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Relativistic Rocket

  1. Nov 21, 2009 #1
    This is actually a problem in Goldstein.

    1. The problem statement, all variables and given/known data
    A rocket that ejects stuff at a speed a in its rest frame. Demonstrate that
    [tex]m\frac{d v}{dm} + a\left(1 - {v^2 \over c^2}\right) = 0[/tex]
    in which m is the invariant mass of the rocket and v is the velocity of the rocket viewed in earth frame.

    2. Relevant equations
    3. The attempt at a solution
    First work in rocket rest frame. At some time, the 4-momentum of the rocket is
    [tex]p_0^\mu = (mc, 0)[/tex]
    After dt time, the 4-momentum of the rocket is
    [tex]p_r^\mu = \left((m-dm)c ,\ m\,dv\right)[/tex]
    I ignored [tex]dv^2[/tex] and [tex]dm\,dv[/tex]. Also the stuff the rocket ejects during this time is
    [tex]p_g^\mu = \left(\delta m(c + \frac{1}{2c} a^2), \ \delta m\, a\right)[/tex]

    Since the equation in the problem has v, I have to boost those vectors into earth frame. And momentum is conserved, therefore
    [tex]\left(\begin{array}{cc}1&-\beta \\ -\beta &1\end{array}\right) p_0^\mu = \left(\begin{array}{cc}1&-\beta \\ -\beta &1\end{array}\right) p_r^\mu + \left(\begin{array}{cc}1&-\beta \\ -\beta &1\end{array}\right) p_g^\mu[/tex]

    But what I get is a very messy and long stuff that has no way to be the same as the equation in the problem. So where did I do wrong?
     
    Last edited: Nov 22, 2009
  2. jcsd
  3. Nov 21, 2009 #2

    Looks like from the beginning. My understanding of this problem is to use [itex]p=mv\gamma[/itex] and not the 4-momentum, take the time derivative and set it equal to zero (there are no forces applied in this problem), some algebra ensues and then you get your result.
     
  4. Nov 21, 2009 #3
    You can never solve this problem correctly as the formula you are asked to prove is wrong. The correct formula reads:

    dv = -a(1-v^2/c^2) dm/m


    To obtain the incorrect formula, you have to confuse the gamma factor of the rocket w.r.t. Earth with the gamma factor of the rocket after it expels some amount of fuel w.r.t. the frame it was in before the fuel was expelled. To see that this formula cannot be correct, integrate the diff. equation and compute how much mass the rocket will have to lose in order to reach lightspeed.
     
  5. Nov 21, 2009 #4
    Complete solution in three steps. The parts become visible if you mouse over the whitened parts.


    Part 1:


    In your own attempt you implicitely assumed that the velocity of the fuel is non-relativistic. You don't need to make any such assumptions. You can write the conservation of momentum equation in the original rocket's frame as:

    m dv' + a dm_f gamma_f = 0

    where subscript f denotes that the quantity refers to the fuel. The gamma factor of the rocket after the fuel is ejected can be set to 1.

    Part 2:


    To relate dm_f to dm, you write down the conservation of energy equation. This yields:

    -dm + dm_f gamma_f = 0

    So, you find that:

    m dv' = -a dm


    Part 3:

    You then use the addition for velocities formula to find dv (Or alternatively, you could perform a Lorentz transform, but in this case that is more work). The new velocity w.r.t. Earth is given by:

    [v + dv']/[1 + v dv'/c^2] = v + (1 - v^2/c^2) dv'

    So, the change in velocity, dv, is given by:

    dv = (1 - v^2/c^2) dv' = -a dm/m (1 - v^2/c^2)
     
  6. Nov 21, 2009 #5
    I don't mean to be a nudge, but shouldn't you have let him try working out the problem from the help I gave plus the help you first gave before giving him the answer? It seems counter-intuitive to try educating people when you spoon-feed them the answers.

    Also, from the PF rules itself (emphasis mine): "On helping with questions: Any and all assistance given to homework assignments or textbook style exercises should be given only after the questioner has shown some effort in solving the problem. If no attempt is made then the questioner should be asked to provide one before any assistance is given. Under no circumstances should complete solutions be provided to a questioner, whether or not an attempt has been made."
     
