The momentum of a photon is proportional to its frequency. But the radiation pressure is independent of frequency (for a monochromatic plane wave, average of poynting vector is, = 1/2 epsilon c A^2. Where A is the Amplitude of the Electric field..). Why is this so? After all, I expect that the pressure is due to momentum transfer?
One has to be a little careful in the translation of classical (e.g. amplitude of electrical field) and quantum (e.g. notion of photons) quantities. What you can equate is the electromagnetic energy flux density (related to the Poynting vector, as you state) to the photon flux density. Once you have the photon flux density, you can connect it with the radiation pressure.
A single photon state has vanishing expectation values for the classical field quantities. But a radiation field of same energy flux density (intensity) has less photons at shorter wavelengths (higher photon energies) compared to longer wavelengths.
So it is like, photon desnity has an exact inverse proportionality with frequency..
Classically I expected the other way... Higher wavelength will have a lower photon denity.. Of course it was a blind intuition, thinking that larger wavelength occupy larger 'space'.. Anyway it could be wrong...
A key point is that A^2 is proportional to the product of the photon flux n (number of photons passing through a unit area per second) and the energy per photon (\hbar \omega). The Poynting vector, S, therefore is proportional to n \omega. It's possible to increase the wavelength (decrease \omega) while maintaining the same intensity, and thus the same radiation pressure, by reducing the photon flux. What matters is the power in the beam of light, not the color.
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