electromagnetism - How do permanent magnets manage to focus field on one side?

The actuator of a hard drive head consists of two very strong neodymium magnets, with an electromagnetic coil between them. If you take that apart, the magnets attract each other very strongly. There's no doubt the field between them is very strong.

But if you flip them back to back, there is no repulsion force - there is pretty much nothing. While the magnets are very strong on one side, they seem completely inert on the other.

I understand how a U-shape of a shaped magnet allows it to focus field near the two exposed poles, but how does a flat rectangular plate manage to keep the field over one flat, wide surface and nearly none near the other?

[sorry about the heavy edit but it seems the question got totally muddled with irrelevant discussion about unipolar magnets and possibility or impossibility to remove magnetic field from one side. Only heavy rephrasing may help.]

Edit: some photos:

1: The magnets stuck together. They hold really hard, I'd be unable to just pull them apart but I can slide one against the other to separate them.

The magnets in "inert position" - they don't act on each other at all, just like two inert pieces of metal.

The magnets seem to have two poles located on the surface off-center. Here, a normal magnet placed against the hard disk magnet, centering itself on one side, then flipped - on another.

The metal shield seems to act like completely unmagnetized ferromagnetic. I can stick the "normal magnet" any way I like to it, and it doesn't act on another piece of ferromagnetic (a needle) at all.

When I apply a small magnet to it, it becomes magnetic like any normal "soft" ferromagnetic - attracts the needle weakly. It behaves as if the (very powerful) neodymium magnet glued to the other side wasn't there at all.

Unfortunately the neodymium magnets are glued very well and so fragile I was unable to separate any without snapping it, and then the "special properties" seem to be gone.

Answer

The metal plates the magnets are glued to are an Iron and Nickel alloy that has a very high magnetic permeability called a Mu-metal.

I don't understand all of the details of magnetism or how Mu-metal works but that should get you started.

homework and exercises - I wonder why I cannot charge a capacitor with alternating current?

Why can't I charge the capacitor with AC? How do the plates block the flow of electrons with DC but not with AC.

Somebody told me that the DC is blocked by the capacitor, so the capacitor gets charge, but I could not get the actual concept about it.

quantum mechanics - Why is $langle x| x' rangle=delta(x-x')$?

I've tried to find any solution or proof for $$\langle x| x' \rangle=\delta(x-x'),$$ but I only came to this post: Wave function and Dirac bra-ket notation

So I got the information, that the vector $|x\rangle$ form a dirac-normalized basis for the Hilbert Space.

I know that the dirac-delta distribution is defined like this: $$\delta(x-x') = \begin{cases} 0 &\mbox{if } x\neq x' \\ \infty & \mbox{if } x=x' \end{cases},$$ this means that my x' is a point on my x axis where I have my infinite high peak. And also $$\int_{-\infty}^{\infty}dx\cdot \delta(x-x')=1.$$

But how actually correlate this with the scalar product of vectors x, x' in the Hilbert Space that form a so-called 'diracl-normailzed" basis of it?

Can you give me some tips on this please? Or maybe you actually know a link, where this is explained.

Answer

Isn't it just from the sifting property?

$$f(x) = \int\mathrm{d}x'\;f(x')\,\delta(x - x')$$

That is, if you accept the above and if you accept that

$$|\psi\rangle = \int \mathrm{d}x'\,\psi(x')\,|x'\rangle$$

then

$$\psi(x) = \langle x|\psi\rangle = \langle x| \int \mathrm{d}x'\,\psi(x') \,|x'\rangle = \int \mathrm{d}x'\,\psi(x')\,\langle x|x'\rangle$$

$$\Rightarrow \langle x|x'\rangle = \delta(x - x')$$

Noether's theorem in general relativity

Noether's theorem yields a conservation law for every symmetry. Is that independent of the Lagrangian i.e. when $\mathcal{L}\neq T-V$? In general relativity the integral that is minimised will be the geodesic: $$S=\int ds$$ What form would Noether's theorem take? I am also looking for a proof of this. All the proofs I've seen assume $\mathcal{L}=T-V$.

thermodynamics - How to interpret phase diagrams?