  7. Nov 21, 2009 #6
    Wat you should do next is solve the differential equation. You can also consider the special case of a = c. It is also a good exercise to solve that case of the photon rocket directly using four momentum algebra, by considering the quantity:

    P_in - P_fin = P_f

    P_in is the initial four momentum of photon rocket, P_fin the final four momentum and P_f the four momentum of all the expelled photons. What can you say about P_f^2 (the Lorentz inner product of P_f with itself)?
     
  8. Nov 21, 2009 #7

    The PF rules do not always apply. In this case the student was trying to solve a problem that was flawed from the start (flawed solution in the textbook, which is highly unusual). The student in fact had done some serious work to solve the problem. I did not spot any real errors, up to one unwarranted approximation (and some typos in the Lorentz transform matrix). The student got stuck simply because he thought that doing the Lorentz transformation would not yield the correct answer.

    Also, I chopped the solution in three parts beind separate spoiler tags.
     
  9. Nov 21, 2009 #8
    The problem the OP was trying to solve is Goldstein's Problem 7.23 (Page 331), in that problem there is no square-root term, but the form is otherwise the same. The only flaw was in his re-writing the problem on the forum here, placing the square-root instead of parenthesis (a large sized mistake, but one nonetheless).

    I did notice you separated the solution into three parts with spoiler tags on each, but that's not the point. You should have waited patiently for a response from the OP before just giving him a solution (or really before offering more advice); let him work on the problem with the insight you and I had given him. (And he got stuck because he was working with 4-vectors, rather than 3-vectors that he should have used)
     
  10. Nov 22, 2009 #9
    Sorry guys. That was not a square root... My bad. I don't have that book right now, sorry.
    I haven't read Count's solution yet. :-p I'll try jdwood983's method first.
     
    Last edited: Nov 22, 2009
  11. Nov 22, 2009 #10
    That should work. But --- maybe my understanding is wrong --- shouldn't these two methods be equivalent?
     
  12. Nov 22, 2009 #11
    Sorry, that was a typo.. When I was trying to solve it, my goal WAS the correct equation.:blushing: Thanks for your reply!
     
    Last edited: Nov 22, 2009
  13. Nov 22, 2009 #12
    I see! Yes, textbooks indeed rarely contain mistakes...

    I'm not sure about jdwood983's method. If you only consider the momentum m gamma v of the rocket w.r.t. Earth, then the information about the fuel moving away from the rocket with velocity a (which therefore has its own gamma factor) is not accounted for at all.
     
  14. Nov 22, 2009 #13
    This is what I am thinking now. I still need two equations: one for momentum, the other for mass-energy, right?
     
  15. Nov 22, 2009 #14

    That's correct. The only thing you should do different relative to what you wrote down in you first post is that you should simply write the momentum of the expelled fuel as

    a dm_f gamma_a

    and the energy as:

    dm_f gamma_a c^2


    You should not expand the gamma factor by assuming that the velocity of the fuel is small.
     
  16. Nov 22, 2009 #15
    Ok I get it. I made two mistakes: I forgot the momentum I used was an expansion; and dv in rocket rest frame is different from dv on earth.

    So let me call the velocity in rocket rest frame u.

    Initially the 4-momentum is (mc, 0). After dt time, the 4-momentum of the rocket is ((m - dm)c, m du), and the 4-momentum of the "gas" the rocket emits is [tex](\delta m\gamma_a c,\ \delta m \gamma_a a)[/tex].

    Boost these vectors into earth frame with [tex]\beta=v/c[/tex], and let the momentum conserve. This results in

    [tex]m \frac{du}{dm} + a = 0[/tex]

    Also boost du into earth frame

    [tex]v + dv = \frac{v+du}{1 + \frac{vdu}{c^2}} \approx v+du - \frac{v^2}{c^2}du[/tex]
    [tex]\Rightarrow dv = \left(1-\frac{v^2}{c^2}\right)du[/tex]

    Therefore

    [tex]m\frac{dv}{dm} + a\left(1-\frac{v^2}{c^2}\right) = 0[/tex]

    Thank you guys! Your help is very valuable to me!:tongue:
     
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