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kordic
2007-Aug-23, 04:57 AM
Hi all, sorry if this has been asked before. I remember hearing somewhere that a neutron star has a "solid" surface. What exactly is it comprised of? I thought a neutron star was only made of neutrons? Are the neutrons the "solid" in the surface?

Neverfly
2007-Aug-23, 06:35 AM
I should post here so that someone more knowledgable will feel constrained to correct me:D

http://www.astro.umd.edu/~miller/nstar.html
This is a quick link to a long read. This website discusses Neutron Stars in Pretty good detail.

http://www.astroscu.unam.mx/neutrones/INT/Workshop.html

This is a little heavier handed reading and you may need to fiddle with the page a bit to get it to where you can comfortably read.

A neutron Star is Dense.
At such a high compression of gravity, we "could" refer to it as solid or hard.
Think past our notions of what is solid if you can imagine it. A solid counter top has a HUGE amount of space between the elements its made of on the atomic level. Now imagine if you took that counter top and squeezed on it so hard that you push the atoms closer together. The counter top takes up less space, but retains the same mass. Keep compressing until there is little space left. You will have a tiny countertop that you would need a microscope to see. But good luck finding a way to pick it up to place it on the scope- it still has the same mass as it did before you squeezed it:p

For a bit of science fiction, try reading Robert L Forwards "The Dragon's Egg".
The science isnt too bad and he discusses the conditions on a neutron star as he speculates about life on one.

Tim Thompson
2007-Aug-23, 07:26 AM
Surface
A neutron star is made of neutrons only in the core, where the material is dense enough. The surface of a neutron star has a density about 106 gm/cm3, and is made mostly of close packed 56Fe nuclei, because that is the most stable nucleus. But the details are sensitive to the strength of the magnetic field, which varies from star to star.

Outer Crust
The layer beneath the surface, called the outer crust is dense enough (~106 - 1011 gm/cm3) to become electron degnerate, and so strongly resembles the material of a white dwarf. Inverse beta decay will generate a lot of nuclei other than 56Fe, such as 62Ni, 80Zn, 118Kr and maybe even heavier nuclei.

Inner Crust
Beneath that is the inner crust, with densities between ~1011 and 1014 gm/cm3. In the lower density region there is a mix of neutron rich nuclei, degenerate free neutrons and degnerate electrons. In the higher density regions, individual nuclei become increasingly lost in a neutron liquid, with very few free electrons or protons.

Core
The core region lies below the inner crust. Here the density exceeds 1015 gm/cm3, and we can no longer say with any real certainty what happens. It all depends on the equation of state for high density matter, and that is a matter of considerable research. The physics here is also complicated by the fact that neutrons are fermions, which can become superconducting and/or superfluid under core conditions. Theree may also be a mixture of exotic particles (called hyperons, I believe), which are implied by some particle theories & high energy equations of state. The possibility of free quarks has also been suggested (even to the extent of suggesting that there are quark stars at higher densities than neutron stars).

I can't say at the moment how thick these layers are, but since a neutron star is about the size of a county, they obviously don't amount to much. And don't forget that even neutron stars have atmospheres, which can be as complicated as their interior structures.

I used the book Theoretical Astrophysics, volume II: Stars and Stellar Systems (http://www.amazon.com/Theoretical-Astrophysics-II-Stellar-Systems/dp/0521566312/ref=sr_1_3/104-0340796-5591110?ie=UTF8&s=books&qid=1187853199&sr=1-3) by T. Padmanabhan as a reference here. The 3 volume set is an excellent collection for advanced astrophysics. There is a huge literature on the topic. My copy of Norman Glendenning's Compact Stars: Nuclear Physics, Particle Physics, and General Relativity (http://www.amazon.com/Compact-Stars-Relativity-Astronomy-Astrophysics/dp/0387947833/ref=sr_1_2/104-0340796-5591110?ie=UTF8&s=books&qid=1187853351&sr=1-2) is in my office, or I would likely have used it instead, but I keep Padmanabhan around the house for those astrophysical moments. Also see Dense Matter in Compact Stars: Theoretical Developments and Observational Constraints (http://adsabs.harvard.edu/abs/2006ARNPS..56..327P), Page & Reddy, Annual Review of Nuclear and Particle Systems 56(1): 327-374, November 2006, for a recent review on the current state of research on dense matter equations of state.

kzb
2007-Aug-23, 12:22 PM
This is from the first link posted by neverfly:

"We'll talk about neutron star evolution in a bit, but let's say you take your run of the mill mature neutron star, which has recovered from its birth trauma. What is its structure like? First, the typical mass of a neutron star is about 1.4 solar masses, and the radius is probably about 10 km. By the way, the "mass" here is the gravitational mass (i.e., what you'd put into Kepler's laws for a satellite orbiting far away). This is distinct from the baryonic mass, which is what you'd get if you took every particle from a neutron star and weighed it on a distant scale. Because the gravitational redshift of a neutron star is so great, the gravitational mass is about 20% lower than the baryonic mass."

It's that last sentence that is interesting to me. The gravitational pull is reduced by redshift. What does this say about black holes, is the gravity redshifted to zero?

