New answers tagged

1

it also appears that no particular form of $H$ was assumed. This is not the case. In page $290$ Sakurai says: If the motion obeys symmetry under time reversal, we expect the preceding ket to be the same as $$\Theta|\alpha,t_0;t=-\delta t\rangle$$ So we are beginning with the assumption of time reversal invariance. So if the condition $[H,\Theta]=0$ ...


0

Assuming physics solves its equations in symmetric way, e.g. QM with Feynman path integrals instead of Schrödinger equation, do Bell's assumptions hold - are local realistic "hidden variables" still disproven? Yes. To check this for spins, you can take a spin to a point $\hat{n}$ in the group manifold $SU(2)$. Then take wave function to be defined by the ...


0

(1) is Lorentz transformation, while (2) is similarity transformation. Lorentz transformation includes rotation and boost. Similarity transformation is performed upon a square matrix that leaves invariant its characteristic polynomial, trace, and determinant. The transformed matrix is similar to the original matrix in the sense that they represent the same ...


0

The proper statement is that the electric dipole moment of a neutral object restricted to a single irreducible representation of the rotation group must vanish, if $T$ symmetry holds. A typical neutron in the lab is in such a state, because the rotational energy levels of nucleons are widely separated, by much greater energies than are available at room ...


0

Your question could be "is CP (Charge reversal parity reversal) obeyed by water molecules?", because it is the CPT theorem that is used for getting a time reversal asymmetry check for the neutron. A permanent electric dipole moment of a fundamental particle violates both parity (P) and time reversal symmetry (T). These violations can be understood by ...


4

If you start with the action of Maxwell theory in four dimensions [I will denote by lowercase letters dynamical gauge fields and by uppercase, background gauge fields] $$S[a] = \frac{1}{2}\int f\wedge\star f = \frac{1}{2}\int \mathrm{d}a\wedge\star\mathrm{d}a. \label{1}\tag{1}$$ This has a $\mathrm{U}(1)_\mathrm{e}^{[1]}$ one-form symmetry, dubbed "electric"....


0

This isn't justified a priori. Instead, this is a useful heuristic that we use to build a solution of the Maxwell equations that we're interested in. Once that solution has been constructed, the method for finding it becomes irrelevant: it is a solution, and that is all we need to know.


2

In the chemistry dialect polar means having a non zero electric dipole moment. That is, if you put your molecule in a uniform electric field, it will orient itself aligning its dipole moment to it. Let's consider an arbitrary molecular structure with no symmetry: it is likely that it will show some dipole moment. But if we stick that to its mirror image in ...


2

In $\text {CO}_2$ the central carbon atom is $sp$ hybridised and hence has a linear shape. Now the $ \text O$ atoms on both side have higher electronegativity and hence the electron density is partially shifted towards $\text O$ atoms. This leads to two separate dipoles of equal magnitude and opposite direction in parts of the atom but the molecule as a ...


11

The difficulty might be due to nomenclature : How is polarity defined? In physics the electric field of any distribution of charges can be decomposed into a series of electric multipole moments. The most fundamental is the monopole which means that the resultant electric charge is non-zero. Chemists use the term ionic. If the distribution is overall neutral ...


15

Yes it is about the partial charges on the atoms but the dipole is a vector not just the charge distribution which is positive in the center and negative on the O's but the vectors cancel


-1

Basic covalent/chemical bonding, the two oxygen atoms are completing the valence shell of the carbon atom, this is stable due to the repulsive forces of the opposite charges being weaker.


5

Linear B-A-B molecules cannot be polar. i.e. cannot have a permanent electric dipole for symmetry reasons. Stated plainly, the electronic displacement from A atom towards the two B atoms (or the other way around if A is more electronegative than B atoms) must be symmetric with respect to the central atom, ending up with a molecule without a net electric ...


1

I’m answering the question you asked at the end: Assuming physics solves its equations in symmetric way, e.g. QM with Feynman path integrals instead of Schrödinger equation, do Bell's assumptions hold - are local realistic "hidden variables" still disproven? The short answer is “Yes.” Solving QM with path integrals is equivalent to using Schrödinger’s ...


0

The premise is only true for substances that can be described with a single refractive index: isotropic substances or cubic crystals. It is not true for uniaxial crystals etcetera like quartz or calcite. Those are birefringent. There will be two rays, with different polarizations. At normal incidence, the extraordinary ray is bent. The ordinary ray ...


