Poof goes the solar sail ?

  • Thread starter drag
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  • #1
drag
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Greetings !

Please read this short story about solar sails' physics:
http://www.spacedaily.com/news/rocketscience-03zg.html

Can the experts here provide their opinions, please !

I personally thought that was all simple and basic stuff
long since solved by physics. I mean, a photon has
energy and hence momentum which it transfers to the
sail and then it is reflected - reemmited thus again the
momentum exchange occurs only that this time the photon
and the sail push against each other (actualy kin'na smells
fishy in terms of energy if I put it this way).

But the light pressure experiment with the reflective and
absorbing surfaces clearly draws a different picture. I've
heard of it before but no details were given about the
surface (or, in fact, I think it was said that a reflecting
surface would move, which is clearly a book's or my memory's
mistake ?).

So, what actualy happens here ?!

The only possible conclusion here as I see it is that
the reflected - reemmited light "drags" the reflective
material back, but how can that be ?!

Is that really so complex that nobody thought about it before ?!

Does this actually doom the whole idea of solar sails - they'll
just evaporate if they're absorbing or produce no thrust
if they are reflecting ?!

HAS ANYBODY EVER DONE AN EXPERIMENT, EXCEPT THE ABOVE MENTIONED,
BEFORE AN ENTIRE SOLAR SAIL SPACE MISSION WAS READY FOR LAUNCH ?!

Thanks ! :smile:

Live long and prosper.
 
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Answers and Replies

  • #2
Tyger
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I don't believe Mr. Gold

Anyone want to place any bets on this one?

Photons will reflect off the whole mirror surface and transfer momentum and energy to the sail. And a reflecting sail will work twice as well as an absorbing one.
 
  • #3
drag
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Originally posted by Tyger
Photons will reflect off the whole mirror surface and transfer momentum and energy to the sail. And a reflecting sail will work twice as well as an absorbing one.
So the experiment brought as an example in the story is a lie ?
(SpaceDaily ussualy doesn't post false info.)
 
  • #4
Tyger
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Originally posted by drag
So the experiment brought as an example in the story is a lie ?
(SpaceDaily ussualy doesn't post false info.)

The Crooke's radiometer has been around a long time, and versions have been tested with a high vacuum and bright light on the reflecting surface. They are the relevant experimants.
 
  • #5
Hurkyl
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I wonder if he considered the doppler effect.
 
  • #6
HallsofIvy
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I don't know this "Thomas Gold" of Cornell but it makes you wonder what the physics department at Cornell thinks of this.

Yes, of course, a photon (or anything else) reflecting perfectly from a "sail" maintains the same energy. However, momentum, unlike energy, is a vector quantity. If the photon has momentum m intially and reflects off the sail in the opposite direction, it imparts momentum 2m to the sail. I believe that's the way it was explained to me when I was in high school (more years ago than I care to remember) and don't see anything in the article that contradicts that.
 
  • #7
I don't think conservation of momentum can be applied to massless particles such as photons. While it is true that they do have momentum associated with them, it is a different type of momentum, originating from the photons frequency.

I guess Quantum mechanics must be able to explain this.

I assume that when a photon strikes a reflective surface, it is absorbed and excites the atom absorbing it. That atom will return to its rest state, releasing a new photon having the same energy.

Since conservation of energy will hold, even at the quantum level, no energy can be imparted to the mirror, and thus it can't be accelerated.

(Maybe I'm wrong, but that seems to be the only way to make sense of the problem) [?]
 
  • #8
jcsd
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Originally posted by Ace-of-Spades
I don't think conservation of momentum can be applied to massless particles such as photons. While it is true that they do have momentum associated with them, it is a different type of momentum, originating from the photons frequency.

I guess Quantum mechanics must be able to explain this.

I assume that when a photon strikes a reflective surface, it is absorbed and excites the atom absorbing it. That atom will return to its rest state, releasing a new photon having the same energy.

Since conservation of energy will hold, even at the quantum level, no energy can be imparted to the mirror, and thus it can't be accelerated.

(Maybe I'm wrong, but that seems to be the only way to make sense of the problem) [?]

No your wrong, a photons momentum is given by p= h/wavelength. Therfore as Planck's constant is always constant an inelastic collision results in a change of wavelength.
 
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  • #9
I didn't deny that photons have momentum.
I only said that I don't think the law
of conservation of momentum,
(which I know is supposed to be universal)
would apply in this case.

Once again I said I *think*
So I might be wrong!
 
  • #10
jcsd
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What actually happens is that from the reflectors point of view the photons momentum stays the same but from a 'stationery' point of view the phtons are redshifted and thus lose momentum.
 
