Poof goes the solar sail ?

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  • #26
Ace-of-Spades


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 realise 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
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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
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 travelling away from the sun)
 
  • #32
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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 travelling 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 exaclty 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?
 
  • #36
drag
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Greetings !
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?)
Are you talking about the frequency that is relative to the sail ?
What about the energy transfered then ? Is this the energy that has to do with the momentum exchange ?

AoC, like russ explained there should be an exchange of momentum.

wimms, Solar Wind propulsion is completely different. It
envolves using EM fields to capture the energetic particles
and use their momentum. I don't know about the energy relation
but I thing the potential momentum relation is higher. Further
more, there's now way (that we know of for now, at least) to
reflect all of the Sun's light. The solar sail only reflects
several frequencies out of the whole spectrum that's emmited by
the Sun (though the area of visual frequencies is the main - more enegized area of the spectrum).

Live long and prosper.
 

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