4
$\begingroup$

What will be temperature inside an isolated box in empty space?

By isolated I meant light from stars cannot enter the box and there is no matter inside the box.(no vibrations or movement of molecules or atoms inside),

Will there be a temperature due to quantum fluctuation?

$\endgroup$
  • 3
    $\begingroup$ about 2.7 Kelvin $\endgroup$ – hft Feb 20 '15 at 3:05
  • 2
    $\begingroup$ Closely related: physics.stackexchange.com/q/91232 $\endgroup$ – Kyle Kanos Feb 20 '15 at 3:07
  • $\begingroup$ Very closely related: physics.stackexchange.com/q/163769/23473 $\endgroup$ – Jim Feb 20 '15 at 16:52
  • $\begingroup$ Question: When you say "box in empty space", do you mean a physical box made of cardboard or metal or whatnot? Or did you mean an arbitrary cube-shaped volume of space, not necessarily contained within a material shell? $\endgroup$ – Jim Feb 20 '15 at 17:14
  • $\begingroup$ @Jimdalf the Grey :Unfortunately I dont know how to isolate a region,By a box in empty space I really meant a isolated region where energy from nowhere can enter,I was thinking of the temperature due to the quantum fluctuating particles. I am not very experimentallist type person,so theoretically I have assumed thats possible. $\endgroup$ – Paul Feb 20 '15 at 18:13
4
$\begingroup$

Even if light from stars cannot enter the box, eventually the box and the surrounding space will come to thermodynamic equilibrium and the box will emit blackbody (or, more likely, graybody) radiation both inside and out and the equilibrium temperature attained.

If you take the box far from any starlight there is still the sea of photons that make up the Cosmic Microwave Background, which has a nearly blackbody distribution peaking around 2.7 K. So, if your box is nearly a blackbody the inside of your box will be 2.7 K since your box will be in thermal equilibrium with the CMB photons.

$\endgroup$
  • $\begingroup$ What about quantum fluctuations? you didnt mention that. $\endgroup$ – Paul Feb 20 '15 at 5:13
  • $\begingroup$ @Paul, I don't have an answer for the quantum fluctuation contribution to the temperature. I was mainly pointing out that even if you isolate a portion of space with a box the photon sea outside the box will still cause there to be a photon sea inside the box. $\endgroup$ – NeutronStar Feb 20 '15 at 15:03
  • $\begingroup$ Note that it would not be a blackbody nor a graybody. A black(gray)body is defined as having an absorptivity/emissivity of 1/1 (0.5/0.5). However, without anything at all in the box, it would have a transmittance of 1 and an absorptivity of 0. Therefore, it completely fails the requirements of being a blackbody. It also means it would not emit radiation. Without anything inside it, there can be no associated temperature. $\endgroup$ – Jim Feb 20 '15 at 16:50
  • $\begingroup$ @JimdalftheGrey, I don't think that a graybody has to have absorptivity/emissivity of 0.5/0.5, it just has to have them being less than 1. I was saying that the photons outside the box would heat it up; at that point it would start emitting photons both inside and outside the box. $\endgroup$ – NeutronStar Feb 20 '15 at 17:03
  • $\begingroup$ Photons emitting thermal photons? I'm pretty sure it doesn't work that way. And yes, a gray body is $0<\epsilon<1$ $\endgroup$ – Jim Feb 20 '15 at 17:05
1
$\begingroup$

There's several things involved in your question.

The first is what would realistically happen if you set up this system and you managed to measure its temperature. Here there's no easy answer: you have to consider in detail how you set up the system and how it will interact with its surroundings. Simple idealizations like "isolated" aren't adequate here because the temperature you can measure in empty space is mostly due to the cosmic microwave background. This means that, for example, if you use box made of matter, microwaves would impinge on the outside and transfer heat so that in practice it's impossible to place a truly isolated box in empty space.

If you did manage to do it through magical unphysical means and you also made sure to remove all the radiation inside the box, then the temperature inside it would be zero due to the absence of any excited degrees of freedom. There would be quantum fluctuations, yes, such as the Casimir effect, but no thermal fluctuations.

The second one is whether or not it's possible to define temperature for a system with no matter in it. And the answer is yes. In quantum field theory we can tune the temperature of a system completely independently of the density. To make a system at finite temperature, we consider a theory in a Euclidean spacetime (that is, the distances inside it are given by $ds^2 = dx^2 + dy^2 + dz^2 + d\tau^2$, where $\tau$ is the "Euclidean time" rather than the usual $ds^2 = dt^2 - dx^2 - dy^2 - dz^2$) and make the time direction compact, that is, we make it finite and impose boundary conditions. Spacetime then becomes like an infinite cylinder whose circumference is $1 \over T$ in natural units.

To make a system have a finite density, we can tune the chemical potential. You can think of the chemical potential as a "bias" that allows you to tune the average particle number per unit volume, much like the temperature allows you to tune the average energy per unit volume. The way to actually implement this in field theories is a bit technical so I'll leave it out, but the gist of it is that it's perfectly reasonable to leave the chemical potential at zero and still have a system e.g. a gas at a finite temperature.

The interpretation here, considering QED at finite temperature would be that the average electron number is zero because you have a gas with equal numbers of electrons and positrons flying around, plus photons and whatever messy interactions they may have.

Perhaps this isn't exactly what you wanted, but to define a finite temperature you really need excitable degrees of freedom. At best, you have this zen like idea where equal numbers of electrons and positrons means the average electron number is zero, but this isn't to say that the box is filled with nothing. There's stuff in it.

Disclaimer: this is the particle physics perspective. Other branches of physics may have subtly different approaches to things which may lead to confusion. Just be aware.

$\endgroup$

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.