  • #11
YES!
Ive solved the problem for once and for all...
Here it goes...

Momentum is always conserved yes? - NO!
Momentum is always conserved in a collision? - YES! :smile:

When a photon strikes a reflective surface, or any
other surface for that matter (could be a brick wall)
It does not undergo your ordinary everyday classical
collision. Instead it is absorbed and annihilated by an
atom of the surface it strikes. This excites the atom
which will then jump back down to a more stable
state releasing a DIFFERENT photon.

Since this is not a collision, Momentum need not be
conserved, and thus no energy needs to be transfered.

So... Therefore the Mirror sail thing will NOT work!
 
  • #12
jcsd
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No, momentum is ALWAYS conserved in some form or other, when the photon is absorbed (tho' in the case of a perfect reflector though it's specifically NOT absorbed but reflected) the electron acquires the photon's momentum. Of course in this case were talking about free electrons and collison not absorbtion.

Rememberit is only the iew of one scientist that the solar sail won't work and from what I know I'm not convinced he is correct.
 
  • #13
I don't know, I think you're wrong!
Photons are not classical particles, and
so will not behave like pool balls which conserve
momentum when they collide.

Photons are quantum particles, which behave
quite differently. Photons do not collide with
things, which bounce them back.

When you look at a red sheet of paper, what is
hapenning to the photons hitting the paper??

QM says...
- The photons hitting the paper are absorbed,
they will excite atoms within the paper, the
atoms in the paper will have energy levels which
correspond to a red frequency.

The atoms in the paper will then drop back down
to their ground state emitting a NEW photon
having a red frequency.
The same applies for all matter, including
reflective surfaces.
 
  • #14
jcsd
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In QM mometum is still conserved.

Compton scattering is the QM process of the scattering of phtons by free electrons, there is no absorbtion in this. The idea of a perfect refelctor is that it doesn't absorb any of the light as opposed to it's polar opposite the perfect absorber.

The electrons in a reflective surface are free electrons so they cannot absorb the photons, they scatter them.
 
  • #15
Oh yeah... he he he
My bad (I forgot about Compton Scattering)
I guess youre right!:smile:

...

So, would your ordinary household
mirror not be perfect reflector?
 
  • #16
jcsd
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Actually the sail being a perfect refelector would, I believe use Thompson scattering but still no absorption.
 
  • #17
Whats the difference between Thompson
and Compton scattering?
Ive never heard of Thompson!
 
  • #18
jcsd
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Thompson scattering is classical and elastic, Compton scattering is non-classical (i.e. QM) and inelastic.
 
  • #19
Then why on Earth would you
call this a classical collision?
 
  • #20
Tyger
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Ace, if the photon

is reflected it will be from the whole surface of the mirror and impart momentum to the whole mirror. If it's absorbed it will only be absorbed by a little part of the mirror and it will heat that part up.
 
  • #21
drag
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Originally posted by HallsofIvy
Yes, of course, a photon (or anything else) reflecting perfectly from a "sail" maintains the same energy. However, momentum, unlike energy, is a vector quantity. If the photon has momentum m intially and reflects off the sail in the opposite direction, it imparts momentum 2m to the sail. I believe that's the way it was explained to me when I was in high school (more years ago than I care to remember) and don't see anything in the article that contradicts that.
Precisely my thoughts, but the fact that an absorbing surface
will receive momentum while a reflective one will not in
the experiment mentioned does appear to contradict this.
What's up ?

Live long and prosper.
 
  • #22
jcsd
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It's a simple elastic collision:



(p= 1m) @--> £ (at rest) total mometum = 1m


(p= -1m) <--@ £--> (p= 2m) total mometum = 1m
 
  • #23
jcsd
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btw the above diagram is not meant to be an ilustration of how the solar sail works (for a start the diagram contravenes the conservation of energy) it's just to show how the conservation of momentum works.
 
  • #24
drag
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O.K. But again, what about the experiment when the
absorbing surface is pushed and the reflective surface
is not (that is not supposed to happen then) ?

Thanks ! :smile:

Live long and prosper.
 
  • #25
jcsd
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There must be an exchange of momentum when something is reflected.
 
  • #26


Originally posted by Tyger
is reflected it will be from the whole surface of the mirror and impart momentum to the whole mirror. If it's absorbed it will only be absorbed by a little part of the mirror and it will heat that part up.

Yes, I knew that thanks! I just didn't realize that a perfect reflector is any different from your every day normal mirror that absorbs photons.
 
