Zero-point energy + Thermal runaway

[Grimstone]

IF I am in error, forgive me. but I had this thought cross my mind today.
IF absolute 0 is the stopping of all atomic (or molecular) movement.

and you induce a thermal reaction, i.e. splitting the nuclei. would the reaction be slowed down at all?

(to make a slow burn)
Yea. I understand the effects of splitting the atom and the chain reaction associated with said event.

Yes I understand that I am not allowed to do such a experiment.
The better half has made this perfectly clear. (I"m not bailing you out)
please, Kis'is.
http://en.wikipedia.org/wiki/Absolute_zero
http://en.wikipedia.org/wiki/Thermal_runaway

 

[kurros]

IF I am in error, forgive me. but I had this thought cross my mind today.
IF absolute 0 is the stopping of all atomic (or molecular) movement.

and you induce a thermal reaction, i.e. splitting the nuclei. would the reaction be slowed down at all?
(to make a slow burn)
Yea. I understand the effects of splitting the atom and the chain reaction associated with said event.

I am not entirely clear on what you are asking, but it sounds like you are considering something like taking a lump of uranium and somehow cooling it to say a few nanoKelvin? And then perhaps shooting some neutrons into it to induce some fission?

If so, well, you don't need any neutrons because the uranium will naturally fission and, if your lump is big enough, sustain a chain reaction. And this natural fission rate will be entirely unaffected by the fact that your uranium is extremely cold; the induced fissions from the natural decay neutrons will also be practically unaffected. Your uranium atoms would have to moving very fast before their speed made much difference to their collisions with the neutrons, since the neutrons coming from the uranium decays are already going very fast, something like 10,000 km/s.

Even if you have some moderating material in there to slow the neutrons down to thermal velocities I am not sure that the uranium being ultra-cold would make a noticable difference.

 

[Drakkith]

IF I am in error, forgive me. but I had this thought cross my mind today.
IF absolute 0 is the stopping of all atomic (or molecular) movement.

This is incorrect. Read this definition from wiki:

Absolute zero is the theoretical temperature at which entropy reaches its minimum value. The laws of thermodynamics state that absolute zero cannot be reached using only thermodynamic means. A system at absolute zero still possesses quantum mechanical zero-point energy, the energy of its ground state. The kinetic energy of the ground state cannot be removed. However, in the classical interpretation it is zero and the thermal energy of matter vanishes.

As you can see, absolute zero is not the absence of ALL movement, it is the point at which no more energy can be removed from the system. I also think that absolute zero applies only to a bulk material, IE a system of particles that we can assign a temperature to. It doesn't make much sense to try to assign a temperature to a single atom or molecule.

Simply put you cannot get energy out of a system at absolute zero by thermodynamic means. I don't think nuclear fission is a thermodynamic process, as you are not working with heat.

 

[Crazymechanic]

I would understand that , while being at absolute zero , still the neutrons of uranium have to go somewhere and while they are going and creating a new chainreaction all the time the byproduct of this would be more and more heat , so if you would want to keep it cold all the time you would just remove the heat energy but not the particles doing their job just the byproduct.It's like putting a candle in a ventilated but very cold fridge as long as there is oxygen you can have whatever temp you can achieve the candle would still burn until the fuel ends...
Am I right on this one?

 

[Drakkith]

I would understand that , while being at absolute zero , still the neutrons of uranium have to go somewhere and while they are going and creating a new chainreaction all the time the byproduct of this would be more and more heat , so if you would want to keep it cold all the time you would just remove the heat energy but not the particles doing their job just the byproduct.It's like putting a candle in a ventilated but very cold fridge as long as there is oxygen you can have whatever temp you can achieve the candle would still burn until the fuel ends...
Am I right on this one?

First and foremost, we cannot reach absolute zero.
Now, let's say, for the sake of argument, that you have a block of uranium at 0k. One atom decays and releases energy and neutrons. The block of uranium absorbs at least part of this energy and is no longer at 0k. It's that simple.

 

[Grimstone]

Thank you Kurros.

I was asking about (a slow burn) so then the proximity to absolute 0 would have no effect on the rate of the reaction.

 

[Grimstone]

Drakkith
thank you for using the same info I used to pose the question.
You sadly missed my question.
I do thank you for your response tho.

 

[Drakkith]

Drakkith
thank you for using the same info I used to pose the question.
You sadly missed my question.
I do thank you for your response tho.

Yes, I think I must have misunderstood what you were asking. I think I was very tired yesterday, as I just moved.

 

[kurros]

Thank you Kurros.

I was asking about (a slow burn) so then the proximity to absolute 0 would have no effect on the rate of the reaction.

Pretty much, yeah. I think the neutrons would also have to be ultra cold before one started to notice the impact the "ultra-coldness" was making. I could believe that say an ultra-cold gas of uranium atoms and neutrons might have some odd properties, but I don't imagine any method is currently known for cooling neutrons by that much, especially if they are originating from gaseous uranium, from which they whiz out at extremely high speed. Not to mention it may be impossible to trap neutrons... also probably they don't cause any fissions if they are so very cold, I guess they just slowly "bounce" off the uranium atoms... but I am just being silly now I think :).