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2

The midpoint is the zero frequency component of the image. That is, it will be the sum all pixels in the image (or the amplitude of a wave with no frequency, a flat field across the image). I did the FFT of the image you posted above (using Python, sorry), and found the midpoint value to be 12598085. The image is 336 x 205 pixels, so the midpoint pixel ...


0

Theorem. If $\hat{K}$ is Hermitian then the eigenvectors of $\hat{K}$ can be chosen so as to form an orthonormal basis $|\xi_i\rangle$. (The "can be chosen" wording completely handles the cases of degenerate eigenvalues, and the fact that an eigenvector times a constant is still an eigenvector) (the converse is not true. In fact replace the word ...


1

When talking about string theory they say how there is 12 dimensions and I am assuming they didn't pull that figure out of no where. Actually they start with string theories that have 10 dimensions but there do exist theories with higher ones. They do not pull these numbers out of their hat but they come from the constraints imposed in formulating ...


0

This book contains some letters from Landau to young physicists: If you have enough will, you can study theoretical physics independently - after all, it requires nothing but books and paper. (Mind you, Landau lived in Soviet Russia) 't Hooft has similar advice: It should be possible, these days, to collect all knowledge you need from the ...


2

Strange question! I assume that you are thinking about doing theoretical physics in your spare time. In this case, you a need a relative low amount of money. What you really need is a pc, an internet connection and some books. You can find books and articles online. It can be useful to attend seminars and schools in universities, so maybe you need some ...


0

Start by noting the velocities ($v$ and $V$) of both blocks (mass $m$ and $M$) after a collision, noting the shorthand $\delta = M-m$ and $\mu = M+m$: $$\begin{align} v_\textrm{after} &= -\frac{\delta}{\mu}v_\textrm{before}+\frac{2m_2}{\mu}V_\textrm{after} \\ V_\textrm{after} &= \frac{2m_1}{\mu}v_\textrm{before}+\frac{\delta}{\mu}V_\textrm{after} ...


1

An alternative method to Schur's lemma is to use the orthogonality relations of the characters. Let $\chi_R(g)$ be the trace of the matrix of $g$ in the representation $R$. If you have a finite group $G$, with $|G|=n$, there is an inner product between representations $$\langle \chi_R,\chi_{R'} \rangle = \frac{1}{n}\sum_{g\in G} \chi_R(g)\chi_{R'}(g)^*$$ Now ...


0

As you describe, let $v_n$ and $V_n$ be the velocities of masses $m$ and $M$ just before the $(n+1)$th collision (so both move to the right). $V_0$=0. In each collision, you have two conservation laws you can use: energy and momentum. Being careful about directions, you get $$ m v_n + M V_n = -m v_{n+1}+M V_{n+1}\\ \frac{1}{2}m v_n^2 + \frac{1}{2}M V_n^2 = ...


1

Schur's lemma asserts that a representation is irreducible if its commutant is trivial, that is it only contains multiple of the identities. Let $G$ be your group and let $\pi$ be any representation of $G$ over the representation vector space $V$ (usually a Hilbert space). Then $\pi$ is a group homomorphism between $G$ and $\text{GL}_n(\mathbb C)$. The ...


-1

I think this would not work for perfectly elastic collisions, the small mass $m$ would transfer all of its momentum and kinetic energy to $M$, so that after the collision $m$ is stationary and $M$ would have all the momentum and kinetic energy.


1

$\newcommand{\ro}{\mathbf{r}_o} \newcommand{\zo}{z_o} \newcommand{\rs}{\mathbf{r}_s} \newcommand{\zs}{z_s}$My first answer was wrong, but I think it was wrong in an instructive way so I will leave it. Here is my second attempt. I couldn't find a clever way to do it, so I did it the straightforward way. I hope it is right this time, but there may be algebra ...


1

Update: This solves the wave equation for the wrong initial conditions. One way of seeing the solution is to see that a spherical wave emanating from a point is a solution, and if you take a picture of that wave at any time, you will see exactly your initial condition. This means one solution is an outgoing spherical wave. Also, since the wave equation has ...


1

The main thing that goes bad in nonseparable Hibert space is the loss of Stone-von Neumann theorem. Loosely speaking, the Stove-von Neumann theorem assures us that Schroedinger representation of the canonical commutations rules is irreducibile, and it is unique modulo unitary equivalence. Hence Schroendinger and Heisenberg pictures are physically equivalent, ...


