# Lightsail | Integrity under thrust

To inform the study, a beamer in the 100 GW class was considered. If, for example, 10^{-5} of the energy is absorbed by a 4mx4m sail, it will be heated by about 60kW per m^{2}, which is roughly 60 times more than sunlight illumination on Earth. This will heat the material but not melt it. Using fully dielectric sails, we can reduce the absorption to less than 10^{-9} for optimized materials.

Two possible approaches to mitigate the heating challenge have been identified:

**1. High reflectivity**

Use a material with better than 99.999% reflectivity. Usually, highly reflective surfaces are dielectric mirrors, which are composed of ‘sandwiches’ of material, with each layer reflecting back a modest fraction of the total. Each layer needs to be at least a quarter of a wavelength thick. The weight can be reduced by using a monolayer with high reflectivity at the correct wavelength. Based on recent research, this could be achieved by a ‘hole-pocked’ layer, highly reflective for very specific angles where reflectivity caustics arise. (These caustics occur for wavelengths of light that are actually longer than the sheet thickness.) Adding the holes serves a dual purpose; it reduces the weight of the sail and it could greatly increase reflectivity. This is but one possibility being explored. Modern materials research will explore new materials such as graphene; Breakthrough Starshot aims to take advantage of this rapidly advancing field. The basic Starshot system allows a wide range of options for nanocraft masses and capabilities, all using the same array. This gives it great flexibility in optimizing the science and technology roadmaps.

**2. Low absorption**

Use a material (such as glass) that has a very low absorption coefficient even when not highly reflective. Such materials are used in fiber optics systems with high power applications. Without the protection of a highly reflective sail, the StarChip electronics would need to be protected from the incoming flux. But this could be accomplished by a combination of geometry (orienting the electronics ‘sideways’ with a low cross-section) and placing a very highly reflective coating only on the sensitive components. These can use the multi-layer dielectric approaches mentioned above, which have already been demonstrated in the lab. Using low absorption sail material, together with a limited use of high-reflectivity shielding for critical electronics, would protect the StarChip without increasing its mass beyond the gram scale. There are a number of high-reflectivity, low absorption materials in existence. For possible fabrication and verification, a demonstrable design of silicon microcubes on a silicon dioxide substrate is under consideration.

As demonstrated by the Japanese IKAROS mission, spinning the sail can reduce wrinkles on its surface. Special attention is needed to avoid impurities and non- uniformities in the sail composition - for example, near mechanical attachments - or accumulation of dust particles on its surface, which could otherwise lead to a localized deposition of energy. There are a wide variety of options allowing optimization of the sail design.

**Research:**

- SOLAR SAILING: Technology, Dynamics, and Mission Applications
- Hsu, C.-W., et al. “Observation of Trapped Light Within The Radiation Continuum”, Nature, Vol. 499, pp. 188-191 (2013)
- Slovick, B., Gang Yu, Z., Berding, M., and Krishnamurthy, S., “Perfect Dielectric-Metamaterial Reflector”, Physical Review B, Vol. 88, pp. 165116-1 – 165116-7 (2013)

Posted on: Breakthrough Initiatives

RE:

"Jan 09, 2017 16:49 michael.million@sky.com Posted on: Centauri Dreams

If we use silicon we can go a way into the micron wavelength region before absorption due to the Doppler effect starts to become a problem.

http://www.pveducation.org/pvcdrom/materials/optical-properties-of-silicon";

Answer:

Thank you for your consideration. Silicon is heavy in that we have difficulty building sheets of Silicon much thinner than several hundred adams thick. Which will make the material to heavy also the sail needs to withstand 60,000 g’s of acceleration and silicon is very brittle. With effort I think we can solve all of these problems.

- Pete Klupar, Breakthrough Starshot

Posted on: Breakthrough Initiatives

The plan is to use a 100 GW laser beam to propel the lightsail. To avoid heating the lightsail, the material must have over 99.999% reflectivity to reduce the energy absorbed to 10^-5 of the laser energy. I will show here that this is not possible in principle. The physical laws governing the generation of thrust in the lightsail by light are the conservation of energy and momentum in inelastic collision between the lightsail and light photons.

