# Is the *measured* temperature inside a hollow black body the same if we place a thermometer inside it or let it make contact with the body itself?

Imagine the inside of a black body shell with non-zero thickness. Inside the shell, there is photon gas produced by the shell [at a temperature much higher than zero, say $$1000(K)$$]. Is there is a temperature difference (which objectively there is not, but according to the thermometer) between placing the thermometer somewhere in the shell and bringing the thermometer in direct contact with the shell? Obviously, the two means of heat transfer (radiation and heat conduction) are different from each other. Is the temperature of the photon gas (as indicated by the thermometer) lower because not all energy that hits the thermometer (by means of the photons) is absorbed by it, while all the energy transferred by conduction to the meter is absorbed?
Please correct me if I'm wrong.

## 2 Answers

It really depends on the specifics of how you would make your measurements and how accurate you are being.

In an ideal scenario, where the thermometers are allowed to reach equilibrium with the surroundings, both will measure the same temperature. This makes sense, and follows how we define temperature (see also zeroth law of thermodynamics). If they are at thermal equilibrium with each other, they are by definition the same temperature, and it's that concept that allows us to use thermometers in the first place.

The problem will be actually reaching that equilibrium. In theory, equilibrium is only approached over infinite time between the two systems (the less temperature difference, the less heat transfer, leading to slower change in temperature, creating an approach to an equilibrium you theoretically never reach). Depending on the properties of your thermometer, conduction+radiation may reach a near equilibrium temperature far faster than radiation alone, but the final equilibrium temperature which they approach(which is ideally what you measure with a thermometer) will always be the same.

Basically, the radiation method might look like it's measuring a lower temperature; but if you left both experiments running long enough, the difference between the two temperatures would be completely insignificant, and if you waited infinite time (as you should when using a theoretical perfect temperature measurement) they would be completely identical temperatures.

• This. Thermalization is always the bug-a-boo for practical handling of these kinds of questions (as opposed to blackboard physics where it we can mostly just wave our hands at the issue). – dmckee Jan 16 at 3:21

Eventually the thermometer will read the 1000C, it will emit and receive radiation at a rate that is the same as the walls assuming they are both back body materials.