particle physics - $Delta^+$ decay in GZK process

The dominant channels in the GZK process are

$$p+\gamma_{\rm CMB}\to\Delta^+\to p+\pi^0,$$ $$p+\gamma_{\rm CMB}\to\Delta^+\to n+\pi^+.$$

According to the pdg, $\Delta\to N+\pi$ makes up essentially 100% of the branching ratio (BR). It doesn't, however, say which process is favored: the proton and neutral pion or neutron and charged pion. My instinct is that they should each contribute about 50%, but I am not sure. So my question is, what are the BRs for each of the processes described above?

Answer

[...] $\Delta^+ \rightarrow p + \pi^0$, [...] $\Delta^+ \rightarrow n + \pi^+$,

which process is favored: the proton and neutral pion or neutron and charged pion [?]

Since the kinematics (and corresponding "phase space" factors) for the two final states are presumably as good as equal, the evaluation of the branching ratio

$$\text{BR} := \frac{\Gamma[ \Delta^+\rightarrow p+\pi^0 ]}{\Gamma[ \Delta^+\rightarrow n+\pi^+ ]}$$

simplifies to determining the ratio of "state constituent" transition probabilities

$$\text{BR} := \frac{\Gamma[ \Delta^+\rightarrow p+\pi^0 ]}{\Gamma[ \Delta^+\rightarrow n+\pi^+ ]} \simeq \frac{\left\lvert \langle p; \pi^0 \mid \Delta^+ \rangle \right\rvert^2}{\left\lvert \langle n; \pi^+ \mid \Delta^+ \rangle \right\rvert^2}.$$

Analyzing (or defining) the initial state $\Delta^+$ and the two distinct final states in terms of isospin leads to the expressions

$$ \lvert \Delta^+ \rangle \equiv \big\lvert \left(3/2, 1/2\right)_i \big\rangle, $$

where the first value represents the magnitude of $\mathbf I$, and the second value represents the magnitude of $I_3$, along with

$$ \lvert p; \pi^0 \rangle \equiv \big\lvert (1/2, 1/2)_f; (1, 0)_f \big\rangle \equiv \sqrt{ \frac{2}{3} }~\big\lvert (3/2, 1/2)_t \big\rangle - \sqrt{ \frac{1}{3} }~\big\lvert (1/2, 1/2)_t \big\rangle, $$ and

$$ \lvert n; \pi^+ \rangle \equiv \big\lvert (1/2, -1/2)_f; (1, 1)_f \big\rangle \equiv \sqrt{ \frac{1}{3} }~\big\lvert (3/2, 1/2)_t \big\rangle + \sqrt{ \frac{2}{3} }~\big\lvert (1/2, 1/2)_t \big\rangle, $$

where

• 
  

  the coefficients of the linear combinations on the right-hand sides are Clebsch-Gordan coefficients (specificly those values listed in table "$1/2 \otimes 1$"),

  
  


• 
  
  

  all states are normalized, and

  
  


• 
  

  the indices $f$ and $t$ are to distinguish final states and "state representations to evaluate transition probabilities"; such that in particular the states $(1/2, 1/2)_f$ and $(1/2, 1/2)_t$ are (meant to be) distinct; and both are distinct, and indeed disjoint, from the initial state $\lvert \Delta^+ \rangle \equiv \lvert (3/2, 1/2)_i \rangle$.

  
  

Now identifying

$$\big\lvert (3/2, 1/2)_t \big\rangle \equiv \big\lvert (3/2, 1/2)_i \big\rangle $$

we can evaluate

\begin{align} \langle p; \pi^0 \mid \Delta^+ \rangle & \equiv \bigg\langle \sqrt{ \frac{2}{3} }~ (3/2, 1/2)_t - \sqrt{ \frac{1}{3} }~ (1/2, 1/2)_t \bigg\vert (3/2, 1/2)_t \bigg\rangle \\ & = \bigg\langle \sqrt{ \frac{2}{3} }~ (3/2, 1/2)_t \bigg\vert (3/2, 1/2)_t \bigg\rangle \\ & = \sqrt{ \frac{2}{3} } \end{align}

and

\begin{align} \langle n; \pi^+ \mid \Delta^+ \rangle & \equiv \bigg\langle \sqrt{ \frac{1}{3} }~ (3/2, 1/2)_t + \sqrt{ \frac{2}{3} }~ (1/2, 1/2)_t \bigg\vert (3/2, 1/2)_t \bigg\rangle \\ & = \bigg\langle \sqrt{ \frac{1}{3} }~ (3/2, 1/2)_t \bigg\vert (3/2, 1/2)_t \bigg\rangle \\ & = \sqrt{ \frac{1}{3} } \end{align}

obtaining the sought branching ratio value as

$$\text{BR} := \frac{\Gamma[ \Delta^+\rightarrow p+\pi^0 ]}{\Gamma[ \Delta^+\rightarrow n+\pi^+ ]} \simeq \frac{ (\sqrt{ 2/3 })^2 }{ (\sqrt{ 1/3 })^2} = 2.$$

cosmology - Cosmological Inflation: If Photon expands, why not other matter?

