Hamiltonian_mechanics

    [tex]
    \frac{dp}{dt} = -\frac{\partial H}{\partial q} \quad
    v = \frac{dq}{dt} = \frac{\partial H}{\partial p}
    [/tex]

    The first equation just says dp/dt = 0, that momentum is constant.
    The second equation just says force * velocity = rate of work

    force = dp/dt
    rate of work = dH/dt

    (dp/dt) v = dH/dt

    multiply both sides by dt, then you get

    dp v = dH

    Substituting, we get:

    [tex]v =\frac{p}{\sqrt{p^2+m^2}}[/tex]

    We can invert this to find the possibly more familiar p(v), and E(v), for a free particle
    [tex]p =\frac{mv}{\sqrt{1-v^2}} \quad E = H = \frac{m}{\sqrt{1-v^2}}[/tex]

    So we get energy and momentum as a function of velocity. And we're done. If we want the motion of the particle in a conservative force field, we just add V(q) to the energy.
     
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