What is the paradox in Planck's Law?

[lightarrow]

I see the sun's light as white, because the human visual system adjusts to whatever light source is present in determining the color of objects.

Ok, forget the Sun. You have two light sources (lamps) which you can approximate as blackbodies; one peaks at 502 nm, the other at 885 nm. What colour do you see them? Do you see them of the same colour?

 

[Redbelly98]

Are you referring to the peak of the radiant flux per unit wavelength, or per unit frequency?

 

[lightarrow]

Are you referring to the peak of the radiant flux per unit wavelength, or per unit frequency?

Exactly. But is it always specified in the books of physics, when they talk about light sources?

 

[Almanzo]

With quantum mysteries, it is sometimes a good idea to see if one can state a similar problem in the classical domain.

Suppose we have a bag full of ball bearings in many different but macroscopic sizes. (A few millimeters to a centimeter.) We proceed to measure their diameters, using a screw-micrometer, and their volumes, using the displacement of some liquid. Two histograms may be created, each of which we will assume to have just one peak.

Now, for every individual ball, the diameter has a certain relationship to the volume. This is the same relationship (the same formula) for each of the balls, provided that they are exactly spherical. The question is now whether that same relaionship must exist between the peak-diameter and the peak-volume of our two histograms.

The cat paradox does now translate into a device which accepts the ball bearings one by one, and counts those who fall within two specified categories, one corresponding to the peak in the diameter-histogram, the other to the peak in the volume-histogram. The cat dies, or is set free according to the first category into which a thousand balls have been counted.

 

[cesiumfrog]

Nice analogy, Almanzo, I like how you incorporated quanta.

 

[Redbelly98]

Exactly. But is it always specified in the books of physics, when they talk about light sources?

If it's not, I assume it's per whatever variable is used for the x-axis.

 

[lightarrow]

If it's not, I assume it's per whatever variable is used for the x-axis.

Of course, when there is a diagram of the spectrum.

 

[Redbelly98]

Please let me ask, what is your point? (I am losing track of where our conversation is going.)

 

[lightarrow]

Please let me ask, what is your point? (I am losing track of where our conversation is going.)

In wikipedia, for example, it's written:
http://en.wikipedia.org/wiki/Wien's_displacement_law

Light from the Sun and Moon. The surface temperature (or more correctly, the effective temperature) of the Sun is 5778 K. Using Wien's law, this temperature corresponds to a peak emission at a wavelength of 2.89777 million nm K/ 5778 K = 502 nm = about 5000 Å. This wavelength is (not by accident) fairly in the middle of the most sensitive part of land animal visual spectrum acuity.

Now, while it's quite clear from the context that they are talking about the radiance as a function of wavelenght, sometimes in the books of physics or in scientific magazines, what I quoted from wiki, is written without specifying the spectrum domain and this has no physical meaning.
I just wanted to point it.

 

[Redbelly98]

Valid point. And even more so when you realize the human visual response curve does not have the "per-wavelength or per-frequency?" issue like an emission spectrum does. Change the sun' spectrum to the "per frequency" variety, and it's peak will be somewhere else relative to the visual response curve.

 

[Almanzo]

The problem with a histogram is the arbitraryness of the division of parameter space into different slots. This arbitraryness can be removed -- for the classical case, at any rate -- by using a accumulation curve.

In the ball-bearing experiment this curve would have height zero for diameter or volume zero, and increase by one step at every diameter or volume which is found. (If two or more balls have exactly the same diameter or volume, the curve would increase by two or more steps.) At infinite diameter or volume the height of the curve would be equal to the number of balls tested.

Now, instead of finding a peak in the distribution, one has to find the region of greatest steepness in the accumulation curve. Finding this region for the volume-curve and the diameter curve, we find a volume and a diameter. The question is now whether the relation between this diameter and volume are the same relation which we would expect in any individual ball.

In the quantum case, one would have to test two equivalent samples, measuring wavelength for one sample and frequency for the other one. In principle one of them should be measurable exactly if one forgoes measuring the other one. But if exact measurements are not possible, the accumulation curve will change from a sharp curve consisting of little steps into a more fuzzy, "S-shaped" band. (Not really S-shaped, of course; more like a hillside.) A region of greatest steepness should still be there to be found, though not to be pinpointed exactly.

 

[apolski]

\dot n(\nu) = \frac{2\pi\nu^2}{c^2} \frac{1}{e^{h\nu/kT}-1}

How you found that expression?

 

[apolski]

\dot n(\nu,T) = \frac{J(\nu,T)}{h\nu}

But why?