# Absolute Zero

1. Oct 5, 2004

### Matrixman13

Am i right when i say its impossible to reach absolute zero? Because of the uncertainty principle, you would need an infinite amount of energy to keep a particle completely still. Thanx

2. Oct 5, 2004

### chroot

Staff Emeritus
You are correct, absolute zero is unobtainable.

- Warren

3. Oct 5, 2004

### geometer

Yes, you are correct. In fact, the impossibility of reaching absolute zero is sometimes referred to as the Third Law of Thermodynamics. You can approach arbirtrarily close, but you can't ever reach exactly absolute zero.

4. Oct 7, 2004

### primal schemer

ab zero

Hi,

Just wondering why you need an infinite amount of energy to keep the particles still. How would you use energy to restrict the movement of particles??

Thanks,

PS

5. Oct 7, 2004

### Atheist

Assuming with absolute zero you talk about temperature :

Uncertainty has absolutely nothing to do with temperature:

- In QM particles are in definite states and T=0 would simply mean that the total system is in the (energetically) lowest state possible. The fact that in this definite state you can measure position and momentum of a particle only to a certain degree of accurancy (<- uncertainty) doesn´t make the state any less definite.

- Thermodynamics on the other hand deals with systems that you don´t know the state of, but only certain macroscopic properties. In isolated systems, for example, you then assume that the system is in all QM-states that have these properties with equal probability. Note that this is a completely other type of "uncertainty".
For systems in thermal contact with the surrounding the relative probabilities are not equal anymore but depenant on the state´s energy and a distribution-variable called Temperature.

The inability to reach T=0 is a bit tricky. Most people refer to the x-th law of thermodynamics (x depends on where you start to count) which sais that you cannot reach T=0 with any finite thermodynamical process. I´ll also leave it there because I had to look up quite a few things to give a detailed correct explanation. Note however that Thermodynamics only applies where it applies (a single atom can very well be in it´s ground state which would be T=0 if threatened by TD).

6. Oct 7, 2004

### HallsofIvy

Staff Emeritus
That's simply an incorrect statement. It has nothing to do with energy, an infinite amount or not- because of the uncertainty principle, you CAN'T have an collection of completely immobile particles.

7. Oct 7, 2004

### Gara

I've heard that IF atoms got to 0°K, they would cease to exsist. Is this true, and why?

8. Oct 7, 2004

### Atheist

Not true, although there are people saying "you can´t know that because you can´t reach T=0". But those
a) Don´t understand what temperature is and
b) Can´t say the atoms stop to exist for the same reason.

9. Oct 7, 2004

### mathfeel

This has absolutely nothing to do with uncertainly principle. Are you trying to argue that $$\Delta E \Delta t \le \hbar$$, so you must have some flutuation in energy, and hence zero temperature is impossible? This principle has nothing to do with temperature. What this says is that you can violate conservation of energy temperarily, provided that you pay back what you borrow in short enough amount of time. (The more you "borrow", the shorter you can have it).

Definition of temperature can be seen from the thermodynamic identity:
$$dS=\frac{dE}{T}+\frac{pdV}{T}+\frac{\mu dN}{T}$$
So, $$\frac{1}{T}\equiv\left(\frac{\partial S}{\partial E}\right)_{V,N}$$. In fact, I believe, you can in principle define temperature to be any (nice) fuction of this derivative. (In particular, you can definte temperature to be just this partial deriative, instead of the inverse of it, problem being: temperature then goes inverse as energy, contrary to what we are used to). If you hold everything else constant (volume and what not), and change energy slightly, what do you pay in term of entropy is, by definition, temperture (or the inverse of it). With this definition, temperature can indeed be positive, zero, or even negative. But the latter cases can only happen in nonequilibrium system or in phase transition (and it does--I just a qual. exam problem regarding a magnetic system that does just that!)

Since $$S=\log \Omega$$ where $$\Omega$$ is the number of state must be finite no matter how you change your energy, This partial derivative is also finite. In equilibrium, increasing energy (holding everything else constant), always means increasing entropy. So, the partial must be strickly positive...

In the situation, where your system is very very close to the ground state. Giving the system a little bit of energy makes it a much more flavorable for one or few particles to jump to a higher state, which greatly increases the entropy. This corresponds to when $$\frac{1}{T}$$ is a huge number, i.e. low temperature.

In the situation, where particle of your system is already pretty randomized. Increasing energy doesn't improve your randomness anymore (kind of a measure of entropy). This derivative is small, i.e. high temperature.

Last edited: Oct 7, 2004
10. Oct 7, 2004

### PMASwork

OK, going on memory of 1 thermo class 10 years ago...

Heat energy flows in one direction only - from a "hotter" object to a "colder" object.

The colder object must be at a temperature strictly less than the hotter object (cannot be equal) in order for this energy transfer ("cooling" of the hotter object) to occur.

Another way to look at this: to cool an object from T1 to T2 (T2<T1), you need a second object in which to dump the energy, and the second object's temperaure T3 must be strictly less than T2. T3<T2.

So, to cool an object from T1 = 1degK to T2 = 0degK (absolute zero), you must already have an object that is BELOW absolute zero on hand (at, for example, T3 = -1E-100 degK). ...but objects with temperatures less than absolute zero do not exist!