Let:

dp = change in momentum of lightsail = change in momentum of photon

m = mass of lightsail = 1 g = 0.001 kg

dv = change in velocity of lightsail

Conservation of momentum

dp = m dv

dv = dp/m

Let:

dKEp = change in kinetic energy of photon

dKEs = change in kinetic energy of lightsail

dQ = change in thermal energy

c = speed of light

Conservation of energy

dKEp = dKEs + dQ

Conservation of momentum and energy

dp c = 1/2 dp dv + dQ

From the above equation, it is evident that a change in momentum is not possible in an elastic collision where dQ = 0 because to obey the conservation laws, dv = 2 c or the lightsail must be traveling twice the speed of light. Hence, it will always be an inelastic collision where dQ > 0

Solving for dQ and substituting dv = dp/m

dQ = dp c - 1/2 dp (dp/m)

Ratio of thermal energy to lightsail kinetic energy

dQ/dKEs = 2 mc/dp - 1

Let:

h = Planck constant

df = change in light frequency = 10^14 hz

Change in momentum of photon

dp = h/c df = 10^-28

Calculating the ratio

dQ/dKEs = 10^33

This ratio shows that almost all the loss in kinetic energy of the photon is converted to heat (dQ). Why? A familiar analogy is useful. Imagine the molecule of the lightsail is a steel ball and the photon is a hammer. When the hammer strikes the steel ball, the kinetic energy of the hammer is transferred to the steel ball accelerating it. But when you hold the ball, it cannot move so the energy is converted into heat. When a photon collides with a molecule, it should accelerate and eject out of the lightsail. But the molecular bond prevents it from moving like your hand holding the steel ball. Thus the energy is converted into heat. This will increase the temperature of the lightsail and emit infrared and/or light photons. This will generate thrust in the lightsail via conservation of momentum. However, since photons are emitted on both sides of the lightsail, net thrust occurs only if there's a temperature difference between the two sides.

We can calculate the acceleration of the lightsail given the temperature difference between the two sides. Assume a graphite lightsail with boiling point of 4,000 K (it doesn't melt at low pressure) and temperature difference of 100 K

Let:

P = net power emitted by lightsail

o = Stefan-Boltzmann constant

A = area of lightsail = 16 m^2

Ta = temperature at side A = 4,000 K

Tb = temperature at side B = 3,900 K

Net power emitted by lightsail

P = o A (Ta^4 - Tb^4) = 2 x 10^7 W

We can convert this into acceleration of the lightsail.

Let:

P = net power emitted by lightsail

E = net energy of photons

dp = change in momentum of lightsail/photons

dv = change in velocity of lightsail

a = acceleration of lightsail

t = time

c = speed of light

m = mass of lightsail = 1 g

P = E/t = dp c/t

dp/t = P/c

dp/t = m dv/t

a = dv/t

P/c = m a

a = P/(m c)

Calculating for the acceleration

a = 75 m/s^2 or 7.6 g's

Hence, the max. acceleration of the lightsail at max. temperature and 100 K temp. difference is 7.6 g's. This is 10,000 times lower than the planned acceleration of Breakthrough Starshot, which aims to accelerate the lightsail to 0.2 c within the distance of two million km. This is a fundamental limit by the boiling/melting point of the material and the temperature difference between the two sides of the lightsail.

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RE:

"Feb 24, 2017 00:20 ENRICO BELMONTE Posted on: Breakthrough Initiatives"

Thank you for the very detailed discussions here. We welcome and encourage all such thoughts and you have given this a lot of your time.

The interaction of photons with matter is largely a photon-electron process. The sail is NOT driven the differential flux on the sides of the reflector FROM temperature differentials. This is a very small effect compared to laser photon-reflector electron interaction. Other examplea of photon-electron interaction can be seen in Compton scattering where photons can drive electrons to relativistic speeds. The actual laser photon-electron physics is quite interesting due to both thermal and quantum fluctuations (including zero point fluctuations) leading ultimately to Doppler shifting in the reflected photons which is the basic mode by which momentum conservation is achieved.

Please see the technical paper "A Roadmap to Interstellar Flight":

A Roadmap to Interstellar Flight – Lubin – submitted April 2015

JBIS Vol 69, Pages 40-72, Feb 2016

A Roadmap to Interstellar Flight

http://arxiv.org/abs/1604.01356

See the table of contents as the paper is quite long. See section 3 for the mathematics behind the reflector absorption and reflection as well as discussions on material issues.

There is also on online photon calculator that implements the equations from the above "roadmap" paper you find useful:

http://www.deepspace.ucsb.edu/wp-content/uploads/2015/04/Laser-Propulsion-Classical-1D-Standalone.html

Many more papers and discussions are here:

http://www.deepspace.ucsb.edu/projects/directed-energy-interstellar-precursors

- Phil Lubin, Breakthrough Starshot

Posted on: Breakthrough Initiatives

Prof. Lubin,

Thank you for your reply. I understand what you're saying and it does not contradict my scientific argument. Compton scattering occurs in free electrons or when the photon energy is higher than the binding energy of atomic electrons. Please note that if this occurs in the lightsail, its atoms will be ionized and the ejected electrons will carry the momenta transferred by the photons. It will not accelerate the lightsail. On the other hand, if the electrons are not ejected out, they will jump to a higher quantum energy level in the atom. Eventually they will revert to a lower energy level and emit a photon. This will generate thrust in the lightsail but also increase its temperature according to Stefan-Boltzmann law.

