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Relativistic rocket equation

  1. Sep 20, 2005 #1


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    I'm trying to derive the relativistic rocket equation, and I'm getting two different answers depending on the method I use. I think the first method is right, and they would agree if not for one sign in the second equation, but I can't seem to get rid of it. Can anyone tell me what I'm doing wrong?

    Problem: A rocket ejects mass at a constant speed u in its rest frame. Find the ratio of the initial and final rest masses of the rocket in terms of u and the final speed v. (Note: In both methods I'll be using c=1 units)

    Method 1:

    Let O be the rest frame of the rocket at some time. The rocket ejects a small mass dM at velocity u. By conservation of energy, the measured mass in this frame will equal the initial mass of the rocket (its rest mass) minus dM, and since it is moving at an infinitessimally small speed, this is equal to its new rest mass. Then the new momentum of the system is (M-dM)dv'-dMu, which must be zero by conservation of energy. So Mdv'=udM, and dv'=u dM/M. I use dv' because this is the change in velocity in the rockets rest frame (ie, the frame it was at rest in before it ejected the mass). The change in speed in the frame where it was initially at rest (eg, Earth) is found by assuming it had a speed v in this frame. Then v+dv=(v+dv')/(1+vdv')=v+(1-v^2)dv'. So dv/(1-v^2)=u dM/M, and integrating this gives:

    [tex]\frac{1}{2} \mbox{ln}\left(\frac{1+v}{1-v}\right)=u \ \mbox{ln}\left(\frac{M_f}{M_i}\right)[/tex]

    Where v is now it's final velocity, and Mf and Mi are its final and initial rest masses.

    Method 2:

    For this method, always stay in the rest frame of Earth. At some point it has a velocity of v and a measured mass of M. It ejects a measured mass of dM at a speed of u in the rocket's frame, so at a speed (v-u)/(1-vu) in this frame. Again, conservation of energy says the measured total mass is the same, so conservation of momentum gives: Mv=(M-dM)(v+dv)+dM(v-u)/(1-vu). Rearranging and ignoring second order differentials, Mdv=dM(1-(v-u)/(1-vu)), or dv=u dM/M (1-v^2)/(1-vu). This gives:

    [tex] \int_0^{v_f} \frac{dv (1-vu)}{1-v^2} = u \int_{M_i'}^{M_f'} \frac{dM}{M}[/tex]

    integrating both sides (I've checked the integral on the left and it's definitely right, but you can check if you want):

    [tex]\frac{1}{2} \mbox{ln}\left(\frac{1+v}{1-v}\right)+\frac{u}{2} \mbox{ln}\left(1-v^2\right)=u \ \mbox{ln}\left(\frac{M_f'}{M_i'}\right)[/tex]

    But remember that these M's are the measured M's, equal to M[itex]\gamma[/itex]. This is the same for the initial mass, but for the final mass there is an extra factor of gamma that must be removed:

    [tex]\frac{1}{2}\mbox{ln}\left(\frac{1+v}{1-v}\right)+\frac{u}{2} \mbox{ln}\left(1-v^2\right)=u \ \mbox{ln}\left(\frac{\gamma M_f}{M_i}\right)[/tex]

    [tex]\frac{1}{2} \mbox{ln}\left(\frac{1+v}{1-v}\right)+\frac{u}{2} \mbox{ln}\left(1-v^2\right)=u \ \mbox{ln}\left(\frac{M_f}{M_i}\right)-\frac{u}{2}\mbox{ln}\left(1-v^2\right)[/tex]

    And now you see my problem. Hopefully that was all clear, and I can answer any questions if it's not.
    Last edited: Sep 20, 2005
  2. jcsd
  3. Sep 20, 2005 #2


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    When the rocket ejects a small mass dm at velocity u, it's mass decreases not by dm, but by dE/c^2

    where E is dm*c^2 / (1 - sqrt(1-(v/c)^2) as long as m>0.

    Consider the case of an ideal photon drive, for instance, and you'll see why you need to decrease the mass by dE and not dm - dm will be zero with a photon drive, but the rocket's mass will still be decreasing by E/c^2. (Imagine an idealizied matter/antimatter drive, one that emitted photons and not pions, to see this).

    To double-check, use the equations in

    http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html [Broken]

    (I haven't done this yet, I'm feeling lazy).
    Last edited by a moderator: May 2, 2017
  4. Sep 20, 2005 #3


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    I'm using energy and (relativistic) mass interchangeably, and units where c=1. Also, I've seen that page, and the only equation they give is for the case where u=c, and their formula matches the one you get from my first method, but not the second.
  5. Sep 20, 2005 #4
    I think your Mf and Mi are fliped. Your gamma then would appear in the denominator of the second part accounting for the sign. I would also suggest not using mass to mean energy for the second part. That leads students to confusion and may have contributed to your problem. Leave it invariant.
    Last edited: Sep 20, 2005
  6. Sep 20, 2005 #5
    Compare your solutions after some manipulation (with the mi and mf flipped) to equation3.3.6 a b or c at http://www.geocities.com/zcphysicsms/chap3.htm#BM3_3 This solution is consistent with the Baez faq.
    Last edited: Sep 20, 2005
  7. Sep 20, 2005 #6


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    I was *not* suggesting the use of relativistic mass in my previous comment.
    The _invariant mass_ of the rocket will decrease by E/c^2 by the conservation of energy (where the exhaust energy, E, is measured in the rest frame of the rocket) in the limiting case of an ideal, perfectly efficient rocket. This shows up most clearly in the limiting case of a photon rocket. It has nothing at all to do with relativistic mass!

