Suppose there is a container (filled with vacuum) in which there is a particle (only one).Let the initial temperature be $0K$ so that the particle will have $0$ Thermal energy. Now as we increase the temperature the particle will start to accelerate from rest. I searched the internet and my book which points out that this is due to the fact that it has some Thermal Energy. But according to Newton's First Law the particle should've been acted upon by a force to accelerate it ,so which is the force here? Also if the particle was given some initial velocity (such that it moves in a straight line),does the particle deflect from its path due to the temperature increase? if so, why?


1 Answer 1


From quantum physics, we know that even at 0K, particles would still have motion. But let's assume a classical view for a moment, where particles could be motionless at absolute zero. In addition, it really doesn't make sense to talk about the temperature of a single particle. We would instead talk about its speed.

As we increase the temperature of the walls of the container, the molecules that make up the wall vibrate more quickly. This has two effects.

The first is that if the particle in question approaches the wall, it can get a kick from a fast-moving molecule there. The interaction takes energy from the wall and gives it to the particle, which rebounds with a higher speed. The motion of the particles in the wall give the force required.

The second is that the warming walls are jostling the electrons around them. This jostling causes the electrons to radiate energy. This radiation will move through the cavity and may interact with the particle. If it scatters off the particle, it can transfer energy, increasing the particle's energy. Here we say the momentum of the photon is responsible for the force felt by the particle.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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