Welcome back @JohnDuffield

:-)

what is the O(2,4) group like? I have never studied a pseudo-Riemann group besides the Lorentz group.

it looks like a group to operate on a 4-dimensional spacetime with 2 time directions.

Hallo, in a post they mentioned gating detectors to an energy level (MeV), how would you usually gate a detector to a given energy level? Or what is the method name I could google to learn about it? Thanks

Here is their post (physics.stackexchange.com/questions/577859/…)

@antimony In this case, a thallium-activated sodium iodide NaI(Tl) scintillation detector was used.

You could better use a lithium-drifted germanium detector Ge(Li) or even better a high-purity germanium (HPGe) detector.

For this particular experiment, you need two detectors.

The energy levels are found using pulse height analysis as usual.

What's special in this case is that the two detectors work in anticoincidence mode.

The rejection is carried out by passing the pulses from the first detector through an electronic gate that is closed if a coincident pulse is detected from the second detector.

That's why we don't see the big 1.33249 MeV peak in the given Co-60 spectrum.

morning

huomenta

morning

dusk

dawn

midnight at full moon, invoking the spirits by swinging a dead cat three times around my head

I have a question for all veterans here. How did you deal with feeling dumb while in grad school? The last month has been extremely unproductive and I am not able to understand what I am doing and it almost makes me want to quit

It doesn't go away :p

I've had a lot of unproductive months :-/

"unproductive" sounds normal; "not able to understand what I am doing" is something you might want to try to change

Just to clarify a bit. I understand the physical picture of what is happening, but the actual calculation that I need to do is something that I am finding difficult to understand

Are you talking about a single set of calculations you need to do or more broadly you don't understand how to do the calculations you're being assigned? @xcodeking

Single set of calculations

You will forever be chasing a "better" version of yourself, if you don't ever turn around and appreciate how much progress you've made it's easy to feel like you're not moving. If your issue is a single set of calculations that's at least a good sign that the misunderstanding is relatively small

@skullpatrol Hi!

xcodekin: you should ask a question about it.

Question on the calculation that I am stuck on?

Yes.

I already did, albeit on math SE. I'll post the link. Give me a sec

Since this is a physics audience, the background might be appreciated. I am trying to do inverse double Mellin transform, and that is how I end up with a similar looking integral

Ouch. That's a reminder of just how much I don't know. I can't help you with that, xcodeking. I hope somebody else can. Sorry.

Nice thanks @Fad

* Nice, thanks @FadedGiant

So it is mostly material bandgap?

Do you mean the better energy resolution of Ge detectors?

Christopher Nolan be reusing the same actors for 20 years lol. A lot of his movies use the same actor/s from a previous movie

Just so I'm sure I have the "big picture" idea in my head. Once we find a representation of the Lorentz/Poincare group, the elements of the underlying representation space are not "Lorentz invariant" objects, but we can build Lorentz invariant objects from them. So a general spinor $v\in\Bbb C^4$ is not manifestly Lorentz invariant, but we can construct objects like the bispinors used in the Dirac field which then are lorentz invariant geometric objects (or Lorentz covariant with a basis)

Also "bilinear" and "bispinor" in this context are used interchangeably I assume

Oh, in general the energy gating of the detector types you mentioned

Well most objects aren't lorentz invariant

They are lorentz covariant

Scalars are, and pseudo-scalars also (to some degree)

The objects the representations act on usually have some sort of inner product

for spinors it's the product $$\psi^A \xi_A = \psi^A \varepsilon_{AB} \xi^B$$

Where $\varepsilon$ is the spinor metric

@Slereah This is something that I only really gave real thought to today, the point of finding a representation of the Lorentz group isn't that the resulting vector space contains only lorentz invariant vectors, but that we can use them to build lorentz invariant objects which we can then use to build lorentz invariant actions

yeah people tend to be pretty loose with their use of "invariant"

In my head lorentz covariance is just what you get when you make an arbitrary choice of coordinates (or components etc) for a lorentz invariant object

hmm I guess in that sense they don't have to be distinguished if it's clear from context but meh

Covariance literally means "vary with"

yeah that makes sense

Roughly speaking, if you vary the coordinates, the object will vary accordingly

it's invariant if the object doesn't vary at all (beyond being shifted)

I mean every object that is lorentz covariant is technically "underlyingly" lorentz invariant right

the covariance just appears which a choice of coordinates/components with which to "vary" with

unless I have made a horrific mistake in my understanding

A "mnemonic" that I find useful is to just scale my coordinates and see what happens to the object. If it scales by the same amount, it is covariant. If it scales in the opposite way (1/k or something) then it is contravariant. If nothing happens, it is invariant

Just an easy way to think about it

I know scaling is not a part of Poincare group. It is just an easy way to check

yeah I get the distinction, but the coordinates are arbitrary so they're describing an object whose underlying geometric properties are invariant

@Charlie No, invariant means the object does not transform at all. A vector does transform - think about ordinary rotations: Certainly vectors are not "underlyingly" invariant under rotations!

