energy carried in and the momentum is E = pc where c is the
velocity of light. It is also true that a change in momentum gives rise
to a force. It’s as simple as F(force) = delta p / delta t. (I'm 80 years
old and don’t have the “tech” to put these equations in the right
way.) Anyway, consider what happens on the black side.
The light comes in and is absorbed. The change in momentum
is just whatever momentum the radiation had and this results in a
force. Now consider what happens on the shiny side. Remember
that momentum is a vector quantity. You have twice the change in
momentum on the shiny side as on the black side and twice the
force.
Remember the change from going in to going out is twice that
of going in and stopping. The radiometer vanes should rotate
towards the black side. I don't know that this has ever been
observed. The forces are very small.
Well, thanks for letting this old man have his say.
Dean Kaul
Response: Thanks for the note, Dean.
One reason I have the radiometer on my desk is because there
are several theories as to why it works or shouldn't work the way it
does.
We could look at the wave nature of light or the particle nature
— as you have — or delve into quantum physics. I like your argument,
as well. Yet another reason to keep the bulb where it is!
Bryan Bergeron
I decided to delve into how a Crookes’ radiometer (light-mill)
works a bit more.
I think I came across a definitive paper on the subject. Just go
to Google and type in "How does a light-mill work?" It mentions that
you need a partial vacuum.
It also mentions that the explanation I gave that the warmer air
near the dark side of the vane is warmed, expands, and pushes the
vane forward is wrong. However, I don’t feel too bad because the
article says that the Encyclopaedia Britannica still gives this
erroneous explanation.
Out of curiosity, I did a calculation of the radiation forces in a
perfect vacuum.
I took the intensity of the input light to be 10 solar constants.
(I'm sure an arc lamp could do it a lot better.) Anyway, with an input
intensity of 13. 6 kilowatts per square meter, the force on an
absorbing 1 cm square would be 16. 3 billionths of an ounce.
I believe it when they say, "The experiment is very difficult.”
Dean Kaul
Response: Yes, it is a partial vacuum.
I actually built a few of these things in high school physics.
We had a mercury vacuum pump — things didn't start happening
until most of the atmosphere was evacuated. The most difficult part
of construction is balancing the vanes so the contraption sits level on
the needle bearing.
Glad you find the 'simple' device fascinating.
Bryan
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