# Rocket propelled by a giant monochromatic laser

I am preparing for my quals and stumbled across the following problem, and although it only requires undergraduate-level physics, I feel I can't piece everything together.

"A rocket of mass $m_0$ is propelled by a giant monochromatic laser mounted on the back of the rocket. The laser emits a beam with a power of $P_0$ watts and a frequency $f_0$, both measured in the rest frame of the rocket. When the beam is turned on, the rocket is driven in the opposite direction by the recoil.

(a) At $t=0$, the laser is turned on, with the rocket's velocity initially at rest in the Earth's reference frame. Calculate the instantaneous acceleration of the rocket.

(b) If the rocket is moving at velocity $v$, what is the instantaneous beam power as measured in the Earth's reference frame?

(c) The laser is kept going until the speed of the rocket reaches $v=0.9c$. What is the rest mass of the rocket at this point?"

I work in units where $c=1$. For (a), in the Earth's reference frame, I get a momentum $p(t) = \gamma m_0 v + h f$, where $\gamma = \frac{1}{\sqrt{1-v^2}}$ and $f = f_0 \sqrt{\frac{1-v}{1+v}}$ because of the Doppler shift in frequency as the rocket travels further away. Differentiating w.r.t time, I get

$$\frac{dp}{dt} = \gamma^3 m_0 a - \frac{\gamma a}{(1+v)} h f_0$$

This is as far as I get. Equaling $dp/dt=0$ for conservation of momentum make the $a$ cancel out, and there isn't any other time dependency to differentiate $v(t)$.

For part (b), we can say that in the frame of the rocket, $P_0 = r h f_0$, where $r$ is the rate of emission of photons. In the frame of the Earth, we can therefore write

$$P(t) = \frac{dE}{dt} = \frac{d}{dt} \left( \gamma m_0 + r h f \right) = m_0 \gamma^3 v a - \frac{\gamma a}{(1+v)} P_0$$

but since I don't have $a$, there's not much I can do. Also, I am not sure if $r$ needs to be corrected with a proper time factor when we boost between frames.

Also, for part (c), my first intuition was to say that the rest mass was $m_0$, but now I am thinking that the rocket's total mass decreases because we need to account for the energy lost from the laser. I am slightly confused as to what I need to take into account.

• I think you are complicating things too much. For instance, the thrust from the photons is $P_0/c$ and therefore the acceleration asked under a) is just $P_0 \ / m_0 c$. Aug 25, 2013 at 18:14
• If photons have non-zero effective mass, then why wouldn't a mirror on the rocket and on earth give perpetual free momentum as the photons bounce back and forth? Mar 20, 2014 at 6:54
• @Ehryk if you post your comment above as a phys.se question, I can try to answer it there. Mar 20, 2014 at 7:22
• Mar 24, 2014 at 19:53
• en.wikipedia.org/wiki/Photon_rocket Jul 6, 2014 at 20:33

This is a fun, high-quality qualifying exam question. The algebra is not hard; the physical insight takes some real thought; there are many ways to be partly right. Here's my take on it.

## Acceleration

From Einstein's equation $$E^2 = p^2 + m^2$$ we have for each photon $$E = p = hf_0$$ (in the reference frame of the laser). We can use the power of the laser $$P_0$$ to find the rate at which individual photons are emitted: $$P_0 = \dot N_0 hf_0.$$ (Pardon me for writing $$\dot N$$ rather than $$dN/dt$$.) The total force exerted by the laser is the time derivative of its momentum, which is the time derivative of its energy, which is just the power, $$F = \dot p = \dot E = P_0,$$ and since the laser's momentum must be coming from the rocket we have its acceleration is $$a = P_0/m_0$$. You'll have to stick some $$c$$s back in to make the dimensions come out right.

If the rocket is moving away from Earth at constant speed $$\beta = v/c$$, with corresponding relativistic factor $$\gamma = 1/\sqrt{1-\beta^2}$$, there are three factors that affect the power received at Earth:

1. Each photon's frequency is red-shifted to $$f = f_0\sqrt{\frac{1-\beta}{1+\beta}}$$.
2. Time on the rocket will be dilated, reducing the rate at which photons appear to be emitted to $$\dot N = \dot N_0 / \gamma$$.
3. Each photon emitted by the rocket has a little bit further to travel back to Earth. If the time between photon emissions (in Earth's frame) is $$\Delta t = 1 / \dot N$$, each photon has to go $$\Delta x = v \Delta t$$ further than the last, adding an extra delay $$\Delta x / c$$ to its trip. So the interval between photons arriving at Earth is $$\Delta t' = \Delta t \cdot (1 + \beta) .$$ The rate at which photons reach Earth is therefore $$\dot N' = 1 / \Delta t'$$.

