Radiation pressure has been hypothesized since the early days of special relativity and quantum mechanics. It is a simple consequence of the mass-energy equivalence and the photon nature of light. Quantum mechanics states that light is made of photons, and that each photon is a discrete bundle of energy. The energy \(E_{pho}\) carried by each photon is proportional to its frequency \(\nu=c/\lambda\), and the proportionality constant is Planck's constant \(h\).Abstract:The Breakthrough Starshot project proposes using radiation-pressure propulsion to send multiple spacecraft in the gram-range mass to interstellar targets. There are numerous engineering challenges to accomplishing this mission, but their project discusses speeds and accelerations previously only discussed for cannon projectiles and particle physics experiments. This article examines the problem from first principles to see if the project is even orders-of-magnitude possible. A preliminary check on their numbers seems to confirm that the project is possible in principle.

\[E_{pho}=h\nu=\frac{hc}{\lambda}\]

Everyone knows the famous mass-energy equivalence equation \(E=mc^2\). However, that is only a special case of the momentum equation:

\[E=\sqrt{p^2c^2+m_0^2c^4}\]

We can solve this for the momentum of a photon, considering that its rest mass is zero.

\[\begin{eqnarray}E_{pho} & = & \sqrt{p_{pho}^2c^2+0^2c^4}\\

& = & \sqrt{p_{pho}^2c^2} \\

& = & pc\\

\frac{hc}{\lambda}&=&p_{pho}c\\

\frac{h}{\lambda}&=&p_{pho}\end{eqnarray}\]

Note that momentum is a vector quantity, but this just deals with the magnitude. The direction of the momentum vector is the direction the photon is traveling.

Now, we know the energy and momentum of each photon, so we know that if we throw so many joules of photons at a target, it will transfer so much momentum. The irradiance is defined as the power \(P\) (energy \(E\) per unit time \(t\)) of the light hitting each unit area of the target, so the units are W/m^2 or J s^-1 m^-2. We can calculate it from the amount of energy striking the target of known area \(A\) per unit time:

\[I=\frac{P}{A}=\frac{E}{tA}\]

So given a certain amount of irradiance with a known photon wavelength, how much momentum does that light carry? First, figure out how many photons per time per area \(I_{pho}\) that irradiance represents:

\[\begin{eqnarray}I_{pho}&=&\frac{I}{E_{pho}}\\

&=& \frac{I\lambda}{hc}\end{eqnarray}\]

The dimension of photon irradiance is photons per time unit per area unit (photon s^-1 m^-2 in SI units).

Now a bit about momentum. Again, everyone knows Newton's second law \(\vec{F}=m\vec{a}\), but again this is a special case. Force is defined as the change in momentum per unit time:

\[\vec{F}=\frac{d\vec{p}}{dt}\]

For massive objects at sub-relativistic speeds, momentum is defined as the mass of the object times the velocity of the object. For relativistic conditions, we use the energy relation above, where energy still has the customary units J=kg m^2 s^-2. Working through the units of the energy equation in the special case of a photon, we find that momentum still has the same units as it does in sub-relativistic conditions.

So from this, a given amount of energy of photons carries a certain amount of momentum. A certain power of photons carries a flow of momentum per unit time, or in other words

*Force.*A certain amount of momentum flow impinging on a certain area is the same as a force exerted on that area, or a

*Pressure*\(\rho\). So, we can calculate the pressure exerted by a given irradiance of light from the intensity in photons/time/area, multiplied by the momentum of each photon, which gives momentum/time/area which equals force/area which equals pressure:

\[\begin{eqnarray}\rho&=&I_{pho}p_{pho}\\

&=&\left(\frac{I\lambda}{hc}\right)\left(\frac{h}{\lambda}\right)\\

&=&\frac{I}{c}

\end{eqnarray}\]

So the pressure exerted by the light is the irradiance of the light divided by the speed of light. That is amazingly simple, and notice that both Planck's constant and the wavelength cancel out. Now we can start plugging numbers.

First, let's do one of Clarke's space yachts. It weighs on the order of 1000kg and has on the order of 1km^2=1,000,000m^2 of sail, and it uses natural sunlight. How much force does it collect and what is its acceleration? Note that since the sails are reflective, the light is reversed in direction and has a momentum change of twice its original momentum, so we collect twice as much force as the momentum suggests.

\[\begin{eqnarray}I& & &\approx&1400\mbox{W/m}^2\\

c& & &=&299,792,458\mbox{m/s}\approx 3\times 10^8 \mbox{m/s}\\

\rho&=&2\frac{I}{c}&=&2\frac{1400}{3\times 10^8}=9.4\mu \mbox{N/m}^2\\

A& & &=&1,000,000 \mbox{m}^2\\

F&=&P_{pho}A&=&9.4\mbox{N}\\

m& & &=&1000 \mbox{kg}\\

a&=&\frac{F}{m}&=&\frac{9.4}{1000}=9.4\mbox{mm/s}^2\approx1.0\mbox{mg}

\end{eqnarray}\]

So the whole sail exerts a measly couple of newtons. Pulling on a ton of mass, we get almost a full milli-g of acceleration. Not a whole lot, but in order of magnitude of that described in the story.