Neverfly
2007-Aug-23, 12:48 PM
This is from the first link posted by neverfly:

"We'll talk about neutron star evolution in a bit, but let's say you take your run of the mill mature neutron star, which has recovered from its birth trauma. What is its structure like? First, the typical mass of a neutron star is about 1.4 solar masses, and the radius is probably about 10 km. By the way, the "mass" here is the gravitational mass (i.e., what you'd put into Kepler's laws for a satellite orbiting far away). This is distinct from the baryonic mass, which is what you'd get if you took every particle from a neutron star and weighed it on a distant scale. Because the gravitational redshift of a neutron star is so great, the gravitational mass is about 20% lower than the baryonic mass."

It's that last sentence that is interesting to me. The gravitational pull is reduced by redshift. What does this say about black holes, is the gravity redshifted to zero?

As I understand SR, the gravitational redshift must be finite.

R_\mu\sigma=0


ETA: I'm looking into it and doing a little math, so not trying to keep you hanging.
See if anyone else posts with bigger brains:p

Hornblower
2007-Aug-23, 01:47 PM
Because the gravitational redshift of a neutron star is so great, the gravitational mass is about 20% lower than the baryonic mass."

It's that last sentence that is interesting to me. The gravitational pull is reduced by redshift. What does this say about black holes, is the gravity redshifted to zero?
If the gravity was redshifted to zero, reputed black holes would not give us any evidence of their existence.

I think what we have here is some lousy technical writing by Dr. Miller. He understands mathematically what is going on, but his attempt at putting it into plain English may be garbled and/or incomplete. What we need is a mathematical walk-through of the transition from the uncompressed state to the neutron star state, showing whatever transformation which leads to the calculated amount of whatever it is that he is calling "baryonic mass".

Tim Thompson
2007-Aug-23, 03:44 PM
This is from the first link posted by neverfly:

"... This is distinct from the baryonic mass, which is what you'd get if you took every particle from a neutron star and weighed it on a distant scale. ...
When neutrons & protons bind together into a nucleus, some of their mass is converted into binding energy, which holds the nucleus together against the proton electrostatic repulsion (that's where the strong force energy comes from). Hence the phenomenon that the mass of the nucleus is less than the sum of the masses of its constituent particles, where those masses are the free particle masses. Likewise, in a neutron star, some of the mass of the particles is converted into binding energy. So, the mass of the neutron star, which looks like a giant nucleus, should be less than the sum of the free particle masses of its constituent particles. The conversion is just E = mc2, where the mass m is converted into binding energy E.

In both cases the binding energy does not show up as an equivalent inertial mass, as in m=E/c2, and so by the equivalence principle, it cannot show up as gravitational mass either.

Ken G
2007-Aug-23, 06:57 PM
The gravitational pull is reduced by redshift. What does this say about black holes, is the gravity redshifted to zero?

No, I believe the mass-energy of the black hole would be the same as if you took all the matter that fell into it and arranged it in a hollow sphere just outside its event horizon. Then you'd incorporate the correct amount of negative gravitational binding energy into the mass-energy, and it wouldn't be that much different from the neutron star result-- maybe it's a 30% or 40% reduction instead of 20%, I don't know, but neutron stars aren't that far from black holes in terms of their gravitational effects on other things.

Ken G
2007-Aug-23, 07:01 PM
I think what we have here is some lousy technical writing by Dr. Miller. He understands mathematically what is going on, but his attempt at putting it into plain English may be garbled and/or incomplete.I think it's pretty accurate, the use of the term "redshift" instead of "gravitational self-energy" might be a bit confusing.
What we need is a mathematical walk-through of the transition from the uncompressed state to the neutron star state, showing whatever transformation which leads to the calculated amount of whatever it is that he is calling "baryonic mass".
It sounds like you are worried about the rest energy changes that go on in these transitions, but I don't think that's actually the issue-- it's the gravitational self-energy as the object contracts. Whether it is neutrons or iron nuclei is very much a second-order correction, the internal binding energies of these things is much smaller than the gravitational self-energy we are talking about. That may not be true for hyperons or quark stars, and even the binding of the neutrons may be uncertain to a large degree, but I'll bet that 20% number he quotes is essentially all just mundane gravitational self-energy that has nothing at all to do with nuclear transitions. Where you need the equation of state is in figuring out what that gravitational self-energy will be-- essentially, to know what the radius will be.

Ken G
2007-Aug-23, 07:06 PM
Likewise, in a neutron star, some of the mass of the particles is converted into binding energy. So, the mass of the neutron star, which looks like a giant nucleus, should be less than the sum of the free particle masses of its constituent particles. But again, I believe most of that binding energy is gravitational not nuclear-- and that's what Miller is saying as well.


In both cases the binding energy does not show up as an equivalent inertial mass, as in m=E/c2, and so by the equivalence principle, it cannot show up as gravitational mass either.
Actually, I would say it does show up as equivalent mass, but it is negative-- that's the point. A binding energy always reduces the energy of a system by introducing a negative component, which also shows up in the equivalent mass. You of course have to extract the resulting excess energy to get the binding to occur, presumably that means neutron stars are very bright while they are forming (can you say, supernova?).

