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Determining relativistic velocity from work done from rest with a given rest mass

  1. Feb 25, 2012 #1

    jaketodd

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    I am wondering how to determine relativistic velocity from a given amount of work done on a body starting from rest with a given rest mass.

    This is not homework.

    Thanks in advance,

    Jake
     
  2. jcsd
  3. Feb 25, 2012 #2

    Mentz114

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    It depends where the work is being done. If the accelerating body is a rocket then the relativistic rocket equations apply. See for instance

    http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html [Broken]

    If the work is being done from outside the accelerated body ( such as in a particle accelerator ) the relativistic expression is the same as the Newtonian one but for a factor of γ or γ2
     
    Last edited by a moderator: May 5, 2017
  4. Feb 25, 2012 #3

    jaketodd

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    Thanks for the link, but I still could use some help...

    Here's the equation it provides for velocity:
    v = at / sqrt[1 + (at/c)2]

    How do we get 'a' and 't' from a given amount of work done on a given mass from rest?

    Maybe I shouldn't have said "relativistic velocity." I just want to know how to get the velocity it has compared to when it was at rest, due to the given amount of work done to the given mass.

    Thanks,

    Jake
     
    Last edited by a moderator: May 5, 2017
  5. Feb 25, 2012 #4

    Mentz114

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    Use conservation of energy. The work done is equal to the increase in kinetic energy.
     
  6. Feb 25, 2012 #5
    You could set up an integral of the force dp/dt over dx, where p is the relativistic momentum. This would be unnecessary work, however, because when you evaluate it you just end up with the relativistic kinetic energy formula (as per Mentz's suggestion).
     
  7. Feb 25, 2012 #6

    Dale

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    The total energy is given by
    [tex]\frac{mc^2}{\sqrt{1-v^2/c^2}}[/tex]

    It is also given by
    [tex]mc^2+w[/tex]

    Set those two expressions equal and solve for v which gives:
    [tex]v=\pm\frac{c \sqrt{w \left(2 c^2 m+w\right)}}{c^2 m+w}[/tex]
     
  8. Feb 25, 2012 #7

    robphy

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    In terms of rapidities,
    DaleSpam's expressions would be
    [itex]mc^2 \cosh\theta[/itex] and [itex] mc^2+w [/itex],
    where velocity [itex]v=\tanh\theta[/itex].

    So, [itex]v=\tanh\left(\cosh^{-1}\left(1+\displaystyle\frac{w}{mc^2}\right)\right)[/itex].
     
  9. Feb 26, 2012 #8

    jaketodd

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    It's getting over my head, but thanks all.

    Jake
     
  10. Feb 27, 2012 #9
    If we rewrite Dalespam's expression slightly differently as:

    [tex]v=\pm c \sqrt{1- \left(\frac{mc^2}{ mc^2+w}\right)^2}[/tex]

    it is slightly easier to see that the work required to achieve v=c is infinite.
     
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