Well, why does this answer get so many downvotes? Because it mentions a WORD file? But the first paragraph does answer my question straightforwardly: physics.stackexchange.com/a/591468/18778

@JohnRennie Can anyone help me with this physics.stackexchange.com/questions/590968/…

?

Hey guys, is it possible for highly energetic (e.g. 22 MeV) gamma rays to travel for thousands of miles in air? Mostly I found on the Internet gamma rays will be absorbed by air after only several meters for example 500 meters.

I am concerned mainly with the second part of the question . Please help...!!!

@Ankit hi :-)

Hi All,

There are several reasons why we don't see the properties of solids oscillating. The most obvious one is that the oscillation frequency is of order E/h, where E is the interaction energy and h is Planck's constant. In solids the interaction energies are of order one eV so the frequencies are in the terahertz range.

That means any oscillation in the properties happens on timescales far too short to be observed.

Pls explain if you have time. In linear track experiment (as in K&K book) if we apply a force on rider by standard rubber band and release it. Why it is accelerating because once we released rider there is no force which in contact.

@JohnRennie Pls see my question.

@123 the rider only accelerates when the rubber band is exerting a force on it.

In K&K and saw a video on youtube for inear track. Once they released they released the rider it is moving on linear track but continuously increaseing it velocity.

What is the K&K book?

My question is that once we released the rider there is no force on it. It just moved with constant velocity not acceleration.

K&K Kleppner and Kolenkow book on mechanics

And you'll need to link the YouTube video

Pls read newton's second law page-78 1st edition of K&K.

I share you the link

@123 I've watched the video, but I don't see why you say it shows the slider accelerating when no force is applied. It only accelerates while it is in contact with the rubber bands.

@JohnRennie My question is that when force is applied it should not accelerate because the force is not in contact after release.

It just give initial velocity as in projectile motion if we kick the ball it just give initial velocity and i am applying continuous force on the ball.

I don't understand what you are asking. There is a force on the slider only for the time that the slider is in contact with rubber band, and during that time the slider accelerates due to the force the rubber band is exerting on it.

what about after release the slider.

After the slider leaves the rubber band there is no force on the slider and it moves at constant velocity.

Aaah.. Okay Let me check video one more time.

When you were at Cambridge sir @JohnRennie were chemistry majors required to take one full year of an organic chemistry course?

@JohnRennie my main doubt was with the second part of the question. Can you help me with that ?

@skullpatrol everyone would do some organic chemistry as part of their first and second year courses. For the final year you specialise and you don't have to do an organic chem course.

@Ankit do you mean:

> One thing which I think for the above question is that we need a repulsive force to separate two closer molecules and this is the reason why we don't notice oscillations . Am I right ?

@JohnRennie I can not upload the image. I will share you the text of K&K book

@123 I have the book.

Newton's Second Law (Page-78 1st edition)

@123 this page?

We now turn to how the rider on the air track behaves when it is no longer isolated. Suppose that we pull the rider with a rubber band. Nothing happens while the rubber band is loose,

but as soon as we pull hard enough to stretch the rubber band, the rider start to move. If we moves our hand ahead of the rider so that the rubber band is always stretched to the same standard length, we find the rider moves in wonderfully simple way; its velocity increases uniformly with time. The rider moves with constant acceleration.

@JohnRennie above where second law started first paragraph. (I shared you the text above)

That section?

Yes @JohnRennie this one

That describes the situation when we pull the rider along with the rubber band i.e. one end is tied to the slider and you pull the other end.

The reason for using a rubber band in the example is that by pulling the rubber band so it stays at a constant length we can apply a constant force to the slider.

So in that case we applying a constant force to the slider, and therefore it has a constant acceleration.

Ooooops.. Superb @JohnRennie

:-)

Why they don't say this in book. This is very very important point.

Let me think again the scene and complete situation. Thanks @JohnRennie

Note the text I have underlined.

I see. I skipped this information or did not pay attention to it.

Now I pay attention to the diagram. Is it possible for us keep constant stretch of rubber band on air track?

