cosmology - Is Poincare recurrence relevant to our universe?

If the theory of everything indicates a singularity-free and finite universe, will Poincare recurrence be relevant to the universe? If so, is there any interesting physical consequence, e.g. in superstring or quantum gravity?

Answer

Yes, our Universe is approaching the empty de Sitter space – we are already pretty close to it, actually, because the cosmological constant dominates the vacuum energy (68% of it). It has a cosmic horizon (the boundary behind which we can't see) and the degrees of freedom are formally living on that surface.

Via the holographic principle (in a somewhat less tested context), one may claim that this means that de Sitter space has a finite entropy so a finite-dimensional Hilbert space is enough to describe everything that happens in it (including the matter in not-yet-empty mostly de Sitter space, like the present Universe). If that's so, the Universe we inhabit behaves much like any system with finitely many degrees of freedom, and it has Poincaré recurrences.

The Poincaré recurrence time is extremely long, something like $\exp(10^{120})$ billion years – it is because the entropy of the de Sitter horizon is $10^{120} k_B$ (the cosmological constant is $10^{-120}$ in Planck units or so). After this superlong time, approximately, events start to repeat themselves. At least in some sense, it is fair to say that the time is literally periodic.

This very long timescale is the ultimate "maximum duration" that may be discussed in physics. For example, if a calculation in field theory or string theory implies that the lifetime of a vacuum is longer than this Poincaré recurrence time, the decay is considered unphysical because "it cannot happen in time", anyway.

The timescale is vastly longer than anything we have a chance to experimentally test. Each 10 billion years or so, the linear distances between galaxies double and the density of normal matter decreases by an order of magnitude. In hundreds of billions of years, a vast majority of the currently active stars will be inactive and even "new generations" of the stars will already be gone or dying. In trillions or certainly quadrillion years, there will be nothing left to energize star-powered life as we know it, and similar "local, more modest" sources of useful energy will be diminishing in similar ways.

It is hard to imagine that there will be any intelligent beings in a quadrillion years. This is still vastly smaller than the Poincaré recurrence time. And this result of the comparison is no coincidence. Of course that things must have a chance to "destroy any pattern" before the chaos has a chance to reassemble itself into the patterns again.

thermal field theory - Matsubara frequencies as poles of distribution function

Is there any deeper meaning to why the bosonic/fermionic Matsubara frequencies appear as poles of their corresponding distribution functions (with an additional $i$)?

For example in the bosonic case we have: $$\omega_n=\frac{2n\pi}{\beta}$$ which are the poles of $$n_B=\frac{1}{e^{\beta \hbar \omega}-1}$$ if we say that $\omega = i\omega_n$

I know this is sometimes exploited when evaluating sums of functions of $\omega_n$.

See Matsubara frequency.

Answer

The point is that to achieve the sum over Matsubara frequencies $$\sum_{n} g(i\omega_n)$$ we can use a contour integral $$\oint_C g(z) f(z)$$ with the contour described in fig 1 here, so long as we choose an $f(z)$ with simple poles exactly at the Matsubara frequencies $\omega_n$. This determines $f(z)$ to be proportional to the Bose-Einstein distribution. In fact, if we think about the sum as computed in a correlation function, we could have derived the integral above instead by considering computing the expectation value in the thermal density matrix $\int dE n(\beta,E) |E\rangle\langle E|$ determined by the Bose-Einstein distribution $n(\beta,E)$. I think this is an equally good starting point of the logic, then deriving the Matsubara frequencies as the poles of the distribution function.

I'd say the whole point of the Matsubara frequencies is that they are the poles of the distribution function.

quantum mechanics - Uncertainty Principle Intuition

So, as your usual physics undergrad, I read Griffiths's derivation of the general uncertainty principle. I understood it but there was no physical intuition given behind it in the book. It was basically a bunch of manipulations and inequalities used to get just the right result.

Other than "It's because of the statistical interpretation." is there a more intuitive reasoning behind uncertainty. Like commutators, behavior of particles, etc.

Thank you for the replies and as always, much obliged.

