Tuesday, April 19, 2016

quantum field theory - Why do neutrino oscillations imply nonzero neutrino masses?



Neutrinos can pass from one family to another (that is, change in flavor) in a process known as neutrino oscillation. The oscillation between the different families occurs randomly, and the likelihood of change seems to be higher in a material medium than in a vacuum.


Why does the oscillation of neutrinos directly imply that they must have a nonzero mass, since the passage of one flavor to another can only occur in massive particles? And What are the cosmological implications of these oscillations?




Monday, April 18, 2016

quantum field theory - Definition of Casimir operator and its properties


I'm not sure which is the exact definition of a Casimir operator.


In some texts it is defined as the product of generators of the form: $$X^2=\sum X_iX^i$$


But in other parts it is defined as an operator that conmutes with every generator of the Lie group.



Are these definitions equivalent? If the answer is yes, how could I prove it (I'm thinking in using Jacobi's identity)?



Answer



I'll give you enough hints to complete the proof yourself. If you're desperate, I'm following the notes by Zuber, which are available online, IIRC.


Let's start with some notation: pick some basis $\{t_a\}$ of your Lie algebra, then $$ [t_a,t_b] = C_{ab}{}^c t_c$$ defines the structure constants. If you define $$ g_{ab} = C_{ad}{}^e C_{be}{}^d,$$ then this gives you an inner product $$(X,Y) := g_{ab} x^a y^b, \quad X = x^a t_a \text{ and } Y = y^b t_b.$$ Indeed this "Killing form" is related to the adjoint representation, as $$(X,Y) = \text{tr}(\text{ad } X \text{ ad} Y)$$ (exercise!). Similarly, $$g_{ab} =\text{tr}(\text{ad } t_a \text{ ad } t_b).$$ In this language, the Casimir $c_2$ is given by $$ c_2 = g^{ab} t_a t_b, \qquad \text{ so}$$ $$[c_2,t_e] = g^{ab} [t_a t_b,t_e].$$ Now you need to do some basic work (expand the first factor of the commutator, work out the resulting brackets) and you'll see that this gives you $$ \ldots = g^{ab} g^{dk} C_{bek} \{ t_a,t_d \}.$$ This vanishes (why?), so you're done!


Edit (regarding Peter Kravchuk's remark): when you write $c_2 \sim t_a t_b$, it's not really part of the Lie algebra. The only multiplication that "works" in Lie algebras is the commutator $[t_a,t_b]$. So these guys live in some richer structure, which is called the "universal enveloping algebra." Indeed you often hear that "the Casimir is a multiple of the identity matrix," but the identity matrix is seldom part of the Lie algebra (the identity in a Lie algebra is 0). In practice everything is self-evident, because you do calculations in some vector space.


Gauge-fixing of an arbitrary field: off-shell & on-shell degrees of freedom


How to count the number of degrees of freedom of an arbitrary field (vector or tensor)? In other words, what is the mathematical procedure of gauge fixing?




quantum mechanics - Turning a finite difference equation into code (2d Schrodinger equation)


I am trying to convert the following finite difference equations into code (taken from the bottom of page 12 of this thesis by Maike Schulte Numerical Solution of the Schrodinger Equation on Unbounded Domains (2007)) (NB: it's a PDF). Please be aware I don't know a lot about this sort of thing. $$ i\hbar D_t^+\psi^n_{j,k}=-\frac{\hbar^2}{2m^*}\left(D_x^2\psi^{n+1/2}_{j,k}+D_y^2\psi^{n+1/2}_{j,k}\right)+V_{j,k}^{n+1/2}\psi^{n+1/2}_{j,k}\tag{1.3} $$ where \begin{align} D_t^+\psi^n_{j,k}&=\frac{\psi^{n+1}_{j,k}-\psi^n_{j,k}}{\Delta t} \\ D_x^2\psi^{n}_{j,k}&=\frac{\psi^{n}_{j-1,k}-2\psi^n_{j,k}+\psi_{j+1,k}^n}{\Delta x^2} \\ D_y^2\psi^{n}_{j,k}&=\frac{\psi^{n}_{j,k-1}-2\psi^n_{j,k}+\psi_{j,k+1}^n}{\Delta y^2} \\ \end{align} denote the standard finite difference operators and $$ V_j^{n+1/2}=V\left(x_j,t_{n+1/2}\right)\\ \psi^{n+1/2}_{j,k}=\frac12\left(\psi^{n+1}_{j,k}+\psi^{n}_{j,k}\right) $$