  • #27
FZ+
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Can someone sort this out then... if as a perfect reflector, the photons are reflected at the same relative speed (c, of course) and the safe frequency, where does the additional KE of the sail come from?
 
  • #28
Hurkyl
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They're not at the same frequency; don't forget the doppler effect. If the sail is moving towards the sun, the photons act to slow the sail's motion... but the photons get blueshifted upon reflection. If the sail is moving away from the sun, the photons act to speed up the sail, but the photons are being redshifted by the reflection.
 
  • #29
Tyger
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The sail KE

Originally posted by FZ+
Can someone sort this out then... if as a perfect reflector, the photons are reflected at the same relative speed (c, of course) and the safe frequency, where does the additional KE of the sail come from?

comes from the photons. But each photon only imparts a tiny amount of KE to the sail compared to the momentum. And we can always choose to be in the rest frame of the sail by riding on it, so we would see each photon as being reflected with almost the same energy and almost the exact opposite momentum.
 
  • #30
drag
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Originally posted by Hurkyl
They're not at the same frequency; don't forget the doppler effect. If the sail is moving towards the sun, the photons act to slow the sail's motion... but the photons get blueshifted upon reflection. If the sail is moving away from the sun, the photons act to speed up the sail, but the photons are being redshifted by the reflection.
What do you mean ?
Aren't the reemited photons at the same frequency (total
reflection) ? Their original frequency as they hit the
sail is different of course from that of the source - the
Sun, but why should that effect the sail in any way (if
indeed perfect reflection produces no thrust) ?

Live long and prosper.
 
  • #31
Since the photons impart momentum to the sail,
energy must be imparted to the sail as well.

E = h[nu] , so in order to impart energy, they
must experience a drop in frequency, hence red-shift.
(assuming the sail is traveling away from the sun)
 
  • #32
wimms
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In stable orbit around sun there is zero effect then?

Photons between powerful laser mirrors bounce back millions of times, mirrors don't get bent by light pressure, or do they?

Is the sail photon-energy based at all? Solar wind is full of all sorts of stuff much heavier than photons. Perhaps its reflective surface is made to avoid overheating from light and main 'fuel' is matter blowing from the sun? So maybe 'discovery' isn't exactly relevant?

http://www.genesismission.org/science/module4_solarmax/SolarWind.html [Broken]

edit: oops, apparently I'm way off. Solar wind is supposed to be mere 1% of energy compared to light pressure..

I'm confused by redshift. In frame of sail, isn't it simply incoming photon that becomes outgoing photon without any freq change? If photon would impart energy into a mirror, then lasers should show some redshifting, shouldn't they?
 
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  • #33
russ_watters
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Originally posted by Ace-of-Spades
Since the photons impart momentum to the sail,
energy must be imparted to the sail as well.

E = h[nu] , so in order to impart energy, they
must experience a drop in frequency, hence red-shift.
(assuming the sail is traveling away from the sun)
Wrong. This has been explained before: (-1)^2=1. A perfectly reflected (by definition) photon will have exactly the same energy as before it was reflected.

And as stated before, an absorbed photon emparts exactly half of the momentum of a reflected one because the initial velocity is v for both while the ending velocity is 0 instead of -v.

v-0=v
v-(-v)=2v

e=mv^2=m(-v)^2

What that equation also describes is a perfectly elastic collision between a ball and a wall. It works pretty much the same way. The ball leaves the wall with exactly the same speed and the opposite direction (-v). And since energy is v^2, the negative goes away, giving the exact same energy as it had before.
 
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  • #34
Hurkyl
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What do you mean ?
Aren't the reemited photons at the same frequency (total
reflection) ? Their original frequency as they hit the
sail is different of course from that of the source - the
Sun, but why should that effect the sail in any way (if
indeed perfect reflection produces no thrust) ?

Nope! The frequency of the photons just before they hit the sail is different from their frequency just after they're reflected. The proof is identical to the derivation of the Doppler effect. Numerically, the result is identical to the situation where the mirror absorbs then re-emits the photon. (Isn't that what perfect reflection is, anyways?)
 
  • #35
russ_watters
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Originally posted by Hurkyl
Nope! The frequency of the photons just before they hit the sail is different from their frequency just after they're reflected. The proof is identical to the derivation of the Doppler effect. Numerically, the result is identical to the situation where the mirror absorbs then re-emits the photon. (Isn't that what perfect reflection is, anyways?)
Different issue that I hadn't considered. Doppler effect. In my example, the wall was stationary, so there is no doppler effect. So this basically means that the faster the sail goes, the less effecient it gets?
 

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