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I obviously do not understand how a wavelength greater in length than a cell can be detected by a cell.

As I understand it, infrared ranges from 300GHz to hundreds of THz.

As an example, if an infrared wave is 1 THz, it is 0.3mm in length. How does this get even noticed by a cell?

I would think it doesn't see the wave simply because the wave is longer (larger?) than the cell.

Qmechanic
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Gary La
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The cells don't react directly to radiation in that wavelength range. Temperature sensitive cells can react to the heating caused by this radiation, though. It doesn't take a lot of power to set off our temperature detectors. I would estimate that one can easily "feel" 1W per square decimeter of skin or less in an otherwise "quiet" thermal environment. The reason for that sensitivity is biological: the human body produces $100W$ of heat and has approx. $2m^2$ of surface area. It is of vital importance to us to maintain our body temperature with that heat loss of $100W/200dm^2=0.5W/dm^2$. A few tenths of a $W/dm^2$ of heat loss more or less will lead to hypo- or hyperthermia.

FlatterMann
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You can very well ask how does an infrared photon get absorbed by a hydrogen atom - which has a size of just about 0.05 nm.

The condition you used: $\lambda > L$ where $L$ is the cell's length, accounts for just scattering alone (which you seem to know). But there's just one photon. If you send more photons, there will be some absorption or scattering by the same cell. And there's a multitude of scattering effects out there; elastic (Rayleigh) and inelastic scattering (Raman effects) - which transfers energy to the cell.

Karthik
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@hyportnex: The secret are optical phonons. They are oscillations, in any lattice cell with at least two different ions oscillating against eachother. The very stiff coupling yields frequencies in the infrared range of light, only weakly k-dependent, and the dell dipole moments in a line are coupling to light perfectly, hence the name optical phonons.

Even if they don't play a role energetically at 37C by the Boltzmann factor $e^{- \hbar \omega/kT}$, they are perfect signal transmitters.

Roland F
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