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Change in kinetic energy between reference frames

  1. Dec 13, 2013 #1
    I was trying to do some studying regarding the definitions of energy, and I've hit a road block.

    I know that since velocity is frame-dependent, an object's kinetic energy as observed in two different frames will yield two different results. My initial expectation was that changes in kinetic energy of an object should be the same regardless of the reference frame (as a change is kinetic energy is a scalar and therefore invariant under coordinate transformations). I tried to work this out however in the following manner:

    Say you have an object that is considered "moving" in two frames: S and S'. The object is initially moving at
    v1 in S and v'1 in S'. It experiences a force which changes the velocity to v2 in S and v'2 in S'.

    v is the velocity of the object moving in frame S
    v' is the velocity of the object moving in a frame S' which is moving with respect to frame S. This
    movement is characterized by the constant velocity vector vc.

    I assume v'=v+vc.

    Now the "generic" differential I'm working with here is that d(mu2) (I have omitted the 1/2 factor out of the diffferential while I work for simplicity's sake and u here is just a "generic" velocity that is not related to the velocity of the scenario). Now, assuming that the mass of the object does not change, I get that the differential in frame S is md(v2) and the differential in frame S' is md(v'2). What I tried to do next was to expand the magnitude of v' and got that v'2=v2+vc2+2(dot_prod(v,v')). Applying the differential over this I get:

    d(v'2)=d(v2)+2(d(dot_prod(v,v'))

    (Note: The vc2 contribution is zero since this value is constant)

    Does the dot product contribution zero-out? Is my math correct? Did I do something wrong? When I do the integration for these terms from the first set of velocities to the second set, the dot product term will not cancel out and thus indicates that the change in kinetic energy as seen in one reference frame differs from that a reference frame moving at uniform velocity with respect to the former (which to me physically makes no sense).
     
  2. jcsd
  3. Dec 13, 2013 #2
    Nope. Consider an object of mass 1 kg. Calculate its change in kinetic energy if it accelerates from 0 m/s to 1 m/s. Now calculate its change in kinetic energy if it accelerates from 1000 m/s to 1001 m/s.

    It's invariant under rotations and translations, but not under boosts.
     
  4. Dec 13, 2013 #3
    I've been trying to picture this on my own, but is there a physical justification for that (the mathematical one seems apparent between your example and what I've worked out so far)?
     
  5. Dec 13, 2013 #4
    1. Your calculation is right.
    2. Work is defined as the inner product of the force vector and the displacement vector.

    Assumption:
    (1)The object we're discussing is a mass point.
    (2)The net force on the object is the same in both frames. Let it be Fnet.
    (3)v+vc=v', so dr+drc=dr'

    In frame S
    [itex]F_{net}\cdot dr=d(\frac{1}{2}mv^{2})[/itex]

    In frame S'
    [itex]F_{net}\cdot dr'=d(\frac{1}{2}mv'^{2})[/itex]

    Now, I'll show these two equations are equivalent. Thus, you'll understand why your calculation are right and the changes in kinetic energy is not the same in these two different inertial frames.

    (Obviously, [itex]F_{net}\cdot dr[/itex] is not equal to [itex]F_{net}\cdot dr'[/itex])

    [itex]d(\frac{1}{2}mv'^{2})=\frac{1}{2}md(v^{2}+v^{2}_{c}+2v\cdot v_{c})[/itex]

    [itex]=d(\frac{1}{2}mv^{2})+\frac{1}{2}md(v_{c}\cdot v_{c}+2v\cdot v_{c})[/itex]

    [itex]=d(\frac{1}{2}mv^{2})+\frac{1}{2}md[(v_{c}+2v)\cdot v_{c}][/itex]

    [itex]=d(\frac{1}{2}mv^{2})+\frac{1}{2}m[d(v_{c}+2v)\cdot v_{c}+(v_{c}+2v)\cdot dv_{c}][/itex]

    [itex]∵v_{c}\;is\;a\;constant\;vector[/itex]

    [itex]∴dv_{c}=0[/itex]

    [itex]∴d(\frac{1}{2}mv^{2})+\frac{1}{2}m[d(v_{c}+2v)\cdot v_{2}+(v_{c}+2v)\cdot dv_{c}]=d(\frac{1}{2}mv^{2})+\frac{1}{2}m(2dv\cdot v_{c}+0)[/itex]

    [itex]=d(\frac{1}{2}mv^{2})+mdv\cdot v_{c}[/itex]

    [itex]=d(\frac{1}{2}mv^{2})+m\frac{dv}{dt}\cdot v_{c}dt[/itex]

    [itex]∵v_{c}dt=dr_{c}[/itex]

    [itex]∴d(\frac{1}{2}mv^{2})+m\frac{dv}{dt}\cdot v_{c}dt=d(\frac{1}{2}mv^{2})+m\frac{dv}{dt}\cdot dr_{c}[/itex]

    [itex]∵All\; above\;ploynomials\;are\;equal\;to\;d(\frac{1}{2}mv'^{2})[/itex]

    [itex]∴d(\frac{1}{2}mv'^{2})=d(\frac{1}{2}mv^{2})+F_{net}\cdot dr_{c}[/itex]

    [itex]∴d(\frac{1}{2}mv'^{2})-d(\frac{1}{2}mv^{2})\neq0[/itex]

    Finally,

    [itex]F_{net}\cdot dr' = F_{net}\cdot dr+F_{net}\cdot dr_{c}[/itex]

    [itex]∴F_{net}\cdot dr' = F_{net}\cdot (dr+dr_{c}),\;indeed.[/itex]
     
    Last edited: Dec 13, 2013
  6. Dec 13, 2013 #5
    How would this result be interpreted physically? (I have gleaned pages and pages of Google information that keep either contradicting the math that we have already established as correct, or just do not touch on the subject of my question to a rigorous enough extent)
     
  7. Dec 13, 2013 #6
    And thank you both by the way (Ethan and The_Duck)
     
  8. Dec 13, 2013 #7
    Well,
    What do you mean by "interpreted physically"?
    Isn't those derivation sufficient to interpret it physically?
     
  9. Dec 13, 2013 #8
    Actually nevermind. Looking at your derivation closer, I see that I can interpret the term dot_prod(Fnet,drc) as the difference in work done by the force along dr'
    and the work done by the force along dr.

    I want to thank you again. Your derivation was very straightfoward and rigorous enough that leaves no doubt in my mind with regards to the fact that "the math don't lie mun"(even if you start doing the math at midnight and become perprlexed by your result. :) ).
     
  10. Dec 13, 2013 #9
    When I say "difference in work done" I of course mean difference in work from the POV of an observer in S' and the work done from the POV of an observer in S.
     
  11. Dec 13, 2013 #10
    Okay, you're welcome^^
     
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