This is basically my argument. Reflection of light is essentially an elastic collision of photons with matter without a change in momentum of photons. Hence, no thrust is generated in the matter. When thrust is generated in an inelastic collision, heat is also produced. It is true for photon-electron collision.

Posted on: Breakthrough Initiatives

By the way, it does not matter if we use photons, electrons or protons as projectiles to hit the lightsail, they will all heat it up before it attains high acceleration. Electron-electron collision will also lead to electron quantum jumping in the atoms. Proton-electron collision will ionize the atoms since protons are 1836 times more massive/energetic than electrons. It will vaporize the lightsail into plasma.

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Enrico

Lets break down your concerns. I will answer as a physicist as you seem comfortable with this.

1) Lets consider a system that has no conduction band (no free electrons - no metals) and thus there is no Thomson scattering (free electron scattering) but only bound Rayleigh scattering. My reference to Compton scattering was to indicate other examples that free electrons can exchange energy with photons but this is in the limit of free systems and not important for our system.

Back to our system. Lets assume there is no internal loss mechanism in the material and hence zero energy loss internally - ie a perfect reflector. This leads to zero temperature increase.

We also assume the reflector is initially at rest relative to the photon source (laser array).

The photons thus have the same magnitude of momentum (but opposite direction) upon reflection and hence the photon energy is the same after reflection as before.

In this case we thus conclude there is no energy transfer from the photons to the reflector and hence it does not move. Seems counter intuitive but correct. The "perfect" reflector is analogous to an infinite mass . The kinetic energy (non rel limit) for photons that transfer momentum 2p (incident p = -reflected p) KE(2p) = (2p)^2/2m BUT if m=infinity then KE(p)=0 --> no movement EVER.

This is analogous to pushing on a car with the brakes on. You exert a force but the car does not move and hence there is no energy transfer. The car exerts an equal and opposite force on you. This is logical. --> no energy transfer and no heating and no movement..

a) There is an important lesson here. We enter a realm where the classical regime and the quantum mechanical regime often get swept under the carpet. Lets consider a single photon impact on a single electron in the "perfect" sail. Since it is a perfect (stationary) sail the incident photon and the reflected photon have equal and opposite momentum (assuming normal incidence) and thus the incident and reflected photon have the same energy and thus NO energy is transferred and thus NO movement. If NO energy is transferred for a single photon then NO movement is EVER possible. How can this be? The problem is that there is a "loss mechanism" here but a very subtle one that is not obvious at all. It is the key to understanding the photon electron interaction at a micro physical regime. The "loss mechanism" here is NOT the usual loss mechanism you might imagine. It is not a normal "damping mechanism" such as "friction" or conversion to heat. It is NOT this. NO heating takes place in this perfect system. So what is the loss mechanism?

The issue is that a system at any non zero temperature (above absolute zero) has thermal motion and thus the electron is moving during the interaction. This motion during the photon-electron interaction leads to a Doppler shift in the photon and hence the incident and reflected photon DO NOT have equal and opposite momentum and thus DO NOT have equal energy. This IS the "loss mechanism" but NOT a typical loss mechanism. There is NO heating or damping going on.

A more subtle loss mechanism is due to "zero point energy". Even if the system were at absolute zero in temperature BUT was a bound finite system there would be "zero point energy" in the electron and thus we once again DO NOT have equal and opposite momentum and thus the same type of Doppler shift and hence a "loss mechanism" . It is these "loss mechanism" that leads to motion. Note that in this case the reflected photon can actually have increased energy (blue shifted) or decreased energy (red shifted) depending on the electron velocity vector relative to the incident photon. A related effect is used for laser cooling of the lattice for extremely narrow line lasers (much narrower than ours).

b) Classical analogue - lets imagine a perfect spring (no internal degrees of freedom - no damping). Lets attach the spring to a finite mass car. Now throw a ball at the spring. The ball compresses the spring and since car has finite mass the car starts to move DURING the compression of the spring. As the spring reaches maximum compression and the ball comes to a rest relative to the (now moving car) the spring now starts to push the ball back to you BUT since the car is moving the ball that comes back to you has less speed than when you threw it. Where did the kinetic energy of the ball go? The "loss mechanism" in this perfect system is in fact the "Doppler shift" of the ball on the finite mass car. There is NO heating. Only energy transfer from the KE of the ball to the finite mass car through the "perfect spring" during the spring compression and decompression cycle. The finite mass of the car is the quantum mechanical analogue of the "zero point" energy of a finite system.