I keep repeatedly reading in many Stack Exchange and Quora questions that space of universe expands but particles (matter) don't, see e.g. this Phys.SE post. The reason given is that particles are governed by much stronger Electromagnetic and Nuclear forces, where space expansion is relevant to only on much larger inter galactic scale where all natural forces including gravity is weaker.

How come photons in CMB (Cosmic Wave Background) which is also a kind of matter has expanded to radio wave range from gamma range? Is photon not affected by nuclear, Electromagnetic forces?

If the answer is Photons are weightless while subatomic particle like electron has mass, then does it mean space expansion affects other massless particles as well? Have the other massless particles proven to show expansion since big bang?

I am confused why space expansion affects Photons but not other forms of matter.

general relativity - GR as a gauge theory: there's a Lorentz-valued spin connection, but what about a translation-valued connection?

Given an internal symmetry group, we gauge it by promoting the exterior derivative to its covariant version:

$$ D = d+A, $$

where $A=A^a T_a$ is a Lie algebra valued one-form known as the connection (or gauge field) and $T$ the algebra generators.

For GR, we would like to do the same thing with the Poincaré group. But the Poincaré group isn't simple, but rather splits into translations $P$ and Lorentz transformations $J$. I would thus expect two species of connections:

$$ D = d + B^a P_a + A^{ab} J_{ab}. $$

But the covariant derivative of GR as usually found in textbooks is:

$$ \nabla_\mu = \partial_\mu +\frac{1}{2}(\omega^{\alpha \beta})_\mu J_{\alpha \beta}, $$

where $\omega$ is the spin connection. It is defined for any object that has a defined transformation under $J$, i.e., under Lorentz transformations, like spinors or tensors. But it makes no mention of the translation generator $P$. What happened? Shouldn't I have this extra gauge field?

fluid dynamics - Why there's a whirl when you drain the bathtub?

At first I thought it's because of Coriolis, but then someone told me that at the bathtub scale that's not the predominant force in this phenomenon.

Answer

The whirl is due to the net angular momentum the water has before it starts draining, which is pretty much random.

If the circulation were due to Coriolis forces, the water would always drain in the same direction, but I did the experiment with my sink just now and observed the water to spin different directions on different trials.

The Coriolis force is proportional to the velocity of the water and the angular velocity of Earth. Earth's angular velocity is $2\pi/24\ {\rm hours}$, or about $10^{-4}\ s^{-1}$. If water's velocity as it drains is $v$ the Coriolis acceleration is about $10^{-4} v\ s^{-1}$.

The water moves about a meter while draining, which takes a time $1\ m/v$, so the total velocity imparted by Coriolis forces could be at most $10^{-4} v\ s^{-1} * 1\ m/v = 10^{-4} \ m/s$.

So the Coriolis effect is quite a small effect. But this first-order Coriolis effect does not cause the water to rotate.

The direction of Coriolis force depends on your direction of motion. All the water in your tub is moving the same direction, so the Coriolis force pushes it all the same direction. The effect is that if the bathtub starts out perfectly flat and begins draining (and it points north), all the water will get pushed east. The two edges of the tub will have very slightly different depths of water, because the Coriolis force is pushing sideways.

The Coriolis force could create "spinning" on uniformly-moving water, but only as a second-order effect. As you move away from the equator, the Coriolis force changes. This change in the Coriolis force is because the angle between "north" and the angular velocity vector of Earth changes as you move around; as you go further north (in the Northern Hemisphere) the "north" direction gets closer and closer to making a right angle with the angular velocity vector, so the Coriolis force increases in strength. The size of this effect would be proportional to the ratio of the size of your tub to the radius of Earth. That ratio is $10^{-7}$, so this effect is completely negligible.

The Coriolis force could also create some "spinning" if different parts of the water are moving different speeds. If the tub is draining to the north in the northern hemisphere, and water near the drain is moving faster than water far away, then the water near the drain would be pushed east more than water far away is. If you subtracted out the average effect of the Coriolis force, what remained would be an easterly push near the drain and a westerly push far away. This gives a clockwise spin as viewed from above.

We've already estimated the typical velocities as $\omega L$, so the angular momentum per unit mass induced this way would be on the order of $\omega L^2$ (but maybe smaller by a factor of 10). That's only $10^{-4}\ m^2/s$. To get an equivalent effect, in a tub of $100\ L$, you could give just one liter of water on the edge of the pool a velocity of a few cm/s, something you surely do many time over when removing your body from the tub.