2

In spherical coordinates the position is $$ {\bf r} = r \ \hat{r} $$ and the velocity is (see this) $$ {\bf v} = \dot{r} \ \hat{r} + r \dot{\theta} \ \hat{\theta} + r \sin \theta \dot{\phi} \ {\hat{\phi}} $$ The angular momentum is then $$ \begin{eqnarray} {\bf L} = {\bf r} \times m {\bf v} &=& m r \ \hat{r} \times \left(\dot{r} \ \hat{r} + r ...


1

The Navier-Stokes Equations and its variants are some of the most important/relevant partial differential equations in material science.


0

Quantum mechanics is a physical theory. For a given physical system (set up and possible states) you can fix a Hilbert space, and some linear operators. Some linear operators will be unitary (for instance time evolution, to advance a state at one time to a state at a different time), some operators will be self-adjoint (for instance for an observable). An ...


3

No answers yet? So let's take a shot at a (partial) answer: Therefore as my main question, does symplectic geometry underpin thermodynamics? No. In thermodynamics, we're dealing with a Legendrian submanifold of a contact manifold (cf Wikipedia). Thermodynamic variables are canonical coordinates on that manifold. Morally speaking, in case of symplectic ...


3

Since I started writing it, I will complete this answer to add some extra mathematical detail. As stated by lionelbrits, the $d^2-1$ Hermitian generators of $\mathrm{SU}(d)$ are traceless and orthonormal with respect to the Hilbert-Schmidt inner product $$(A,B) = \mathrm{Tr} (A^{\dagger}B).$$ Therefore, along with the identity, these operators form a ...


3

The generators you mention form a complete basis for traceless hermitian matrices. Thus, to express a hermitian matrix that has a non zero trace in this basis, you have to augment the basis with one more matrix that is trace orthogonal to the others, and the identity does the trick. You can verify that the number of matrices in your basis agrees with the ...


0

I think what you are asking is whether your theory is renormalizable. Renormalizable means: all divergences can be absorbed by terms in the original energy functional by rescaling of the coefficients and fields. If you cannot, then your theory is not renormalizable and can therefore not exist as a quantum field theory at all energies. If you persevere and ...


2

Let $\mathscr{H}$ be a separable Hilbert space. Suppose that the Hamiltonian $H$ is a densely defined self-adjoint operator with domain $D(H)\subset \mathscr{H}$. Then for any $\phi\in D(H)$, $i\partial_t e^{-it H}\phi=He^{-itH}\phi$, where $e^{-itH}\phi$ is the unique solution of the Schrödinger equation. Now, $e^{-itH}$ is a unitary operator for any ...


0

It's important to be more precise when you mean 'bound state of finite size'. Recall that bound state solutions to the Schrödinger equation behave at large distances as $\sim\exp(-\kappa x)$, where $\kappa$ is dependent on potential strength. Typically as potential strength weakens, $\kappa$ becomes smaller. Conventionally it is said that $\kappa$ is the ...


10

For a given quantum system, the kernel of the path integral is, in fact, the kernel of an integral transform as you explicitly write down. It is the transform that governs time evolution of the system as is manifest in your first equation. For this reason, it is often referred to as the propagator of a given system. For example, for a single, ...


1

I'm now a theoretical physics PhD student, and I did mathematics as an undergrad. In many ways doing mathematics prepares you very well for theoretical physics, particularly if you want to work in string theory or phenomenology. You'll pick up lots of useful skills for developing theories and doing calculations. You should definitely make sure you at least ...


10

The mathematics of QM seems ad-hoc when one sees it in physics courses. In von Neumann's approach it becomes clear that a good mathematical candidate for the description of quantum phenomena are operator algebras, specifically C*-algebras and von Neumann algebras. Their representation theory then leads to the usual Dirac/Schrödinger/Heisenberg picture of ...


5

We find that the description of quantum systems with the concept of superposition yields predictions highly consistent with experiment. Linearity seems to be central and essential to this highly successful description, so, until someone works out another way of describing the phenomena we measure, the notion of vector space i.e. linear, Hilbert space is here ...



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