Another interesting point here is, how can neutron stars, with relativistic degenerate neutrons, support a mass of 1.4 solar masses, when relativistic degenerate electrons (white dwarf) cannot? Shouldn't relativistic electrons have the same degenerate properties as relativistic neutrons? It's because the nucleons aren't relativistic until the star is much smaller than a white dwarf, and in getting smaller, you bring in all that negative gravitational self-energy that effectively lowers the mass of the star below the Chandrasekhar mass. The process doesn't end until you have the supernova, which happens when you do reach the Chandrasekhar mass in the core after including all that negative gravitational self-energy. By itself, according to Miller, this process would explain why you need 1.7 solar masses of baryon rest masses to get a supernova, not 1.4-- you just end up with 1.4 equivalent solar masses in your neutron star. If you want to turn that into a black hole, you have to make it contract even more, and only then do you really get into all that bizarre nuclear stuff. I don't think you really need that at the Chandrasekhar limit.

trinitree88
2007-Aug-24, 12:12 PM
And, as the neutron star forms, the nucleons change from roughly spheroidal to bean-shapes. It allows for denser packing , and gives the object a radial symmetry. (MIT, Astrophysics colloquia, circa 2004) Pete.

Neverfly
2007-Aug-24, 12:58 PM
And, as the neutron star forms, the nucleons change from roughly spheroidal to bean-shapes. It allows for denser packing , and gives the object a radial symmetry. (MIT, Astrophysics colloquia, circa 2004) Pete.

Wow! I wonder how the neucleons knew that! I wouldn't have thought to do that.

:lol::lol:

neilzero
2007-Aug-24, 01:25 PM
Likely Tim Tompson is correct about iron nuclii on the surface of a neutron star, or the next most stable nuclii if the magnetic field shifts the iron elsewhere. At millions of g gravity, 25,000 degrees k and few electrons to go with the nuclii, usual meanings of gas, liquid and solid may not apply. The analogy that pops in my head is liquid mercury seems like a solid surface to a feather. Neil

mugaliens
2007-Aug-25, 12:42 PM
If the gravity was redshifted to zero, reputed black holes would not give us any evidence of their existence.

I think what we have here is some lousy technical writing by Dr. Miller. He understands mathematically what is going on, but his attempt at putting it into plain English may be garbled and/or incomplete. What we need is a mathematical walk-through of the transition from the uncompressed state to the neutron star state, showing whatever transformation which leads to the calculated amount of whatever it is that he is calling "baryonic mass".


Hmmm... Yes indeedy...

But this is just beautiful, regardless!

Hmmm.. Try this:

http://www.astro.umd.edu/~miller/Images/crabchandra.jpg

trinitree88
2007-Aug-26, 01:42 AM
Wow! I wonder how the neucleons knew that! I wouldn't have thought to do that.

:lol::lol:

Neverfly.:eek: Don't tell anybody....there's a secret society of nucleons...with a da pinci code, for when dey gets squeezed down in a neutron star, youse see?...and dey holds dere annual convention, replete with feathered hats,and tooting horns, in Frankfurt Germany...

so it's dem beans and franks, wid a little bit o salt pork in every neutron star.Pass da ketchup, please, and da conference ain't da only place where it's rootin, tootin...:shifty::D
pete

Neverfly
2007-Aug-26, 05:48 AM
My lips have bean sealed.

SteveA
2007-Aug-26, 07:04 PM
Neutron stars are NON EXISTANT!! They are "apperntly" comprised of atoms with a nucleus of mainly neutrons, which would make them extremly dense, almost 1/10th that of water. However it is well known throughout physics that a neutron based nucleus would be extrmely unstable, as a neutron would decay very quickly (in a matter of minutes) into a proton! An atom needs to have roughly the same number of neautrons as protons (and their associated electorns) and therefore a neutron star cannot be a real world object.

Neverfly
2007-Aug-26, 07:16 PM
Neutron stars are NON EXISTANT!! They are "apperntly" comprised of atoms with a nucleus of mainly neutrons, which would make them extremly dense, almost 1/10th that of water. However it is well known throughout physics that a neutron based nucleus would be extrmely unstable, as a neutron would decay very quickly (in a matter of minutes) into a proton! An atom needs to have roughly the same number of neautrons as protons (and their associated electorns) and therefore a neutron star cannot be a real world object.

Uhhhh... Present evidence please. As well as show your math.
Knowledge in Quantum Physics is a requirement - not an option;)

Ken G
2007-Aug-26, 07:36 PM
Neutron stars are NON EXISTANT!! They are "apperntly" comprised of atoms with a nucleus of mainly neutrons, which would make them extremly dense, almost 1/10th that of water. They are a whole lot denser than that, are you kidding? Water is denser than that, you obviously didn't mean what you said. I'm afraid that casts doubt on everything else you are saying, if there wasn't already plenty of that.

However it is well known throughout physics that a neutron based nucleus would be extrmely unstable, as a neutron would decay very quickly (in a matter of minutes) into a proton! An atom needs to have roughly the same number of neautrons as protons (and their associated electorns) and therefore a neutron star cannot be a real world object.What is "well known" is how to determine the lowest energy of a system of neutrons and protons-- and that comes out fairly equal numbers in small nuclei, but not in neutron stars. We don't know all the physics of a neutron star, but that's the easy part!

Delvo
2007-Aug-26, 08:00 PM
How does a neutron star accumulate so many neutrons and/or lose so many protons and electrons? I do know that one form of radioactive decay of heavy metals is for a neutron to decay into a proton and an electron. Does that mean that density imposed by gravitational compression can cause electrons and protons to have their available space so constricted that they do the opposite of that decay, combining to form neutrons? If so, I guess that would answer my other question, of exactly how the electrons fail to keep the star "propped up" at a larger size anymore: they quit doing that because they aren't there anymore!

