Freeze time at Absolute Zero?

[browndn]

can u freeze[pause] time at 0K? because time is relative to mass and energy and if theoretically particles stop moving at 0K does that mean you can freeze time?

 

i heard people talk about that you cant because gravity is still acting on it. any thoughts?

[thedarkshade]

First you cannot reach 0K, everyone knows that!

 

And at 0K only the rotational movement of molecules stops, and they move in a collective way!

 

But how could you freeze time by freezing particles? Time is not made of particles, it's subjective!

[swansont]

You can't reach absolute zero, but atoms show no special temporal effects from getting them very, very cold, or slow, other than reducing the time dilation they might experience manifesting itself in a longer excited state lifetime.

[timo]

can u freeze[pause] time at 0K? because time is relative to mass and energy and if theoretically particles stop moving at 0K does that mean you can freeze time?

 

i heard people talk about that you cant because gravity is still acting on it. any thoughts?

- Time being relative to mass and energy does not sound true. The statement doesn't even make much sense to me. What should that mean?

 

- Different objects can have a different understanding of the time-direction similarly as the forward-direction of two cars differ when one travels to the north and the other one travels to the north-east. Seen from the perspective of the first car, there is a point where the 2nd car will have travelled 1 km into the forward direction (of the 1st car), i.e. 1 km north. At this point, however, the 2nd car will have travelled [math]\sqrt{2}[/math] km in its forward direction ([math]\sqrt{2}[/math] km north-east = 1 km north + 1 km east). In that sense, objects can disagree on a distance travelled in a certain direction. In the same sense (except for some different signs), objects can disagree on the distance travelled in the time direction.

 

- Consider two objects: Yourself as the observer in the laboratory and some gas molecule bouncing around. As said, you and the molecule can have a different understanding and disagree on the amount of time passed. Neglecting gravity and assuming the molecule to have a constant speed relative to you, the factor by which the two "distances" (=experienced time) differ is [math]\gamma^{-1}= \sqrt{1-v^2/c^2}[/math], where gamma is called the Gamov Factor, v is the velocity and c is the speed of light. You might (but shouldn't) say that time stops for the molecule, if [math]\gamma^{-1}[/math] approaches zero.

 

- Now, let's take a classical gas and assume the velocity of the molecules did approach 0 m/s as the temperature approaches 0 K. In that case, the factor by which their time measurements differ from yours approaches one. In other words: Their time-direction and their time measurements approach yours. That's pretty much the opposite of what you probably meant by time stopping.

 

- I see no good reason why someone would want to include gravity into that question (I see some bad reasons, though ).

 

In short: No, you can't freeze time at 0 K.

[Riogho]

If you were at absolute zero the atoms would still be moving.

[insane_alien]

If you were at absolute zero the atoms would still be moving.

 

then it would not be absolute zero as the atoms have kinetic energy which is proportional to temperature. or rather, temperature is proportional to kinetic energy.

[John Cuthber]

No, it's zero point energy. If they were stationary you would know their momentum and position exactly. That's a breach of the uncertainty principle.

They are moving but, since you can't stop them, they don't have energy that you can take from them. If you cn't take that energy away then you can't "cool" them so they must be at absolute zero.

(Strictly that only applies vibrational movement of atoms in a crystal lattice or a molecule, atoms of a gas could be stationary because you wouldn't know where they were so the uncertainty remains. OTOH, at absolute zero pretty much everythuing freezes)

WTF this has to do with time I don't know.

[swansont]

No, it's zero point energy. If they were stationary you would know their momentum and position exactly. That's a breach of the uncertainty principle.

They are moving but, since you can't stop them, they don't have energy that you can take from them. If you cn't take that energy away then you can't "cool" them so they must be at absolute zero.

 

 

You can't reach absolute zero, so that doesn't strictly follow. However, the third law of thermodynamics is classical, not QM, so this is a bit of apples-and-oranges.

[Riogho]

Everything must be in constant motion, you cannot be at rest. It's known as zero-point energy. Or simply an effect of the uncertainty principle.

[John Cuthber]

Actually, I think a single electron in an infinitely large empty box can be at rest. More importantly, even for a finite box ( and, therefore finite quantisation of translational energy) I think the ground state is the one with zero kinetic energy. The uncertainty principle isn't bothered because you don't know where in the box to find the particle.

[timo]

Actually, I think a single electron in an infinitely large empty box can be at rest. More importantly, even for a finite box ( and, therefore finite quantisation of translational energy) I think the ground state is the one with zero kinetic energy.

Not exactly sure what you mean by "at rest". Here's some thoughts:

 

Ground state of a particle in a box [0,L]: [math]\psi(x) = \alpha \sin (x\pi/L) \Theta(x) \Theta(L-x) [/math], with alpha some non-zero normalisation constant.

 

=>

1: [math]\left< p \right> = \left<\psi | \hat p \psi \right> = | \alpha |^2 \int_0^L dx \ \sin(x\pi/L) i\hbar \partial_x \sin(x\pi/L)[/math]=[math] i \hbar | \alpha |^2 \pi/L \int_0^L dx \sin(x\pi/L) \cos(x\pi/L) = 0[/math], with the zeroness of the integral following from the anti-symmetry of the integrand around L/2.

 

2:[math]\left< E_{\text{kin}} \right> = \left< p^2/2m \right>= \left< \psi| \hat p^2 \psi \right>/2m = |\alpha|^2 / 2m \int_0^L dx \ \sin(x\pi/L) (-\hbar^2 \partial_x^2) \sin(x\pi/L)[/math][math] =|\alpha|^2 \hbar^2 \pi^2/(2mL^2) \underbrace{\int_0^L dx \ \sin^2(x\pi/L)}_{=L/2} =|\alpha|^2 \hbar^2 \pi^2/(4mL) > 0[/math]. Comment for a large box: [math]|\alpha |^2 = 2/L \Rightarrow \lim_{L \to \infty} \left<E_{\text{kin}} \right> = 0[/math], in that case.

 

Time-dependent case: It's easy to see that for the ground state above, [math]\left< x \right>(t) = L/2 \Rightarrow v(t) \approx \frac{d}{dt}\left< x \right>(t) = 0[/math] because the additional term due to time, [math] \exp (-i E t/\hbar) [/math], cancels out for all t. Note that E = Ekin+Epot, where Epot depends on how you set up the value of the potential of the box (and that all of above statements hold true independently of Epot).