Short wavelengths carry no energy?

[dlevanchuk]

short wavelengths carry no energy?

Hey guys (and girls),
I'm reading the book "The Black hole war", by Leonard Susskind and on the page 208 it has written something that kind of surprised me. I ll just quote an entire paragraph:
"The hypothesis that light is composed of indivisible photos whose energy is proportional to their frequency solved the problem (refering to ultraviolet catastrophe). Applying Boltzmann's statistical mechanics to these photons, Eisntein found that the very short wavelengths (high frequency) have less than a single photon. Less that one means none. So the very short wavelengths carry no energy, and the ultraviolet catastropphe ceased to exist."
The part where the author said that high frequency em waves carry no energy gives me a trouble to understand. Was the author referring to the wavelength equal to gamma rays (which I though have extremely high energy) or was the reference made to the wavelengths beyond gamma rays??

 

[Bob S]

, Eisntein found that the very short wavelengths (high frequency) have less than a single photon. Less that one means none. So the very short wavelengths carry no energy, and the ultraviolet catastropphe ceased to exist."

Susskind is saying that the probability of finding a photon is less than one, an an insignificant amount of energy is carried away by UV photons, because their probability is nearly exactly zero. So the integral over wavelengths converges as the wavelength limit goes to zero.

 

[dlevanchuk]

It is the statement "So the very short wavelengths carry no energy" confuses me.

If use Planck-einstein equation E=hc/lambda, where h is Plancks constant, c is a speed of light and lambda is wavelength, I see that as the wavelength goes smaller - the photon energy goes up.

Am I missing something here? :(

 

[jtbell]

As the wavelength decreases, the energy per photon [STRIKE]decreases[/STRIKE] increases, but (in this situation) the (average) number of photons decreases even faster, so the so the total energy decreases with wavelength.

 

[hamster143]

This looks like the discussion of EM spectrum of black-body radiation.

If you look at the electromagnetic field inside a black box, you can break down all energy contained into frequency bands. At low frequencies and low energies, there may be lots of photons flying back and forth in a given frequency range. As you look at higher and higher frequencies, the expected number of photons eventually falls below 1.

So, the statement is that there's no energy in short wavelength bands in such a black box, because there are no photons with that frequency there.

For any given photon, the statement E = hc/lambda still holds, of course.

 

[Bob S]

Here is a short discussion of the ultra violet (short wavelength) divergence problem.
http://www.statemaster.com/encyclopedia/Ultraviolet-divergence
Convergence required the number of UV photons to go to zero.
[Edit] What is really important is for the integral over all emission quanta to converge for high energy (short wavelength) quanta. So the product of N, the number of quanta in an energy interval dE, and the the energy of each photon E go to zero for short wavelengths: N(E) E dE ==> zero.

 

[Halcyon-on]

This means that in the UV limit the EM fields have low intensity with respect to the energy of a single photon. In particular the intensity here is fixed by the thermal noise KT << h v. Thus the Boltzmann law says that the probability of a single photon is low exp(-h v / K T).

 

[Dadface]

As the wavelength decreases, the energy per photon decreases, but (in this situation) the (average) number of photons decreases even faster, so the so the total energy decreases with wavelength.

Just to clarify this did you mean that the energy per photon increases?

 

[jtbell]

Just to clarify this did you mean that the energy per photon increases?

Oops, yes. I'll go back and fix my post. Thanks for pointing it out, even if a bit late.