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Kinetic energy in SR.

  1. Feb 25, 2006 #1
    In normal mechanics, the kinetic energy is the integral of the momentum with respect to the velocity ( [tex] \int m_0v.dv [/tex]. So, why is the kinetic energy not given by [tex] \int \frac{m_0v}{\sqrt{1 - v^2/c^2}}dv[/tex]?

    Sorry it took a few edits to get the maths looking right.
     
    Last edited by a moderator: Feb 25, 2006
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  3. Feb 25, 2006 #2

    Hurkyl

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    Because the actual definition of work is:

    [tex]
    W = \int \mathbf{F} \cdot d\mathbf{x}
    [/tex]

    and your expression for work in the Newtonian case is derived from the particular form that F takes in that theory.
     
  4. Feb 25, 2006 #3

    jtbell

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    Working in one dimension for simplicity, non-relativistically we have

    [tex]\Delta K = W = \int {Fdx} = \int {\frac{dp}{dt}dx} = m \int {\frac{dv}{dt}dx}[/tex]

    whereas relativistically we have

    [tex]\Delta K = W = \int {Fdx} = \int {\frac{dp}{dt}dx} = m_0 \int {\frac{d(\gamma v)}{dt}dx}[/tex]
     
  5. Feb 25, 2006 #4

    pervect

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    Momentum is always the derivative of the Lagrangian, L, with respect to velocity. Thus the intergal of momentum with respect to velocity is always the Lagrangian.

    Thus your intergal gives not the kinetic energy, but the relativistic lagrangian. In Newtonian theory the Lagrangian is equivalent to the kinetic energy when the potential energy is zero, as it is in this example (L=T-V and V=0), but not in relativisitc mechanics.
     
    Last edited: Feb 25, 2006
  6. Feb 25, 2006 #5
    The correct definition of kinetic energy in classical mechanics is as follows
    (note: Work done equals change in kinetic energy)

    In SR the same definition is used. I.e. dK = Fdx which yields [itex]K =(\gamma - 1)m_0c^2[/itex].

    Pete
     
    Last edited: Feb 25, 2006
  7. Feb 25, 2006 #6
    That is the canonical momentum, not the (linear mechanical) momentum. In any case the canonical momentum is the positive derivative of the lagrangian.
    ... is something not mentioned here.
    Huh? Since when? In classical mechanics the Lagrangian, L, is given by L = T - V, not as you gave.

    Pete
     
  8. Feb 25, 2006 #7

    pervect

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    In this particular problem, the canonical momentum is equal to the mechanical momentum. If the system had fields, one might have to make the distinction. Since the canonical momentum is the only sort that's conserved, one would generally be interested in the canonical momentum of a system even in the case where the system has fields (for instance a moving charge).

    You may have missed the point. If the derivative of the Lagrangian is the momentum, then the intergal of the momentum is the Lagrangian.
    I'll revise and expand my original post to make my meaning more clear.
     
    Last edited: Feb 26, 2006
  9. Feb 25, 2006 #8

    Hans de Vries

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    To give the Lagrangian explicitly:

    [tex]-L\ = \ \sqrt{1-v^2/c^2}\ m_o c^2 \quad \approx \quad m_o c^2\ -\ \frac{1}{2}m_o v^2 \qquad . [/tex] (for v << c)

    Which corresponds to the time dillation of an object moving at v.
    In QM words: the number of phase changes (ticks) over the trajectory
    of the particle (the t' axis) is less by a factor gamma.

    While for the Hamiltonian we have:

    [tex]H\ =\ \frac{1}{\sqrt{1-v^2/c^2}}\ m_o c^2 \quad \approx \quad m_o c^2\ +\ \frac{1}{2}m_o v^2 \qquad .[/tex] (for v << c)

    In QM words: the number of phase changes (ticks) over the t axis is
    higher by a factor gamma.

    That's why -L = V-T while H = V+T in the non-relativistic limit.


    Regards, Hans
     
    Last edited: Feb 25, 2006
  10. Feb 26, 2006 #9
    I neglected to comment on he questoners statement. I must have misread it since I didn't see it the first time. I don't understand why he thinks that kinetic energy is defined as the integral of the momentum with respect to the velocity. I guess some people confuse definitions with equalities.

    Pete
     
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