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## Main Question or Discussion Point

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 M

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.

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 M

_{f}and M_{i}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.

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