Tim Thompson
2007-Aug-26, 11:15 PM
How does a neutron star accumulate so many neutrons and/or lose so many protons and electrons? I do know that one form of radioactive decay of heavy metals is for a neutron to decay into a proton and an electron.
That's the beta decay process. In the case of a free neutron, the mean lifetime against beta decay is 885.7 ± 0.8 seconds, according to the 2006 version of the Particle Data Group (http://pdg.lbl.gov/) summary tables. The timescale of neutron beta decay in bound nuclei, however, depends on the particulars of the nucleus. You can find that information in a table of nuclides (http://atom.kaeri.re.kr/ton/index.html).


Does that mean that density imposed by gravitational compression can cause electrons and protons to have their available space so constricted that they do the opposite of that decay, combining to form neutrons? If so, I guess that would answer my other question, of exactly how the electrons fail to keep the star "propped up" at a larger size anymore: they quit doing that because they aren't there anymore!
Exactly correct. The pressure drives the electrons into the protons and they beconme neutrons. Electron degeneracy is what holds up white dwarf stars. Neutron stars are supported by a similiar pressure, but in this case provided by neutrons instead of electrons.

Ken G
2007-Aug-26, 11:57 PM
If so, I guess that would answer my other question, of exactly how the electrons fail to keep the star "propped up" at a larger size anymore: they quit doing that because they aren't there anymore!

No, that puts the cart before the horse. You don't lose the electrons until you get the superhigh density, and you don't get the superhigh density until electron degeneracy pressure has already failed to support the star. The reason for the latter is that if the mass is above 1.4 solar masses, electron degeneracy pressure cannot support the star against gravity, nor can degeracy pressure in any ideal gas. Neutrons fare a bit better expressly because the additional contraction has released a lot of gravitational energy that effectively lowers the mass below 1.4 solar masses by about 20%.

Fortunate
2007-Aug-27, 07:56 PM
From an entirely different angle: Here is a press release about relevant recent activities of the experimentalists.

"Knowing...allows physicists to describe the 'stiffness,' or 'equation of state,' of matter...inside these [neutron stars]."

http://www.physorg.com/printnews.php?newsid=107433430

trinitree88
2007-Aug-30, 11:57 AM
That's the beta decay process. In the case of a free neutron, the mean lifetime against beta decay is 885.7 ± 0.8 seconds, according to the 2006 version of the Particle Data Group (http://pdg.lbl.gov/) summary tables. The timescale of neutron beta decay in bound nuclei, however, depends on the particulars of the nucleus. You can find that information in a table of nuclides (http://atom.kaeri.re.kr/ton/index.html).


Exactly correct. The pressure drives the electrons into the protons and they beconme neutrons. Electron degeneracy is what holds up white dwarf stars. Neutron stars are supported by a similiar pressure, but in this case provided by neutrons instead of electrons.


Tim Thompson. Nitpik. An electron plus a proton will not make a neutron, so the pressure can not drive them together. When a neutron decays, it produces a proton, an electron, and an electron type antineutrino. That's beta decay, and it's energetically favorable.
If you want to reverse the process, you must have all three particles come together at the same point in time, and that's particularly unfavorable with regards to the antineutrino fluxes, and the energetics.So, How does it happen?
The key thing to visualize is the weak interaction mediated by a weak boson. There are three...W+, W-, and W0.When normal beta decay occurs, a down quark in the neutron (2 downs, 1 up)..converts to an up quark, as it emits a W-. The W- disintegrates into an electron, and an electron-type antineutrino. The (up,up,down) combination now constitutes a proton. In space, the free neutron would disappear, and an atom of hydrogen forms from the decay. Each up quark has fractional electric charge of 2/3, so two of them are 4/3, and the down is minus 1/3, for a net charge of plus 1 for the proton.
When inverse beta decay occurs, the proton uses a W+, emitted by an up quark, as it converts to a down....making a neutron (2 downs, 1 up....2 minus 1/3 charge, 1 plus 2/3 charge for zero electrically)
the W+ disintegrates into a positron (which annihilates a local electron into two or three gamma rays, 180 degrees or co-planar..), and an electron-type neutrino. All conservation laws are upheld. The conversion of kinetic energy to mass is endothermic, and the escape of the neutrinos where opacity is not present, is endothermic to the region. Coupled with the lack of electric charge, the neutrons can gravitationally aggregate, and bind into a giant nucleus. The "strong" nuclear force is actually a shadow of the color force exerted between the quarks, which is mediated by colored gluons in quantum chromodynamics. So, even electrically neutral neutrons still want to bind if in close contact.
As KenG has said... mass is effectively converted to gravitational and binding energy. Pete.

mugaliens
2007-Aug-30, 02:47 PM
What is "well known" is how to determine the lowest energy of a system of neutrons and protons-- and that comes out fairly equal numbers in small nuclei, but not in neutron stars. We don't know all the physics of a neutron star, but that's the easy part!

Generally, charge is conserved, so compress a molecule enough and bingo - a bunch of neutrons.

When these "unstable" neutrons decay, where does the electron go? Well, somewhere until it hits another proton...

I don't understand the "instability" comment, though, as neutrons and protons are strongly bound in the nucleus, and most nuclei are relatively stable. In the immense gravitational gradient of the neutron star I would simply assume the gravity is able to overcome the instability of large-scale neutron-only "molecules" or that the compression factor is such that it causes the strong nuclear forces to somehow align and become superconduction, much like metallic hydrogen.

trinitree88
2007-Aug-30, 08:31 PM
Generally, charge is conserved, so compress a molecule enough and bingo - a bunch of neutrons.