I find quite difficult to interpret phase diagrams in general, for example I see people discuss them along the following lines:

• Here we see the coexistence line between liquid-solid phases..


• a tricritical point..


• this section describes the stable solid phase..


• in between these lines we have the metastable zone..

In other words, these diagrams seem to be telling us everything about a given system in a very compact way.

As an example, let us take the following phase diagram [source] shown in terms of temperature vs density for a Lennard-Jones fluid in three dimensions:

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Questions:

1. How does one interpret such a diagram in terms of the drawn curves and the outlined regions? Do all the points along the drawn lines correspond to phase transition points for different $T,\rho?$


2. Is the region between the two near-vertical lines depicting a coexistence region? That is, any point would correspond to a Lennard-Jones fluid in a coexistence phase comprised of solid and liquid.


3. What about the stable regions? For example, how can one know which sets of density values (and temperatures) correspond to stable solid phase of the fluid?


4. I admit these are rather naive questions, but I really don't have a good grip on how to read such diagrams which are oh so important. Additionally, if you happen to know of good lecture notes teaching how to understand these diagrams together with explanations of the different kinds of points (tricritical, bicritical...), it would be very helpful.

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Source: Mastny, Ethan A., and Juan J. de Pablo. "Melting line of the Lennard-Jones system, infinite size, and full potential." The Journal of chemical physics 127.10 (2007): 104504.

Answer

I agree with the commenters that this is a very broad question, and that you should start with some background reading, e.g. a textbook. Many standard Physical Chemistry texts give a good introduction to the phase diagrams of simple substances, and the Lennard-Jones system (although an idealized model) is fairly typical. The links provided by Jon Custer may also be helpful, but they are mainly concerned with systems of more than one component, so I would recommend starting with the simpler, one-component, case first.

I think that there's some value in doing what you wanted, and using the very specific example you have picked out from a simulation paper, to answer your questions. That paper is looking at the solid-liquid coexistence line: the "melting line". Plotted as a function of $T$ and $P$, it would indeed be a line (not, in general, a straight line, but a curve): along that line, the chemical potentials $\mu$ of the two phases would be equal, and the equation $\mu_\text{solid}(P,T)=\mu_\text{liquid}(P,T)$ will define a line in $P$-$T$ space. As you cross such a line, properties such as density $\rho$ change discontinuously (it's a first-order transition). Think of the lines in the $P$-$T$ diagram as marking these discontinuities. Like a crude topographical map, except that we don't mark out the contours of height, just the locations of cliff edges. A typical phase diagram in $P$-$T$ variables (but plotted, as is most common, with the temperature along the horizontal axis) can be found on the Wikipedia page.

If the phase diagram is plotted in temperature-density variables, the melting "line" becomes a coexistence region. In your picture, you'll see the dots on the two near-vertical lines come in pairs, which could be connected by horizontal lines. These are called "tie lines". They will be horizontal because the temperatures of the coexisting phases must be equal. The densities corresponding to the dots at the end of each tie line are those satisfying $\mu_\text{solid}(\rho_\text{solid},T)=\mu_\text{liquid}(\rho_\text{liquid},T)$. The general rule is, any state point $(\rho,T)$ in the two-phase region does not correspond to a stable phase, but to a mixture of the two phases $(\rho_\text{solid},T)$ and $(\rho_\text{liquid},T)$ , whose densities can be read off from the points at either end of the tie line. When plotted in other variables, in more complicated situations, it may be that the tie lines are not horizontal, and are actually drawn on the phase diagram, to help people make this construction.

