Wednesday, 2 October 2019

particle physics - What is $p_T$? (transverse momentum?)


I've been looking at a few papers in experimental physics (from the ATLAS collaboration, for example) and I've often run across phrases such as "high-$p_T$ electron." What exactly is $p_T$? Is it simply momentum, but with the component parallel to the main beam projected off?


Also, why is $p_T$, as opposed to $p$, an important characteristic of a particle in a collision process?




Answer



The component of momentum transverse (i.e. perpendicular) to the beam line.


It's importance arises because momentum along the beamline may just be left over from the beam particles, while the transverse momentum is always associated with whatever physics happened at the vertex.


That is, when two protons collide, they each come with three valence quarks and a indeterminate number of sea quarks and gluons. All of those that don't interact keep speeding down the pipe (modulo Fermi motion and final state interaction).


But the partons that react do so on average{*} at rest in the lab frame, and so will on average spray the resulting junk evenly in every direction. By looking at the transverse momentum you get a fairly clean sample of "stuff resulting from interacting partons" and not "stuff resulting from non-interacting partons".


There are also advantages related to the engineering of the detector.




{*} Only on average. Indivudual events may involve high Bjorken $x$ particles and be a long way from at rest in the lab frame.


general relativity - Relationship between Coordinate Time and Proper Time


While I was reading Ta-Pei Cheng's book on relativity, I was unable to derive the correct relationship between coordinate time $dt$ (the book defined it as the time measured by a clock located at $r=\infty$ from the source of gravity) and proper time $d\tau$ from the definition of metric.


The book states that for a weak and static gravitational field, $g_{00}(r)=-\left(1+\frac{2\Phi(r)}{c^2}\right)$ (with the metric signature $(-1,1,1,1)$ and $\Phi(r)$ is the gravitational potential) and the proper time $d\tau=\sqrt{-g_{00}}\,dt$.


From the gravitational redshift result I know that the above result is correct (in a more unambiguous form $d\tau=\sqrt{-g_{00}(r_\tau)}\,dt$).


However, if I simply use the formula for spacetime interval $ds^2=g_{\mu\nu}dx^\mu dx^\nu$ (assuming two clocks that measure proper time and coordinate time are at rest relative to each other), I have


$$ ds^2=g_{00}(r_\tau)c^2d\tau^2=g_{00}(r_t)c^2dt^2=-c^2dt^2\\ \implies \sqrt{-g_{00}(r_\tau)}\,d\tau=dt$$ This suggests that time flows faster with a lower gravitational potential which is incorrect.


I'm not sure why the above method lead to a wrong conclusion, did I misunderstood the the definition of proper time, coordinate time or spacetime interval?





Update:



  1. One mistake I've made is letting $ds^2=g_{00}(r_\tau)c^2d\tau^2$, which should be $ds^2=-c^2d\tau^2$ by definition. However, I'm confused about two definitions of $ds^2$ now. $ds^2=-c^2d\tau^2=g_{\mu\nu}dx^\mu dx^\nu$, this suggests that $g_{00}$ is always $-1$ for the frame that measures proper time, but in my problem $g_{00}$ is a function of $r$ which is only equals to $-1$ if $r=\infty$, how could two both be true at the same time?

  2. Assuming $ds^2=-c^2d\tau^2$ is true, as all the answers pointed out that $d\tau=\sqrt{-g_{00}}\,dt$. But by the definition of $g_{00}$ and $ds^2$ the $g_{00}$ used here must be $-(1+2\Phi(r_t)/c^2)=-1$, but I want $g_{00}$ here to be $-(1+2\Phi(r_\tau)/c^2)$ so that $$d\tau=\sqrt{-g_{00}}\,dt=\sqrt{1+2\Phi(r_\tau)/c^2}\,dt\approx (1+\Phi(r_\tau)/c^2)\,dt\\ \implies \frac{d\tau-dt}{dt}=\frac{\Phi(r_\tau)}{c^2}=\frac{\Phi(r_\tau)-\Phi(r_t)}{c^2}$$


Please correct me if I've made any mistakes!




history - How did ancient physicists around 500 BC decide that the earth is spherical?


