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GNU/Linux EFI boot for Lenovo ThinkPad x230
The EFI file should be located in <EFI partition>/EFI/boot/bootx64.efi
, so copy the EFI file installed by the GNU/Linux distribution (for exemple, <EFI partition>/EFI/ubuntu/grubx64.efi
in Ubuntu 16.04) to the proper location, and your system will boot.
Essay Contest: How has social media enhanced your research?
Welcome to Authorea!
Hey, welcome. Double click anywhere on the text to start writing. In addition to simple text you can also add text formatted in boldface, italic, and yes, math too: \(E = mc^{2}\)!
Add images by drag'n'drop or click on the "Insert Figure" button.
Calibrate your display in GNU/Linux with Gnome 3 + Argyll and a ColorVision/DataColor Spyder2
The Spyder2 from ColorVision/DataColor is a display calibration device which aims at being used with Microsoft Windows or Apple MacOS X operation systems. I had bought it a long time ago to calibrate my display on my computer under Windows, at the time.
Regnskapsfører
Bookkeeper er ditt lokale regnskapsbyrå som tilbyr et bredt spekter av tjenester som regnskap, lønn, HR, rådgivning blant andre. Vår regnskapsførere vil ta seg av regnskap og finans jobb å sørge for at du lever et bekymringsløst liv. Vi sørger for at regnskap er enkelt, tilgjengelig og morsomt for hver kunde som bruker våre tjenester.
Secure Research Funding With Visuals
and 1 collaborator
Numerki
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- Równania liniowe
Równanie Laplace
Równanie to wyraża następującą własność pola potencjalnego: dywergencja (rozbieżność) wektorowego pola potencjalnego (czyli gradientu potencjału), pod nieobecność źródła jest równa zeru. Opisuje ono zatem wiele procesów zachodzących w przyrodzie, np. potencjał grawitacyjny poza punktami źródeł pola (czyli bez punktów materialnych), potencjał prędkości cieczy przy braku źródeł. Równanie Laplace’a jest szczególnym przypadkiem równania Poissona, wyrażającego analogiczny związek w przypadku istnienia źródeł pola.
$ \bigtriangleup u = \sum_{i=1}^{n} u_{x_ix_i} =0$
$ u_{x_i} = \frac{du}{dx_i}$
$ u_{x_i x_i} = \frac{d^2u}{dx_i^2}$
Równanie Poisona
Równanie Poissona opisuje wiele procesów zachodzących w przyrodzie, np. rozkład pola prędkości cieczy wypływającej ze źródła, potencjał pola grawitacyjnego w obecności źródeł, potencjał pola elekrostatycznego w obecności ładunków, temperaturę wewnątrz ciała przy stałym dopływie ciepła.
△u = f
Liniowe równanie transportu
$u_t + \sum_{i=1}^{n}b^iu_{x_i} = 0$
przewodnictwa / dyfuzji
ut − △u = 0
Schrodingera
ut + △u = 0
falowe
utt − △u = 0
- Równania nieliniowe
nieliniowe równanie Poissona
− △ u = f(u)
p-harmoniczne
dir(|Du|P − 2Du)=0
dir - dywergencja
$dir v = \sum_{i=1}^n \frac{dv_i}{dx_i}$
Du = ▽u
Food resources - info about healthy eating, habits and more
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How To Submit Your Essay
Essay Contest: How has social media enhanced your research?
Welcome to Authorea!
Reinventing Peer Review
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PHYS6562 F2 Daily Quaetion
First observe that the probability density at the origin at any time is zero.
\begin{equation} \rho(0,t) = 0 \end{equation}
And the probability density on x = x′ at t = 0 is
\begin{equation} \rho(x,0) = \delta(x-x') \end{equation}
We can assume that there is a image probability density at t = 0 on x = −x′. Therefore the total probability at t = 0 should be
\begin{equation} \rho(x,0) = \delta(x-x')-\delta(x+x') \end{equation}
The Green’s function for each delta function is a time-dependent Gaussian distribution,
\begin{equation} G(x,t) = \frac{1}{\sqrt{4\pi Dt}} e^{-x^2/4Dt} \end{equation}
with the particle originally at the origin. For this case, both delta functions diffuse, yielding the same Gaussian distribution.
\begin{equation} G(x,t) = \frac{1}{\sqrt{4\pi Dt}} \left( e^{-(x-x')^2/4Dt} - e^{-(x+x')^2/4Dt}\right) \end{equation}
The time evolution probability distribution is shown in fig.1.
PHYS6562 M3 Daily Question
Most of the volume of high dimensional spheres is near the surface. As we can approximate the volume as V ≈ r3N in 3N dimensions, near the center, r ≈ 0 and V ≈ 0. Near the surface, where r becomes larger and contributes to most of the volume.
