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Solar Sailing

  1. Mar 12, 2008 #1
    The photon comes in with momentum p and leaves with the same momentum in the opposite direction. The sail picks up a momentum of 2p in the original photon's direction. Energy conservation states that:

    [tex]h\nu_1=h\nu_2+\frac{(2p)^2^}{2m}=p_{1}c=p_{2}c+\frac{(2p)^2^}{2m}[/tex]

    Isn't |p1|=|p2| thus violating energy conservation?
     
  2. jcsd
  3. Mar 12, 2008 #2

    dst

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    I remember reading about this and it turns out it was XKCD's blag:

    http://blag.xkcd.com/2008/02/15/the-laser-elevator/

    The Laser Elevator
    February 15th, 2008

    Solar sails suck.

    In a 2002 paper, Laser Elevator: Momentum Transfer Using an Optical Resonator (available at your local school/library, possibly electronically — J. of Spacecraft and Rockets 2002), Thomas R. Meyer et. al. talk about a neat way to get a lot more speed out of light reflection than with a regular solar sail. The basic physics are pretty simple, and it’s a fun subject to think about.

    When a photon hits a solar sail, it gives the sail momentum. If the photon has momentum P and bounces off a stationary sail, it looks like this:

    Think of where the energy is in this system. Before it hits, the photon has energy E. After it bounces, the photon still has roughly energy E. But the sail’s moving, so where did it get its kinetic energy? (Remember, energy — unlike momentum — has no direction.)

    The answer lies in the word “roughly”. The photon loses a tiny fraction of its energy to Doppler shifting when it’s reflected, but only a tiny fraction. It is this tiny fraction that goes into pushing the sail. This is a phenomenally small amount of energy — far less than a percent of what the photon has. That is, not much of the photon’s energy is being used for motion here.

    This is why solar sails are so slow. It’s not that light doesn’t have that much energy, it’s that it has so little momentum. If you set a squirrel on a solar sail and shone a laser on the underside, do you know how much power would be required to lift the squirrel? About 1.21 gigawatts.

    This is awful. If we were lifting the squirrel with a motor, railgun, or electric catapult, with 1.21 gigawatts we could send it screaming upward at ridiculous speeds.

    This is where Meyer and friends come in. They’ve point out a novel way to extract momentum from the photon: bounce it back and forth between the sail and a large mirror (on a planet or moon, perhaps).

    With each bounce, the photon loses a little more energy and adds another 2P to the sail’s momentum. The photon can keep this up for thousands of bounces — in their paper, Meyer et. al. found that with reasonable assumptions about available materials and a lot of precision, you could extract 1,000 times the momentum from a photon before diffraction and Dopper shifts killed you. This means you only need 1/1,000th the energy to levitate the squirrel — a mere megawatt.

    This isn’t too practical for interstellar travel. It requires something to push off from, and probably couldn’t get you up to the necessary speeds. It may, they suggest, be useful for getting stuff to Pluto and back, since (somewhat like a space elevator) it lets you generate the power any old way you want (a ground nuclear station, solar, etc). But more importantly, it’s kind of neat — it helped me realize some things about photon momentum that I hadn’t quite gotten before. It’s like Feynman says, physics is like sex — it may give practical results, but that’s not why we do it.

    Now we’ll let things get sillier. I spent a while trying to brainstorm how to use this with a solar sail (that is, using the sun). I imagined mirrors catching the sun’s light and letting it resonate with a sail.

    But you really need lasers for this — regular light spreads out too fast. Maybe a set of lasing cavities orbiting the sun …

    Supplemented by a Dyson sphere …

    And since by this point we’ll probably have found aliens …

    Why settle for interstellar communication when you can have interstellar war? And we could modulate the beam to carry a message — in this case, “**** YOU GUYS!”
     
  4. Mar 12, 2008 #3
    But is the momentum of the photon before and after truly equal.

    Isn't it true that [tex]p=\frac{h}{\lambda}[/tex], thus making the momentum depend on wavelength which changed due to a Doppler shift.
     
  5. Mar 13, 2008 #4

    pam

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    It's not a Doppler shift, but the photon just loses E and p like any other projectile.
    This also causes a change in its wave length.
     
  6. Mar 13, 2008 #5
    Energy Conservation:[tex]h\nu_1=h\nu_2+\frac{(2p)^2^}{2m}=p_{1}c=p_{2}c+\frac{(2p)^2^}{2m}[/tex]

    Momentum Conservation:
    [tex]p_1=p_2+p_{sail}[/tex]

    [tex]\frac{h\nu_{1}}{c}=\frac{h\nu_{2}}{c}+p_{sail}[/tex]

    If we assume that it is a totally elastic reflection then the photon must have momentum p_1 in the opposite direction and the sail must have momentum 2p_1 in the initial direction. This means that the wavelength of the photon should stay the same (the magnitude of the momentum is the same). However, from energy conservation we see that the photon should lose some energy and thus change its wavelength in order to give to the sail as kinetic energy. Where am I going wrong?
     
  7. Mar 13, 2008 #6

    pam

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    If the sail recoils, the reflected photon will not have the incident momentum.
    Your equations would show that if you didn't assume a mistake.
    Trust your equations over your intuition.
     
  8. Mar 13, 2008 #7
    So then the sail momentum gain isn't really exactly 2p right? This is in order to maintain momentum conservation.

    Can you explain why this isn't a simple mechanical reflection where the incident beam gets reflected with exactly the same momentum backwards and the target gains 2 times that momentum?
     
  9. Mar 13, 2008 #8

    pam

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    If the sail RECOILS, the reflected photon will not have the incident momentum.
     
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