Thus it is not possible to cool any object to absolute zero.

There is an equation (not in my head) for the amount of energy that must be transferred from object 1 to object 2 in order for a specific temperature reduction in object 1 to occur. I believe this equaiton will yield an infinte amount of energy should the object of lower temperature be at absolute zero.

(as I stated above, this is purely based on a dusty memory, so feel free to poke at this idea and correct it if needed!)

11. Oct 11, 2004

### Matrixman13

o im sorry. I failed to explain what i meant by energy. i was speaking of the method used to try to produce temperatures nearing absolute zero. Lasers are aimed at a particle from many different directions, in an attempt to "pin" the particle to one spot.

12. Oct 11, 2004

### Atheist

Laser cooling is just a method of cooling. Assuming your laser takes finite power to run, then, by interpreting "you cannot reach absolute zero by a finite thermodynamical process" as "it takes an infinite time to cool them down" you could construct a statement like "it takes infinte energy". However, this still has nothing to do with uncertainty and also doesn´t imho tell me anything interesting (not to speak of the relatively vague interpretations used that would need some backup).

If you´re interested in cooling there is a page about experimental realization of bose-einstein condensation I often like to recommend:

Keep klicking "next" and play with the applets a bit.
The page should be very easy to understand even for non-scientifically interested persons but is still very accurate (to the point I can judge that) and should also be interesting for physicists working in other fields (it was for me, at least).

Last edited by a moderator: May 1, 2017
13. Oct 12, 2004

### Chronos

I side with the dissenters. HUP has everything to do with absolute zero. All atoms have a zero point energy state that is above absolute zero. This means they still radiate [albeit in the extremely long wavelengths] and still have a blackbody temperature. This is one instance where quantum theory and GR do agree.

14. Oct 12, 2004

### Atheist

@Chronos:

Three questions and one suggestion:
q1) Where is the connection between uncertainty and temperature in your post? Uncertainty never appears except at the point where you claim that "it has everything to do with absolute zero".
q2) If an atom in it´s energetically lowest state emits a photon (radiates): What state does it change to?
q3) Could you please show from my 1st post where you think I am wrong? Simply saying "I disagree" is ok, but doesn´t help anyone, i think.

s1) Please take a bit more time to explain what you want to say. Here, you jump from uncertainty over Thermodynamics (very well knowing that I, for example, don´t see any connection) over to Relativity (why?!?!) within four sentences. Also, your posts (this mainly reffers to one of your posts that raised my attention just yesterday) sometimes are very vague and leave room for a lot of interpretation.

Last edited: Oct 12, 2004
15. Oct 13, 2004

### Chronos

Very good questions.
q1] If you define temperature as a measure of the intensity of motion [kinetic energy], and absolute zero as the point at which all motion ceases, then absolute zero is forbidden by the uncertainty principle. If absolute zero is defined as the ZPE state, then it is not forbidden by the uncertainty principle. We do, however, have a problem with the definition of temperature as a measure of motion. If you apply the state equation for an ideal gas PV=nRT, you clealy have a dilemma at T=0. P cannot be zero if there is motion and V obviously cannot be zero without a singularity arising or a violation of energy conservancy.
q2] It cannot radiate unless a lower energy state is available. It must radiate per Planck's Law, if there is [ie, it has a temperature].
q3] I only disagreed with the assertion uncertainty has nothing to do with temperature.
s1] The connection to relativity involves potential violations of energy conservancy at absolute zero. It is rather technical, but not terribly important.

Vague posts? Just because I sometimes feel like I am the only one who understood what I meant doesn't mean it's vague, does it? [you might have a point].

16. Oct 13, 2004

### Bystander

you've got a definition that is incongruent with any in accepted use.

17. Oct 13, 2004

### Chronos

In kinetic theory it customary to substitute (mv^2/2) for temperature, where (mv^2/2) is the average kinetic energy of each atom in a gas.

18. Oct 13, 2004

### Atheist

Well, I think the problem is that your view on temperature derives from Classical Thermodynamics. Classical Termodynamics more or less directly compares to the more modern Quantum Statistics as Classical Mechanics compares to Quantum Mechanics - obviously becaue they are based on Classical Mechanics and QM, respectively.
Both Classical Thermodynamics and Classical Mechanics are only valid withing a certain (yet the most important) range and both Quantum Mechanics and Quantum Statistics converge to their classical analogons within that range.

In Classical Thermodynamics T=0K would indeed mean stop of all motion - to stay with your example of an ideal gas. However, as Classical Thermodynamics is based upon Classical Mechanics there´s no Uncertainty that would prevent that. Classically, there´s no problem if all atoms were at rest (inobtainability let aside - this has different origins even in Classical TD). But Classical Thermodynamics is simply not a good approximation for T->0K anymore.

Plugging Uncertainty into a classical problem does not seem to make much sense. Actually, I couldn´t spontaneously think of any problem where Uncertainty plays a role that could be threatened classically. Indeed, the very common assumption that uncertainty and the inobtainability of T=0K were related (I can very well understand that this assumption is alluring) sounds like a good reason not to introduce Uncertainty to Classical Mechanics to me.

Last edited: Oct 13, 2004
19. Oct 14, 2004

### Chronos

I think you have made a very good case pointing out there are problems between the two approaches.