Proving the Mass Ratio of a Rocket using Logarithmic U-Substitution

[MathematicalPhysicist]

problem statement
citation: introduction to SR by wolfgang rindler, second edition, page 96-97 question number 6.
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 veclocity V relative to its initial rest frame.
prove that the ration of initial to the final rest mass of the rocket is given by:
$$\frac{M_i}{M_f}=(\frac{c+V}{c-V})^{\frac{c}{2U}}$$

attempt at solution
well I am given the hint that:
$$(-dM)U=Mdu'$$
where: M is the rest mass of the rocket and u' is its velocity realtive to its instantaneous rest frame.
well i think that the equation for the mass is: $$M(t)=M_i-dM\frac{dt}{d\tau}$$
where: $$\frac{dt}{d\tau}=\gamma(u')$$
so the equation (erroneously) transforms to:
$$-UdM\gamma(u')=Mdu'$$ adn then one only needs to integrate between the masses: M_i upto M_f and from 0 to V in the velocities side of the equation, but i don't get the required answer, any pointers here on where i got it wrong?

thanks in advance.

 

[MathematicalPhysicist]

still waiting for help in this issue.

 

[D H]

well I am given the hint that:
$$(-dM)U=Mdu'$$

More on this later.

well i think that the equation for the mass is: $$M(t)=M_i-dM\frac{dt}{d\tau}$$

Why would you think that?

What you need to do is to relate your du', the change in velocity in the instantaneous rest frame with dv, the change in velocity in the initial rest frame. The hint gives you an the relation between du' and the given exhaust velocity U and the vehicle's mass in its instantaneous rest frame. All you need is gamma. There is no need to invoke the concept of relativistic mass in this problem.

Now to the hint: This is not quite right. Consider the case of a rocket that emits massless photons for propulsion. Even though photons are massless, the rocket will still lose mass because the ejected photons have kinetic energy. It would be more accurate to say relate the power of the exhaust (change in energy per spacecraft time) to the change in mass. The hint assumes a non-relativistic exhaust velocity.

 

[MathematicalPhysicist]

but where does gamma comes?
perhaps it's the relation between du' and dv, so it should be something like this:
-dMU=Mdu'=Mdv*gamma(v)
or the gamma should be of v_rel between v and U, i.e of (v-U)/(1-vU/c^2) or something else?

thanks in advance.

 

[D H]

Something else. This gamma: http://scienceworld.wolfram.com/physics/RelativisticGamma.html

 

[MathematicalPhysicist]

i see you didn't understand me, the gamma should be computed with regard to v or v_rel which is what i wrote as (v-U)/(1-vU/c^2)?

 

[MathematicalPhysicist]

i.e v_rel=(v-U)/(1-vU/c^2)

 

[D H]

Why are you trying to involve U here?

The relation $-UdM = M dv'$ pertains to the vehicle's instantaneous rest frame.
U is a constant here. Moving the M to the left-hand side yields $-U dM/M = dv'$. Integrating the left-hand side yields $-U(ln(M_f)-ln(M_i))$. Integrating the right-hand side as-is is meaningless.

All you have on the right-hand side is $dv'$. You can make it meaningful by changing that $dv'$ to $dv$. What is the relationship between $dv'$ and $dv$? Hint: You only need $v$ and $c$.

 

[MathematicalPhysicist]

so dv'/dv=gamma(v)?
but still the integral you get is of arcsin(v/c)/c, which i don't think suits here.

missing another thing?

 

[D H]

so dv'/dv=gamma(v)?
but still the integral you get is of arcsin(v/c)/c, which i don't think suits here.

missing another thing?

Yep. Your missing something. Since velocity is length/time, both time dilation and length contraction come into play. You only addressed one part.

 

[MathematicalPhysicist]

you mean it should be:
dv'=d^2x'/d^2t'*dt'
when dx'=gamma(v)(dx-Udt)
dt'=gamma(v)(dt-(U/c^2)dx)
so v'=dx'/dt'=(v-U)/(1-vU/c^2)
and then i need to differentaite it with respect to t, something like this?

 

[learningphysics]

You can use the velocity addition formula

$$u' = \frac{u-v}{1-uv/c^2}$$

Note, that this is not the same u as the "U" used for this problem... it's just that the velocity addition formula commonly uses u as the velocity of the object, so I'm just sticking with that here...

Now calculate $$\frac{du'}{du}$$. Then set u = v to get $$\frac{du'}{du}$$ at u = v. But this answer is also $$\frac{du'}{dv}$$ since du = dv in this problem.

 

[D H]

you mean it should be:
dv'=d^2x'/d^2t'*dt'
when dx'=gamma(v)(dx-Udt)
dt'=gamma(v)(dt-(U/c^2)dx)
so v'=dx'/dt'=(v-U)/(1-vU/c^2)
and then i need to differentaite it with respect to t, something like this?

You don't need U. This is the relative exhaust velocity; as such it is no longer a part of the vehicle. Since you are really struggling here I am going to bring you within a small epsilon of the solution.

You can use the velocity addition formula

$$u' = \frac{u-v}{1-uv/c^2}$$

Better as

$$s = \frac{u+v}{1+uv/c^2}$$

What happens when $u$ is an infinitesimally small change $dv'$ in the vehicle velocity, as measured in the vehicle's instantaneous rest frame? The answer:

$$v+dv = \frac{v+dv'}{1+vdv'/c^2}$$

Since $dv'$ is infinitesimally small,

$$v+dv = (v+dv')(1-vdv'/c^2) = v + dv' -v^2/c^2dv' = v + (1-v^2/c^2)dv'$$

Now apply this to the previous result:

$$-U\frac {dM}{M} = dv' = \frac{dv}{1-v^2/c^2}$$

Integrate between the initial and final states to get the result in the book.

 

[MathematicalPhysicist]

ok, thank you both.

 

[D H]

Can you get from the differential form to the answer in the book?

 

[MathematicalPhysicist]

you mean can i integrate this?
yes, why not? it's a simple integral substitution sin(w)=v/c
which gives us c*(integral(dw/cos(w)))
the integral of 1/cos(w) is resolved by substituing s=tg(w/2) and you get a rational integral.
the maths isn't where iv'e gone bad, just the physics.

 

[D H]

That looks like too much work. You will get an ugly mess. Why not use some logarithmic u-substitution instead?