# Special Relativity - Rocket problem (particle mechanics)

• Thales Castro
In summary: Calculate that the ratio of the initial to the final rest mass of the rocket is given by$$\frac{m_{i}}{m_{f}} = \left( \frac{c+v}{c-v} \right) ^{\frac{c}{2u}}$$-Evaluate that this is equal to$$\frac{m_{f}}{m_{i}} = \left| \frac{v+c}{v-c} \right| ^{\frac{c}{2u}}$$
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?

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## traveling 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##.

Thales Castro
Orodruin said:
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## traveling 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##.

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.

Thales Castro said:
The will give me the correct answer, but are there any other mistakes in what I did?
Apart from the energy equation and the definition of ##dm## such that it is the actual change in the mass, the approach is correct.

Thales Castro

## 1. What is the Special Theory of Relativity?

The Special Theory of Relativity is a scientific theory developed by Albert Einstein in 1905. It explains the relationship between space and time and how they are affected by the motion of objects in the universe. The theory states that the laws of physics are the same for all observers in uniform motion and that the speed of light in a vacuum is constant.

## 2. How does the Special Theory of Relativity apply to the "Rocket problem" in particle mechanics?

The "Rocket problem" in particle mechanics is an example used to illustrate the concepts of the Special Theory of Relativity. It involves a rocket moving at a constant velocity in space and the effects of time dilation and length contraction on the rocket and its passengers. The theory helps to explain how the observations of time and space differ for observers in different frames of reference.

## 3. Can the Special Theory of Relativity be tested or proven?

Yes, the Special Theory of Relativity has been extensively tested and proven through experiments and observations. One of the most famous experiments is the Michelson-Morley experiment, which showed that the speed of light is constant and does not depend on the motion of the observer. Other experiments, such as the Hafele-Keating experiment, have also confirmed the predictions of the theory.

## 4. How does the Special Theory of Relativity impact our understanding of the universe?

The Special Theory of Relativity has greatly impacted our understanding of the universe by providing a framework for understanding the behavior of objects moving at high speeds. It has also led to the development of other theories, such as the General Theory of Relativity, which explains the effects of gravity on space and time. The theory has also been applied to various fields, such as astrophysics and particle physics, to make accurate predictions and advancements in these areas.

## 5. Are there any practical applications of the Special Theory of Relativity?

Yes, there are many practical applications of the Special Theory of Relativity. One of the most well-known is the Global Positioning System (GPS), which relies on the theory to accurately determine the positions and times of satellites and receivers. The theory also has applications in nuclear energy, telecommunications, and particle accelerators, among others.

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