If I was to measure a spectrum of my palm, how would it look like? WIll it be continuous? Will it be possible to identify the components the palm is made of by digesting the spectrum?

  • $\begingroup$ Hi and welcome. What makes you think your palm shows a temperature spectrum? $\endgroup$
    – Gert
    Nov 14 '21 at 18:55

To a first approximation the palm shows a blackbody radiation spectrum of a 37C surface, which would be a smooth continuous function. It peaks around 10 micrometers.

However, there are deviations from this curve due to different emissivity at different wavelengths. If the emissivity is low (it is more reflective and releases less radiation on its own) the thermal energy will be released at wavelengths that have higher emissivity (it is "blacker" and absorbs more but can also radiate more). The result is peaks and valleys on the curve.

Figure from (Folbert & Heim 1984) Figure from (Folbert & Heim 1984), showing measured emissions from skin. Note the somewhat messy "blackbody" curve - this setup seems somewhat unstable.

In practice the emissivity over the infrared range is 0.990 to 0.999, essentially a blackbody (and independent of skin colour). When we go down towards the visible range emissivity does go down, again in a somewhat uneven manner. Emissivities from (Ginesu et al. 2004):

(Ginesu et al. 2004)

NIST has been measuring reflectance of skin (reflectance is the opposite of emissivity, if there is no significance energy transfer), finding a fairly high variability in the visible range because of different amounts of melanin, hemoglobin and other components.

NIST measurements

So, yes, there is a continuous spectrum. Yes, some components like oxy- and deoxy-hemoglobin affect the shape of the spectrum. But no, it is not a particularly neat and clean spectrum.

  • $\begingroup$ Why aren't the dashed blackbody curves smooth? $\endgroup$
    – ProfRob
    Nov 14 '21 at 21:11
  • $\begingroup$ I suspect their experimental setup wasn't very good. It is not a recent paper. $\endgroup$ Nov 15 '21 at 14:00
  • $\begingroup$ But the blackbody curve is the Planck function? $\endgroup$
    – ProfRob
    Nov 15 '21 at 14:01

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