This effect is too small to affect your bathtub, but it's still observable under the right conditions. According to Wikipedia, Otto Tumlirz conducted several experiments in the early 20th century that demonstrated the effects of the Coriolis forces on a draining tub of water. The tub was allowed to settle for 24 hours in a controlled environment before the experiment began. This was enough to damp out the residual angular momentum left over from filling the tub up to the point where Coriolis effects were dominant.

general relativity - Is it possible to escape from within event horizon?

I always think that it is not possible to escape from within event horizon. However, some one recently told me with deep conviction that it is possible with sustained energy output. I countered with calculations in Schwarzschild metric showing that any objects below event horizon in the simplistic black hole will hit the singularity within a finite proper time, and was met with the response that Schwarzschild metric was a bad choice for coordinate system.

Now I understand that the singularity at $r=\frac{2GM}{c^2}$ in Schwarzschild metric is an artifact of the coordinate choice, and thus it seems that he could have a point. So my question is, can one prove that it is (or not) possible to escape from within the event horizon of a simple static black hole using a better metric?

Answer

See Why is a black hole black?

It's quite true that the Schwarzschild coordinates misbehave near the horizon, but there are plenty of other coordinate systems we can use. My answer to the question I've linked above uses Gullstrand-Painlevé coordinates, but you can also use Eddington-Finkelstein or Kruskal-Szekeres coordinates. The conclusion is the same in all coordinate systems - once you have crossed the event horizon there is no way back.

Classical and quantum correlations in bipartite system

I would like to know how to answer following questions: Is there classical/quantum correlations in given bipartite pure/mixed state?

I have gathered several definitions. Some of them (it seems) lead to counterintuitive conclusions. I have marked them from 1 to 4. I would be glad if someone could point out mistakes (if there are any) in my logic.

$\underline{\text{Classical correlations}}$ are correlations which can be created by LOCC (local operations and classical communication).

1. Do local operators acting on bipartite Hilbert space have following form: $U_{A} \otimes I_{B} \text{ and } I_{A} \otimes U_{B} $? If so then locality of these operators have nothing to do with spacial locality. Is that correct?

$\underline{\text{Quantum correlations}}$ are correlations which can not be created by aforementioned way.

2. Would that mean that quantum correlations will be created by nonlocal(?) operators of form: $U_{A} \otimes U_{B}$? Is there any other way to create quantum correlations? Can classical correlations be created by action of this operator on a given state as a byproduct(at list in case of mixed state)?

I have heard that entanglement is not the only form of quantum correlations. Yet since at the moment I am not interested in exotic states but rather in general concepts I will ignore this knowledge. If this information can't be neglected please let me know.

Here are some statements which seem to be correct yet counterintuitive:

$\underline{\text{Pure states:}}$

3. 
   

   No pure (bipartite) state has classical correlations between its subsystems. Proof: If we start from direct product state $|A\rangle\otimes |B\rangle$ no local operator will be able to create any correlations between subsystems. Since classical communication will lead to creation of mixed state we are out of options.

   
   
   
   
   • Schmidt decomposition tells us whether given pure state is entangled(has quantum correlations) or separable.
   
   
   
   
   

$\underline{\text{Mixed states:}}$

4. 
   

   Every mixed bipartite state has classical correlations between its subsystems. Proof: Any mixed state can be written as a mixture of pure states. As I understand it this means that any density matrix can be diagonalized. Diagonalized matrix will have eigenvalues on its diagonal. Due to properties of density matrix, eigenvalues $\lambda_{i} \in [0,1]$ and their sum always equals 1. Hence these eigenvalues may be interpreted as probabilities corresponding to some pure states in a mixture. Presence of at least two nonzero eigenvalues would indicate presence of classical correlations between subsystems.

   
   
   
   
   • Separability tells us whether given mixed state has entanglement(i.e. quantum correlations) or not. For bipartite system one can use so called Peres-Horodecki criterion.
   
   
   
   

Answer

Let's go through your list.

$\underline{\text{Classical correlations}}$ are correlations which can be created by LOCC (local operations and classical communication).

Yes, that is correct.

1. Do local operators acting on bipartite Hilbert space have following form: $U_{A} \otimes I_{B} \text{ and } I_{A} \otimes U_{B} $?

Yes, that is correct, though this does not take into account classical communication, so the LOCC set is bigger than that. A full expression is rather clunky (since it needs to account for an arbitrary number of round trips for the classical communication, with a local unitary and projective measurement at each end), but there's good details on Wikipedia if you want them.

If so then locality of these operators have nothing to do with spacial locality. Is that correct?

It doesn't need to have anything to do with spatial locality. You normally enforce this by stipulating that the A and B subsystems are at remote locations and that their measurements occur sufficiently fast that they're within spacelike-separated regions of spacetime. If you don't enforce that, then 'locality' just becomes a marker of the different tensor factors of the state space.