Combining these we have a power recieved at Earth of $$P' = \dot N' hf = \frac{P_0}{\gamma(1+\beta)}\sqrt\frac{1-\beta}{1+\beta} % = \frac{P_0}{\gamma^2(1+\beta)^2} = P_0 \frac{1-\beta}{1+\beta} .$$

It's always possible in relativity problems to get identical results using classical EM fields for light instead of photons, with Poynting vectors carrying momentum, etc. I would not know how to go about that in this case.

## Final rest mass

This part was not immediately obvious to me. The messy option is to try and integrate the expression from the previous section; that probably requires assumptions about the time profile of the acceleration. Usually when you only know the initial and final conditions for a problem, conservation of energy is a good strategy. I wasted some time before I remembered to use conservation of momentum, too.

We know that the final momentum of the rocket is $$p_f = \gamma_f v_f m_f$$, and that the combined momentum of the rocket and its laser exhaust, in the initial rest frame, is zero. Then, using Einstein's equation again, we have a bunch of backwards-going photons with total energy $$E_{\text{exhaust}} = \left|-p_f\right|$$ and a rocket with total energy $$E_{\text{rocket}}^2 = p_f^2 + m_f ^2.$$ If we assume that none of the laser power was wasted as heat, these components must add up to the initial rest mass of the rocket,

\begin{align} p_f + \sqrt{p_f^2 + m_f^2} &= m_0 \\ m_f \left( \gamma_f v_f + \sqrt{\gamma_f^2 v_f^2 + 1} \right) &= m_0 \\ m_f \left( \gamma_f v_f + \gamma_f \right) &= m_0 \\ m_f &= m_0 \sqrt{\frac{1-v_f}{1+v_f}} \end{align}

Actually this result also holds if the laser power supply is inefficient, so long as the rocket is thermally insulated such that the thermal waste photons are all emitted in the same direction as the exhaust --- tail of the rocket hot, head of the rocket cold. If there's forward-going thermal radiation, it will be focused in the forward direction in a complicated way, and the problem becomes much more difficult.

# Edit

It seems from the comments I was taking a naive classical perspective on momentum, and this is an incorrect answer. However, I'm leaving it for anyone tempted to think the same.

# Original

The recoil? From massless photons? How do you figure?

a) Instantaneous Acceleration: $$0\ m/s^2$$

b) Power: $$P_0$$

c) It will never reach $$0.9c$$, and if it did it would have the same rest mass, $$m_{rocket}$$

Let's use the Tsiolkovski Rocket Equation: $$\Delta v = v_\text{e} \ln \frac {m_0} {m_1}$$ https://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation

Where: $$v$$ = exhaust velocity (c), $$m_0$$ = initial mass, $$m_1$$ = final mass.

So $$\Delta v = c \ln \frac {m_{rocket}} {m_{rocket}} = c \ln 1 = c * 0 = 0$$

Or how about conservation of momentum?

$$\Delta Momentum_{laser beam} = \Delta Momentum_{rocket}$$

$$m_{laser}*\Delta v_{laser} = m_{rocket}*\Delta v_{rocket}$$

$$0 * c = m_{rocket}*\Delta v_{rocket}$$

$$\Delta v_{rocket} = 0$$

So if it was initially at rest, it would stay at rest.

Power does not equate to thrust - in this case, the laser would just be producing heat as far as the rocket is concerned.

• Photons carry momentum even though they don't have mass. So the laser could indeed accelerate the rocket. Mar 20, 2014 at 1:49
• From answers.yahoo.com/question/index?qid=20100521204409AAk1lkq, $m_{photon} = \frac {h*f} {c^2}$? Seems bizarre, and the conservation of momentum could be used to calculate it. How do you account for the Tsiolkovski Rocket Equation, based on the (rest) mass and velocity of the exhaust? Does it not apply in this situation? Mar 20, 2014 at 6:12
• The rocket equation to which you refer is, at heart, a statement of momentum conservation. For non-relativistic exhaust (moving at speeds that are small compared to $c$), it takes the form you have quoted. However, that form of the equation does not apply to relativistic exhaust like photons. The resulting rocket equation in the relativistic case is the subject of this question, and is slightly more complicated than merely replacing the mass in the classical formula with $h f / c^2$. Mar 20, 2014 at 7:28