For statistical mechanics, instead of taking the whole E sphere into account, which might involve high-dimensional integral, it’s easier to focus on the shell of E sphere. Then the momenta can be considered to be $p = \sqrt{2mE}$ instead of doing
\begin{equation} \int_0^{\sqrt{2mE}} (....)dp \end{equation}
PHYS6562 W2 Daily Question
a. The density of tin atoms ρSn(x) is locally conserved since there is no tin atom created or annihilated.
b. It depends. Under the constraints that one cannot change his/her political belief, yes, the density of democrats is locally conserved. However, if changing of political belief is allowed, the density of democrats is not locally conserved. There might be democrats changing to republicans or vise versa, which implies that the democrats or republicans can teleport, created or annihilated.
Suppose there is no external force. We can write the current as
\begin{equation} J = -D(\rho)\nabla\rho \end{equation}
And apply
\begin{equation} \frac{\partial\rho}{\partial t} = -\nabla\cdot J \end{equation}
\begin{equation} \frac{\partial\rho}{\partial t} = \frac{\partial D(\rho)}{\partial\rho} \nabla\rho\cdot\nabla\rho + D(\rho)\nabla^2\rho \end{equation}
\begin{equation} \frac{\partial\rho}{\partial t} = \frac{\partial D(\rho)}{\partial\rho} (\nabla\rho)^2 + D(\rho)\nabla^2\rho \end{equation}
This is the diffusion equation with density-dependent diffusion constant.
PHYS6562 W3 Daily Question
First of all, Ω1(E1)δE1 and Ω2(E2)δE2 are the phase space volume for these two subsystems. Since most of the volume in phase space is near the surface, we can instead write Ω1(E1) and Ω2(E2) to denote the two volumes in phase space. Second, it’s clear that the probability density of subsystem 1 having energy E1 is proportional to the ’allowed’ volume product of the two subsystems.
\begin{equation} \rho_1(E_1) \propto \Omega_1(E_1)\Omega_2(E_2) \end{equation} where E2 = E − E1. And finally, we impose the normalization condition:
\begin{equation} \int \rho_1(E_1)dE_1 = A \int \Omega_1(E_1)\Omega_2(E-E_1)dE_1 = A \Omega(E) = 1 \end{equation} A is the proportional constant and equal to 1/Ω(E). Therefore,
\begin{equation} \rho_1(E_1) = \frac{\Omega_1(E_1)\Omega_2(E_2)}{\Omega(E)} \end{equation}
If we maximize the probability at E1:
\begin{equation} \frac{d\rho_1}{dE_1} = 0 \end{equation}
\begin{equation} \frac{1}{\Omega_1}\frac{d\Omega_1}{dE_1} - \frac{1}{\Omega_2}\frac{d\Omega_2}{dE_2} = 0 \end{equation}
By plugging the definition of entropy S = kBlog(Ω),
\begin{equation} \frac{1}{k_B}\frac{dS_1}{dE_1} - \frac{1}{k_B}\frac{dS_2}{dE_2} = 0 \end{equation}
\begin{equation} \frac{dS_1}{dE_1} + \frac{dS_2}{dE_1} = \frac{d}{dE_1}(S(E_1)+S(E_2)) = 0 \end{equation} ,which implies the maximization of the sum of entropies of both subsystems.
For (6), we see that if the probability density ρ(E1) is maximized,
\begin{equation} \frac{1}{T_1} = \frac{1}{T_2} \end{equation}
PHYS6562 F3 Daily Quaetion
For N distinguishable particles, if at certain instant they occupy N states in the phase space, there are N! different choices of such configuration. The total volume for this system in phase space is the product of the volume for each particle in its own sub phase space.
\begin{equation} \Omega_D = N! \space \omega_1 \omega_2 \omega_3 ... \omega_N \end{equation}
If the particles are now undistinguished, there is only one choice of such configuration without permutation.
\begin{equation} \Omega_U = \omega_1 \omega_2 \omega_3 ... \omega_N \end{equation}
Therefore, $\Omega_D = N!\space \Omega_U $.
For the mixing problem of black and white particles of the same pressure on both sides, if you cannot tell the difference between black and white ones, you cannot extract work from the mixing. Actully, you won’t know if they’ve mixed already or not. However, suppose the black and white particles are of different pressure, and you still cannot tell them apart. If there were a membrane which allows particles of lower pressure to penetrate through, this membrane then can tell the difference between the two sides without telling the particles apart. Then work can be extracted.
Now if a door can distinguish the black from the white particles and let through white ones, after very long time, the white particles will distribute themselves equally in the two sides, i.e. there will be $\frac{1}{4} N\space white + 0 \space black$ in the left and $\frac{1}{4} N \space white + \frac{1}{2} N \space black$ in the right. And we can extract work from it due to the pressure difference between the two sides.