$\underline{\text{Quantum correlations}}$ are correlations which can not be created by aforementioned way.

Generally speaking, yes.

2. Would that mean that quantum correlations will be created by nonlocal(?) operators of form: $U_{A} \otimes U_{B}$?

Yes.

Is there any other way to create quantum correlations?

It depends on what you count as "ways". There's a bunch of quantum channels that create quantum correlations which are not of the form $U_{A} \otimes U_{B}$; as an example take a unitary channel of the form $U_{A} \otimes U_{B}$ and follow it up with some limited decoherence. Does that count?

Generally speaking, if you have a given quantum channel and you know that it creates quantum correlations, you'll typically be able to decompose it in such a fashion, but that decomposition does not necessarily speak to what is "really going on" inside the system (and you will typically not have any access to that).

I would therefore say that the answer to the question is morally yes, but that it's too ill-defined to say anything concrete.

Can classical correlations be created by action of this operator on a given state as a byproduct (at list in case of mixed state)?

Frankly, this is completely unclear to me.

I have heard that entanglement is not the only form of quantum correlations.

This is correct. You probably want to look at quantum discord as the core example of such correlations.

Yet since at the moment I am not interested in exotic states but rather in general concepts I will ignore this knowledge. If this information can't be neglected please let me know.

That depends on what you mean by "can't be neglected". There are many important conceptual settings in which the information can't be neglected. There are many important conceptual settings in which it can. Some examples of the former are places where quantum contextuality is an important consideration, with the Kochen-Specker theorem taking a role similar to that of Bell's theorem in the study of entanglement. Some examples of the latter are the study of entanglement. Which side you want to listen to is a personal choice.

$\underline{\text{Pure states:}}$

3. No pure (bipartite) state has classical correlations between its subsystems. Proof: If we start from direct product state $|A\rangle\otimes |B\rangle$ no local operator will be able to create any correlations between subsystems. Since classical communication will lead to creation of mixed state we are out of options.

Yes, that is correct. You may be interested in purity as a quantum resource theory.

$\quad \cdot$ Schmidt decomposition tells us whether given pure state is entangled (has quantum correlations) or separable.

That is correct.

$\underline{\text{Mixed states:}}$

4. Every mixed bipartite state has classical correlations between its subsystems.

No, this is incorrect. To get a counter-example, simply take any two mixed density matrices $\rho_A$ and $\rho_B$, and set $\rho = \rho_A\otimes \rho_B$ as a mixed separable state with no classical correlations between its subsystems.

Proof: Any mixed state can be written as a mixture of pure states. As I understand it this means that any density matrix can be diagonalized. Diagonalized matrix will have eigenvalues on its diagonal. Due to properties of density matrix, eigenvalues $\lambda_{i} \in [0,1]$ and their sum always equals 1. Hence these eigenvalues may be interpreted as probabilities corresponding to some pure states in a mixture.

So far so true, but

Presence of at least two nonzero eigenvalues would indicate presence of classical correlations between subsystems.

This doesn't follow from anything. Study the counter-example above to see why this fails.

$\quad \cdot$ Separability tells us whether given mixed state has entanglement (i.e. quantum correlations) or not.

Entangled states are, by definition, those states that are not separable. So yes. But it tells you very little.

For bipartite system one can use so called Peres-Horodecki criterion.

That is one possible criterion, but it is not infallible at detecting entanglement. In other words, you get the implication $$ \rho\text{ is separable} \implies \text{its partial transpose is positive semidefinite} $$ and its contraposition $$ \text{the partial transpose of }\rho\text{ is indefinite} \implies \rho\text{ is entangled}, $$ but you don't get the converse, which would be more useful, i.e. it's possible that $\rho$ is entangled but you've just chosen a basis in which the partial transpose is still positive semidefinite.

classical mechanics - Types of circular acceleration?

To my knowledge there are three types of acceleration when a body (e.g. a rod) is moving in a circle about an axis. These are:

1. 
   

   Angular acceleration : this is the rate of change of angular velocity.

   
   
   


2. 
   

   Tangential acceleration : this is the linear acceleration of the system in a tangential direction to the circle and equals the radius times the angular acceleration.

   
   


3. 
   

   Radial/centripetal acceleration : this is the linear acceleration of the system that is directed inwards towards the center of the circle.

   
   

I also think there are two types of velocity:

1. 
   

   Linear velocity: this is its velocity in the tangential direction and is constantly changing

   
   


2. 
   

   Angular velocity or angular frequency: this is the rate of change of angle.

   
   

Is the above correct? And have I missed anything?

Answer

No. Frequency is defined as 2π*θ/t where theta is the angle rotated for a time t. You maybe tempted to equate frequency to angular velocity. But it is not so. Angular velocity = dθ/dt. Angular frequency= 2*pi*(Integral of x over time interval t)/t