Solar Sail will not work?

[GAHD]

*sigh* ok, I'll try a different approach:

 

Solar sails do not gain velocity from bouncing photons; they gain velocity from bouncing massive particles, and by absorbing a portion of that massive particle's velocity.

 

A photon, being a massless waveform, cannot impart velocity via 'bounce' unless some of it's energy is absorbed and converted into kinetic energy. If such an absorption/conversion/reemission phenominan were to occur, the photons would all red-shift due to the loss of energy from the bounce.

 

Momentum *is* energy, *is* heat, and *is* rays; just in various diferent structures. Conservation of energy applies to *all* of these at once, they *all* obey the same basic laws. For a sail to get "double energy" from reflecting a photon, that would imply 'free energy' and break the laws of conservation.

 

Follow now?

 

Edit: P.S. I hope the explanation is useful if not ask specific questions. :surprise:

[Kayra]

*sigh* ok, I'll try a different approach:

 

Sorry GAHD, last thing I wanted was to increase your frustration level, but I fear you still have not convinced me. :surprise:

 

Solar sails do not gain velocity from bouncing photons; they gain velocity from bouncing massive particles, and by absorbing a portion of that massive particle's velocity.

 

Nope. Light is energy. Energy and mass are interchangeable. It just takes a lot of energy to show up as a small amount of mass.

While light is a massless particle, and therefore can travel at the speed of itself (sorry, could not resist), at the moment of reflection, the energy is absorbed by an atom, and then retransmitted (the covalent thing you were speaking of earlier). For that instant that it was absorbed, it increased the mass of the atom slightly, and when transmitted back in the opposite direction, the atom got slightly lighter (this is my own visualization, and the physicists in the forum should be able to straighten me up if I am wrong).

 

http://www.science.ca/askascientist/viewquestion.php?qID=699

 

A photon, being a massless waveform, cannot impart velocity via 'bounce' unless some of it's energy is absorbed and converted into kinetic energy. If such an absorption/conversion/reemission phenominan were to occur, the photons would all red-shift due to the loss of energy from the bounce.

 

Why would it red shift? No energy is lost in reflection except due to imperfections in the mirror or quantum tunneling of the energy (both of wich result in the energy being converted to heat). Someone help me out here.

 

Momentum *is* energy, *is* heat, and *is* rays; just in various diferent structures. Conservation of energy applies to *all* of these at once, they *all* obey the same basic laws. For a sail to get "double energy" from reflecting a photon, that would imply 'free energy' and break the laws of conservation.

 

Follow now?

 

True, momentum, heat, radiant are all forms of energy, and they do obey the law of conservation of energy.

 

The sail gets double the energy from reflecting the photon for exactly this reason. I only wish I could explain this properly.

 

Lets try another thought experiment

 

You and I are in space and playing catch with a rock

I throw the rock at you and you catch it.

You begin to move farther away from me as a result (action and reaction). Now you decide to throw the rock back to me. as a result, you have just doubled your speed. No "Free energy" involved here, and no laws broken.

 

The first law of thermodynamics is based upon the law of conservation of energy as it applies to heat only.

 

Does that help?

[GAHD]

If we're playing catch in space, you throw that rock at me, I catch it, both of us are pushed away from each other, and if I throw it back just as hard and you catch it...But (And this is the big thing) the energy in our muscles at that point has been converted into kinetic energy. That is why in the rock thowing example it *does* work; with a solar sail there is no such imput of chemical energy. If a photon is reflected, it simply reflects and that is the end of the reaction; unless it red-shifts.

 

Why would the photons red-shift? Red-shifting implies a loss(transfer) of energy, while blue shifting indicates and increase(transfer) in energy, because high-frequency waves have more energy than low-frequency waves. In order for a photon to apply *any* kinetic force it has to loose that much energy itself. If this is *not* true you could put 2 mirros facing eace other, trap a photon in between the two, and hook them up to 2 spring-loaded slide-generators to create a perpetual motion macheine (because the force DOUBLES each time the photons reflect!) which violates conservation of energy laws.

