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Author Topic: Do neutrons in a neutron stars bend Heisenberg's Uncertainty Principle?  (Read 20660 times)

Offline lightarrow

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That's exactly what I'm getting at, though. You can compress & compress until you get to the quarks. You can't compress quarks as they're fundamental.

You see, this is 1 of the things I don't get about Heisenberg's principle. If you could stop the neutrons moving by squeezing them tightly enough together the quarks would still have as much freedom of movement as ever. But you could know both the positions (stationary) and momenta (zero) of the neutrons. So where does HUC stop? At fundamental particles or with composite particles?

I only used the marbles as a very simplistic analogy.
There are two things I don't understand:
1. why you don't ask about a simple atomic nucleus? There's no much difference between nuclear matter and neutron star matter.
2. why do you think that neutrons cannot move in a neutron star?
 

Offline DoctorBeaver

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How do you propose to measure the position or velocity of a neutron, considering its made up of quarks?  You'd probably have to somehow measure some property of the quarks.  Therefore, if all the quarks obey the uncertainty principle, shouldn't the neutron position necessarily obey the uncertainty principle?

I think that depends on how neutrons are thought of. Are they considered entities in their own right, albeit composite entities? Is the quantum wave function of a neutron a function of the combination of the wave functions of the quarks? Or is its wave function independent of the quarks? I don't know enough about QM to answer that. However, I believe that the uncertainty principle only applies if the object is smaller than its wave function; am I correct in that belief?
 

Offline DoctorBeaver

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There are two things I don't understand:
1. why you don't ask about a simple atomic nucleus? There's no much difference between nuclear matter and neutron star matter.
2. why do you think that neutrons cannot move in a neutron star?

1. A simple atomic nucleus has electron orbits. A neutron in a neutron star doesn't.
2. I don't think that because I don't know. That's why I'm asking this question.
 

Offline lightarrow

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1. A simple atomic nucleus has electron orbits. A neutron in a neutron star doesn't.
Ok, it has electrons around; so? Protons and neutrons in an atomic nucleus are not more free than neutrons in a neutron star.
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2. I don't think that because I don't know. That's why I'm asking this question.
Sincerely I don't know too ;). However there's an old model of the nucleus called "liquid drop model". Just the name makes me think that a nucleus couldn't be thought of as a "clump of marbles", and the same (following this logic) for a neutron star.
 

Offline DoctorBeaver

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Ok, it has electrons around; so? Protons and neutrons in an atomic nucleus are not more free than neutrons in a neutron star.

That may or may not be. In an atomic nucleus there is not 31012 times Earth's gravity pressing them together. And that's just the surface gravity. I don't know what it is near the centre of a neutron star. That's what my question is about - what is the effect on the neutrons of that amount of gravity. Does it squeeze them together in such a way that their position & momentum could be known simultaneously? I don't see that talking about hydrogen atoms or neutrons in an atom addresses that question.
 

Offline JP

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How do you propose to measure the position or velocity of a neutron, considering its made up of quarks?  You'd probably have to somehow measure some property of the quarks.  Therefore, if all the quarks obey the uncertainty principle, shouldn't the neutron position necessarily obey the uncertainty principle?

I think that depends on how neutrons are thought of. Are they considered entities in their own right, albeit composite entities? Is the quantum wave function of a neutron a function of the combination of the wave functions of the quarks? Or is its wave function independent of the quarks? I don't know enough about QM to answer that. However, I believe that the uncertainty principle only applies if the object is smaller than its wave function; am I correct in that belief?

The uncertainty principle appears because of the way in which measurements effect a quantum system.  Its actually quite a natural phenomenon in dealing with waves, so it arises naturally if you think of particles as represented by waves.  If you want to go for the most fundamental widely-accepted theory, you'd treat the neutron as a composite of quarks interacting via the strong force.  Then you'd have to figure out what your measurement does to the quarks.  Then you'd have to define the "position" of a neutron in terms of the quark measurements.  However, fundamentally each quark should obey an uncertainty relation, and this uncertainty relation should carry over to the neutron.

If you're willing to go with the most basic model of QM, which is "good-enough" in a lot of cases, you can treat the neutron as a single particle, and because any single particle has to obey an uncertainty relation, it should as well.

For your final question about the uncertainty relation only applying when the wave function is smaller than the particle--I'm not sure I've ever heard of an object being bigger than its wave function.  Fundamental particles are thought of as point-like, so they're smaller than any wave function.  The "size" of a more complicated object is tricky, since most of it is empty space, with forces holding these point-like particles apart.  
 

Offline DoctorBeaver

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JP - I understand what you're saying, but particles also behave like particles. I take your point about the uncertainty associated with the constituent quarks being carried over to the neutron as well. In which case, what about quarks in a quark star (theoretical, I know)? I can't help getting the feeling that there must come a point where everything is squeezed together so tightly that nothing can move about at all.
 

lyner

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particles also behave like particles.
But you can't predict what they're going to do without doing some wave calculations.

What is so attractive about the notion of everything being squashed to tight that it can't move?
It's like wanting to get to absolute zero temperature and traveling at the speed of light. The energy involved in getting things closer
and closer together just increases without limit.
 

Offline DoctorBeaver

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It's not that I find the idea attractive. It's just something I was wondering.

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The energy involved in getting things closer and closer together just increases without limit.

Ah, at last an answer. So squeezing things together that closely is impossible because of energy constraints. That makes sense. Thank you.
 

lyner

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The Noddy answers are always the best.
I'm just waiting for lightarrow to point out something wrong with such a simple answer.
 

Offline DoctorBeaver

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I am a Noddy when it comes to this sort of stuff.
 