You just need to move your hand ahead of the slider so that you keep the rubber band stretched to a fixed length.

Thanks a lot @JohnRennie

For the help.

:-)

If you have time one more think I don't clear Atwood machine tension force.

I'm a little busy at the moment but I will be around later.

Oookay... :-)

@Ankit two reasons:

Firstly the forces oscillate on a timescale of picoseconds so we only see their average values not their instantaneous values.

Secondly the solid contains of order Avagadro's number of atoms, and the forces between each pair of atoms oscillates independently of all the other pairs. So at a moment when the force between one pair of atoms is weak the force between another pair of atoms is strong. Again what we see is the average over all Na atoms.

@Azmuth if you're being serious, those statistics have been debunked. Please don't spread fake news.

Deleted

Hey guys, is it possible for highly energetic (e.g. 22 MeV) gamma rays to travel for thousands of miles in air? Mostly I found on the Internet gamma rays will be absorbed by air after only several meters for example 500 meters.

@MohamedObeidallah This paper has equations for gamma ray absorption in air.

It's behind a paywall. I don't know if your institution has access, but if not some determined Googling will probably find the content.

@MohamedObeidallah there is a table here

At 20 MeV μ/ρ = 0.017 cm²/g and the density of air is about 1.25 mg/cm³ so that gives μ = 2 x 10⁻⁵ cm⁻¹ or 2 x 10⁻³ m⁻¹

So if you measure distance in kilometres the intensity at a distance of x kilometers would be exp(-2x). Even at only 1km that's about 0.14. At 10km it's about 0.000000002.

You aren't going to sending gamma rays even ten km in air, let along a thousand km.

Wow, Georgi's Lie Algebras in Particle Physics is really comprehensive

@ACuriousMind Check (37) here: theorie.physik.uni-muenchen.de/lsfrey/teaching/archiv/sose_09/… If I understand it well, in a rotated Higgs field, the fermion masses would be different. There is a matrix for all the fermions, describing how their yukawa lagrangians depend on the higgs components. Which also answers my this question here: physics.stackexchange.com/q/264577/32426

But other say, rotating the higgs components would result the same theory. Maybe... it would not be so same.

@JohnRennie phew, that means we can still be safe from very energetic gamma rays on Earth just by going further from them. Anyway thank you so much John! Much appreciated. Thanks for the link too.

@MohamedObeidallah actually no :-)

@JohnRennie can you explain why? So that means shielding is very necessary

Gamma rays can create muons and muons are much less strongly absorbed by air. So the gamma ray could create a muon and the muon could then hit you.

Although muons are not very dangerous. You get hit by several muons a second generated by cosmic rays, and they don't do you any harm.

Actually, ignore me. A 22MeV gamma ray isn't energetic enough to create any muons.

@JohnRennie so in order to be truly safe, we still need a shielding then. Correct?

@JohnRennie can metals for example steel stop them?

Muons have a mass of about 106 MeV, and a gamma ray would have to create a muon/antimuon pair. So it would need an energy of at least 212 MeV.

@MohamedObeidallah I have seen a cloud chamber on the ground floor of a tall building detect muons that passed right through the building to reach the ground floor. I don't know what the absorption of muons in steel is like, but they go straight through metres of concrete.

But as I said, muons aren't dangerous. Life on Earth has been exposed to muons from cosmic rays for billions of years.

@JohnRennie gotcha. Thanks for the detailed responses John.

@JohnRennie Wow, if there is a beta decay which releases at least 106MeV, maybe it could create also muons

Hmm, I've never heard of a radioactive decay that produces muons.

I suspect that the probability of producing an electron is so much greater that in practice muons are never observed.

A neutron decay cannot create a muon since there isn't enough energy. Pion decays create muons but that's a very different process.

@JohnRennie Yes, but if the mass of the product nucleus is smaller by at least 106MeV, then it might by energetically possible.