I would like to make a simulation of the general behaviour of the Schrodinger equation in 2d, and scales are unimportant so I figure I can ignore $\hbar$, $m$, $\Delta t$, $\Delta x$ and $\Delta y$, setting them all $=1$. I am also dropping the potential term setting $V=0$ for now.


Presumably from the definition of $D_t^+$ the direction of time is positive thus the aim is to calculate $\psi^{n+1}$. So, the confusing thing, $\psi^{n+\frac{1}{2}}$ is used on the RHS hence $\psi^{n+1}$ appears in what I'm trying to compute. How can the formula for $\psi^{n+1}$ be expressed purely in terms of $\psi^{0


Answer



The issue is with short-hand notation. The term $\psi^{n+1/2}_{j,k}$ being operated on by $D_x^2$ really means \begin{align} D_x^2\psi^{n+1/2}_{j,k}&=D_x^2\left[\frac12\left(\psi^{n+1}_{j,k}+\psi^n_{j,k}\right)\right]\\ &=\frac12D_x^2\psi^{n+1}_{j,k}+\frac12D_x^2\psi^{n_{j,k}} \\ &=\frac1{2\Delta x^2}\left[\psi^{n+1}_{j-1,k}+\psi_{j+1,k}^{n+1}-2\psi_{j,k}^{n+1}\right] \\ &+\frac1{2\Delta x^2}\left[\psi^{n}_{j-1,k}+\psi_{j+1,k}^{n}-2\psi_{j,k}^{n}\right] \end{align} and similarly for $D_y^2\psi^{n+1/2}_{j,k}$. The definition of $\psi^{n+1/2}_{j,k}$ is given in that document. Using this, you'll find something along the lines of $$ \alpha\psi^{n+1}_{j+1,k}+\beta\psi^{n+1}_{j,k+1}+\gamma\psi^{n+1}_{j-1,k}+\delta\psi^{n+1}_{j,k-1}+\epsilon\psi^{n+1}_{j,k}=f\left(\psi^{n}\right) $$ where the term on the right side is all the $t=n$ terms and $\alpha,\,\beta,\,\gamma,\,\delta,\,\epsilon$ are constants of parameters $dt,\,dx^2,\,dy^2,\,m^*,\,\hbar$ and $i$. This is a band-diagonal matrix and can be computed in a number of manners (e.g., sparse matrix solvers or iterative methods).


It was also pointed out in this CompSci post that it might be easier to solve the Schrodinger equation in two pieces, one real and one imaginary ($\psi=R+iI$) and solve them at alternate half-time steps. Though it might be a little more complex, the method should be extensible to 2D.


In one dimension, you can solve the Crank-Nicolson method with a tri-diagonal matrix algorithm. In 2D, you get a penta-diagonal matrix that is a bit more complicated to solve (cf. this FORTRAN routine by Dr Kevin Kreider at the University of Akron). If you sweep along the dimensions individually (i.e.g, solve $x$ for each $y$ cell and vice versa), you can the aforementioned tri-diagonal solver to solve 2D problems (though it's typically referred to as "alternating direction implicit" when done this way).


Alternatively, you can use some sparse solvers in BLAS routines or try an iterative solver like the Gauss-Seidel method.


Sunday, April 17, 2016

thermodynamics - Why the temperature is getting lower when the universe is expanding


As we know, if an ideal gas expands in vacuum, as its energy is unchanged, the temperature remains the same. An ideal gas's energy does not depend on volume. In general, the energy is $kT$ times the total degrees of freedom, like in an ideal gas, the total degrees of freedom is $N$ particles plus three dimensions, $3N$.