While the above seems like a long complex analysis of a single photon electron interaction it show where the fundamental "loss mechanism" comes from even in "perfect "lossless" systems.

When we add Thomson scattering with metals, and other loss mechanisms and even relativistic effects the above analysis still holds the key to why a system ultimately moves.

I hope this helps.

Philip Lubin

Posted on: Breakthrough Initiatives

Philip

I completely agree with your classical analogue. In the quantum mechanical system, when relativistic effect is considered, I agree there's a change in momentum in photon-electron collision. My energy-momentum equation is Newtonian since the lightsail is not traveling near the speed of light unlike an electron where relativistic effect must be considered.

Sorry but I don't see how your good explanation disproves my argument. Yes the electrons will gain momentum and energy. This does not immediately translate to a gain in momentum of the lightsail. The electrons will simply jump to a higher energy level in the atom. Then return to a lower energy level and emit a photon. That's the time the lightsail gains momentum accompanied by an increase in temperature. The radiative flux is proportionate to the 4th power of temperature (Stefan-Boltzmann law) My computations relate the increase in temperature to the increase in momentum of the lightsail.

Posted on: Breakthrough Initiatives

Here’s an example to show the conservation of energy and momentum mandates that the energy loss of reflected photon is converted to thermal energy.

Frequency of incident photon = 600 Thz

Frequency of reflected photon = 599 Thz

Momentum loss = h/c (600 – 599) x 10^12 = 2 x 10^-30 kg-m/s

Energy loss = h (600 – 599) x 10^12 = 6.6 x 10^-22 J

The photon collided with an electron and transferred its momentum and energy to the electron

Momentum loss of photon = momentum gain of electron = 2 x 10^-30 kg-m/s

p = momentum gain of electron

m = mass of electron = 9.1 x 10^-31 kg

Gain in kinetic energy of electron = p^2/2m = 2.6 x 10^-30 J

Missing energy = energy loss of photon – energy gain of electron = 6 x 10^-22 J

Almost all the energy loss of photon is missing! Energy gain of electron is negligible. Where’s the missing energy? It’s converted to thermal energy, the random motion of atoms. This increases the temperature of the material and radiation is emitted according to Stefan-Boltzmann law. The emitted photons have momenta that generate thrust in the lightsail.

Momenta of emitted photons = gain in momentum of lightsail

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Enrico

Good discussion here. Thanks again for thinking about this.

The photons are too low in energy to cause atomic electronic transitions. For a 1 micron wavelength laser the energy per photon is about 1 eV. This is too low for typical atomic electronic level transitions but even if we were using higher energy photons that would cause atomic transitions the basic physics would not change. The term "elastic" interaction needs to be carefully understood. The classical definition is that in an elastic interaction no energy is translated into internal degrees of freedom in the local system (ie no heat is generated) BUT due to the finite mass of the system the photon transfers some of its energy into translational degrees of freedom (motion) without heating (assuming no local dissipation - ie a perfect reflector) and the energy "loss" in the reflected photon goes into the translation (motion) of the (in this case) bound electron in the lattice. If you are in a metal with a conduction band the analysis is similar. The reflected photon has less energy than when incident with the reduction due to "Doppler" shifting not electronic transitions. This is also why the efficiency of the energy transfer between the photons and the reflector increases as the speed increases. The instantaneous efficiency is 2 v/c and the average efficiency is v/c assuming the reflector starts from rest relative to the laser . See the "roadmap" for a discussion of this point.

In your analysis a perfect reflector would not move as no heating would occur internally. This is not correct.

BTW - In the case of complete absorption of the photons on a reflector that absorbs all photons with no reflection the reflector will obviously heat and some additional thrust can come from the emitted (thermal) photons (assuming you can maintain a temperature differential between the front and back of the reflector) but this is typically a small effect compared to the the direct momentum transfer of the photons being absorbed.

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Philip

1 micron wavelength is near infrared. Electron transition can occur up to microwave wavelengths by Lamb shift. Electron transition to lower atomic energy levels requires emission of photon to satisfy the conservation of energy. In classical radiation theory this is interpreted as increase in temperature of the material as all matter above absolute zero temperature emit radiation.

Doppler shifting is an effect not the cause of energy loss of reflected photon. It is an effect of observations in different frames of reference. A stationary observer will see the reflected light redshifted because it sees the lightsail moving away - a receding light source. An observer on the lightsail will not see the redshift because the lightsail is stationary as far as the observer is concerned. But it will see the light from the laser on earth redshifted. Anyway, the Doppler effect is not an explanation of how the photon lost energy. What is important is a satisfactory explanation (theory with calculations) of photon-electron interaction that conserves both energy and momentum. This is what I tried to show and it seems heat is an essential part of the interaction. I'd like to see the calculations without heat that obey the conservation laws.