@JohnRennie Maybe beta decays where unstable nucleus decays to a very stable one, could do that. I would first check some bismuth->lead beta decay

@JohnRennie Say that to that man en.wikipedia.org/wiki/Anatoli_Bugorski

who keeps starring John's comments about muons? I mean, I love muons to an unhealthy degree, but even I find this weird

I am not sure if this question makes a lot of sense, but I'll have a go. Characters of irreducible representations are in general complex numbers. Are there any cases or requirements that give purely real characters?

I was considering some representations of some space group, and noticed all characters were real, and then for another space group they were complex. Now I wonder if this is just a coincidence, an arbitrary choice or something physical

Remember remember the today of November...

@B.Brekke The characters of a group are real-valued (I assume that you meant this, but note that mathematicians often call a character "real" if it corresponds to a real representation, which is not the same thing) if and only if the inverses for all group elements are conjugates of them (i.e. $g^{-1} = k_ggk_g^{-1}$ for all $g$ and some $k_g$)

(because the trace is cyclic and so $\mathrm{Tr}(\rho(g^{-1})) = \mathrm{Tr}(\rho(g))$ for elements where $g^{-1}$ is conjugate to $g$.)

Hmm. I achieved 1/1000th of John Rennie's rep

seems like a cause for celebration

@ACuriousMind Thanks! That is very helpful

Heyo

sorry

wrong room

Hi all! I am shamefully confused about the Bose-symmetry of states of spinning particles. The state of two spin-j particles (of the same kind) with definite momenta $p_1$ and $p_2$ is $|p_1,m_1> \otimes |p_2,m_2>$, where m_{1,2} are the spin-z projection quantum numbers. Does Bose symmetry simply act irrespectively of the values of $m_1$ and $m_2$, so $|p_1,m_1> \otimes |p_2,m_2> =|p_2,m_2> \otimes |p_1,m_1>$? What does mathematically change if these two particles are not of the same kind?

For example, if these two particles were not of the same kind, then the state with inverse order, would still be equal to the previous one?

If I understand correctly, Bose symmetry acts on states transforming under the same representation of the Lorentz group, so I cannot strictly conclude that $|p_1,m_1> \otimes |p_2,m_2> =|p_2,m_2> \otimes |p_1,m_1>$ if these two states belong to two different representations. But maybe we can conclude it by a different argument?

@JohnRennie Pion decay produces muons, because angular momentum and parity non-conservation forbid $\pi\to e\nu$. That's where cosmic ray muons come from, mostly.

There are no nuclear decays which produce muons. All nuclear excited states above about 10 MeV can decay by evaporating a proton or neutron. You can find a table of "nucleon separation energies."

How can a group of authors post basically the same thing in two different journals?

is that not allowed or something?

@Secret are you sure it's the "same thing" or are you just basing that on the abstracts?

many of the references on the latter are more recent than the publishing date of the first, so I'd suspect the second one is at least an updated version of the first

yo

Good Night everyone :)

Sweet Dreams

@Azmuth I'm fairly sure it's only night time for us two :)

Hi, is anyone here well versed in consensed matter? I've been trying to formulate a question for a while but it's coming out as a rambling flow of confusion rather than a question, so I thought I might ask here.

Literally every paper on topological phases starts by saying that once upon a time we thought all phase transitions were described by symmetry breaking signalled by a local order parameter suddenly becoming nonzero, or something like that

then they go on saying that this definition does not capture topological phases, and introduce the definition of continuous gapless paths between Hamiltonians to define phases. If you require Hamiltonians in the same phase to have the same symmetry (and the gapless path as well), you get symmetry protected topological order

now, some SPTO phases apparently have an order parameter, a nonlocal one, see Haldane phase

so what's the deal? Would it have been enough to just remove the world local from Landau's definition to recover SPTO? What's the relation between nonlocal order parameters and topological phases?

@user2723984 it does not sound like rambling confusion to me. seems worthy of posting but with the caveat one can never anticipate reception on SE. my other idea is that theres actually now a lot of study of phase transitions in CS (a few decades now) and it might lead to other (helpful?) povs.