One of the standard stories that is given for this (and is mentioned in Aristotle) is that the hull of a ship disappears first as it sails towards the horizon on a calm sunny day. Is this a myth, or is it possible? After all, at the time no telescopes were available.


Scenario: Assume, for instance the hull of the ship rides 1 metre above the surface of the sea, with a sail whose height is 3 metres (I am going by 'Min of the Desert', a replica of a 3800 years old oceangoing Egyptian ship, so about the right time period.)




  1. How far must it travel before the curvature of the earth completely hides the hull?

  2. How far must it travel (assuming a flat earth) before visual acuity of a normal naked eye can no longer discern it?


I get an answer for question 1 of around 120km, this seems too distant (by my unaided intuition) for the naked eye to resolve. Also it seems wrong from general observation of the horizon from the coast, as one of the commenters has pointed out it looks like only a few kilometers. And surely at that kind of distance atmospheric efects begin to kick in, especially in the mediterrean where its generally warm.


Its been a very long time since I've done geometrical optics, so the answer to question 2 is beyond me. Can someone help. Also a verification of 1 would be nice (I've used a radius of the earth as 6000km).




fluid dynamics - Calculational method for determining surface tensions from photograph of menisci?


How can I get from a photograph of a liquid surface to a value for the surface tension.




nuclear physics - How can a proton be converted to a neutron via positron emission and yet gain mass?


The mass of a neutron is greater than mass of a proton so how is it possible in positron emission for a proton to form a neutron and a positron?




Tuesday, 1 October 2019

Does quantum reversibility require many worlds?


The source S sends a photon into the beam splitter below. There is a 50% chance that it will be detected at A and a 50% chance it will be detected at B.


S -----------\---------> A
|
|
|

v
B

Now if we assume that physics is (generally) reversible we should be able to time-reverse this process. This would imply that if we send in a photon at A or B it should always appear back at S.


How can this happen? We know that if we send in a photon at A (or B) we would expect it to appear 50% of the time at S and 50% at C as below.


             C                                         C
^ ^
| |
| |
| |

S <----------\--------- A S <------------\
|
|
|
B

In order that the photon always appears back at S and never at C we need a superposition of the two situations above such that the paths to S constructively interfere and the paths to C destructively interfere.


Thus it seems that in order to retain time-reversibility we need to assume a many-worlds view. As a photon is sent in at A (or B) it must be assumed that another photon is simultaneously sent in at B (or A) and both states weighted with the correct amplitude such that a photon appears with certainty back at S.


What do people think?




quantum mechanics - Why are SU(2)-generators interpreted as *spatial* components of spin?


The generators of the unitary representation of $SU(2)$ on the internal spin Hilbert space of (say) a spin-$1/2$-particle are typically said to represent components of spin along various spatial axes. For example, if $|+\rangle$ is the eigenvector of $S_z$ with eigenvalue $1/2$, then (according to the standard story) if the system is in the state represented by $|+\rangle$, the component of the system's spin along the $z$-axis is $1/2$.



In general, textbook wisdom has it that the component of spin along the spatial axis $\vec{n}$ is given by $\vec{n}\cdot \vec{S} = n_xS_x + n_yS_y + n_zS_z$, where $\vec{n}$ is a unit vector.


I find this puzzling, for at least two reasons. (1) The generators of $SU(2)$ are self-adjoint operators on an internal spin Hilbert space that has no spatial degrees of freedom at all. (In the case of spin-$1/2$-particles, this is the two-dimensional Hilbert space of Pauli spinors.) It is hard to see what connection these generators have with axes in physical space. (2) One may think that the coincidence between the number of spatial dimensions and the number of basis elements of the $SU(2)$ Lie algebra is suggestive of interpreting the latter as representing components of spin along spatial axes. However, it is my understanding that the fact that there are three $SU(2)$-generators is a result of the structure of $SU(2)$, and not of the fact that the relevant physical space is three-dimensional. Indeed, the "components" of spin of a spin-$1/2$-particle in \textit{four} spatial dimensions would also be represented by triples of linearly independent $SU(2)$-generators.


