Nuts and Volts - February 2013

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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