[Kayra]

If we're playing catch in space, you throw that rock at me, I catch it, both of us are pushed away from each other, and if I throw it back just as hard and you catch it...But (And this is the big thing) the energy in our muscles at that point has been converted into kinetic energy. That is why in the rock thowing example it *does* work; with a solar sail there is no such imput of chemical energy. If a photon is reflected, it simply reflects and that is the end of the reaction; unless it red-shifts.

 

Analogies bite :surprise: I had only meant to show why you would get twice the energy from a reflected as opposed to an absorbed photon.

 

Why would the photons red-shift? Red-shifting implies a loss(transfer) of energy, while blue shifting indicates and increase(transfer) in energy, because high-frequency waves have more energy than low-frequency waves. In order for a photon to apply *any* kinetic force it has to loose that much energy itself. If this is *not* true you could put 2 mirros facing eace other, trap a photon in between the two, and hook them up to 2 spring-loaded slide-generators to create a perpetual motion macheine (because the force DOUBLES each time the photons reflect!) which violates conservation of energy laws.

 

Proof again that learning is a difficult process, but always a rewarding one.

 

You are quite correct. There must be a loss.

The laws of thermodynamics DO have to apply here as well, including the Carnot and 2nd law. If not, we could convert heat to radiant energy and get all the free power we would ever want.

 

Nor sure why I never saw this before GAHD, but it makes sense.

 

The amount of redshift wold have to be pretty darned small or else it would have been noticed before.(it would be pretty damned obvious). This means that the amount of momentum transfered would have to be MUCH smaller then what is predicted.

 

I am going to need to think about this for a while GAHD, thanks for the headache ;)

[Kayra]

http://www.newscientist.com/article.ns?id=dn3895

[Kayra]

I keep re-reading this part from the origional article

 

http://arxiv.org/html/physics/0306050

 

"From a formal point of view, it is clear that one could not equate radiative momentum content with Newtonian momentum. Newtonian momentum is Mv, clearly a vector, while the momentum attributed to radiation is E/c, a scalar, since E is a scalar and c is a universal constant of nature. When an amount of energy E is captured as heat energy in a body from the light, this amount is thereby converted into a vectorial quantity, moving with the velocity of the body. It is only at this stage that this vectorial quantity can be compared with Newtonian momentum. How much of the radiant energy is absorbed depends not only on the amount of radiation that is directed towards the body, but also on the temperature of the body and the difference of that to the average radiation temperature striking it. This is defined as the temperature a black body would have when equilibrated in the radiation environment to which it is exposed. It is this consideration that brings the radiation result into compliance with Carnot's rule and thus with the amount of free energy that can be obtained from a source of heat. The mass added to the body is given by the equivalent relativistic mass of the energy absorbed, and the radiation pressure is the force we would deduce as necessary to change the momentum of the body by the observed amount. If the body is a perfect mirror or reflector of all incident energy, instead of a black body, then the energy absorbed is zero and so the radiation pressure is zero also. The same is true for any body, when it has reached temperature equilibrium with the radiation to which it is exposed.

"

 

It is starting to make sense.

[GAHD]

Too bad it died before anything happened.

 

Hope the headach was worth it. :surprise:

 

Edit:P.S. Arxiv is freakin awesome, I've loved that place for a long time

[Kayra]

Too bad it died before anything happened.

 

Hope the headach was worth it. ;)

 

Yep, it really is.

 

Headaches give me an excuse to crack a beer open and ponder life.

Bring it :surprise:

[InfiniteNow]

Hmm

I thought he was trying to say that if you could reflect 100% of the radiation, you get work with no losses.

 

Old thread, but I just realized I missed a point. :evil:

 

 

You are not getting work with no losses. The "losses" occur in the sun itself... all that gurgling, bumping, and grinding is what gives off the radiation energy which pushes the sail. Before the radiation went out, there would be "work" done local to the sun. So, although the solar sail is very close to 100% reflective, there is work being done to push it, and that comes from the sun itself.