Offline LeeE

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Ok, it has electrons around; so? Protons and neutrons in an atomic nucleus are not more free than neutrons in a neutron star.

That may or may not be. In an atomic nucleus there is not 31012 times Earth's gravity pressing them together. And that's just the surface gravity. I don't know what it is near the centre of a neutron star. That's what my question is about - what is the effect on the neutrons of that amount of gravity. Does it squeeze them together in such a way that their position & momentum could be known simultaneously? I don't see that talking about hydrogen atoms or neutrons in an atom addresses that question.

Are you not mistaking pressure for gravity here?  The gravity will be strongest at the surface and will be zero at the center of the neutron star.

The estimated max density of a neutron star (at it's center) is only up to around three times the density of an atomic nucleus and, when you remember that the neutron has a slightly higher mass than the proton anyway, neutron stars are still in the same league of density as atomic nuclei.  Sure, two or three times isn't insignificant but it's not really an order of magnitude of difference, so I wouldn't expect radically different behaviour.
 

Offline DoctorBeaver

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LeeE - My mistake about the gravity. Yes, I was referring to density & pressure.

Is it really on 3 times greater than an atomic nucleus? I'm sure I read it was many orders of magnitude greater. Maybe I got that confused too.
 

Offline LeeE

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I wasn't sure before, but that's the figure given by the wiki article...

http://en.wikipedia.org/wiki/Neutron_star

which states:

"Neutron stars have overall densities predicted by EOS APR of 3.7 1017 (2.6 1014 times Solar density) to 5.9 1017 kg/m (4.1 1014 times Solar density), which compares with the approximate density of an atomic nucleus of 3 1017 kg/m."

...and which includes the upper figure you stated.  The next sentence says:

"The neutron star's density varies from below 1 109 kg/m in the crust increasing with depth to above 6 or 8 1017 kg/m deeper inside."

...so we're looking at around a max of 3x atomic nuclei density.  Heh - I think I remembered it because I think I expected it to be higher too.
 

Offline DoctorBeaver

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Thanks for that. I was under the impression that the EOS for neutron stars isn't known. Is my information out of date?
 

Offline lightarrow

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That may or may not be. In an atomic nucleus there is not 31012 times Earth's gravity pressing them together. And that's just the surface gravity. I don't know what it is near the centre of a neutron star. That's what my question is about - what is the effect on the neutrons of that amount of gravity. Does it squeeze them together in such a way that their position & momentum could be known simultaneously? I don't see that talking about hydrogen atoms or neutrons in an atom addresses that question.
I don't know if it's possible to treat a singol neutron in a neutron star as individually distinct from all the others and so if it's correct to say that its wavefunction there is as more spatially localized as the pressure increases. It could even result that its wavefunction becomes less spatially localized (and so its position indeterminacy increases instead of decrease), going to take all the star's volume. For example, if you compress hydrogen atoms very very much (don't remember, maybe millions of atmospheres) it becomes a metal, because the atom's electrons becomes delocalized over all the bulk of material.
 

Offline DoctorBeaver

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It could even result that its wavefunction becomes less spatially localized (and so its position indeterminacy increases instead of decrease), going to take all the star's volume.

My poor little brain is having trouble with that.
 

Offline LeeE

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Is my information out of date?

Heh - I think that unless you're actively working in a field, the answer is usually yes  ;D
 

Offline DoctorBeaver

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Is my information out of date?

Heh - I think that unless you're actively working in a field, the answer is usually yes  ;D

I can't recall the last time I worked in a field apart from clearing up horse poo.
 

Offline lightarrow

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It could even result that its wavefunction becomes less spatially localized (and so its position indeterminacy increases instead of decrease), going to take all the star's volume.

My poor little brain is having trouble with that.
Microscopic objects like neutrons are not tiny balls; they are particles and waves. Imagine an electron in an atom. Do you think it's a small ball rotating around the nucleus, there? Or do you think it's a cloud smeared on the entire orbital? If you compute its position indeterminacy with HUP (Heisenberg Uncertainty Principle) you find that it equals the entire orbital. Now think to what happens if you clump together many atoms of Iron, for example. When they are still not very close, the volumes pertaining to the electrons are still the same, but as soon as the atoms bind together, the external electrons take the volume of all the external orbitals of the atoms, becoming so much more free. For this reason Iron is a metal, that is electrons are free to move inside of it.
 

Offline DoctorBeaver

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I think I understand that. Thank you.
 

lyner

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I can't recall the last time I worked in a field apart from clearing up horse poo.

A farmer is an expert: A man out_standing in his field.
 

Offline DoctorBeaver

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I can't recall the last time I worked in a field apart from clearing up horse poo.

A farmer is an expert: A man out_standing in his field.



It will be a farmer who invents the tractor beam  :D
 

Offline syhprum

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I allways thought that temperature was a measure of the kinetic energy of the particles making up a body, what happens to the concept of temperature when they are packed so tight that they cannot move ?
 

Offline lightarrow

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I allways thought that temperature was a measure of the kinetic energy of the particles making up a body, what happens to the concept of temperature when they are packed so tight that they cannot move ?
It's an interesting question and I sincerely cannot state to have a complete answer. However, if you take a gas inside a cylinder and you reduce the cylinder's volume isothermically (constant temperature), what you get is to reduce the particle's average free path, not their speeds, so you will have a constant increase of the collision's frequency; if you have a liquid or a solid it's quite the same, with a solid the only difference is that now you will talk of particle's vibrations around their equilibrium points, not of free movement, but the concept is quite the same.

I don't expect a much different behaviour from neutrons in a neutron star.
 

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