Was the relation between E and f known BEFORE quantum theory?

[nonequilibrium]

Hello,

In classical EM, would one have said that red light carries less energy than blue light? (if both have the same intensity)

 

[granpa]

classically, the energy density of the electric field is proportional to the square of the field
http://en.wikipedia.org/wiki/Electric_field#Energy_in_the_electric_fieldh was first used in placks law

[tex]
I(\nu,T) =\frac{2kT}{\lambda^2} \frac{hf}{kT} \frac{1}{e^\frac{hf}{kT} - 1}
[/tex]

http://scienceworld.wolfram.com/physics/PlanckLaw.html

http://en.wikipedia.org/wiki/Planck's_law

 

[nonequilibrium]

I know that, but that's not really my question...

 

[Hootenanny]

Hello,

In classical EM, would one have said that red light carries less energy than blue light? (if both have the same intensity)

Intensity is the power per unit area. Thus, by definition, the answer to your question is no. Moreover, this is also the case in quantum mechanics - the definition of intensity does not change in going from EM to QM.

 

[nonequilibrium]

Hootenanny: if I have one photon of red light and one photon of blue light, doesn't QM say they are of the "same intensity", because intensity is a measure for the amount of photons? Yet a red photon carries less energy than a blue photon, so I don't see how you can say that QM implies that if two light sources have the same intensity, they give the same energy.

 

[Hootenanny]

Hootenanny: if I have one photon of red light and one photon of blue light, doesn't QM say they are of the "same intensity", because intensity is a measure for the amount of photons?

Nope. Intensity is a measure of power per unit area, as it is in the classical case. Light intensity is not exclusively determined by the number of photons - it depends on both the photon flux and the energy of each photon.

 

[tiny-tim]

Hello,

In classical EM, would one have said that red light carries less energy than blue light? (if both have the same intensity)

hello mr. vodka!

red light by definition has a longer wavelength (lower frequency) than blue light … this can be determined from diffraction gratings, Young's fringes etc

if you move away from a wave, its frequency lessens, and its energy also lessens (Doppler shift), so between the same amount of blue light and of red light, the red light has less energy (without knowledge of photons, I don't see any other way of comparing "the same amount" of different colours)

 

[nonequilibrium]

tiny-tim: hello :) indeed, but what I was wondering about: did they also know this in classical physics? (EDIT: woops, didn't see your last line between the parenthesis, but Redbelly (next post) seems to have replied to that already)

Hootenanny: but isn't intensity classically determined by the amplitude of the EM-wave? (if not, what is the correct formula?) In the case I am right about that: the amplitude then gives the probability the photon being in a certain place (cf Born), and that probability is determined by the flux, and not by the energy of each photon

 

[Redbelly98]

(without knowledge of photons, I don't see any other way of comparing "the same amount" of different colours)

Classically you could compare different color waves that have the same electric field amplitude, and they would have the same intensity. Then if you want to bring photons into the discussion, you would say the same intensity of red light has more photons (per m², per second), so each red photon has less energy. It takes fewer blue photons to produce light of the same intensity as red, so the blue photons must be more energetic than red ones.