Rocket propelled by a beam of photons

  • #1
ruairilamb
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Homework Statement
A rocket of initial mass, ##m_{i}##, starts at rest and propels itself by an engine ejecting photons backwards. After the rocket has reached a speed of v, it switches off its engine and its final mass is ##m_{f}## . By using four-momenta and three-momenta for the rocket and photon beam, calculate the ratio of the initial to final mass for the rocket as a function of β only.
Relevant Equations
##β = \frac{v}{c}##
conservation of four momentum
conservation of 3 momentum
conservation of energy
Before the engine is switched off: $$P_{Initial rocket} = (m_{i}c, 0)$$

$$P_{photons} = (E/c, -p_{photons})$$

where E is the energy of the photons and ##p_{photons}## is the 3 momentum of the photons.
Rocket after engine switched off: $$P_{Final rocket} = (m_{f}c, m_{f}v)$$
By conservation of four momentum:

$$P_{Initial rocket} + P_{photons} = P_{Final rocket}$$
So, ##(m_{i}c, 0) + (E/c, -p_{photons}) = (m_{f}c, m_{f}v)##
Now, ##E = p_{photons}c## so,
$$(m_{i}c, 0) + (p_{photons}, -p_{photons}) = (m_{f}c,m_{f}v)$$
##p_{photons}## can be written as ##p_{photons} = m_{i}β##
Hence, $$(m_{i}c, 0) + (m_{i}β, -m_{i}β) = (m_{f}c, m_{f}v)$$
I am not sure how to proceed from here, or even if my solution so far is remotely right. If I equate the components of each of the terms I get:
$$m_{i}c + m_{i}β = m_{f}c$$
This implies a ratio of $$\frac{m_{i}} {m_{f}} = \frac{1} {1+β/c}$$
 
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  • #2
##\beta## is dimensionless number and c has dimension of LT^-1 so something wrong in dimension in your result. Your formula seems to allow rocket to have light speed c "getting" half of its initial mass. Is that reasonable ? Not + sign in your formla but - seems reasonable, though I have not investigated the procedure.
 
  • #3
anuttarasammyak said:
##\beta## is dimensionless number and c has dimension of LT^-1 so something wrong in dimension in your result. Your formula seems to allow rocket to have light speed c "getting" half of its initial mass. Is that reasonable ? Not + sign in your formla but - seems reasonable, though I have not investigated the procedure.
The minus sign is due to the photons being ejected downwards. Thanks for the reply.
 
  • #4
A number of issues with your attempt:

Looking at just the momentum part of your conservation equation, it says that the rocket's final momentum is equal to the rocket's initial momentum, which is 0, plus the momentum of the photons, which points in the. negative direction. So the rocket's final momentum is equal to the photon beam's momentum, which is in the negative direction—that is, the rocket moves backwards. Does that make sense?

Similarly, the energy component says the rocket's final energy is its initial energy plus the energy of the photons, so the energy of the rocket is increasing with time. Does that make sense?

I'm not sure where you got ##p_{\rm photon} = m_i \beta##, whatever ##\beta## is defined as.

Finally, the four-momentum of the rocket when it's moving is ##(\gamma m_f c, \gamma m_f \vec v)## where ##m_f## is its rest mass. It looks like you may be using the idea of relativistic mass, which you should avoid.
 
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  • #5
Hi @ruairilamb. Welcome to PF. I'd like to add a few comments in addition to what @anuttarasammyak and @vela have said.

ruairilamb said:
Before the engine is switched off: $$P_{Initial rocket} = (m_{i}c, 0)$$
Are you sure? Do you mean 'Before the engine is switched on'?

ruairilamb said:
$$P_{photons} = (E/c, -p_{photons})$$
where E is the energy of the photons and ##p_{photons}## is the 3 momentum of the photons.
The minus sign is wrong IMO. You are basically saying that if the photons move (say) left, you will take their 3-momentum as acting to the right!

(It may be that one or more of the momentum's components have a negative value, But any minus signs would then appear in later working.)

ruairilamb said:
Rocket after engine switched off: $$P_{Final rocket} = (m_{f}c, m_{f}v)$$
Are you sure? ##m_f## is (almost certainly) intended to be the rocket's final rest mass. See @vela's reply.

ruairilamb said:
By conservation of four momentum:

$$P_{Initial rocket} + P_{photons} = P_{Final rocket}$$
Conservation of momentum requires that (in the absence of any external forces): total initial momentum = total final momentum. There are no photons initially.

I'm stopping at this point. The question is (IMO) quite a difficult one for someone not already familiar/comfortable with 4-momentum.

I recommend taking a step back and reading/practicing with some simpler problems until you are more confident with the basics.
 
