Do hot metals radiate? (Thermography) I was looking into thermography which talks about emissivities of metals and other materials. Polished metals which have low emissivity appear to be colder in thermal imaging cameras even if they are actually hot (because they have low emissivity). Refer to the image below:

^ Source: http://www.flir.com/science/blog/details/?ID=71556
My understanding of thermal radiation says that a material should be radiating when hot. So, a coffee mug made of two materials (ceramic mug with say metal decorative leafs on the outside) should be radiating equally? If they are not radiating equally, then how is metal leafs part moving towards thermal equilibrium? 
Thermal radiation is the radiation generated by the thermal motion of charged particles in matter. So if the metal part if actually hot, why is not radiating? If it does radiate, why doesn't it show up in the IR cameras (until you adjust the region of interest and manually enter emissivity of material in camera)?
 A: A hot material will radiate heat to a colder, that is to say it will radiate
more heat outward than it absorbs from the colder object.   The problem is
only that the radiation RATE, as well as the absorption rate, is not
determined by temperature alone, but by the coupling of the material to
light of any given wavelength.  Metals are electrically conductive, and
if they haven't any oxide or other nonmetal surface layers, that makes them
reflective.  Reflection means the absorption of radiation is very small, and implies that the emission of radiation by a hot metal, while larger
than absorption, can be also very small.
An easily treated case is that of a perfectly absorbing material, a
'black body'.   In that case, the material property of absorption
versus wavelength turns into a factor of '1', and simplifies the
equations wonderfully.   Such effects as greenhouse warming are
not  present in the simple model where all bodies are black, though.
A: There is a relationship between emissivity and absorptivity, wherein materials that don't absorb well are also poor emitters.  Therefore, shiny metals typically don't emit thermal radiation at as high a rate as other objects, and that's what you see in the image. 
This can all be reasoned out by considering the need for conservation of energy in thermal equilibrium.  Kirchoff has done this for us, and has nabbed another law named in his honor (aside from the more well known circuit laws) , Kirchoff's law of thermal radiation:

For an arbitrary body emitting and absorbing thermal radiation in thermodynamic equilibrium, the emissivity is equal to the absorptivity.

A: The mug is made up of two materials, ceramic (an insulator) and metal (a conductor). These two types of materials certainly have different heat coupling coefficients, and metals in general both emit and absorb heat faster than insulators. So it definitely makes sense that metals, by virtue of more efficient radiative heat transfer with the environment, are almost in a state of thermal equilibrium with the environment in this picture shown. On the other hand, the ceramic has poor thermal coupling with the environment for radiative heat transfer, so the thermal energy cannot be transferred to the environment quickly. This results in the mug "containing" more energy and appear hotter in this picture. 
It would be interesting to take a time series photographs of this mug from the moment you pour hot liquid into it to the moment where the metal reaches thermal equilibrium with the environment (primarily via radiative lossses). If you wait long enough, you'd also see the ceramic reach thermal equilibrium but it would most likely do this via convection (heat loss to the air through the surface of the liquid rather than radiative losses from the ceramic to the air). https://en.wikipedia.org/wiki/Heat_transfer
Hope this helps!