    Reading over your response, I realize that you use the term "measured mass", so it is was very unclear to me whether or not you were using invariant mass or relativistic mass. Since I hardly ever use relativistic mass, I assumed you meant invariant mass.

    Anyway if your first approach is giving the same answer as the FAQ for the limiting case of a photon rocket, and your second approach is NOT getting the same answer as the FAQ for this case, it would appear likely that the problem is in the second approach.

    This means that I was probably wrong about finding the error :-(. I'll take another crack at it, meanwhile I would encourage you to look closely at the second approach. I'll take another look at it, now that I'm aware that there might be some confusion about which mass you were using.
    Last edited: Sep 20, 2005
  8. Sep 20, 2005 #7


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    First off, yes, I did mean relativistic mass when I said measured mass. Sorry for the confusion. Now, it seems I have flipped Mi and Mf, and looking again at the first link pervect gave, I actually was wrong up to this flip. So flipping them in the last step gives the right answer in the first part (matches both this link and the one Trilairian gave) and leads to the same answer in the second part. The only problem is that I'm not seeing why these should be flipped. Clearly M=Mi when v=0 and M=Mf when v=V, so I think the bounds of the integral are correct, and ln(Mf)-ln(Mi)=ln(Mf/Mi). Can anyone explain why these should be flipped? Because that certainly seems to be the mistake.
  9. Sep 20, 2005 #8
    In your first set up
    dv/(1-v^2)=u dM/M
    should be
    dv/(1-v^2)=-u dM/M
    as a decrease in M leads to an increase in v. This accounts for the Mf, Mi flip. Its a matter of a positive element of exause mass corresponding to a negative element in ship mass change.
    Last edited: Sep 20, 2005
  10. Sep 20, 2005 #9


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    Since you apparently *are* using relativistic mass, (something I hardly ever do), you should be aware that the mass ratio in your second equation is different from the mass ratio in your first equation. The relativistic mass of an object depends both on the object and on the frame in which it is measured. The relativistic mass is not a property of the body in and of itself. (The invariant mass is - it is the same for all observers).

    Because the invariant mass does not depend on the frame in which it is measured, it is probably the invariant mass ratio that you want to use, which is the one used in the first set of equations.

    Let's clarify this.

    If a rocket starts out with a mass of fuel = mass of payload, and burns until it has only payload left, the invaiant mass ratio (full)/(empty) is always equal to 2, for all observers.

    The ratio of "relativistic masses" depends on the final velocity of the rocket, so if the rocket starts out stationary in the Earth frame, the mass ratio will not in general be 2.

    [end add]

    In addition, starting with

    dv/(1-v^2) = u*dm/M

    I get

    arctanh(v) = u*ln(M)

    rather than your result.

    [add] silly me! It's the same! Nevermind!
    I'll leave the rest in, it might be useful, sort-of
    [end add]

    Solving for v, I get


    v = \frac{M^{2u}-1}{M^{2u}+1}

    M is a number greater than 1, the ratio of the initial (invariant!) mass of the full rocket to the final (invariant) mass of the empty rocket.

    u = v/c is a dimensionless number proportional to the "exhaust velocity".

    We can solve for gamma = 1/sqrt(1-v^2)

    [tex] \gamma =\frac{1}{2}(M^{2u}+M^{-2u})[/tex]

    This seems to agree with the FAQ when u=1, it satisfies

    [tex]\gamma(1+v) = M[/tex]

    note that M here is (M+m)/m in the FAQ due to different notation
    Last edited: Sep 20, 2005
  11. Sep 20, 2005 #10
    I don't like the concept either and don't use it at all, but I think he was aware of that as indicated by the primes on the second method's masses.
    Aside from the missed sign, in his, thats equivalent.
  12. Sep 20, 2005 #11


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    Right, I just thought of the dM problem. I should have realized that my answer had an initial mass smaller than the final mass. Anyway, thanks both of you for your help.
  13. Sep 20, 2005 #12


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    Re: arctanh
    Yeah, I realized this (eventually), it's an alternate form of the same function.

    Re: relativistic mass
    The point I was making, though, is that if you do use this concept, the ratios of the relativistic masses are not in general the same as the ratios of the invariant masses. This is why I think the results are different with the two approaches.
    Last edited: Sep 20, 2005
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