PHYS6562 W4 Daily Question
It’s not true that neighboring initial conditions will converge to the fixed state after long time. As we saw from the Jupiter problem, there are some regions where the initial conditions for chaotic and periodic trajectories lies infinitesimally close but lead to different results. The chaotic initial conditions will not make the system converge to certain state. Therefore, there is no attractor in this 3-body problem.
Stable Hamiltonian systems can withstand a minor perturbation in the process and the system will still end up where it’s supposed to.
Unstable Hamiltonian system will diverge if perturbed in the process and the system will not come to the original equilibrium regardless of whether the system becomes chaotic or not.
PHYS6562 F4 Daily Quaetion
The system of classical particles bouncing back and forth is not ergodic since each particles can only occupy certain close volume in phase space, with fixed x, y position, Px, Py momenta and ±pz. Therefore, the given volume in phase space of the ensemble will not evolve with time.
(a) True.
(b) True, since the work in done on the gas.
(c) False, it’s a refrigerator.
(d) True.
\begin{equation}
W = (P_0)(3V_0) + \int_{4V_0}^{V_0} \frac{4P_0V_0}{V} dV = 3P_0V_0 - 4P_0V_0\log4
\end{equation} (e) True, a part does no work so ΔU = Q. \begin{equation}
\Delta U = \frac{3}{2} nRT_c - \frac{3}{2} nRT_h = -\frac{3}{2} (3P_0V_0) = -\frac{9}{2}P_0V_0
\end{equation} where n is the number of particles in mole and R = 8.317 is the ideal gas constant. (f) True. The whole cycle ΔU = 0, the heat input and the net work done onto the gas must equal the heat output. \begin{equation}
Q_h + W = Q_c
\end{equation}
PHYS6562 M5 Daily Question
(a) \begin{equation} \Omega = \frac{V^N}{N!} \end{equation}
\begin{equation} S = k_B \space \log(\Omega) = k_B(N\log(V) - B\log(N) + N) = Nk_B\left(1+\log\left(\frac{V}{N}\right) \right) \end{equation} The average volume of a region for single particle is v = V/N with the entropy of s = kBlog(v). Therefore, the configuration entropy is related to N times the entropy of a particle in its region with average volume.
(b) \begin{equation} S = Nk_B\left(\frac{5}{2}-\log(\rho\lambda^3)\right) = Nk_B\left(\frac{5}{2}+\log\left(\frac{V}{N}\lambda^3\right)\right) \end{equation} The configuration entropy for an ideal gas particle is also related to the small average volume v. \begin{equation} S_c = Nk_B\log(v) \end{equation}
But the ideal gas particles also occupy the momentum space and each particle has the average length of λ in each dimension. So
\begin{equation} S_p = 3Nk_B\log(\lambda) = Nk_B\log(\lambda^3) \end{equation}
Combine the configuration and momentum entropy for one particle, \begin{equation} S = S_c+S_p = Nk_B\log(v*\lambda^3) = Nk_B\log\left(\frac{V}{N}*\lambda^3\right) \end{equation} The total entropy is related to the N times the space occupied both in configuration and momentum space.
PHYS6562 W5 Daily Question
For a child’s room containing 100 things in the room, we only need to consider the configuration space since those things are stationary in the room. The room before tightening up, the number of configurations is
\begin{equation} \Omega = V^N = 5^{100} \end{equation}
Let’s consider the center-of-mass location accuracy of 1 cm. First of all, the center-of-mass can be located itself at any point of the room. Therefore,
\begin{equation} \Omega_{CM} = 5 \end{equation}
Then we can assume that the center of mass in x direction is at 0. For each randomly located thing A1 at x, there would be the another thing A1′ located at the −x. We now have only 50 things we can throw at will. For the A50′, since the deviation of 1cm is acceptable, A50′ can have roughly 1m of freedom in the x direction. Therefore,
\begin{equation} \Omega_{item} = (L^{50}*1)^3 = 5^{50} \end{equation}
where L = 51/3.
\begin{equation} \Omega' = \Omega_{CM}\times\Omega_{item} = 5^{51} \end{equation}
Entropy decrease can be found to be
\begin{equation} \Delta S = k_B\log{\Omega'} - k_B\log{\Omega} = -49k_B\log{5} \end{equation} \begin{equation} \Delta S_{total} = 10^9\times\Delta S = 1.09\times10^{-12} \end{equation}
For 1 liter baloon contracting by 1% in volume, \begin{equation} \Delta S = Nk_B\log{0.99V} - Nk_B\log{V} = \frac{PV}{T}\log(0.99) \end{equation} where P = 1atm, V = 1L and T = 300K. \begin{equation} \Delta S = -0.0034 (J/K) \end{equation}