My question is therefore: what is the basis for the identification of the generators of the unitary $SU(2)$-representation on the internal spin Hilbert space with the components of spin along spatial axes?


(Note that I am aware of the Bloch sphere illustration of the Hilbert space of a spin-$1/2$ system. I have also read this thread: Do spins have spatial directions?. Neither contains an answer to my question.)



Answer



An $SU(2)$ symmetry, a priori, has absolutely nothing to do with spatial rotations. For instance, the symmetry between protons and neutrons is described by isospin, an $SU(2)$ symmetry that treats the proton and neutron just like the spin up and spin down states of a spin $1/2$ particle. Similarly there is weak isospin, also described by $SU(2)$, which relates the electron with the electron neutrino. Neither of these are related to rotations, except for the math being the same; you cannot do a physical rotation to turn a proton into a neutron. (However, the analogy with spin is so useful for conceptualizing what's going on that they all have the word 'spin' in their names.)


If a textbook says "we've identified an $SU(2)$ symmetry, so it must physically correspond to rotational symmetry", then that book is being sloppy. The argument should be phrased in reverse. Below I'll show how that goes very explicitly.




We start with experimental data that we want to understand. For example, suppose we want to model the precession of the magnetic dipole moment of a nucleus in a time-varying magnetic field. (It turns out this dipole moment is proportional to the spin, so this is exactly what you're asking about.) To make a quantum mechanical model we must define a Hilbert space and a Hamiltonian, as well as operators $\mu_i$ corresponding to the components of the magnetic dipole moment.


Of course, there is no unique mathematical prescription for this. For example, you may choose the Hilbert space to be zero-dimensional, but then it clearly wouldn't fit the data. It turns out that for some nuclei the model works if we choose the Hilbert space to be two-dimensional, with the states $|\uparrow\rangle$ and $|\downarrow\rangle$ corresponding to the dipole moment pointing vertically up and down. In other words, this defines $\mu_z$ as $$\mu_z |\uparrow \rangle = \mu_0 |\uparrow\rangle, \quad \mu_z |\downarrow\rangle = - \mu_0 |\downarrow \rangle.$$ For convenience we'll ignore spatial degrees of freedom. You can think of the nucleus as nailed down at the origin, if you want.



Next, we define rotation operators that physically rotate the system around. We know that classically, the rotations of space form a group $SO(3)$. Because of issues with quantum phases, this means the rotation operators must necessarily be a representation of the group $SU(2)$. This is a bit weird, but something you get used to after a while -- there is generally a turn-the-crank mathematical procedure for figuring out which group to use in the quantum case.


However, we still don't know which representation of $SU(2)$ it is. For example, the rotational operators could conceivably all do nothing -- that is exactly the right choice for isospin symmetry (ignoring the spins of the proton and neutron) because you can't rotate a proton into a neutron. But it's not the right choice for the magnetic moments, because we can observe that tilting the magnet and running the experiment again gives a different result.


We also know that a rotation about the $z$ axis should fix $|\uparrow \rangle$ and $|\downarrow \rangle$, again by observation, while $180^\circ$ rotations about the $x$ and $y$ axes should interchange these states. (Everything here is just up to phases.) We can't proceed further and get explicit expressions because they will vary depending on the phases of the states. But in the standard phase conventions, one can continue with this logic to show that the $\mu_i$ operators must all be proportional to $\sigma_i$, and the rotation operators are exponentials of $\sigma_i$. This is described in detail here.




This is a pretty long and very explicit argument to make one simple point: in physics, we do not blindly do mathematics and put the physical interpretation in at the end. We start with a physical system we want to describe and define mathematical objects accordingly. We know that rotations must be described by $SU(2)$ by general principles, so we define a Hilbert space with a representation of $SU(2)$. Which one? Whatever works.


Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...