Not true. Without the addition of the electron-type antineutrino, spin is not conserved, nor Electron Family number....never been seen in a particle physics lab. pete

When these "unstable" neutrons decay, where does the electron go? Well, somewhere until it hits another proton...

I don't understand the "instability" comment, though, as neutrons and protons are strongly bound in the nucleus, and most nuclei are relatively stable. In the immense gravitational gradient of the neutron star I would simply assume the gravity is able to overcome the instability of large-scale neutron-only "molecules" or that the compression factor is such that it causes the strong nuclear forces to somehow align and become superconduction, much like metallic hydrogen.

In the lighter nuclei...elements in the periodic table...the electron is emitted as a high speed beta particle. Even in the gravitational grip of a neutron star, neutrons decay at the poles due to the fields there. pete see the "pulsar with a tail" thread.:shifty: pete

mugaliens
2007-Aug-30, 10:19 PM
Understand there will be some losses. Thanks.

Northwind
2007-Aug-31, 01:30 AM
SteveA wrote
Neutron stars are NON EXISTANT!! They are "apperntly" comprised of atoms with a nucleus of mainly neutrons, which would make them extremly dense, almost 1/10th that of water. However it is well known throughout physics that a neutron based nucleus would be extrmely unstable, as a neutron would decay very quickly (in a matter of minutes) into a proton! An atom needs to have roughly the same number of neautrons as protons (and their associated electorns) and therefore a neutron star cannot be a real world object.

So have "we" observed a Neutron star yet? 'cos afaik "Current understanding of the structure of neutron stars is defined by existing mathematical models" and "A neutron star is one of the few possible conclusions of stellar evolution." and if stellar evolution is wrong?? LINK (http://en.wikipedia.org/wiki/Neutron_stars) :whistle:

So so many holes in the mathematical model, when it's referenced against the real world observations that like the BB model it's been adjusted, tampered, changed and rewritten that it makes it a complete farce.

Cos the types of Neutron stars out there all hold one other common property! :shifty:

Magnetic fields :doh:

No electric field, no magnetic field, it's as simple as that! (Don't tell the existing mathematical models this though)

Charges do not only produce electric fields. As they move, they generate magnetic fields, and if the magnetic field changes, it generates electric fields. This "secondary" electric field can be computed using Faraday's law of induction

And the easiest way to make radio noise? X-rays? Gamma rays?... A strong electric field (http://en.wikipedia.org/wiki/Electric_field) :doh:


Radio emitting
X-ray pulsar
Soft gamma repeaters

So I'm at a complete loss as to why we need such exotic particles and interactions, when KISS will do, after all simple lightning can do this, terrestrial gamma ray flashes (TGFs) (http://www.physorg.com/news3959.html).

Also check out the Vela SNR http://www.astro.psu.edu/users/green/Main/vela_all2.gif

Intense plasma focus?

http://img126.imageshack.us/img126/9620/image2bq2.jpg


Couldn't be!

Ken G
2007-Aug-31, 05:14 AM
You seem to be of the opinion that modern neutron star modelers are unaware of Maxwell's equations and the connections between magnetic and electric effects. I'd like to assure you that is not the case.

trinitree88
2007-Aug-31, 03:00 PM
Understand there will be some losses. Thanks.Mugs. You're welcome. pete

mugaliens
2007-Aug-31, 06:06 PM
Intense plasma focus?

http://img126.imageshack.us/img126/9620/image2bq2.jpg


Couldn't be!

You putting candles into the microwave again, Northwind?

Tim Thompson
2007-Aug-31, 09:56 PM
No electric field, no magnetic field, it's as simple as that! (Don't tell the existing mathematical models this though)
This is a false statement. The existence of magnetic fields without corresponding electric fields is standard, and was in fact described by Alfven. You should know your own sources well enough to not have to be told.

It is what happens when a magnetic field is trapped by a plasma which has a higher energy density than does the field. Energy wins in physics, and in this case the magnetic field becomes "frozen" into the plasma (a process described by Alfven in his book on cosmic plasmas). So when the neutron star forms, the trapped field is compressed along with the trapped matter, and the neutron star acquires & maintains a strong magnetic field.


Couldn't be!
Actually, it could be, but is unlikely. The comparison between your plasma focus images abd the Chandra X-ray image is qualitative, but not quantitative. See, for instance, that the Chanrda image has a double bow, but your images do not. And also note that there is no significant plasma outside the bow of your images, while there is in the Chandra images. How would the additional plasma affect the form of the focused bow? We also note that, as the Chandra webpage (http://chandra.harvard.edu/photo/2000/vela/) tells us, there is a perfectly acceptable, and more realistic interpretation: "The rings are thought to represent shock waves due to matter rushing away from the neutron star. More focused flows at the neutron star's polar regions produce the jets. The origin of this activity is thought to be enormous electric fields caused by the combination of the rapid rotation and intense magnetic fields of the neutron star." We do know that shock waves will do that. So why is your ATM concept better than the M concept?

