Tag Info

59

The crucial part is that earth's outer core is fluid, and that it's conductive. That the material happens to be iron which we know as ferromagnetic is actually rather unimportant, because the geomagnetic field is not created as a superposition of atomic spins like in a permanent magnet. Rather, it's generated via Ampère's law from macroscopic currents,...

26

The core of the Earth isn't a giant bar magnet in the sense that the underlying principles are different. A bar magnet gets its magnetic field from ferromagnetism while Earth's magnetic field is due to the presence of electric currents in the core. Since the temperature of the core is so hot, the metal atoms are unable to hold on to their electrons and ...

23

For an unmagnetized iron needle to align with an external magnetic field, the field would need to be able to induce a magnetization in the needle. This is definitely possible with a large enough field. If a naturally ferromagnetic material is unmagnetized, it still contains small magnetic domains inside. However, the sum of the magnetizations of all the ...

10

Indeed, one of the definitions of spontaneous symmetry breaking is in terms of its susceptibility: Suppose we add a symmetry breaking perturbation $h \; \delta H$ to our Hamiltonian (as you do), if $$\lim_{h \to 0} \lim_{N \to \infty} \langle m \rangle \neq 0$$ then we say our system has spontaneous symmetry breaking. (Note: $N$ is the number of ...

9

Consider the simple ferromagnetic system shown below. The instantaneous power consumed by the system at any time $t$ is simply: $$p(t) = v(t)i(t)$$ Where $v(t)$ and $i(t)$ are simply the voltage and current in the terminal of our coil. Now the total energy dissipated from time $t_A$ to $t_B$ in the system is: W_{A-B} = \int_{t_A}^{t_B} p(t)dt =\int_{t_A}^{... 8 First, I strongly advise you to read sections 24 and 25 of Tolman's excellent book on statistical mechanics. My answer will mainly go along the lines of what is in the book. The ergodic hypothesis states that a system will eventually in some time interval T visit all the states compatible with a given energy constraint. As you said, this implies an ... 8 A magnetic dipole would be induced in the iron bar and the iron bar would try and align itself along the magnetic field lines because of the torque applied on it by the interaction of the induced dipole and the Earth’s magnetic field. However since the torque which was applied on the iron bar would be very small the chances are that there would not be an ... 7 Currently our best model of the electron is that it is a point particle. Measurements give an upper bound of 10^{-22} meters on its radius. A point particle has no radius, and thus cannot rotate. Therefore we do not consider an electron with spin as a "spinning" electron in any classical sense. The spin and the magnetic moment of the electrons are instead ... 6 Landau free energy is just an approximation to the real free energy in the thermodynamic limit. For that reason, Landau free energy can be analytic, while the real one is not. Let me show how the approximation works. As you may know, the Landau free energy is defined in the following way, assuming the Ising model:Z\left(h,T\right)=\sum_{\left\{ s_{i}\...

5

It turns out that if you work out the physics behind the magnetism of a material, the magnetisation $M$ obeys a transcendental equation of the form $$M=\tanh\left(\frac{\alpha M+\gamma H}{T}\right)$$ where $\alpha$ and $\gamma$ are some constants, $H$ is the magnitude of the external magnetic field and $T$ is the temperature. What this says is that at a ...

4

The Wikipedia page is being sloppy. They mean that the free energy density is an analytic function of the mean-field order parameter, whereas at a thermal phase transition the free energy density is a non-analytic function of the temperature (or for a zero-temperature phase transition, of the external parameter being tuned across the transition). The order ...

4

The magnetic dipole dipole interaction is just by far too weak. In Stephen Blundell: "Magnetism in Condensed Matter" this is estimated to be on an energy scale of about $10^{-23}$J which may play a role at temperatures of about 1K or so. But when we are talking about magnetism we typically talk about much higher temperatures. In detail the energy due to ...

4

It is coincidence. For example not all forms of iron are ferromagnetic, even though they all have the same nuclei. Iron exists in three slightly different crystal structures called ferritic, martensitic and austenitic. The ferritic and martensitic are ferromagnetic but the austenitic form is not. So iron can switch between a ferromagnet and paramagnet just ...

4

$\vec{H}$ is sometimes called the magnetic field strength (I prefer "H-field"). More commonly $H$ will confusingly be called the magnetic field -- we will return to this. As you likely know $\vec{B}$ is just the typical magnetic field. The key of this is $\vec{M}$. $\vec{M}$ is the magnetization, which represents the amount of magnetic dipoles in material. ...

