astrophysics - Plasma and Stars

I have read that most stars are made mostly of plasma.

My questions in this statement are:

1. 
   

   Are there stars not made of plasma?

   
   


2. 
   

   In what percentage stars are made of plasma?

   
   

Answer

Are there stars not made of plasma?

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Plasma is an electrically neutral medium of unbound positive and negative particles (i.e. the overall charge of a plasma is roughly zero). It is important to note that although the particles are unbound, they are not ‘free’ in the sense of not experiencing forces. When a charged particle moves, it generates an electric current with magnetic fields; in plasma, the movement of a charged particle affects and is affected by the general field created by the movement of other charges.

For more details see this link too.

A basic effect of the motion of charges is that electromagnetic radiation is created, i.e. light and thus stars certainly have plasma because they are called stars for being stationary sources of light in the night sky, in contrast to planets. The sun in the center of the solar system is a star and allows us to study the composition of stars, including the evident plasma.

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Our Sun, and all of the other stars, are made of plasma, much of interstellar space is filled with a plasma, albeit a very sparse one, and intergalactic space too.

[note that "all other stars is not really correct in this wiki link. see below]

Stars that are not wholy plasma are neutron stars:

A neutron star is the collapsed core of a large (10–29 solar masses) star. Neutron stars are the smallest and densest stars known to exist.1 With a radius on the order of 10 km, they can, however, have a mass of about twice that of the Sun. They result from the supernova explosion of a massive star, combined with gravitational collapse, that compresses the core past the white dwarf star density to that of atomic nuclei.

....

Neutron stars that can be observed are very hot and typically have a surface temperature around 6×10^5 K.

They are complex stars.

Also very large stars that become supernovae, and in general the whole spectrum in the evolution of stars has stars which are not wholly plasma.

To be visible stars, they have to emit light so that their outer shell must be plasma.

So plasma in the outer atmosphere is necessary for a star to be visible in the night sky, but there do exist stars that are not wholly plasma. Thanks to DrunkenCodeMonkey for catching it.

you ask:

In what percentage stars are made of plasma?

They are mostly plasma, i.e neutral ionized matter, even the core, because of the very large kinetic energies acquired in the formation from the primordial plasma due to the gravitational attraction .

The core of the Sun extends from the center to about 20–25% of the solar radius. It has a density of up to 150 g/cm3 (about 150 times the density of water) and a temperature of close to 15.7 million kelvins (K). By contrast, the Sun's surface temperature is approximately 5,800 K.

This very high temperature does not allow nuclei and electrons to stabilize into neutral atoms, and even at that high density the core is a plasma. The temporary formation of neutral nuclei gives spectral lines detectable in the star's spectrum, but the temperatures are so high that no solid core can result. The amount of neutral atoms in a plasma is very small, and is controlled by the relevant equations, as was pointed out in the comments.

The planetary masses cooled off enough to acquire a solid core.

quantum field theory - How does QFT account for localization (into a finite volume of space), in practice?

In Quantum Field Theory there is until yet no agreement (as far as I know) on the issue of localization of particles. When one talks about a 'particle' in QFT, one usually means a single-particle state of definite momentum, or a wavepacket made out of such states. It is not clear, however, what (if any) are the states that correspond to something that is localized in space, or even something that is localized into a finite region of space.

Some textbooks on QFT (e.g. Peskin and Schroeder, page 24) suggest that (at least in the case of the free Klein-Gordon theory) the field operator $\phi(\vec{x})$ creates a particle at position $\vec{x}$, i.e., the state \begin{equation} |\vec{x}\rangle := \phi(\vec{x})|0\rangle \end{equation} would correspond to a particle localized at $\vec{x}$. However, it can be easily shown that such states are not mutually orthogonal, i.e., $\langle \vec{y}|\vec{x}\rangle\neq 0$ if $\vec{y}\neq \vec{x}$. So these states cannot possibly correspond to localized particles.

This bothers me and I would gladly hear other people's views on this. Still, I can imagine, for instance, that these states do actually correspond to effectively localized states, by which I mean that in practice it makes sense to regard them as localized states, even if they technically aren't. But this is only a shot in the dark; I have no idea whether that makes any sense. And if this is the case, then what is the justification for this view?

Other references advocate that one should use the eigenstates of the so-called Newton-Wigner position operator, which is explained in detail in this excellent answer. Although these states also have their peculiarities, they seem to be preferable over the states $\phi(\vec{x})|0\rangle$.

So theoretically it is not clear how we should describe localized particles. Nevertheless, in collider experiments, for instance, the particles (or perhaps I should say the quantum fields) clearly are effectively localized into a finite region of space. And there the theory really works! So apparently we are able to describe localized particles. So how does one describe this spatial dependence, in practice? I imagine one uses some kind of wavepackets? And does this give any insight into the theoretical problem?

quantum mechanics - Conceptual difficulty in understanding Continuous Vector Space

I have an extremely ridiculous doubt that has been bothering me, since I started learning quantum mechanics.

If we consider the finite dimensional vector space for the spin$\frac{1}{2}$ particles, I guess it is nothing but $\mathbb{C}^2$. Each vector has two components (which is why it is two dimensional right ?), each of which can be any complex number.