Ken G
2007-Sep-01, 12:31 AM
It is what happens when a magnetic field is trapped by a plasma which has a higher energy density than does the field. Energy wins in physics, and in this case the magnetic field becomes "frozen" into the plasma (a process described by Alfven in his book on cosmic plasmas).Nitpick for clarity-- what freezes the field in is not an energy comparison, that just tells you whether the frozen field will tell the plasma where to go or the plasma will tell the frozen field where to go. What freezes in the field (and eliminates electric fields, which is what Tim is driving at) is the extreme mobility of the charges in a plasma. If every time you try to induce an electric field into the local fluid frame of the plasma, the charges move so as to neutralize that electric field, then you have a hard time maintaining an electric field at all and the results are primarily magnetic.

mugaliens
2007-Sep-01, 10:46 AM
Ok, I'd always been taught that:

1. Associated with any charge (even an electron) is a field.

2. Move the field and you produce a magetic field.

3. Magnets work because more electrons are orbiting one way than the other about a given axis.

4. Magnetic fields associated with plasma exist because plasma contains charged particles that are moving.

Is this what you mean by "the extreme mobility of the charges in a plasma?" and perhaps the "eliminates electric fields" is referring to removing electric fields external to the plasma?

Tim Thompson
2007-Sep-01, 03:56 PM
1. Associated with any charge (even an electron) is a field.
2. Move the field and you produce a magetic field.
4. Magnetic fields associated with plasma exist because plasma contains charged particles that are moving.

Any charged particle has an electric field (http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elefie.html#c1), and if you move the electric field, a magnetic field (http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magfie.html#c1) is generated (Bio-Savart Law (http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/biosav.html#c1) and Faraday's Law & Lenz's Law (http://hyperphysics.phy-astr.gsu.edu/hbase/electric/farlaw.html#c1)). Magnetic fields generated in a plsama arise from the motion of the charged particles (magnetohydrodynamics (http://mysite.du.edu/~jcalvert/phys/mhd.htm)). But note that in most cases, the plasma has the same number of positive charges as it does negative charges. So the electric fields cancel each other out. The result is that the plasma has no net electric field; the electric field will vary due to local conditions such as variable density in the plasma, but there is no net, global electric field. But there will be a net global magnetic field. The fans of the ATM approach do not believe this, rather choosing to believe that there must be a large electric field. Physics does not allow that, but the ATM idea does not care much for the mundane limitations of physics (a significant internal or external electric field will destroy a plasma very quickly by separating the charges).

Once the magnetic field has been generated, it can be maintained & amplified by processes in the plasma which never include a net, global electric field. That's what happens when the neutron star forms from the collapsing core, and drags the magnetic field with it. The result is a strong, compressed magnetic field, with no associated global electric field.



3. Magnets work because more electrons are orbiting one way than the other about a given axis.

That's close enough for a general explanation. The process is called ferromagnetism. See the Wikipedia ferromagnetism page (http://en.wikipedia.org/wiki/Ferromagnetic) or the Hyperphysics ferrromagnetism page (http://hyperphysics.phy-astr.gsu.edu/hbase/solids/ferro.html).



Is this what you mean by "the extreme mobility of the charges in a plasma?" and perhaps the "eliminates electric fields" is referring to removing electric fields external to the plasma?

As I described above, its all about canceling the charges internally, not externally. An external electric field will destroy a plasma by pulling the charges apart, whereas an external magnetic field will drive the plasma, but not separate the charges (in both cases the field does have to be strong enough).

Ken G
2007-Sep-02, 03:09 AM
It would require a pretty strong external electric field to "destroy" a plasma-- typically the plasma will simply react (via a small amount of charge separation, so local imbalance but not global imbalance) to neutralize the imposed electric field inside the plasma. It'll still be a plasma, just a slightly "tweaked" one-- but that tweaking involves energy releases that might be quite dramatic (lightning discharges, for example).

mugaliens
2007-Sep-02, 01:06 PM
(plasmas aren't my forte')

So, theoretically, you could use a small electric field to slightly separate plasma charges, then rotating electric fields to induce movement, then rotating magnetic fields to continue inducing movement and confinement. Is that the typical approach?

What about Bussard's Polywell (http://en.wikipedia.org/wiki/Polywell)approach? What's your opinion of the liklihood of his multiple magnetic mirror design overcoming the Farnsworth-Hirsch fusor's limitation, namely, too many nuclei striking the wall?

Ken G
2007-Sep-02, 05:16 PM
There are engineering issues there that require significant personal experience, I would say.

Northwind
2007-Sep-03, 02:24 AM
This is a false statement. The existence of magnetic fields without corresponding electric fields is standard, and was in fact described by Alfven. You should know your own sources well enough to not have to be told.


Reference? Source?

Alfven showed later this was a big mistake of his and MHD theory. i.e. the are mo "frozen" in magnetic fields!


Any charged particle has an electric field, and if you move the electric field, a magnetic field is generated (Bio-Savart Law and Faraday's Law & Lenz's Law). Magnetic fields generated in a plsama arise from the motion of the charged particles (magnetohydrodynamics). But note that in most cases, the plasma has the same number of positive charges as it does negative charges. So the electric fields cancel each other out. The result is that the plasma has no net electric field; the electric field will vary due to local conditions such as variable density in the plasma, but there is no net, global electric field. But there will be a net global magnetic field. The fans of the ATM approach do not believe this, rather choosing to believe that there must be a large electric field. Physics does not allow that, but the ATM idea does not care much for the mundane limitations of physics (a significant internal or external electric field will destroy a plasma very quickly by separating the charges).