The line going from $(\rho,T)\approx(0.84,0.694)$ up to $\approx(0.6,1.15)$ is the right hand boundary of the liquid-gas two phase region. If we extended the plot to lower density, the curve would continue to rise up to the critical point, at about $T=1.3$, and would then come down again, reaching $T=0.694$ at a very low density. There should be tie lines drawn horizontally across this region as well, corresponding to the coexistence densities of liquid and gas.

You'll see the horizontal dashed line at $T_\text{tp}=0.694$. This is the triple point, at which liquid, gas, and solid are all in equilibrium. Nothing much is shown below that temperature (this is not the interest of the authors of that paper). In fact, there will be yet another two phase region: solid-gas. This extends from a near-vertical line at very low density ($\rho_\text{gas}$) across to a near-vertical line extending downwards from roughly $\rho=0.96$ ($\rho_\text{solid}$). To the right of that region would be solid (one phase); to the left of that region would be gas (one phase).

The one-phase regions in the diagram are labelled "solid" and "liquid". The combinations $(\rho,T)$ of points in these regions correspond to a single stable phase. In the "liquid" region, for any temperatures higher than about $1.3$, it would be better to refer to the phase as a "supercritical fluid", since we do not distinguish between liquid and gas above the critical point.

I think there's not much more to be said about that particular diagram, but hopefully it has clarified things a bit. There is no tricritical point (you should not worry about those, until you've studied phase transitions more deeply) and in most circumstances phase diagrams show equilibrium phase boundaries rather than "metastable zones", so I would suggest putting those on the back burner too.

newtonian gravity - Does the Earth gets closer to the Sun?

We know that the sun loses an amount of it's mass equivalent to the amount of energy it produces, according to the $E=mc^2$ equation. so the sun is losing mass every second. Does this affect the space-time curvature it creates. Or does this affect the distance between the Sun and the Earth. Does losing mass affects the gravity of Sun or other planets?

Answer

The answer is here

There exists the effect of the loss of mass and therefore gravitational attraction between the earth and the sun but it is small:

If we assume that the Sun's rate of nuclear fusion today is the same as the average rate over those 10 billion years (a bold assumption, but it should give us a rough idea of the answer), then we're moving away from the Sun at the rate of ~1.5 cm (less than an inch) a year. I probably don't even need to mention that this is so small that we don't have to worry about freezing.

There is also the even smaller effect of the tides induced on the sun by the earth:

It turns out that the yearly increase in the distance between the Earth and the Sun from this effect is only about one micrometer (a millionth of a meter, or a ten thousandth of a centimeter). So this is a very tiny effect.

plasma physics - Vlasov equation why "long range" interactions?

In discussions of the Vlasov equation it is often said (e.g. Cercignani, 1988; pg59) that we require rarefied gas with weak, long range interactions. I understand why we need weak interactions (since we take a scaling of $1/N$ for the force) but I can't see why we need the interaction to be long range.

Does anyone know why this is so? (a source would be helpful).

Answer

I understand why we need weak interactions (since we take a scaling of 1/N for the force) but I can't see why we need the interaction to be long range.

It is one of the assumptions required to create a limit where the Boltzmann collision operator goes to zero. More importantly, the physical reason they state as much is because the fluids handled by the Vlasov equation are typically ionized gases, namely plasmas. In a plasma, the particles are charged and respond to the collective fields of all the charged particles out to infinity (in principle/theory, but generally Debye shielding and quasi-neutrality keep the interactions more local).

Does anyone know why this is so?

The only real difference between the Boltzmann and Vlasov equations are the lack of a collision operator in the latter, namely, that the latter is considered time-reversible. Using things like ensemble averages (e.g., mean field theory) can change the Vlasov equation from a time-reversible to a time-irreversible equation, but that is an additional nuance.

In discussions of the Vlasov equation it is often said (e.g. Cercignani, 1988; pg59) that we require rarefied gas with weak, long range interactions.

If the particle-particle interactions (e.g., Coulomb collisions) are strong, then the Vlasov equation is no longer the appropriate approximation to use and requires a collision operator.