4
$\begingroup$

I found this on Wikipedia article on black bodies:

A black body in thermal equilibrium (that is, at a constant temperature) emits electromagnetic radiation called black-body radiation. The radiation is emitted according to Planck's law, meaning that it has a spectrum that is determined by the temperature alone (see figure at right), not by the body's shape or composition.

Now I'm a bit confused. Isn't the black body radiation given by $\frac{P}{A}=\sigma T^4$

So it doesn't depend on body color, composition or anything else than its size. Shouldn't the radiation be the same for all bodies, but we use the black body because its a convenient model because all of its radiation comes from emission and not reflection or anything else?

$\endgroup$
3
$\begingroup$

A real object will radiate less energy than a black body at the same temperature. The ratio of the emission to the black body emission is called the emissivity. The Engineering Toolbox web site has data on emissivities for a range of materials. For example a polished silver surface has an emissivity of 0.02 i.e. it radiates only 2% of the power that a black body of the same temperature radiates.

The emissivity is related to the reflectivity by E + R = 1. A black body reflects none of the light falling on it so R = 0 and E = 1. Anything that has a non-zero reflectivity must have an emissivity of less than 1 otherwise it could be used to build a perpetual motion machine.

$\endgroup$
1
$\begingroup$

What you've written is the Stefan-Boltzmann equation for a blackbody which relates the flux at a certain distance to the temperature at that distance...

The wavelength of blackbody radiation is actually given by Planck's law. But, it doesn't matter...

You're right. The blackbody is an ideal one-way window. So, it's always at some higher value than the observable. Because, nothing is perfect. This radiation depends only on temperature and not any other physical properties. That's the reason why the lava from volcano looks very similar in color to molten iron. The same reason why we're at 98 F and always emit IR radiation.

The reason why we use blackbody is because, take some star (for instance, Sun). The temperature of Sun-like stars don't necessarily be the same. That's why we approximate (cruel, but necessary for science) them with our blackbodies.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.