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Photon Powered Windmill |
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| Jun23-05, 04:45 PM | #1 |
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Photon Powered Windmill
Okay, I got one for you all.
Years ago, in school, I saw a propellor shaped thing, mirror side up, black side down. It was resting on a pin in, what I assume was, a vacuumed bell jar. It spins in sunlight as per what you'd expect... But what is the physics behind this? As light never slows due to energy loss, and light is massless, it can't be through that mechanism. The light never changes wavelength is no energy is lost that way. Maybe the force which drives it is proportional to the light absorbed by the mirror and blown out by the black surface? I'd like to know for sure thanks. |
| Jun23-05, 04:50 PM | #2 |
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Light moves in little particles called photons which have no mass but momentum < (the part I dont get!) so they basicly hit the blades and give them momentum to move... maybe someone else can give more detail.
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| Jun23-05, 05:08 PM | #3 |
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But if the momentum is transferred, we have loss of energy in the wave. Something that can never happen unless you have a change in frequency. It'd be like saying you bounce a ball that bounces back with the exact energy it hit an object with, and still move the object. Hmm, creation of energy is where I have a problem.
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| Jun23-05, 06:01 PM | #4 |
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Recognitions:
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Photon Powered Windmillsci.physics.faq http://math.ucr.edu/home/baez/physic...ight-mill.html howstuff works http://science.howstuffworks.com/question239.htm the answer turns out to be an effect known as thermal transpiration as mentioned in the FAQ. It takes a more sophisticated instrument to directly measure the pressure of light (but such an instrument is possible - it mostly involved a better suspension and a better vacuum than is found in the typical radiometer). As far as accounting for energy goes, note that light is reflected by a mirror, and absorbed by a black surface. Reflection light from a non-moving mirror does not change the frequency (or energy) of the light, it just transfers momentum. However, reflection from a moving mirror does change the frequency (and hence the energy) of the light via the relativsitic doppler shift. When light is absorbed by a black surface, all of the energy and momentum in the light is absorbed by the surface. |
| Jun23-05, 11:38 PM | #5 |
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since momentum is [tex] p = m v [/tex] if either mass [itex] m [/itex] or velocity [itex] v [/itex] were zero, so would momentum [itex] p [/itex] be zero. what light doesn't have is rest mass and that's because its velocity is [itex] c [/itex]. since [tex] E = m c^2 [/tex] and [tex] E = \hbar \omega [/tex] then the mass of the photon is [tex] m = \frac{\hbar \omega}{c^2} [/tex]. but since relativistic mass is also [tex] m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}} [/tex] where [itex] m_0 [/itex] is the particle's rest mass, and [itex] v [/itex] is the particle's velocity, then [tex] m_0 = m \sqrt{1 - \frac{v^2}{c^2}} [/tex] and because [itex] v = c [/itex], then [itex] m_0 = 0 [/itex]. photons have mass. that's the only way they can have momentum with a finite velocity. r b-j |
| Jun24-05, 01:54 AM | #6 |
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Recognitions:
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See the sci.physics.faq Does light have mass? for a fuller discussion. I'll take the liberty of posting a brief quote: http://math.ucr.edu/home/baez/physics |
| Jun24-05, 09:03 AM | #7 |
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Mentor
Blog Entries: 1
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Momentum is defined as: [tex] p = \frac{m v}{\sqrt{1 - \frac{v^2}{c^2}}} [/tex] only equalling the Newtonian form for small speeds. |
| Jun24-05, 11:42 AM | #8 |
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Edit: as noted by Pervect,
The device is called a radiometer. It doesn't work by momentum transfer of photons, that wouldn't be strong enough with room lighting. There is neearly but not completely a vacuum in the device. The dark sides of the vanes get warmer and there is an increased pressure on that side by the gas. The device spins away from the black side toward the white. If it worked by photon pressure it would spin away from white toward black. Hope that helps... |
| Jun24-05, 12:21 PM | #9 |
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Admin
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In a radiometer, the dark side absorbs photons (in the optical range) which produce thermal energy (atomic vibration) in the dark surface material. The light side is cooler because photons are reflected.
The air molecules strike both surfaces, but on the dark side the vibrating atoms in the dark material transfer some energy/momentum to the air molecules and this causes the radiometer to spin. So it is simply a transfer of energy/momentum involving the air molecules. Photons have momentum, p = E/c = h/[itex]\lambda[/itex], but not rest mass. Refer to the work by Arthur Compton, and the effect known as Compton scattering. This is however a small quantity compared to the momentum of a molecule, even at room temperature. |
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