Answer

Heisenberg's Uncertainty Principle is, in essence, a consequence of two basic facts: de Broglie's relation between a particle's momentum and its associated wavelength, and a mathematical fact known as the bandwidth theorem, which states an equivalent uncertainty relation between a wave's position and its wavelength.

The fundamental physical leap here is of course the first one, the statement that particles have a 'wavelength' of some meaningful sort. This leap takes you directly outside of classical physics and there is nothing within the classical sphere that is really enough to motivate it. De Broglie's relation remains a cornerstone of quantum theory, though it is usually stated in a more sophisticated form as the canonical commutation relation $[\hat x, \hat p]=i\hbar$.

The bandwidth theorem, on the other hand, is a very intuitive statement about waves, and it essentially says that if you want to measure a wave's wavelength $\lambda$ to some uncertainty $\Delta \lambda$, you will need to measure a number of periods $n$ that is inversely proportional to $\Delta \lambda$. Thus, on one end, to have zero uncertainty in the wavelength you need to measure an infinite number of periods, and on the other end, a wave pulse that's so short it only fits one peak and no troughs can hardly be said to have a defined wavelength.

The precise statement is cleanest in terms of the wavevector $k=2\pi/\lambda$, and the theorem states that the size $\Delta x$ of the region occupied by some waveform, and the width $\Delta k$ of the region of wavevectors that appreciably contribute to the waveform's plane-wave decomposition, are related via $$\Delta x\,\Delta k\geq 2\pi.$$ Since de Broglie's relation reads $p=\hbar k$, the extension to the Heisenberg principle is obvious.

mathematics - A Fairly Simple Equation

I'm new, so I thought I'd start with a fairly simple puzzle.(Simple doesn't necessarily mean easy.)

Can you fit the numbers $2, 3, 4, 5, 6,$ and $10$ into this equation, each replacing one letter, so that the result is $93$ (rounded to the nearest whole number)

Equation: $(A+B)(C-D)(E/F)$

I'm fairly certain there is only one solution. (excluding reversing the order on the addition)

Answer

Since rounding is allowed, but the rest are not; I can promise that there are at least 2 solutions:

$(4 + 10)(6 - 2)(5 / 3) = 93$
$(10 + 4)(6 - 2)(5 / 3) = 93$

I was actually curious if there were any further solutions and created a brute force style application in C# to find out. There are actually exactly two solutions.

decimal[] nums = new decimal[] { 2, 3, 4, 5, 6, 10 };
foreach (decimal a in nums)
    foreach (decimal b in nums)
        foreach (decimal c in nums)
            foreach (decimal d in nums)
                foreach (decimal e in nums)

                    foreach (decimal f in nums) {
                        if (a == b || a == c || a == d || a == e || a == f ||
                            b == c || b == d || b == e || b == f ||
                            c == d || c == e || c == f ||
                            d == e || d == f ||
                            e == f) continue;
                            if (Math.Round((a + b) * (c - d) * (e / f)) == 93)
                                Console.WriteLine($"(({a} + {b}) * ({c} - {d}) * ({e} / {f}) = 93");
                    }

---

Without doing any of the following, there are no solutions.

• Combining numbers.


• Reusing numbers.


• Rotating numbers.


• Rounding.

When allowing numbers to be combined, there are thousands of solutions given that:

There are 6 numbers, and this means there are 7 basic representations of combinations for each number, for example:
$2, 22, 23, 24, 25, 26, 210$.

However, if you go even further; you can create roughly 36 combinations for each number if you limit yourself to only a single repetition in a single combination.

When rotating numbers is allowed, this number of solutions increases substantially, especially if combination is also allowed. Throw in rounding on top of all of this and you've got yourself a solution celebration similar to the ball drop in Time Square.

general relativity - How does the string worldsheet affect the space-time in which they live?

I don't understand much about string theory and never really got much further past the Nambu-Goto action and very basic supersymmetry (SUSY) lectures in my undergraduate courses, but the only thing that stuck is that the string worldsheet lives in space-time.

So, if strings are allowed to interact only by string exchange (which amounts just to changing the genus of the worldsheet, that still keeps living in space-time), how do we go from that state of things where we have actions (that is, Polyakov) that depend on the metric properties of the worldsheet, and describe how the worldsheet evolve inside space-time as a fixed background, to strings-exchange-things-that-curve-spacetime-itself, understanding spacetime the very space those strings live within.

