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Special Relativity - Rocket problem (particle mechanics)

  1. Jul 25, 2017 #1
    Problem statement:
    A rocket propels itself rectilinearly by giving portions of its mass a constant (backward) velocity ## u ## relative to its instantaneous rest frame. It continues to do so until it attains a velocity ## v ## relative to its initial rest frame. Prove that the ratio of the initial to the final rest mass of the rocket is given by
    $$ \frac{m_{i}}{m_{f}} = \left( \frac{c+v}{c-v} \right) ^{\frac{c}{2u}} $$


    Attempt at a solution:
    Consider, first, the instantaneous rest frame of the rocket at a given time.
    The rocket ejects a small mass ## dm ## with backward velocity ## u ## and has an increase ##dv'## in its velocity. It's rest mass has a new value ## m' ##, which can be calculated by mass conservation law (considering that ## dv' ## is small and ##\gamma(dv') \approx 1##):
    $$ m = dm \gamma(u) + m' \implies m' = m - dm \gamma(u)$$

    Then, by conservation of mommentum, we have:
    $$ 0 = -dmu + \left( m-dm \gamma(u) \right) dv' $$
    Ignoring second order differentials, we have:
    $$ dm u = m dv' $$

    Now, we need to transform ##dv'## to the velocity measured by the inertial frame in which the rocket had velocity 0 when ##t=0## (lab frame).
    Assume the rocket had velocity ## v ## at ## t = t_{1} ## and had an increase ## dv ## after ejecting material. Then, by relativistic velocity transformation (ignoring second order differentials):
    $$ v + dv = \frac{ v + dv' } { 1+ \frac{vdv'}{c^{2}} } \implies dv' = \frac{1}{1-\frac{v^{2}}{c^{2}} } dv $$

    So we have:
    $$ \frac{u}{m} dm = \frac{1}{1-\frac{v^{2}}{c^{2}} } dv $$
    $$ \int_{m_{i}}^{m_{f}} \frac{u}{m} dm = \int_{0}^{v} \frac{1}{1-\frac{(v')^{2}}{c^{2}} } dv' $$

    Finally:
    $$ \frac{m_{f}}{m_{i}} = \left| \frac{v+c}{v-c} \right| ^{\frac{c}{2u}} $$

    Problem is, the left side should be ## \frac{m_{i}}{m_{f}} ##. The mistake is most likely on the first two equations, since the relation between the differetials shoud have a minus on one side (since velocity gets bigger when mass is released), but I can't find it. Can anyone help me?
     
  2. jcsd
  3. Jul 26, 2017 #2

    Orodruin

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    Your equations are inconsistent. To start off, it is not very illuminating to pick the mass of the ejecta to be ##dm## as this gives you the (false) impression that it is a change in ##m##. In fact, ##m## decreases if your selected ##dm## is positive, not the other way around. I suggest that you instead call the energy of the ejected material ##dE## and reserve ##dm## for a change in ##m## that is positive when ##m## increases and negative when it decreases. This is the only way you can connect ##dm## and ##m## in your integration. Otherwise you are not integrating with ##dm## actually being a differential in ##m##, it just seems that way because of your improper choice of variables. Your energy equation would then be
    $$
    m = dE + (m+dm)\gamma(dv').
    $$
    I leave it to you to linearise this for small ##dE##. Note that ##dm## will be negative when ##dE## is positive!

    Note that with your chosen ##dm##, your momentum equation is wrong as well. The momentum of a mass ##dm## travelling at a speed ##u## is ##u\gamma(u) dm##.

    Note: You do not actually need to know the mass of the ejecta. You can use that the momentum ##p## for an object is related to its energy ##E## via ##p = uE##, where ##u## is its speed. The result will then be valid for ##u = 1## as well as for ##u < 1##.
     
  4. Jul 26, 2017 #3
    Thanks for replying.

    So, from your equation, I have (Since my textbook doesn't assume ## c = 1 ## for this exercise, I'll just keep it there):

    $$ mc^{2} = dE + (m+dm) \gamma(dv') c^{2} \implies dE = mc^{2} - (m+dm) \gamma (dv') c^{2} $$

    Assuming ## p = \frac{u E}{c^{2}} ## and using conservation of mommentum, we have:

    $$ 0 = -u \frac{ mc^{2} - (m+dm) \gamma (dv') c^{2} } {c^{2}} + (m+dm) \gamma(dv') dv' $$

    Assuming ## \gamma(dv') \approx 1 ## and ignoring second order differentials:

    $$ udm = -mdv' $$

    I guess the velocity transformation and the integration are now analogous to what I did on my first attempt, so I'll just stop here.

    The will give me the correct answer, but are there any other mistakes in what I did? I still don't feel very comfortable working with energy in relatvity.
     
  5. Jul 26, 2017 #4

    Orodruin

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    Apart from the energy equation and the definition of ##dm## such that it is the actual change in the mass, the approach is correct.
     
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