So what's your interpretations on the Tim Thompson?
Recent studies suggest that flux-tube-like structures may exist in the solar wind. Flux tubes in the fast and slow solar wind (http://scitation.aip.org/getabs/servlet/GetabsServlet?prog=normal&id=APCPCS000932000001000026000001&idtype=cvips&gifs=Yes)

Interesting? Because you then state
Once the magnetic field has been generated, it can be maintained & amplified by processes in the plasma which never include a net, global electric field. That's what happens when the neutron star forms from the collapsing core, and drags the magnetic field with it. The result is a strong, compressed magnetic field, with no associated global electric field.


No electric field? You sure?

Ken G
2007-Sep-03, 04:02 AM
MHD is an approximation, as is the very concept of a plasma to begin with. It explains why global magnetic fields are generally more important in plasmas than are global electric fields. Locally, life gets complicated and lots of non-MHD effects become important on small scales or on short times. So what matters is the phenomena being modeled, and the scales involved. Comparing descriptive aspects of pictures from very different scales is inadequate for reaching an understanding of the observed phenomena.

Tim Thompson
2007-Sep-03, 05:38 AM
Reference? Source? Alfven showed later this was a big mistake of his and MHD theory. i.e. the are mo "frozen" in magnetic fields!
My source is Alfven's description of frozen flux in his book on cosmic plasma, but I don't have that book here at home & cannot at the moment be more specific. I have no knowledge of Alfven entirely abandoning the idea of frozen flux, and I do not believe that he ever did. What he did say, in the same book, was that he feared the concept was being improperly applied. On that he was quite wrong. There certainly are "frozen in" magnetic fields, and there is no way to avoid them in plasma physics.


So what's your interpretations on the Tim Thompson? Flux tubes in the fast and slow solar wind (http://scitation.aip.org/getabs/servlet/GetabsServlet?prog=normal&id=APCPCS000932000001000026000001&idtype=cvips&gifs=Yes)
My interpretation of the Tim Thompson is likely to be biased. However, I presume you meant "this". I don't know much about flux tubes & magnetic ropes, but what's to interpret anyway? You linked to an abstract, and the abstract pretty well says what they are doing. I do know that a flux tube is a concentrated, rope-like structure with a strong magnetic field aligned along the rope. My understanding is that the magnetic flux is woven into these tubes by the differential rotation of the sun, and driven to the surface by magnetic bouyancy. It would not surprise me that such structures could make it into the solar wind.


Interesting? Because you then state

Once the magnetic field has been generated, it can be maintained & amplified by processes in the plasma which never include a net, global electric field. That's what happens when the neutron star forms from the collapsing core, and drags the magnetic field with it. The result is a strong, compressed magnetic field, with no associated global electric field.
No electric field? You sure?
I said can be, not must be. Of course electric fields begat magnetic fields and vice versa, we all know that. But I know what I meant by "no electric field", and yes, I am sure. I already explained it, how many times do you want me to repeat myself, especially when you already put it in bold type: But note that in most cases, the plasma has the same number of positive charges as it does negative charges.? Which part of "same number" don't you undetstand? of course there are small electric fields throughout the plasma, because the plasma itself is variable. But there is no global electric field unless there is a global charge separation, and we don't normally see that. The whole point is that moving plasma generates large global magnetic fields without also generating equivalent large global electric fields. That happens to be boringly standard plasma physics.

Northwind
2007-Sep-03, 07:51 AM
However, I presume you meant "this". I don't know much about flux tubes & magnetic ropes, but what's to interpret anyway?

???

You're serious? :confused: after all the bagging you do of the EU theory, I'm suprised to hear that from you.

The currents were predicted in 1903 by Norwegian explorer and physicist Kristian Birkeland!!!


A Birkeland current generally refers to any electric current in a space plasma, but more specifically when charged particles in the current follow magnetic field lines (hence, Birkeland currents are also known as field-aligned currents). They are caused by the movement of a plasma perpendicular to a magnetic field. Birkeland currents often show filamentary, or twisted "rope-like" magnetic structure.

1903!! :doh:

These "flux tubes" are also called Birkeland currents! :eek:

So the solar wind and presumable all stellar "winds" would also show this structure, not one "global" electrically neutral "wind", why would a Nuetron star be any different?

tusenfem
2007-Sep-03, 09:24 AM
Reference? Source?

Alfven showed later this was a big mistake of his and MHD theory. i.e. the are mo "frozen" in magnetic fields!



Well, Alfvén never really showed it, he mainly posited it together with Fälthammar in the second edition of Cosmic Electrodynamics, which was published before any large scale magnetospheric missions were started. He was right in what he wrote, that if the plasma is inhomogeneous or changing too fast, that you cannot apply the frozen in and/or MHD.

We will get to that in the next days in the thread Plasma Physics for Dummies (http://www.bautforum.com/general-science/63183-plasma-physics-dummies.html#post1045626), with the latest results from modern experiments on Cluster, where it will be found that frozen-in works pretty good when applied to e.g. the external cusp and the high-latitude magnetostheath, and breaks down on small scales e.g. in the magnetosphere boundary layer.

Stay tuned, it will take me a bit to write it up for the next post.

tusenfem
2007-Sep-03, 09:42 AM
The currents were predicted in 1903 by Norwegian explorer and physicist Kristian Birkeland!!!



A Birkeland current generally refers to any electric current in a space plasma, but more specifically when charged particles in the current follow magnetic field lines (hence, Birkeland currents are also known as field-aligned currents). They are caused by the movement of a plasma perpendicular to a magnetic field. Birkeland currents often show filamentary, or twisted "rope-like" magnetic structure.