1

Through the obvious canonical rescaling, $$ q_1=Q_1, \qquad q_2=Q_2/\sqrt{2}, \qquad q_3=Q_3/\sqrt{3},\\ p_1=P_1, \qquad p_2=P_2 ~ \sqrt{2}, \qquad p_3=P_3~\sqrt{3}, $$ which preserves the commutation relation, $[Q_a,P_b] = \delta_{a,b} ~i I$. It transforms the hamiltonian into the ("unmodified") isotropic 3D SHO, whose symmetry is well-known to be U(3). ...


2

The Hubbard model is described by nearest-neighbor hopping $H_t = -t\sum [c^{\dagger}_{\sigma,i}c_{\sigma,j}+\rm{h.c.} ]$, which preserves the spin-orientation, plus an on-site Coulomb replusions $H_U = U \sum n_{i,\uparrow}n_{i,\downarrow}$. Both of these terms are invariant to a global rotation of the spins. The first term will still describe each flavor ...


5

This is a rather academic discussion, but if you plug the trivial transformation into Noether's theorem, the quantity you get is... zero. Which is indeed conserved.


0

The problem with the right hand diagram is that it doesn't scale properly. Suppose this stress field was uniform over a finite sized region. Now take a rectangular element of size $\Delta x$ by $\Delta y$. You now have a direct force $\sigma_y \Delta x$ balanced by the shear forces $2 \tau_{yx} \Delta y$ and you can't eliminate $\Delta x$ and $\Delta y$ ...


1

There's lot of confusing jargon. Let me define the following four terms - Global symmetry - Continuous symmetry parameterized by a finite number of real numbers (could also be discrete). Local symmetry - Continuous symmetry parameterized by a function. Physical symmetry - A true symmetry of the theory. More precisely, such a symmetry implies existence of a ...


3

This is not true. See section 3 of this article where the full analytic expression is derived. For finite mass, we can write the $x^2\rightarrow 0^+$ fermionic propagator (the $x^2\rightarrow 0^-$ limit gives the same answer, but with Bessel instead of Hankel functions when $x^2\neq 0$): $$ \begin{align} S_F(x)&=(i\gamma^{\mu}\partial_{\mu}+m)G_F(x)\\ &...


0

Parity (aka reflection aka the mirror world) is when you switch coordinates from $x, y, z$ to $-x, -y, -z$. This looks so much like a simple coordinate transform that we would instinctively expect it to lead to the same physics, except it doesn't.


0

The mirror universe is a universe in which the arrow time is pointing opposite to the time arrow in our universe (in reality the mirror should be 4-dimensional). Both arrows start at the big bang. Things happening in that mirror universe are happening at the same time as things over here, but "on the other side" of the big bang. I'm not so sure that things ...


2

Yes, it's a math thing. However, since physics is about describing the world as it is, and since physicists use mathematics to "understand" the laws of nature, the implications matter: If we include this "thing" in our mathematics to describe nature, we alter our predictions about nature. Symmetries and conservation laws are fundamental in physics, because ...


0

As with an infinite electric slab, the field produced by a slice of the slab is independent of distance from that slice. As you move away from an infinite plane of charge (electric or mass, anything with $F\propto \frac{1}{r^2}$), you get more outward components from the far-off charges, which balances the fact that the outward force of the charges below you ...


0

As We know Gauss's Law is given by $$\Phi=\iint _S \mathbf{E}.d\mathbf{A} = \frac{\rho}{\epsilon_0}$$ It mainly talk about the net flux (Total flux pass through the surface).In case of infinite slab When you take a slab within the slab ,The net flux is only due to density inside the slab ,the field line due to outer density come in from onside and get out ...


0

For an infinite sheet of mass the field produced is uniform (going away in both directions). In order to maintain symmetry within a slab, the two Gaussian surfaces must be equidistant from the center of the slab. The fields from outside will enter at one surface and leave at the other, contributing nothing to the flux (out).


1

Change the variable of integration to $\phi’_i=\phi_i-\phi$ and the dependence on $\phi$ will go away. The new limits of integration will look different but you are still integrating over all 360 degrees of the ring so you can change them back to $0$ to $2\pi$.


0

No, the transformation should be complete. "The Lorentz transformation is a linear transformation. It may include a rotation of space; a rotation-free Lorentz transformation is called a Lorentz boost. In Minkowski space, the mathematical model of spacetime in special relativity, the Lorentz transformations preserve the spacetime interval between any two ...


0

Answer from talk On exotic six-dimensional supergravity theories:


Top 50 recent answers are included