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Does kinetic energy vary by frame of reference?

  1. Oct 28, 2007 #1
    Kinetic energy is 1/2mv^2, and velocity varies depending on velocity of the observer, so does kinetic energy also vary depending on the velocity of the observer? for example, someone sitting in a bus who slides a 2 kg brick down the center isle of the bus at 5 m/s could say the has a kinetic energy of 25 joules, while someone on the side of the road who observes the brick's velocity as 55m/s (the bus is doing 50m/s) would say the bricks kinetic energy is 3025 joules. would they both be right?
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  3. Oct 29, 2007 #2

    Chris Hillman

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    Yes, KE is an observer-dependent quantity!

    The relativistic expression for the kinetic energy of a particle with mass m moving at velocity v wrt some inertial observer (in flat spacetime) is
    {\rm KE} = m \, \left( -1 + \cosh(\operatorname{arctanh}(v)) \right) = m \left( -1 + \frac{1}{\sqrt{1-v^2}} \right) = \frac{1}{2} \, m v^2 + \frac{3}{8} \, m v^4 + O(v^6)
    where the first term in the last expression is the Newtonian expression for the kinetic energy (of a particle with mass m moving at velocity v wrt some inertial observer). All the "relativistic correction terms" here are positive, so the relativistic expression is always greater than the Newtonian expression, but they agree pretty nearly when v is much smaller than unity. (I am using "geometric units" in which c=1.) It was from examining the expression [itex]m \cosh{\operatorname{arctanh}{v}}[/itex] that Einstein deduced the famous equivalence of mass and energy, incidently!

    Now, holding m fixed and varying v it is easy to see that the KE varies in both Newtonian and relativistic physics, so KE is an observer-dependent quantity.
    Last edited: Oct 29, 2007
  4. Oct 29, 2007 #3
    kinetic energy transformation

    Please tell me if the kinetic energy of a particle is K in I and K' in I' then how are they related. Do you know a place where the problem is discussed. Thanks in advance.
  5. Oct 29, 2007 #4

    Doc Al

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    The short answer is yes. Kinetic energy, like velocity, is different for observers moving at different speeds.
  6. Oct 29, 2007 #5


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    The total energy [itex]E = K + m_0 c^2[/itex] and momentum [itex]p[/itex] in two different inertial reference frames are related by the Lorentz transformation:

    [tex]p^{\prime}c = \gamma (pc - \beta E)[/tex]

    [tex]E^{\prime} = \gamma (E - \beta pc)[/tex]

    because [itex]E[/itex] and [itex]pc[/itex] form a four-vector just like [itex]ct[/itex] and [itex]x[/itex].

    [tex]x^{\prime} = \gamma (x - \beta ct)[/tex]

    [tex]ct^{\prime} = \gamma (ct - \beta x)[/tex]
  7. Oct 29, 2007 #6
    kinetic energy transformation

    Thanks for your answer. My question is how does transform kinetic energy? It would be correct to approach as
    and to take into account F(x)=F'(x)
    (W(k), W'(k) kinetic energy in I and I', F(x) and F'(x) the OX components of the force acting on the particle.
  8. Oct 29, 2007 #7
    man, i see a lot of symbols i don't recognize (like the upper case beta and the trig functions with the 'h' on the end), but i think i get the jist of it. at the very least, i know that the answer to my question is yes :D thanks for the responses guys
  9. Oct 29, 2007 #8


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    [itex]\beta = v/c[/itex]

    The hyperbolic trig functions referred to are described for instance at http://en.wikipedia.org/w/index.php?title=Hyperbolic_function&oldid=167748987
  10. Oct 29, 2007 #9


    Staff: Mentor

    If you want to transform KE between reference frames I would recommend using the four-momentum. Energy is the timelike part of the four-momentum and momentum is the spacelike part. Together they transform like any other four-vector.
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