Now coming to the case of position space (say one-dimension). I was taught this LVS is infinite dimensional (also continuously infinite, unlike the number operator basis). I am not able to understand this subtle thing that it is infinte dimensional (is it something like $\mathbb{R}^\infty$?). It is quite confusing every time I encounter this kind of space. Also in this each component (of the infinite no. of them) can take any real value (infinite number of them)? I learnt that the way to represent these can be in term of complex-valued functions, I would like have it elucidated.

Answer

Your doubt is not ridiculous, it is probably simply due to the confused way often mathematics is taught in physics. (I am a physicist too and, during my career, I had to bear ridiculous misconceptions, wasting lot of time in tackling non-existent pseudo-mathematical problems instead of focusing on genuine physical issues). There are sensible mathematical definitions, but there is also a practical use of math in physics. Disasters arise, in my view, when the two levels are confused, especially while teaching students.

The Hilbert space of a particle in QM is not continuous: it is a separable Hilbert space, $L^2(\mathbb R)$ which, just in view of being separable, admits discrete countable orthogonal bases.

Moreover, a well known theorem proves that if a Hilbert space admits a countable orthonormal basis, then every other basis is countable (more generally, all Hilbert bases have same cardinality).

In $L^2(\mathbb R)$, a countable basis with physical meaning is, for instance, that made of the eigenvectors $\psi_n$ of the harmonic oscillator Hamiltonian operator.

However, it is convenient for practical computations also speaking of formal eigenvectors of, for example, the position operator: $|x\rangle$. In this case, $x \in \mathbb R$ so it could seem that $L^2(\mathbb R)$ admits also uncountable bases. It is false! $\{|x\rangle\}_{x\in \mathbb R}$ is not an orthonormal basis. It is just a formal object, (very) useful in computations.

If you want to make rigorous these objects, you should picture the space of the states as a direct integral over $\mathbb R$ of finite dimensional spaces $\mathbb C$, or as a rigged Hilbert space. In both cases however $\{|x\rangle\}_{x\in \mathbb R}$ is not an orthonormal Hilbertian basis. And $|x\rangle$ does not belong to $L^2(\mathbb R)$.

As a final remark, I would like to stress that the vectors of $L^2(\mathbb R)$ are equivalence classes of functions: $\psi$ is equivalent to $\phi$ iff $\int| \psi(x)−\phi(x)|^2 dx=0$, so if $\psi(x)\neq \phi(x)$ on a set whose measure vanishes, they define however the same vector of $L^2$. Consequently the value an element of the space assumes at a given $x$ does not make any sense, since each set $\{x\}$ has zero measure.

gravity - Why are our planets in the solar system all on the same disc/plane/layer?

I always see pictures of the solar system where our sun is in the middle and the planets surround the sun. All these planets move on orbits on the same layer. Why?

Answer

We haven't ironed out all the details about how planets form, but they almost certainly form from a disk of material around a young star. Because the disk lies in a single plane, the planets are broadly in that plane too.

But I'm just deferring the question. Why should a disk form around a young star? While the star is forming, there's a lot of gas and dust falling onto it. This material has angular momentum, so it swirls around the central object (i.e. the star) and the flow collides with itself. The collisions cancel out the angular momentum in what becomes the vertical direction and smear the material out in the horizontal direction, leading to a disk. Eventually, this disk fragments and forms planets. Like I said, the details aren't well understood, but we're pretty sure about the disk part, and that's why the planets are co-planar.

newtonian mechanics - What formula do I use to calculate the force of impact of a falling object?

I am trying to calculate the force of impact of a falling object. I did my egg drop project. I dropped the egg from 10m with a mass of 126kg and with the velocity of 14.1 m/s. Which formula should I use to calculate the force of impact? I found the kinetic energy first and later I used the following formula: $W=Fs$ to find the force of impact. Is that the right formula?

How is possible for current to flow so fast when charge flows so slow?

How is it possible for current to flow so fast when charge flows so slowly?

We know electrons travel very slowly while charge travels at ~the speed of light.

Answer

What is being confused here is not the flow of "current" but rather the transmission of energy.

The individual electrons in a wire move very slowly, as they can be modeled as constantly colliding with atoms (yes, this is a naive classical model, no quantum) and bouncing around randomly in the manner of a gas (the term "electron gas" is real and not inappropriate at all). Electric current is the very slow flow of this electron gas through the wire when an electric field is present. The term "flow of current" actually is misleading - there is no such "substance" called "current", current is a flow. "Flow of charges" or more specifically (in this case - in others, it may differ!) "flow of electrons" makes more sense. (After all, we don't talk about "current" as a substance which is contained within a river and which is what does the "flowing", i.e. "flow of current in the river", rather we talk of flowing "water in" the river, and "the current" means the flow of water.) See:

http://amasci.com/miscon/eleca.html#cflow

Energy, however, is not transmitted by one electron moving all the way around the circuit to the load, but rather through waves in the electrons and more importantly, the associated electric field. It's the same way that mechanical energy is transmitted in, say, a pole that is pushed from one end. The pole compresses slightly, and a sound wave thus appears, initially containing all the energy within your "push", and then travels down it, progressively distributing that energy amongst all the atoms within the pole until they are all moving in a single direction (here I imagine the pole pushed in a vacuum, as in interstellar space, with no other forces acting). The same goes with electrons in the circuit - though I should point out the following model is a bit simplistic but is more to convey the point of how the energy is transmitted than to detail the actual behavior of the electrons, which involves quantum mechanics and is subject to many of the same caveats as one sees within in an individual atom or molecule. But in this loose sense, when you throw the switch, now an electromagnetic wave travels down, setting the electrons ahead in motion and thus distributing its energy throughout the circuit. Of course, the core atoms of the metal are relatively fixed despite the electron motion, so the latter will tend to lose that energy to collision with them, unlike the pole where everyone, atoms and electrons together, start going in synchrony, and thus you have to keep supplying energy to them with a power source like a battery or generator which effectively keeps "pushing the pole" and thus keeps energy going into it - now think about a pole that is now not in vacuum but in molasses, and you have to keep pushing it to keep it moving. This pushing on atoms, of course, is how electrical devices can use electrically transmitted energy to do useful tasks.

Electromagnetic waves, and sound waves, thus energy, travel much faster than the electrons and the atoms in both the circuit and pushed pole. Energy is what lights up your light bulbs, and energy is what makes your computer operate. Since energy travels fast, these devices start operating "at the flick of a switch".

hamiltonian formalism - Decoupled physics of the complex scalar field

The canonical commutation relations for a complex scalar field are of the form

$$[\phi(t,\vec{x}),\pi(t,\vec{y})]=i\delta^{(3)}(\vec{x}-\vec{y})$$ $$[\phi^{*}(t,\vec{x}),\pi^{*}(t,\vec{y})]=i\delta^{(3)}(\vec{x}-\vec{y})$$

How can these commutation relations be obtained from the commutation relations for two free real scalar fields?

Answer

I) The complex scalar field comes from two real/Hermitian scalar fields with equal-time CCRs

$$[\hat{\phi}^j(t,\vec{x}),\hat{\phi}^k(t,\vec{y})]~=~0, $$ $$[\hat{\phi}^j(t,\vec{x}),\hat{\pi}_k(t,\vec{y})]~=~i\hbar{\bf 1}~\delta^j_k~ \delta^{3}(\vec{x}-\vec{y}), $$ $$[\hat{\pi}_j(t,\vec{x}),\hat{\pi}_k(t,\vec{y})]~=~0, \qquad j,k~\in~\{1,2\}, \tag{A}$$

and the definitions

$$ \hat{\phi}~=~\frac{1}{\sqrt{2}}(\hat{\phi}^1+i\hat{\phi}^2),\tag{B}$$ $$ \hat{\pi}~=~\frac{1}{\sqrt{2}}(\hat{\pi}_1\color{red}{-}i\hat{\pi}_2),\tag{C}$$

cf. e.g. this Phys.SE post. This leads to OP's mentioned CCRs.

II) If the $\color{red}{\text{minus sign}}$ in eq. (C) seems strange, consider the following classical argument. The Lagrangian density is $$ {\cal L}~=~|\dot{\phi}|^2 - |\nabla \phi |^2 - {\cal V} ~=~\frac{1}{2}(\dot{\phi}^1)^2+\frac{1}{2}(\dot{\phi}^2)^2-\frac{1}{2}(\nabla \phi^1)^2-\frac{1}{2}(\nabla \phi^2)^2 - {\cal V} .\tag{D} $$ Therefore the momenta read

$$ \pi_j~=~\frac{\partial {\cal L}}{\partial \dot{\phi}^j}~=~\dot{\phi}^j, \qquad j~\in~\{1,2\}, \tag{E}$$

$$ \pi~=~\frac{\partial {\cal L}}{\partial \dot{\phi}} ~=~\frac{1}{\sqrt{2}}\left(\frac{\partial {\cal L}}{\partial \dot{\phi}^1}\color{red}{-}i \frac{\partial {\cal L}}{\partial \dot{\phi}^2} \right) ~=~\dot{\phi}^{\ast} ~=~\frac{1}{\sqrt{2}}(\pi_1\color{red}{-}i\pi_2).\tag{F}$$

III) For reference, let us mention that the Hamiltonian Lagrangian density reads

$$ {\cal L}_H~=~\pi \dot{\phi}+\pi^{\ast} \dot{\phi}^{\ast} -{\cal H} ~=~\pi_1 \dot{\phi}^1+\pi_2 \dot{\phi}^2 -{\cal H}, \tag{G}$$

where the Hamiltonian density is

$$ {\cal H}~=~|\pi|^2 + |\nabla \phi |^2 + {\cal V}~=~\frac{1}{2}(\pi_1)^2+\frac{1}{2}(\pi_2)^2+\frac{1}{2}(\nabla \phi^1)^2+\frac{1}{2}(\nabla \phi^2)^2 + {\cal V} .\tag{H}$$