Then if the total energy of the universe is $kT$ times the total degrees of freedoms of the universe, as the universe expands, its energy and entropy should not change, but if the temperature falls, the number of degrees of freedom should grow. It is quite puzzling to me that the universe is having more and more new degrees of freedom. It seems to be contradictory to the entropy argument.




mathematical physics - Physical significance of getting an non-integrable function in an equation


I just found out during my Calculus course in High School, that there exist functions which cannot be integrated.


Then I thought that I come across a lot of integrals while solving Physics questions.




Although I haven't come across such a non-integrable function yet, what physical significance would such an equation hold?



Has anyone come across such functions while solving a physics problem? Some examples would help a lot(High School level examples would be even better).



Answer



First off I'm not sure exactly what you mean be non-integral function. John Dodson discussed two interpretations that commonly appear in physics: non-elementary functions and singularities. The only other possibility I can think of are integrals like $$ \int_{-1}^{1}\mathrm{sin}(\frac{1}{x})dx $$ This oscillates infinitely many times near the origin, such that it is basically undefined at x = 0. As far as I know, these "pathological functions" don't appear in physics.


That being said, I'll try and give a more satisfactory answer by going over an example from electromagnetism: the Green's Function. So that you know where this is going, I'll give an overview: the Green's Function is important for solving Maxwell's equations for arbitrary charge distributions, but trying to evaluate it naively leads to an integration similar to $\int \frac{dx}{x}$ - but for the equations to have any meaning, we need this integral to be finite. To accomplish this we will need to integrate through the complex plane, a strange process that yields physically interpretable results. The math of this example, will be a little sophisticated, but only a vague understanding of the concepts is required to see what's going on.


Simplified Maxwell's Equations


Maxwell's equations talk about the interactions of the electric and magnetic field through differential equations. It turns out we can represent the electric and magnetic fields by introducing a single potential vector $A$ with four components: one "time" component $\phi/c$ and three space components represented by the vector $\bf{\vec{A}}$. Then the electric and magnetic fields are given by $$ \bf{\vec{E}}=\bf{\vec{\nabla}}\phi,\quad \bf{\vec{B}}=\bf{\vec{\nabla}}\times\bf{\vec{A}} $$ and Maxwell's equations are $$ \frac{1}{c^2}\frac{\partial^2\phi(\bf{x},t)}{\partial t^2} - \nabla^2\phi(\bf{x},t) = \frac{\rho}{\epsilon_0} \\ \frac{1}{c^2}\frac{\partial^2\bf{\vec{A}}(\bf{x},t)}{\partial t^2} - \nabla^2\bf{\vec{A}} (\bf{x},t) = \mu_0\bf{\vec{J}} $$ (You can go here to learn more about Maxwell's equations, and here to learn more about the operator $\bf{\vec{\nabla}}$).


Green's Function



Let's focus on the first of Maxwell's equations, the one for $\phi$. One way to find solutions for any given $\rho$ (that is, the density of charge) is to start with the solutions for $\rho=0$ - that is, solutions to: $$ \frac{1}{c^2}\frac{\partial^2\phi}{\partial t^2} = \nabla^2\phi $$ If you're currently in Calculus, this equation may look daunting; however, the solution turns out to be trivial: see wave equation on Wikipedia. If we can relate solutions to Maxwell's equations for any $\rho$ to these known solutions for $\rho=0$, then we're good! The way to do this is to use a Green's function for the wave equation: a function such that $$ \phi_\rho(\bf{x},t) = \phi_0(\bf{x},t) + \int G(\bf{x-x'},t-t')\rho(\bf{x'},t')d^4 x $$ where $\phi_\rho$ is a solution to Maxwell's equation, and $\phi_0$ is a solution to the $\rho=0$ case. The integral is taken over all four dimensions of spacetime (hence the symbol $d^4x$). There is a special differential equation that will give us Green's function, and its solution can be represented as:


$$ G(\textbf{x},t) = \int \frac{e^{i(\omega t - \textbf{k}\cdot\textbf{x})}d^4k}{\omega^2 - \textbf{k}^2} $$


This integral is carried out over four variables: the "frequency" variable $\omega$ and the "wavenumber" vector $\textbf{k}$. These correspond to the various wave solutions to the $\rho=0$ equation.


