# Special Relativity - Rocket problem (particle mechanics)

1. Jul 25, 2017

### Thales Castro

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. Jul 26, 2017

### Orodruin

Staff Emeritus
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 $u < 1$.

3. Jul 26, 2017

### Thales Castro

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.

4. Jul 26, 2017

### Orodruin

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