If strings are allowed to interact only via string exchange, does it mean that...

1. 
   
   

   strings are connected topologically to the space-time via strings? That would be weird, a two-dimensional surface connected to a four-dimensional one, but I guess it would be logically possible. But that suggests that strings basically live outside space-time and only touch it through its boundaries, so closed strings are out of this picture

   
   


2. 
   

   space-time is itself made of interacting strings? that would not make much sense to me. How would that preserve the traditional Lorentz invariance? Besides, string theory assumes the Lorentz invariance is exact

   
   


3. 
   

   strings have a mass density and just behave as classical general-relativity energy-momentum stress tensor densities? That would make sense, but then that would be openly cheating, since that would not explain how strings create gravity, since gravity would be added ad-hoc

   
   


4. 
   

   I'm out of options. How can strings interact with the space-time manifold?

   
   
   

---

Edit

Thanks for the existing answers, but what I want, or what I was hoping is, if there is some diagrammatic/visual insight into the connection mentioned in the answers, say, the one from flat spacetime with graviton strings propagating that turns into an equivalent curved spacetime.

A delve into extraordinary chess problems: Selfmate 1

[This series continues from a harder puzzle here and continues in this puzzle]

Here we have an easier but still interesting selfmate in 5.

This is a puzzle by A. Rubin found in The Book of Extraordinary Chess Puzzles:

From the book: "In a 'selfmate' problem White plays to force Black to checkmate him - a sort of chess suicide. In the diagram above, White has his king nicely trapped in the bottom right-hand corner, but still has to find a way of forcing Black to deliver the mate. Selfmate in five.

Here is an interactive board for you to play on.

Answer

is it as simple as

1. Ba6 forcing axb6
2. Bb5 bxc5

3. Bc4 bxd4
4. Bd3 bxe3
5. Be2 bxf2#

Taking advantage of the fact

that black's knight can't move (pinned by the queen) and blacks king can't move.

quantum mechanics - Can one define wavefunction for Bogoliubov quasiparticle excitation in a superconductor?

Wavefunction is essentially a single particle concept. It is easily extended to multiparticle system as follows- if one has say five electrons the wavefunction of this five electron state is any completely antisymmetric function of five coordinates which is square integrable in the five dimensional space. Given a five electron ket in fock space $|K\rangle$, its wavefunction is denoted as $\langle x_1 x_2 ...x_5|K\rangle$. But for a superconductor its effective Hamiltonian doesn't conserve the particle number. Then can one come up with any reasonable defination of a wavefunction for a single quasiparticle excitation of the superconductor over its ground state denoted by $\gamma_i^{\dagger} |G\rangle$ where $|G\rangle$ is superconducting groundstate composed of Cooper pairs and $\gamma_i^{\dagger}=\sum_k u_i^kc_k+v_i^kc_k^{\dagger}$ is Bogoliubov quasiparticle creation operator and $c's$ being electron operators and $u's$ and $v's$ being some complex numbers.

In Kitaev chain and it's solid state realisation one usually talks about Majorana fermion(Bogoliubov excitation) being localised at the two ends. How can one do that without a reasonable definition of wavefunction for superconducting states? The papers usually interpret eigenvectors of $H_{BdG}$ in coordinate space as representative wavefunctions. Is it justified?

Answer

The localization of Majorana zero modes has a well-defined meaning: consider a Kitaev chain with two ends. Because of the zero modes, there are two nearly degenerate ground states, let us call them $|0\rangle$ and $|1\rangle$, which have opposite fermion parities. They are localized as "single-particle wavefunctions" in the following sense: if one computes the matrix element $\langle 1|c^\dagger(x)|0\rangle$ where $c^\dagger(x)$ is the creation operator for fermions, the result is an exponentially decaying function of $x$ away from the edge. This definition works even when the system is interacting. Intuitively it means that the weight for creating a single fermion excitation is localized near the edge, and in the bulk there is a finite gap to the single particle excitations.