Non-integrable function


So what does this have to do with your question? Let's say we try to integrate over the variable $\omega$. This looks like $$ \int\left(\int_{-\infty}^{\infty} \frac{e^{i(\omega t - \textbf{k}\cdot\textbf{x})}}{\omega^2 - \textbf{k}^2}d\omega\right) d^3\textbf{k} $$


If we partial-fraction decompose, we have $$ \int\frac{e^{i(\textbf{k}\cdot\textbf{x})}}{2|\textbf{k}|}\left(\left(\int_{-\infty}^{\infty} \frac{e^{i\omega t}}{\omega - |\textbf{k}|}d\omega\right) - \left(\int_{-\infty}^{\infty} \frac{e^{i\omega t}}{\omega + |\textbf{k}|}d\omega\right)\right) d^3\textbf{k} $$ Both of these internal integrals are obviously problematic: integrating straight from $-\infty$ to $\infty$, we run over two singularities when $\omega = |\textbf{k}|$ or $\omega = -|\textbf{k}|$!


When we encounter integrals like these, we have to do some tricks to get meaningful information out of them. The trick we use here is not to integrate straight from one end to the other, but to take a curvy path through the complex plane that avoids the non-integrable singularities. The possible paths include but are not limited to:


"Advanced" solution "Retarded" solution


The retarded solution is so named because it accounts for waves that travel forwards in time (which means the charge $\rho$ can only affect the future) and the advanced solution accounts for waves that travel backwards in time (meaning the charge $\rho$ can affect the past). Since we have not yet witnessed any ability to send messages into the past, we typically interpret the retarded solutions as being the only "allowed" ones; the advanced solutions are "banned" (though Feynman-Wheeler absorber theory offers another interpretation).


The retarded solution is: $$ G_R(\textbf{x},t) = \Theta(t)\frac{\delta(t-\frac{r}{c})}{4\pi r} $$ The step function $\Theta(t)$ is zero for negative $t$, while the Dirac delta function $\delta(x)$ is zero whenever $x$ is nonzero. $r$ represents the radius of the point $\textbf{x}$. Thus, this equation corresponds to waves travelling forwards in time with velocity $c$.



Recap


To summarize - there are certain times that non-integrable functions enter into physical calculations, and the attempt to intepret these integrals and find useful solutions can lead us to a broader understanding of the physical situation.


One other significant way this happens is in quantum field theory. The calculation of interacting quantum fields can lead to severely diverging integrals, none of which are as easily avoided as the one we looked at. A class of techniques known as "Renormalization" were created to deal with these infinities, centering around the idea that field theories are not valid on small scales, pointing us towards a more accurate high-energy theory that will eliminate the infinities. Though initially controversial, the modern understanding of such theories is well-founded and has uses beyond infinity-avoidance. (Thanks, Michael Brown!)


Thus one could argue that non-integrable functions not only occur in our understanding of nature, but they augment our understanding of it by leading us to new methods of interpretation.


I hope that I have provided you with a satisfactory example and answer to your question.


EDIT: Initially I called renormalization techniques "controversial" - as Michael Brown pointed out, this is a bit of a mischaracterization. When they were first introduced, many prominent physicists expressed discomfort with these techniques, but a better understanding of the science behind renormalization has led to a well-founded theory and today most quantum field theories are judged by their ability to be renormalized.


nuclear physics - Do some half-lives change over time?


I was recently doing some physics tuition on radioactivity and the student claimed her chemistry teacher had said that radioactive substances can be grouped into two divisions: those whose half-life is constant and those whose half-life changes over time.



I had never heard of this before and can't think of any reason why a half-life should change, so does anyone else know anything about this?


(I know some half-lives can be altered under certain conditions, but I'm talking about a natural change over time).




classical mechanics - Moment of a force about a given axis (Torque) - Scalar or vectorial?

I am studying Statics and saw that: The moment of a force about a given axis (or Torque) is defined by the equation: $M_X = (\vec r \times \...