4

The force exerted by the magnetic field of the earth on a chunk of iron at its surface is far weaker than the force of gravity. (Also, note that because of the shape of the earth's magnetic field, any attractive force between it and a piece of ferrous metal would be concentrated at the earth's north and south magnetic poles.)

4

It seems like you are confusing spatial rotation symmetry and spin rotation symmetry. They are different symmetries (and both are present in nature). In terms of group theory we will say that spin rotation symmetry is manifested by invariance of the hamiltonian with respect to group $SU(N)_{\text{spin}}$ where $N=2S+1$. Spatial symmetry by $SO(3)_{\text{... 3 I'll provide one possible answer to your question. I use the language of the Ising ferromagnet/paramagnet phase transition, but this is just for convenience as the same holds much more generally. Assume that the system is below the Curie temperature. Then, the free energy is an analytic function of the free energy both when$h>0$and when$h<0$. At$h=...

3

Hysteresis is a kind of non-time-reversible transition, so it is associated with entropy (which must, in theory, generate heat). Because magnetic materials have crystal structure, the internal magnetic polarization is not uniform, but aligns in small patches to the crystal, and these patches (magnetic domains) shift in size and shape during a magnetic ...

3

No. The earth's North Pole is a "magnetic south pole". With a compass, the part of the needle that points north is a magnetic north pole. With magnets, opposite poles attract, so what we call North on the earth must be magnetic south to attract the needle's north.

3

Neodym magnets are a lot stronger than iron, copper and nickel.Neodym is a mixture of different chemical elements. What is the reason that this material creates a lot more magnetic field strength than others? How you answer this question depends on what exactly you mean by "stronger." In the colloquial sense of the word, there are a few important material ...

3

The way I see it, your questions are closely related. Imagine a system at high temperature, rapidly exploring a lot of its phase space. You might say the ergodic hypothesis is correct in this situation. Then you start reducing the temperature, and the energy, and it may happen that for low energies there are two regions in phase space with the same energy, ...

3

Can you have a hysteresis curve in a mono domain material? Yes. One example is the Stoner-Wohlfarth system, a model system for single domain particles. In this system hysteretic behavior arises as a result of shape anisotropy energy competing with the Zeeman energy. The shape anisotropy prefers the magnetization to be aligned along an axis defined by the ...

3

The melting of ice is a first order phase transition. First order phase transition involve the release/absorption of a fixed amount of latent heat per unit volume. A finite amount of time is needed for the release/absorption of such heat, and thus for completing the transition. The loss of magnetization of a ferromagnet at the Curie temperature is a ...

3

Probably yes, in a careful experiment. A paramagnetic material like aluminum will align with a strong magnetic field and diamagnetic metal plates align themselves perpendicular to the field. Magnetically weak iron has a relative permeability many orders of magnitude larger than paramagnetic or diamagnetic materials. This probably offsets the much weaker ...

3

No, there is no energy loss. Consider just the upper magnet sitting on the table. Is "energy expended" to hold it there, above the ground, against the force of gravity? No; there is potential energy in the system, which can be converted into kinetic energy if you knock the magnet over the edge and it falls. When the lower magnet is stuck against the table, ...

3

One can provide two characterizations of first-order phase transitions in Ising models at inverse temperature $\beta$ and external magnetic field $h$: Thermodynamic characterization: there is a first-order phase transition at $(\beta^*,h^*)$ if the free energy $h \mapsto f(\beta^*,h)$ is not differentiable at $h^*$. Probabilistic characterization: there is ...

3

That implies a resulting energy deficit in the magnetic field I think all of your questions ultimately boil down to some variation of this statement. So... The problem is that you're ignoring everything before and after "the event". You're creating an artificial line around the system, both physically and in time, and noticing there's a change in energy ...

3

Note that this is rather an opinion than a fully rigorous statement: $\qquad$ For vanishing gradients, i.e. for uniform systems (the case originally considered by Landau), the expansion of the free energy is indeed a Taylor expansion (in even powers of $m$) near the transition. However the addition of gradient terms in the case of non-uniform systems ...

3

It looks to me like you are Taylor expanding $\mathcal{F}$ as a function of $\vec{m}$ without considering that $\vec{m}$ is a function of space. In particular, your $D$ and $D^2$ involve things like $\frac{\partial}{\partial m_x}$ but not $\frac{\partial}{\partial x}$. You are implicitly assuming that $\mathcal{F}(\vec{x})$ only depends on $\vec{m}(\vec{x})$ ...

Only top voted, non community-wiki answers of a minimum length are eligible