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I Morin relativistic rocket example (chapter 12, page 606)

  1. May 11, 2016 #1
    Hello all.

    In Morin's classical mechanics book, chapter 12 (relativistic dynamics), in the axample about the relativistic rocket, we have a rocket that propels itself converting mass into photons and firing them back. Here Morin takes dm as a negative quantity, so the instantaneous mass of the rocket goes from m to m+dm.

    I have tried to solve the same example taking dm as a positive quantity, and assuming that the instantaneous mass goes from m to m-dm, but I do not get the same answer. The correct rocket motion equation should be:

    (dm/m)+(dv/(1-v2))=0

    , but instead I get:

    (dm/m)-(dv/(1-v2))=0

    As you see, the solution is the same, except for one minus sign.
    What could be the problem? Any ideas?

    Thank you so much.
     
  2. jcsd
  3. May 11, 2016 #2
    If dm is a positive quantity in one equation and a negative quantity in the other then it is the same equation.
     
  4. May 12, 2016 #3
    Yes, that is what I thought.

    Nevertheless, in Morin example he integrates this equiation from M to m and from 0 to v to get:

    m=M[(1-v)/(1+v)]1/2

    But if I integrate my equation in the same terms (from M to m and from 0 to v ), what I get is:

    m=M[(1+v)/(1-v)]1/2

    As you see, the signs are exchanged in the quotient, so the result is nor the same.

    What is the problem?
     
  5. May 12, 2016 #4

    robphy

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    This sounds like it might be a confusion between
    the use and/or interpretation of an "infinitesimal change in mass" [whose sign depends on whether one adds or subtracts]
    and as an "increment of the mass variable m" (used in doing an integral) [which is always positive].

    I think it might analogous to this [my reply in an old thread]
    Dot product in the Gravitational Potential Energy formula
     
  6. May 12, 2016 #5

    PAllen

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    Ah, then per robphy's point if the OP reverses the limits of integration, they will get the same answer as the other method.
     
  7. May 12, 2016 #6

    robphy

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    Or is that, when ##dm## changes sign (in the attempted change of variables)
    then ##v## (or ##dv##) changes sign?
     
  8. May 16, 2016 #7
    Confusing, isn't it?
     
  9. May 16, 2016 #8

    robphy

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    It could be, if one isn't careful.

    I took a closer look.
    Let's clarify variable names.
    m is the instantaneous rocket-mass function (which varies with time).
    dm is the incremental change in that rocket-mass function (following what Morin is doing)
    so that m+dm represents the new mass (which is smaller than the old mass since the increment is a negative amount).
    M is the original mass.

    Now if you wish change variables
    so that you can (better?) display the fact that the rocket-mass decreases,
    let ##(d\mu)=-dm##;
    this ##d\mu## is positive (which could be interpreted as the positive photon-energy ejected in the rocket-frame).

    Following Morin,
    the spatial-momentum of the rocket in the ground frame is
    ##(m\gamma v)_{new}=(m\gamma v)_{old}-\gamma(1-v)dm##
    Using ##(d\mu)##,
    ##(m\gamma v)_{new}=(m\gamma v)_{old}-\gamma(1-v)(-d\mu)##,
    which could be written as
    ##d(m\gamma v)= -\gamma(1-v)(-d\mu)##.
    When expanding out the left-hand side, you have ##dm##, but you have ##d\mu## on the right-hand-side.
    If you expand then try to combine terms (in favor of ##d\mu##) to get the diff eq, you get
    ##\frac{-d\mu}{m}+\frac{dv}{1-v^2}=0##.
    However, you can't continue to solve for m unless you use ##d\mu=-dm##.

    So, you are forced to same differential equation for ##m##.
     
  10. May 18, 2016 #9
    All right, I see.

    I will take a deeper look to it, to check if I really get it.

    Thank you so much, I spent a lot of time thinking about it.
     
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