Re: E = mc^2
From: N:dlzc D:aol T:com \(dlzc\) (net_at_nospam.com)
Date: 06/18/04
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Date: Thu, 17 Jun 2004 17:06:57 -0700
Dear sal:
"sal" <believer@nospam.org> wrote in message
news:pan.2004.06.17.20.35.03.809546@nospam.org...
> [Teranews threw away my first response again, which gave me a chance to
> check it before sending...]
>
> On Wed, 16 Jun 2004 19:44:53 -0700, N:dlzc D:aol T:com (dlzc) wrote:
>
> > Dear sal:
> >
> > "sal" <believer@nospam.org> wrote in message
> > news:pan.2004.06.17.01.58.58.906360@nospam.org...
> >> [Apologies if this comes through twice. Having server trouble here.]
> >>
> >> On Wed, 16 Jun 2004 17:18:18 -0700, N:dlzc D:aol T:com (dlzc) wrote:
> >>
> >> > Dear sal:
> >> >
> >> > "sal" <SpamMeHere@foobox.com> wrote in message
> >> > news:d657f887.0406160707.13ea5fd2@posting.google.com... ...
> >> >> You are _assuming_ radiation carries momentum related to its energy
> >> >> in order to assure global momentum conservation.
> >> >
> >> > This is actually a given. Photons are a carrier for momentum
> >> > exchange between charged particles. A radiometer is an example (if
> >> > one could be found that had a "perfect" vacuum inside, anyway).
> >>
> >> Well .... I do have a small quibble with this example.
> >>
> >> Radiometers, you may recall, have a black side and a white side on
> >> their fins, and they rotate _away_ from the black side ... yet, since
> >> the photons are reflected from the white side, they actually pick up
> >> more momentum from each collision with a photon on the white side.
> >>
> >> The reason, of course, is that the rotation of the radiometer has
> >> nothing to do with momentum transfer from photons, and in fact a
> >> typical radiometer would not work if it had a perfect vacuum inside.
> >
> > They do in fact work, as "advertised". The problem is that you cannot
> > make them cheap enough (or permanent enough) to sell as product.
> >
> >> The
> >> radiometer works on the difference in _energy_ absorbed from the
> >> photons, due to a gross difference in the number of photons actually
> >> absorbed by the black surface versus the white surface. The black
side
> >> gets slightly warmer than the white side, as a result of which
> >> collisions with gas molecules in the _partial_ vacuum inside the bulb
> >> transfer more momentum between the blades and the gas when they occur
> >> on the black (warmer)
> > face.
> >> In some sense, the radiometer blades act like little jet engines.
> >
> > Not quite right. The white side gets hotter (think about what black
> > body radiation means), but cannot radiate this heat out well. The
black
> > side can radiate better, therefore is more effective at heating
adjacent
> > gas. Absorptivity and emissivity are very surface dependent.
>
> Now, one moment...
>
> The explanation I gave is something I read a long, long time ago, but I
> just double checked a more up to date "correct" explanation, and it
wasn't
> totally off in left field. A claim that the white side gets _warmer_
> would need some major substantiation!
The amount of energy arriving at a surface is equal to the amount leaving
(since "retained" is not a surface property). Therefore absorptivity +
reflectivity + transmissivity = 1. If reflectance is high, this does not
leave much room to allow heat to transmit into the fin, so the fin surface
is hot. A mirror in sunlight is hotter than a piece of glass (even painted
black). A tin roof is hotter than roof tiles.
> I'm also choking when I try to swallow an explanation that depends
> on radiation from a slightly warm object heating adjacent (transparent)
> gas -- what's it radiating that the gas is opaque to?? Last I heard,
near
> near IR, far IR, and anything else likely to come out of a slightly warm
> object passed through common gasses unmolested. As I understand it,
> common radiometers contain ordinary air (under very low pressure), which
> is quite transparent to such stuff.
Translucent might be better, since all the gasses involved are able to emit
at these frequencies. Since they can emit, they can absorb. There just
isn't a lot of mass in your average volume of air, and the fact that the
binding forces between gas molecules is very small, there is even less
ability to absorb at heat ranges. But not *zero* ability to absorb.
> At to the white side getting "hotter" -- it's in an
> environment where the (small amount) of gas in the bulb is removing heat
> from the fins by _conduction_ while they're absorbing bunches of it from
> the light source which is providing the push. So, sure, black paint
> radiates better than white -- but look, the system's a net radiant energy
> _absorber_ (obviously!) so the fact that black also _absorbs_ better than
> white seems likely to be the dominant effect.
And is equally good at reemitting, but at its black body temperature.
> Since net radiation flux is inward, it would seem surprising if the
> relative temperatures of the various parts were dominated by how well the
> energy was re-radiated.
>
> Finally, I checked out the link pmb posted. I'll take the liberty of
> repeating it here:
>
> http://math.ucr.edu/home/baez/physics/General/LightMill/light-mill.html
>
> Nice link, convincing, talks about both versions, mentions that the one
> using light pressure does turn backwards...
>
> And it also points out that the explanation I originally quoted is
> incorrect but commonly accepted. The correct explanation, first
published
> by Maxwell, is that the black side _does_ get warmer, and the warmer gas
> molecules _do_ provide the driving force, but the actual effect takes
> place entirely at the edges of the fins and beyond that the explanation
> most likely requires more math than I can bring to bear.
I'll visit it.
> >> This would work whether or not light was quantized, and whether or not
> >> light carried momentum -- radiometers only show that light carries
> > energy.
> >
> > If it didn't carry momentum, then why does it move at all? Why doesn't
> > it simply sit still and get hot?
>
> Because the black side warms up faster (radiates better, sure, but absorb
s
> better too and that's dominant in this case). See the above link for a
> more coherent and complete discussion.
>
> Presumably, the gas in the bulb must also be rotating hell-for-leather in
> the opposite direction from the fins, else angular momentum wouldn't be
> conserved...
Not recessary. In fact it is likley that the bulk gas flow follows the
fins around quite closely. Angular momentum does not have to be conserved
if energy is applied.
> >> Radiometers are typically quite crude, and there's no way they'd spin
> >> at macroscopic speeds just as a result of "photon pressure".
> >
> > A gross overgeneralization. They can in fact be made to do so. You
> > just can't buy one at the dollar store.
>
> Apparently so. I've certainly never seen such a device, but then my
> experience with radiometers is pretty limited.
>
> But ... such a radiometer would spin in the opposite direction
> from a "conventional" one, since the white side picks up
> more momentum from each photon than the black side :-)
Correct. We are arguing about "decimal places", for the most part.
David A. Smith
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