Kristian Birkeland posited that the Aurora would be created by electrical currents flowing along the magnetic field of the Earth, most likely driven by the sun in some way. (For a very good biography read "Northern Lights" by Lucy Jago). And indeed, when sounding rockets were send up into the auroral regions, it was found that there were currents aligned with the magnetic field. These specific currents are now called Birkeland currents.

I keep a very restricted view on what Birkeland currents are (a specific current system in a planetary magnetosphere) and what field aligned currents are in general. BCs are a subset of FACs.



These "flux tubes" are also called Birkeland currents!

So the solar wind and presumable all stellar "winds" would also show this structure, not one "global" electrically neutral "wind", why would a Nuetron star be any different?

Flux tubes occur time and again in the solar wind. You happen to quote the abstract of a very complicated paper in which there is turbulence and other stuff. There are many papers of flux tubes of the solar wind enter the Earth's magnetosphere, these are so called FTEs (Flux Transfer Events). Now, flux tubes are never called Birkeland currents, they are called flux tubes and are associated with field aligned currents.

I see that you still think that plasmas with currents flowing in them are not neutral, but I guess we will never be able to explain that to you.

Stars have winds, and neutron stars have also winds, the difference is the object, a "chaotic" magnetic field at stars and a "basically dipolar" magnetic field at a neutron star.

Northwind
2007-Sep-03, 11:10 AM
I see that you still think that plasmas with currents flowing in them are not neutral, but I guess we will never be able to explain that to you.

Whaaa...???

When have I ever said that, goes against the whole EU thing :(

Ya dag!

tusenfem
2007-Sep-03, 11:35 AM
Whaaa...???

When have I ever said that, goes against the whole EU thing :(

Ya dag!

Well, I sort of get that impression from sentences like this one:



So the solar wind and presumable all stellar "winds" would also show this structure, not one "global" electrically neutral "wind", (snip)

Ken G
2007-Sep-04, 06:34 PM
I think by "electrically neutral" he means "zero electric field". Lest the two camps simply talk past each other, I think it is important to stress what tusenfem has already intimated-- physics is about idealizations and approximations, that work in some situations and break down in others. MHD (with its "frozen in" magnetic fields and zero electric fields in the fluid frame) is one such idealization, and it works very well for some things (like the whole flux-tube and FAC concepts to begin with) and not so well for others (like transient discharges or magnetic flux "reconnection"). It is pointless to argue the idealization, what is relevant is how well it applies in any given specific instance. I have no doubt that MHD is used extensively in the very paper that is cited above, but the paper may also refer to breakdowns in MHD. This is how science works, we find unifying principles and then we start looking for when and if they break down, and can we explain neutron stars within those idealizations or not. I still see no evidence in this thread that MHD-inspired models of neutron stars are not useful ways of understanding the global field pattern in those objects, but may not be able to reproduce every transient "burp" that neutron stars might do.

I will also point out the "field-aligned currents" are a perfectly MHD concept and do not require the presence of a sustained global electric field-- that's the whole point of MHD, you can get field-aligned currents without E fields. So "Birkeland currents" in no way suggest we must discard the concept of frozen-in fields or that we need explicit electrical inputs. The generation or dissipation of such currents would have an electrical origin, but can also be modeled in the "frozen-in" picture in terms of "stressing" or "twisting" the magnetic structures. And finally, note that electric fields are not relativistically invariant, so we can get fields appearing in different reference frames. Typically we talk about the electric field in the local fluid frame of the plasma, but even treating the plasma like a fluid is itself an idealization.

William
2007-Sep-06, 09:04 PM
In reply to trinitree88's comment:


An electron plus a proton will not make a neutron, so the pressure can not drive them together. when a neutron decays, it produces a proton, an electron, and an electron type antineutrino. that's beta decay, and it's energetically favorable.

If you want to reverse the process, you must have all three particles come together at the same point in time, and that's particularly unfavorable with regards to the antineutrino fluxes, and the energetics. so, how does it happen?

The neutrino's action in the reaction must be to remove surplus energy from a fixed balanced state. A description of the stellar change and the electron capture process (from an introduction to modern astrophysics, 2nd edition, carrol & ostlie Page 579 Chapter 16 neutron stars) is:


Initially, at low densities the nucleons are found in iron nuclei. ... however, as mentioned in the discussion of the chandrasekhar limit, when the stellar density is approx. 10^9 kg/m^3 the electrons become relativistic. Soon thereafter, the minimum-energy arrangement of protons and neutron changes because the energetic electrons can convert protons in the iron nuclei into neutrons by the process of electron capture.

Where the reaction is: (p + e) combines to form (n + electron type neutrino)

Because the neutron mass is slightly greater than the sum of the proton and electron masses, and the neutrino's rest mass energy is negligible, the electron must supply the kinetic energy to make up the difference in energy .... = 0.78 Mev.

See Wikipedia also:

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

Ken G
2007-Sep-07, 04:25 AM
Note also that the above description is nice as far as it goes, but it skips one pretty important step when it says "soon thereafter"! It is not just that the electron energies become high enough to cause the reaction into neutrons, you still need it to happen at a fast enough rate to win out over the inverse process. That doesn't happen until the density of electrons is very high-- like when the star is down to the size of a city. The electrons become relativistic when the star is still the size of a planet, so the "soon thereafter" rather overlooks the compression of a planet-sized star to a city-sized star (which happens due to gravity). That's not an insignificant "soon thereafter"-- it causes supernovae!