# Is temperature relative?

1. Jun 15, 2009

### Yuqing

Temperature is defined as the average kinetic energy of a system. But would temperature change if some external force was applied to change the total motion? For example, would a ball experience an increase in temperature if the ball was traveling at a high speed as a whole? The average kinetic energy initially is still the same, but the molecules now have another velocity (the velocity of the ball itself). So would we experience this as a change in temperature?

2. Jun 16, 2009

### fatra2

Hi there,

Absolutely.

Instead of a baseball, take a gaseous system (that follows the perfect gas law). This system's temperature is define by the average kinetic energy of its particles. Ok. If, from the outside, I apply a force to this system, I add some pressure to it. Guess what happens, the particles gain some kinetic energy, and the temperature rises.

Ok to make it complete. With an external force applying, my gaseous system has two possibilities, either the volume compresses or the temperature rises. Remember $$PV = nRT$$?

The same happens for solid objects. You let a ball drop from a certain height. The potential energy contained in the ball transforms into kinetic energy. When the ball hits the ground (without a bounce, for simplicity), all this energy is given to the particles. Therefore, heat is produced.

At last, liquids react the same way. Take a cup of water, and stir it with alot of energy, and you will have boiling water after a while. Ok, you might have an armache before having a cup of boiling tee, but still.

Cheers

3. Jun 16, 2009

### Bob_for_short

No. Take a two-body system and separate the variables to the center of inertia and relative motion coordinates. If an external force increases solely the CI kinetic energy, then no change in the internal motion energy occurs. It is also possible to reason without any external force. You just change the reference frame. The relative velocities do not change so the internal kinetic and potential energies do not change either.

Bob_for_short.

4. Jun 16, 2009

### Yuqing

Can you please expand on this? I'm having trouble understanding.

5. Jun 16, 2009

### Bob_for_short

No, I cannot. You have to learn a two-body mechanical problem. After that everything I wrote is clear. I cannot teach here.

Regards,

Bob.

6. Jun 16, 2009

### Count Iblis

In thermal physics, what you do is you describe almost all degrees of freedom of the system (positions and speeds of each particle of the system) statistically and you only keep track of a few degrees of freedom, which we call external variables (like total volume, position of center of mass, velocity of center of mass).

Change in energy of the degrees of freedom that are described statistically is, by definition, heat.

Change in energy associated with changing the external variables is called work.

The relation between temperature and average kinetic energy holds only (under certain conditions) for the average kinetic energy in the degrees of freedom that we choose to describe statistically.

If you were to include the center of mass velocity in the set of variables that you want to describe statistically, then that would amount to adding one degree of freedom to the system. But then the assumption of thermal equilibrium would imply that on average there can only be an energy of 3/2 k T in the center of mass kinetic energy.

You cannot then consider the mass moving (relative to some fixed reference frame) within this framework, as you are then very far from thermal equilibrium. You would have stored an astronomically large amount of energy in the single degree of freedom representing the center of mass motion, while in thermal equilibrium all degrees of freedom contain the same amount of energy.

If you then wait until the mass bumps into something else many times until all the center of mass energy gets dissipated in the environment (assumed to be at some constant temperature so that it acts s a heat bath), you'll have achieved thermal equilibrium.

7. Jun 17, 2009

### physicsnoob93

So basically, you are asking if a system's total energy is altered if it's in motion as compared to when it's at rest? If so, I would think the answer is no. If you are talking about the individual atoms in the system, for gases, you might want to have a look at this: http://en.wikipedia.org/wiki/Maxwell–Boltzmann_distribution.

8. Jun 17, 2009

### ImAnEngineer

The general gas law states: pV = nRT

p= pressure
V= volume
n= number of moles
R= gas constant
T= temperature

Neither p, V nor N or R change when a ball is thrown, so T remains the same as well. (actually, because you apply a force to the ball in order to throw it, pressure inside the ball temporarily increases a little bit, so that could cause a rise in temperature, but that is probably negligible)

9. Jun 25, 2009

### Gerenuk

The reasoning here is OK, in principle. The error is that temperature is not defined as the average kinetic energy. That only works for special systems such as a stationary gas (in general when the density of states is a power law $g(E)\propto E^\alpha$).

Still I could image the temperature increases if all molecules get an additional kick. I will try to figure that out somehow :)

10. Jun 25, 2009

### Andy Resnick

That's *one* way to define a temperature. But yes, an accelerated bolometer will record a different temperature from an unaccelerated bolometer (Unruh radiation).