Rocket propelled by a beam of photons

[ruairilamb]

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 $$\frac{m_{i}} {m_{f}} = \frac{1} {1+β/c}$$

 

[anuttarasammyak]

$\beta$ 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.

 

[ruairilamb]

$\beta$ 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.

 

[vela]

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.

 

[Steve4Physics]

Hi @ruairilamb. Welcome to PF. I'd like to add a few comments in addition to what @anuttarasammyak and @vela have 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'?

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

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.

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.

 

[anuttarasammyak]

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
$$m_i-m_f=\frac{\hbar \omega}{c^2}$$
momentum cocnservaiton
$$m_f\gamma\beta=\frac{\hbar \omega}{c^2}$$
So
$$\frac{m_f}{m_i}=(1+\gamma \beta)^{-1}$$
Please investigate the difference with your result.

 

[Steve4Physics]

Say one high energy photon is generated by mass loss of $m_i-m_f$ with rocket at rest initially
energy conservation
$$m_i-m_f=\frac{\hbar \omega}{c^2}$$

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.

 

[anuttarasammyak]

If so, the rocket's final kinetic energy is missing from the equation.

My bad, thanks. So
$$m_i=\frac{\hbar \omega}{c^2}+m_f\gamma$$

 

[Steve4Physics]

My bad, thanks. So
$$m_i=\frac{\hbar \omega}{c^2}+m_f\gamma$$

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.

Together with momentum conservation

$$\frac{m_f}{m_i}=\sqrt{\frac{1-\beta}{1+\beta}}$$

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.

 

[anuttarasammyak]

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 :
$$\frac{d(m\gamma)}{d\tau}=-\frac{N\hbar\omega}{c^2}=-\frac{N\hbar\omega_0}{c^2}\sqrt{\frac{1-\beta}{1+\beta}}$$
where
$$\frac{dm}{d\tau}=-\frac{N\hbar\omega_0}{c^2}:=-\alpha$$
mass reduced by exhausted photon gas energy per unit proper time. The solution is
$$\sqrt{\frac{1-\beta}{1+\beta}}=\frac{m_0-\alpha \tau}{m_0}$$
I found this ratio relation is hold during this jet acceleration process of any constant fuelling pace.

 

[ruairilamb]

Thanks for all the replies. I think for now this problem is slightly beyond my ability. I will tackle some simpler problems first.