  • #6
ruairilamb said:
This implies a ratio of
Say one high energy photon is generated by mass loss of ##m_i-m_f## with rocket at rest initially
energy conservation
[tex]m_i-m_f=\frac{\hbar \omega}{c^2}[/tex]
momentum cocnservaiton
[tex]m_f\gamma\beta=\frac{\hbar \omega}{c^2}[/tex]
So
[tex]\frac{m_f}{m_i}=(1+\gamma \beta)^{-1}[/tex]
Please investigate the difference with your result.

1697687200317.png
 
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  • #7
anuttarasammyak said:
Say one high energy photon is generated by mass loss of ##m_i-m_f## with rocket at rest initially
energy conservation
[tex]m_i-m_f=\frac{\hbar \omega}{c^2}[/tex]
Presumably ##m_f## is a rest mass - not the (old-fashioned and out-of-favour) relativistic mass.

If so, the rocket's final kinetic energy is missing from the equation.
 
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  • #8
Steve4Physics said:
If so, the rocket's final kinetic energy is missing from the equation.
My bad, thanks. So
[tex]m_i=\frac{\hbar \omega}{c^2}+m_f\gamma[/tex]

1697720962751.png
 
  • #9
anuttarasammyak said:
My bad, thanks. So
[tex]m_i=\frac{\hbar \omega}{c^2}+m_f\gamma[/tex]
Agree. But ##\hbar \omega## is the energy of a single photon; that would be one hell of a photon! I'd probably just define ##E_p## as the total photon energy.

anuttarasammyak said:
Together with momentum conservation

[tex]\frac{m_f}{m_i}=\sqrt{\frac{1-\beta}{1+\beta}}[/tex]
Look good - though the question asks for ##\frac{m_i}{m_f}##, not ##\frac{m_f}{m_i}##. And I'm not sure if, with the rules here, we should give the final answer.

Note that the original question specifically asks for a solution "using four-momenta and three-momenta for the rocket and photon beam". So a solution using 'simple' conservation of momentum and energy equations may not be acceptable - but it provides a useful check if required.
 
  • #10
Steve4Physics said:
Agree. But ℏω is the energy of a single photon; that would be one hell of a photon! I'd probably just define Ep as the total photon energy.
We will need a good technology to control direction of photon gas jet.

I picked up the simplest case that all the generation of a photon or photon gas takes place in an instant. In usual chemical rocket problem it keeps constant jet during its acceleraion. Is the ratio free from these procedures of acceleration ?
Differentiation equaion of more general case :
[tex]\frac{d(m\gamma)}{d\tau}=-\frac{N\hbar\omega}{c^2}=-\frac{N\hbar\omega_0}{c^2}\sqrt{\frac{1-\beta}{1+\beta}}[/tex]
where
[tex]\frac{dm}{d\tau}=-\frac{N\hbar\omega_0}{c^2}:=-\alpha[/tex]
mass reduced by exhausted photon gas energy per unit proper time. The solution is
[tex]\sqrt{\frac{1-\beta}{1+\beta}}=\frac{m_0-\alpha \tau}{m_0}[/tex]
I found this ratio relation is hold during this jet acceleration process of any constant fuelling pace.
 
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  • #11
Thanks for all the replies. I think for now this problem is slightly beyond my ability. I will tackle some simpler problems first.
 
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FAQ: Rocket propelled by a beam of photons

What is a photon rocket?

A photon rocket is a theoretical type of spacecraft that uses photons, or light particles, as the means of propulsion. Instead of expelling mass like traditional rockets, a photon rocket emits a beam of photons to generate thrust according to the principles of momentum conservation.

How does a photon rocket generate thrust?

A photon rocket generates thrust by emitting a beam of photons in the opposite direction of the desired movement. According to Newton's third law of motion, for every action, there is an equal and opposite reaction. When photons are emitted, they carry momentum away from the rocket, which results in a forward thrust.

What are the advantages of using a photon rocket?

The primary advantage of a photon rocket is its efficiency in terms of specific impulse, as it has the potential to achieve extremely high velocities. Since photons travel at the speed of light, a photon rocket can theoretically approach relativistic speeds, making it ideal for long-distance space travel.

What are the challenges associated with photon rockets?

One of the main challenges is the immense amount of energy required to produce a significant amount of thrust using photons. Current technology does not allow for the efficient generation and control of the required photon beams. Additionally, managing the heat and energy dissipation in such a system poses significant engineering challenges.

Are there any practical applications for photon rockets today?

As of now, photon rockets remain largely theoretical and are not used in practical applications. However, research in this area continues, and advancements in laser technology, energy generation, and materials science may eventually make photon rockets a viable option for future space exploration missions.

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