@ACuriousMind Well I have read them in detail. Basically, the newer version is like 2/3 of the original including the exact same equations, and the only new idea is an extra perturbation calculation. I just found it unusual because when there are journal article updates it is usually done as a errata

or they don't look that similar

Maybe I have not read enough papers

My past experience of reading updated versions of journal articles, often the conclusion showed the follow up of the idea in the old version, or there will be some errata, or that the old version is replaced by the new version

These two papers however basically restates the same points, that they concluded there is no universal functional for some electron gases cases

Other things to comment about is that there was a commentary article in 2013 and 2017 that point out they made a mistake in their calculations and hence their conclusion do not hold

perhaps that may explain why 2018 is almost like a restatement, with an extra calculation to show their original conclusion still holds

Still... they are very similar, which is uncommon for me

@vzn thanks, I posted it as a question, let's see what happens. I have the impression that physics SE has a high hep-th representation with respect to the actual distribution of physicists. Is it just my impression?

nah, it's true

I wonder why that is

I think it's a founder effect - for some reason the initial population skewed hep-th, and so we keep attracting a higher-than-average amount of hep-th physicists

well I'm going to skew it to condensed matter just with the 3 papers I'm trying to read

Is the physical notion of "Chern number" the same thing as $2\pi \chi$, where $\chi$ is the Euler characteristic? And this Berry curvature the Pfaffian of the curvature 2-form?

I'm trying to talk to a student who is physically inclined and the notation is making translation difficult, but I have a pretty fair guess as to what everything means. Just wanted confirmation.

@MikeMiller I think the physicists "Chern number" is usually just the first Chern number, i.e. $\int \Omega \mathrm{d}k$, where $\Omega$ is just the ordinary curvature $\Omega = \mathrm{d}A$ of the gauge field/Berry connection. You probably won't find many people talking about the Euler characteristic or Pfaffians because the setting is usually 2D and the first number is all that exists.

Got it. One of the pictures confused me.

What is the line bundle, in physics language?

physicists don't talk about bundles :P

Well then what are you taking the chern number of

A gauge field, with transformation data between coordinate patches specified?

Yeah, I guess the bundle here is defined by its clutching construction

The setting of a Berry connection is that you have a physical state $\phi(x)$ that's "slowly" evolving as $x$ changes and the Berry gauge field is $A(x) = \phi^\ast(x) \nabla_x \phi(x)$, but it's not defined globally because following $\phi(x)$ around a non-infinitesimal path doesn't return to its original state

So you can formalize moving along paths in terms of holonomy or (probably what's better here) in terms of coordinate changes, and the point is that you can encode the coordinate changes of A(x) in terms of the failure of phi(x) to be a function on the whole surface?

Yes (but I've never seen a physicist really do that formalization because all they're interested in is computing $\int \Omega$ and saying it's quantized)

there's probably some physics-y argument for why it is quantized/a topological invariant that I've forgotten

So when a physicist says that some number is topological (or that it measures something topological) they just mean that there are discretely many possibilities

As opposed to meaning "there's some topological+geometric notion that this number is a function of, and the number only depends on the topological part"

Yeah, "topological invariant" does not necessarily mean that the physics viewpoint really cares about the "topology" part

Good to know!

Thanks a lot

@MikeMiller One further point for the dictionary here would be that the holonomy is what the physicist calls the "Berry phase"

That's helpful

I'd only known "Wilson line" (the trace of holonomy)

Yeah, Wilson line is also the holonomy but in an entirely different physical context :D

"Berry phase" is more condensed matter lingo, "Wilson line" is particle physics lingo

Makes sense

I'll finally learn some physics and take a mechanics course next semester. I've been advised to do the senior-level Lagrangian mechanics class so that I don't feel dreadfully bored.

try not to be too harsh with us ;)

not my place as a student, just trying to learn some translation...

that would be next to impossible with a lexicographer like ACM anyway :P

::runs::

if the course involves field theory that could get tricky rather quickly

Fortunately for me, you're quite patient ;)