Now we get into the big numbers. The game Adventure Capitalist teaches us to not be afraid of big numbers, so lets go get some. The Starshot project describes using gigawatts of power on spacecraft weighing grams. Among the enormous engineering challenges are:

- Focusing gigawatts of power onto square-meters sized targets over distances of billions of meters
- Having a reflective enough sail that the GW/m^2 incident power doesn't vaporize the sail
- Having a tough enough sail that the MW/m^2 absorbed power doesn't vaporize the sail
- Having a thin enough sail to stay within the mass budget
- Building a spacecraft with a useful payload, power source, and communications system that will work over interstellar distances and stay within the mass budget
- By the way, the mass budget is about 10g total.

Putting all of these aside, they discuss a spacecraft with a mass of about 10g, a sail area of about 16m^2, and multiple gigawatts of power focused on it. Let's go with 1GW/m^2, or 16GW total, to see what we get:

\[\begin{eqnarray}I& & &\approx&1\mbox{GW/m}^2\\

P_{pho}&=&2\frac{I}{c}&=&2\frac{1\times 10^9}{3\times 10^8}=6.66 \mbox{N/m}^2\\

A& & &=&16 \mbox{m}^2\\

F&=&P_{pho}A&=&16\times 6.66=106\mbox{N}\\

m& & &=&0.01 \mbox{kg}\\

a&=&\frac{F}{m}&=&\frac{106}{0.01}=10.6\mbox{km/s}^2\approx1080\mbox{g}

\end{eqnarray}\]

Well, that's some git-up-and-go, alright. How long does it take this to get to a target speed of a good chunk of the speed of light, say 60,000km/s (0.2\(c\))?

\[v=at\]

\[\begin{eqnarray}t&=&\frac{v}{a} \\

&=&\frac{60,000,000}{10600}&=&5660 \mbox{s}

\end{eqnarray}\]

&=&\frac{60,000,000}{10600}&=&5660 \mbox{s}

\end{eqnarray}\]

\[v=at\]

This is about 10000g. Now we are talking cannon-type acceleration. Since the acceleration is about 10 times greater, it will require 10 times the power, or a couple hundred gigawatts. This is a good fraction of the electricity usage of the United States, so it is a large but doable amount of power. We only need it for 10 minutes. The paper discusses putting the laser in space, which would require a gigawatt-scale power source in space. I suppose a couple of square kilometers of solar cells could do that.

\[\begin{eqnarray}a&=&\frac{v}{t} \\

&=&\frac{60,000,000}{600}&=&100 \mbox{km/s}^2\\

F&=&ma&=&0.01(100,000)=1000\mbox{N}\\

P&=&\frac{F}{A}&=&\frac{1000}{16}=62.5\mbox{N/m}^2\\

I&=&Pc&=62.5(300,000,000)=18.75\mbox{GW}

\end{eqnarray}\]

&=&\frac{60,000,000}{600}&=&100 \mbox{km/s}^2\\

F&=&ma&=&0.01(100,000)=1000\mbox{N}\\

P&=&\frac{F}{A}&=&\frac{1000}{16}=62.5\mbox{N/m}^2\\

I&=&Pc&=62.5(300,000,000)=18.75\mbox{GW}

\end{eqnarray}\]

This is about 10000g. Now we are talking cannon-type acceleration. Since the acceleration is about 10 times greater, it will require 10 times the power, or a couple hundred gigawatts. This is a good fraction of the electricity usage of the United States, so it is a large but doable amount of power. We only need it for 10 minutes. The paper discusses putting the laser in space, which would require a gigawatt-scale power source in space. I suppose a couple of square kilometers of solar cells could do that.

If all of the engineering problems are solved, this could work. There isn't anything physically impossible about it. The engineering challenges are large, but the Starshot project claims that each can be solved incrementally. We don't have to shoot at stars first -- imagine getting back to Neptune in a couple of days, and then being able to orbit once you got there?

References

Lubin, P. A Roadmap To Interstellar Flight

I love your article.you can visit my website: club factory online shopping review

ReplyDelete