🚀 go-pugleaf

In article <4b9tjv$8i9@hg.oro.net>, tessien@oro.net (Ross Tessien) wrote:

> In article <4b6p20$bm7@scotsman.ed.ac.uk>, ajc@reaxp01.roe.ac.uk says...
> >
> >In article <4b6nep$gu3@ixnews2.ix.netcom.com>,
> >Alan \"Uncle Al\" Schwartz  <uncleal0@ix.netcom.com> wrote:
> >>The radiometer is under a light vacuum.  Heating of the black side of 
> >>each vane launches mass convection currents with sufficient orthogonal 
> >>component to drive the vanes.
> >
> >        what is the fine detail behind the phrase `convection currents'?
snip

> Well, all of the previous comments are excellent in this post and the 
> others addressing the question.  Let me take the phenomena one level down 
> in detail to show where the force originates in each of the two kinds of 
> radiometer.
> 
> The first kind of radiometer is a reflective one.  It requires a very 
> good vacuum and is not normally sold as a science toy so I doubt this is 
> the kind the teacher has.  But lets consider it any way.  Normally, to 
> maximize the effect, one has a black surface on one side and a highly 
> reflective surface on the other.  White paint meerly decreases the 
> effeciency of the effect but in space would work fine.
> 
> Now the energy imparted by a photon colliding with the black side of the 
> surface and being absorbed is one times its momentum.  This is given by 
> its relativistic mass times velocity, or, U = mc  where U is momentum and 
> c is the speed of light.  Using the intensity of the light, and the 
> surface area of the absorber (and the effeciency of absorption if you 
> want to get into details, but then you would need to add the radiation 
> and the problem gets ugly.  It essentially all cancels in the end, so 
> don't worry.  I think, though, that the blackbody emissions end up 
> completely canceling the thrust of the absorbed energy since energy in 
> must equal energy out)
> 
> Considering the reflective side, though, we find that the incident energy 
> strikes and reflects which yields a 2U momentum transfer.  The absorbed 
> energy, since reflectors are not perfect, again heats the material, and 
> is again emitted.  A good reflector is synonomous with "poor absorber", 
> and with "poor emitter" in radiation theory.  One is the other.  So the 
> small amount of energy absorbed by the reflective side is again cancelled 
> by its emissions.
> 
> Because Black is synonamous with "good emitter" and "good absorber", it 
> turns out that everything balances.  The black side absorbs and emits the 
> light with which it interacts, and the shiny side does the same.  No net 
> thrust yet.
> 
> But, the shiny side had those pesky photons that dumped 2U momentum and 
> if we want to be picky, the inefficiency of absorption on the Black side 
> did as well.  However, the amount of energy reflected from the shiny side 
> is clearly far greater, and so that is the origin of the momentum 
> transfer that gets the radiometer spinning.  
> 

Wrong: An ideal true vacuum radiometer will not rotate!
While it is true that an absorbed photon imparts momentum E/c,
and a reflected photon 2 E/c, in a true vacuum thermal equilibrium must
be established by radiation. By Kirchhof the white side cannot radiate,
so all the radiating is done by the black side.
This suffers recoil momentum  E/c again,
balancing the torque.
(the photon is massless, so frequency does not matter.)
Of course secondary effects will spoil the complete cancellation.
The only really existing vacuum radiometer is an earth satelite,
which must expend fuel to counterbalance radiation pressure,
(and also solar wind) to maintain its orientation.
True radiation pressure is hard to measure in a laboratory,
precisely because of radiometer effects, but it can be done.

Best wishes,

Jan