 

However, if the sail moved without light or any other source of force, THEN there'd be a problem.

 

 

I wonder what would happen if you put sunscreen on a solar sail. :hihi:

[Qfwfq]

I'd say the emission mechanism is quite aside from the point, actually. The fact is that Thomas Gold doesn't seem to have his physics very straight, as well as that he's knocking down a strawman. I could go into more detail if I had time but I'm not sure it's worth it for such an old thread.

[Pyrotex]

...In order for a photon to apply *any* kinetic force it has to loose that much energy itself. If this is *not* true you could put 2 mirrors facing eace other, trap a photon in between the two, and hook them up to 2 spring-loaded slide-generators to create a perpetual motion macheine (because the force DOUBLES each time the photons reflect!) which violates conservation of energy laws.

BINGO!!

 

Ok, this is the kind of thinking, the kind of "thought experiment", that suffices as "proof". GAHD wins on this one.

[Qfwfq]

OK here come the tricky quiz, suppose you're standing next to the mirror, so it's still in your coordinates. Then the photon comes along, at exactly right angles, and is reflected. Now:

 

• In your coordinates the photon's energy doesn't change.
   
   
  
• The photon reverses its momentum, which means a change equal to twice the modulus, which is equatable to a force applied to the mirror for a brief time, their product giving 2p.
  

 

Is this a contradiction, or is it perfectly alright?

 

:phones:

[GAHD]

it's allright, because it's heat energy, not motion.

[Qfwfq]

because it's heat energy, not motion.

Energy must be conserved, momentum must be conserved, and you can't really talk about heat in the example given.

[CraigD]

OK here come the tricky quiz, suppose you're standing next to the mirror, so it's still in your coordinates. Then the photon comes along, at exactly right angles, and is reflected. Now:

• In your coordinates the photon's energy doesn't change.
  
• The photon reverses its momentum, which means a change equal to twice the modulus, which is equatable to a force applied to the mirror for a brief time, their product giving 2p.
  

Is this a contradiction, or is it perfectly alright?

The first given - In your coordinates the photon's energy doesn't change – isn’t valid. The mirror appears to you accelerated, and energy appears conserved, so the energy (frequency) of the reflected photon is very slightly less than the incident one. Only if the mirror were not accelerated would their energies be the same.

 

The difference is very small, and mirrors complicated, shaky things, so, practically, some reflected photons (the “on the upswing” ones ) would gain a energy, while others lost more than calculated. The energy loss due to the acceleration of the mirror would be an average effect.

 

The calculations sound like challenging fun. The question suggest something interesting: a vibrating mirror should have a slight color pattern as a result of the standing wave of vibration on its surface. Doppler shift, ignoring the momentum of the photons, should dominate this effect, but require a slight correction due to the mirror’s net acceleration.

 

The problem gets really fun when you consider how the mirror is vibrating – via the exchange of virtual photons of magnetic force between protons and electrons within and between atoms - a gigantic exercise of quantum electrodynamics, way beyond my ability to begin calculating.

[Qfwfq]

The first given - In your coordinates the photon's energy doesn't change – isn’t valid. The mirror appears to you accelerated, and energy appears conserved, so the energy (frequency) of the reflected photon is very slightly less than the incident one. Only if the mirror were not accelerated would their energies be the same.

I knew you or some spoil-sport would come along and say that! :rant:

 

It's obvious, if the photon's momentum p isn't small and the mirror's mass isn't large, but it's a second order term and doesn't answer the question at all, certainly not in the [math]\norm p\rightarrow 0[/math] and [math]\norm m\rightarrow\infty[/math] limit.

 

As for the mirror vibrating, this simply contradicts the hypothetical premise of the mirror being still, so it doesn't count in this first tiny step of an analysis. Even less, of course, in the discussion of the formal matter of principle.

 

Another nitpick to address would be the quantum issues for considering it to be a "good reflection" that doesn't heat the mirror, which boils down to it being an elastic collision between the photon and the whole, overall rigid mirror.

 

:cup: