Understanding Photon Collisions with Solar Sails

  • #1
gonzo
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I have some questions about solar sails. Basically, what happens to the photon?

I am under the impression that the "sail" is a giant mirror that reflects photons striking it. Since photons have momentum, they impart some of this momentum to this mirror. Is this correct?

But that must mean that each photon loses momentum/energy. Does this just mean that the reflected photons will always be of a lower frequency?

Does anyone know what determines during a photon collision with another particle how much momentum is transferred as opposed to other collision effects?

Thanks.
 
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  • #2
I don't think it's a mirror intended to reflect photons. It's a material that "captures" that momentum. The photon is absorbed & perhaps re-emited away as heat (lower frequency). (Sure, there's some reflection like anything else.)
 
  • #3
Phobos said:
I don't think it's a mirror intended to reflect photons. It's a material that "captures" that momentum. The photon is absorbed & perhaps re-emited away as heat (lower frequency). (Sure, there's some reflection like anything else.)

the most efficient solar sail would be reflective, like a mirror.
if it bounces the photon directly back then it captures about twice as much momentum as if it absorbs the photon
(this is just conservation of momentum)
 
  • #4
we can do a numerical example, assume we have a square meter of mirror-like mylar that is impacted by a kilowatt of sunlight. perpendicularly.

if it is perfectly reflective, the force in Newtons will be 2000 W/c

2000/3E8 Newtons = 666E-8 = 6.7 E-6 = 6.7 microNewtons.

but if it is perfectly absorbent (black) then the force will be 1000 W/c

3.3 microNewtons.
 
  • #5
gonzo said:
But that must mean that each photon loses momentum/energy. Does this just mean that the reflected photons will always be of a lower frequency?

the doppler effect means that the photon that is reflected has longer wavelength (from the standpoint of a stationary observer who sees the mirror moving and being accelerated) and so the light gives up energy


but if you position yourself on the mirror and say that the mirror is at rest
then the light just bounces off, in a direction depending on how the mirror is oriented, and momentum must be conserved.

I guess the picture is pretty much like a sail. One might tilt it 45 degrees to the sun and use it to gain orbital speed. then the analysis would be slightly more complicated than if the light is striking perpendicular
 
  • #6
I guess all there's left to do is find out which material has the best reflectiveness to density ratio.
 
  • #7
check said:
I guess all there's left to do is find out which material has the best reflectiveness to density ratio.
Mylar gift wrap!
:smile:
 
  • #8
I've been thinking about this now, and I can't help thinking that it still seems like you are getting free energy if the photon doesn't lose energy.

It seems to me that if the photon is just reflected at the same frequency that you could theoretically construct a free energy machine that consisted of a "photon trap" with two mirrors just bouncing photons back and forth between them, and use the slight movement of the mirrors from the photons hitting them to gain energy.

If the photons never lost energy, they could in theory bounce back and forth forever.

What am I missing?
 
  • #9
gonzo said:
I've been thinking about this now, and I can't help thinking that it still seems like you are getting free energy if the photon doesn't lose energy...
...What am I missing?

marcus said:
the doppler effect means that the photon that is reflected has longer wavelength (from the standpoint of a stationary observer who sees the mirror moving and being accelerated) and so the light gives up energy
...

say you are the stationary observer and the light energy (as well as the mirror motion) is being measured on your intruments on your terms

then the energy books are going to balance because every little bit
of kinetic energy that is imparted to the mirror
is balanced by a doppler downshift in the energy of the light after it hits the mirror.

I am not proving this to you with algebra, or even declaring it to you as some kind of "authority". I just want to throw the idea your way so you can see if it works for you.

I think, for me, it works. I don't even have to check the algebra.
as the mirror moves back faster and faster it soaks up more and more energy from the light that is bouncing off it
and that fits the kinetic energy formula with its 1/2 m v2


of course in order to take energy the mirror has to be free to move
(otherwise it is like the wall you bounce tennis balls off of, they keep their energy almost because the wall is connected to the huge mass of the earth
so the wall barely moves when it is hit and the amount of energy it absorbs is negligible)

so it looks like you are considering sometimes a special case where the mirror is almost not moving----maybe because you just begin to accelerate it by shining the light on it. You ask how then can it absorb energy because there is almost no doppler shift?
Well as the mirror is just beginning to move it is also absorbing almost no kinetic energy.

when V is zero (at the very outset) then the Delta E is zero, because the kinetic energy has the V-square term.

I think you should consider a more typical case where the mirror is already moving some.
 
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  • #10
thanks marcus
 
  • #11
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  • #12
I thought the "sail" was powered by solar "wind" {particles} and not light

http://science.nasa.gov/ssl/pad/solar/sun_wind.htm
 
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