# Photon Powered Windmill

1. Jun 23, 2005

### MarsGhost

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.

2. Jun 23, 2005

### JohnnyTheFox

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.

3. Jun 23, 2005

### MarsGhost

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.

4. Jun 23, 2005

### pervect

Staff Emeritus

sci.physics.faq
http://math.ucr.edu/home/baez/physics/General/LightMill/light-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.

Last edited: Jun 23, 2005
5. Jun 23, 2005

### rbj

perhaps the reason you don't get it is because it isn't true. i have read this same non-fact (that photons have no mass, but they have momentum) before on this forum and i wonder how this is taught nowadays.

since momentum is

$$p = m v$$

if either mass $m$ or velocity $v$ were zero, so would momentum $p$ be zero. what light doesn't have is rest mass and that's because its velocity is $c$.

since $$E = m c^2$$

and $$E = \hbar \omega$$

then the mass of the photon is $$m = \frac{\hbar \omega}{c^2}$$.

but since relativistic mass is also

$$m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}$$

where $m_0$ is the particle's rest mass, and $v$ is the particle's velocity, then

$$m_0 = m \sqrt{1 - \frac{v^2}{c^2}}$$

and because $v = c$, then $m_0 = 0$.

photons have mass. that's the only way they can have momentum with a finite velocity.

r b-j

6. Jun 24, 2005

### pervect

Staff Emeritus
This is not correct - photons do not have mass. They do however, have energy.

See the sci.physics.faq

Does light have mass?

for a fuller discussion.

I'll take the liberty of posting a brief quote:

and encourage the interested reader to read the entire FAQ article (and some of the articles referred to by this article, which can be found at the sci.physics.faq homepage:

http://math.ucr.edu/home/baez/physics

7. Jun 24, 2005

### Staff: Mentor

In line with pervect's comments, I think it's become standard to use the term mass (m) to refer to what used to be call "rest mass". (So if you mean relativistic mass, you have to say relativistic mass.)

Momentum is defined as:

$$p = \frac{m v}{\sqrt{1 - \frac{v^2}{c^2}}}$$

only equalling the Newtonian form for small speeds.

8. Jun 24, 2005

### Antiphon

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...

9. Jun 24, 2005

### Staff: Mentor

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/$\lambda$, 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.

10. Jul 16, 2005

### Ouabache

This thread seems like déjà vu ? :uhh: 1, 2, 3