I think I'm going round in circles here but eh

How saturated is the strings community in the US? I would be applying for PhD this year, and being from India, I am worried about the extremely low chances. Would I be better off in the EU or Canada? I would be graduating with a masters, so that should not be a problem in terms of EU admissions

@ACuriousMind I see what you mean but I'm thinking of essentially the coordinate transformations like those found in sr and gr, the vectors themselves are invariant to the coordinate transformations

Ah, that's the "active vs. passive" issue

I guess I didn't make a distinction between changing coordinates and acting on an object

hmm yeah

McDonald's sell 75 burgers per second

if you think that your transformations just change the coordinates (passive viewpoint), then sure, all objects are really "invariant", but then "invariant" doesn't actually mean anything, it's just part of how you chose to think about the transformation :P

if you think that your transformations change the state of the world, then vectors certainly aren't invariant.

In the U.S., 1,700 people become millionaires every day.

I've actually only seen things written in terms of passive transformations like in sr

sometimes it's clear whether a transformation is "active" or "passive" - like when you talk about frames and coordinate changes, but sometimes it's not clear (or rather, irrelevant)

there are more trees on Earth than there are stars in the Milky Way. Today, there are around 3 trillion trees and 400 billion stars.

It's actually really ambiguous in that case because invariant means entirely different things in those two pictures :/

The language of invariance and covariance should not depend on whether or not I'm talking about an active or passive viewpoint, so I'd advise strongly against calling vectors "invariant"

@ACuriousMind diffeomorphism versus tetrad rotation :p

@Charlie no, "invariant" means the same thing in both cases - the object transforms in the trivial representation!

fair enough

ah my bad I meant "different things are labelled invariant" not that invariant literally has a different meaning

i don't know what you mean

anyway don't worry too much about it anyway

it's not like it's extremely important calculation-wise

@ACuriousMind it actually wasn't a particularly interesting or good point, I just meant for instance the vectors themselves are either covariant or invariant depending on whether you're using active or passive transformations

although again if "invariant" is not a good choice of word to use there I'll avoid it

A vector is not invariant in either picture. I'm guessing what trips you up is that we write a vector as $v = v^\mu e_\mu$ and you think that that makes the vector $v$ "invariant", but that's really not what we mean by "invariant". It just means it's a well-defined geometric object

oh

it's tricky because the $\mu$ indices there are "of a different kind" than when you write the contraction of two vector components as $v^\mu w_\mu$ for two vectors $v$ and $w$

active transformations change the $v^\mu$, passive transformations change the $e_\mu$ and hence the basis you're expressing stuff in.

in the passive viewpoint "invariant" would mean "objects that do not change their numerical values after a frame change", not "$v$ stays the same" because, again, everything stays "the same" in the passive viewpoint by definition so that's not a special property

basically every text I read when I was originally learning this essentially says "the basis transforms one way while the coordinates transform another way and they 'cancel out'"

so that is actually not a good way of thinking about it, because you have to change one or the other

I've read that plenty of times, too, and I still find it a weird way to think about this. No proper math text talks about it like that :P

that being said I almost always read that the lorentz transformations for instance are a "coordinate change", not an active change of the vectors

I feel we're getting too much into physical ontology when we quibble about what the transformations really "do"

yeah come to think of it it doesn't sound like a very robust way of thinking about it

it's just math - there's a vector space, and a representation of the Lorentz group. We call the trivial representation "invariant", and everything else co/contravariant depending on whether it transforms like a vector or its dual

ok that makes sense

if there isn't a major difference between the two i'll just live with it for now and think about it later

that we're "doing a coordinate change" rather than "changing the world" doesn't impact the math one bit - it will never happen to you that you accidentally treat something as an active transformation and then get a wrong result at the end because it really was a passive transformation :P

agh but that then brings me back to my original point, that (in more correct language) the spinors aren't covariant themselves, but we build covariant objects out of them

if that statement is correct I'm basically happy

because I don't see any reason why the objects of the representation space are necessarily covariant

I'm not sure what you mean by the spinors not being covariant

"covariant" is just a word for "transforms in a non-trivial representation of the group" :P

ah no actually my analogy with R^3 I was about to give doesn't make sense

ahhh

ok I need think it over more

What is the broadest interpretation of $E=\bar{h} \omega$?

I am not sure if I phrased correctly but how general is the aforementioned equation. What exactly are the terms $E$ and $\omega$?

or, if we're using it contrastive to "contravariant", it means "transforms in a non-trivial representation of the group that doesn't invert the transformation"

@Yashas It's just energy and frequency of a photon. Don't make the mistake thinking this is somehow a general law of quantum mechanics.

@ACuriousMind Somebody showed me a "derivation" (which looked very hacky) of Schrodinger equation which assumes that's true for all particles. He did arrive at the correct differential equation at the end but it puzzled me that this seemly wrong generalization somehow leads to Schrodinger's equation.

The proof starts with the assumption that the waefunction is a plane wave (a photon essentially?) and then takes few derivatives and plugs in $E=\bar{h}\omega$ and de Broglie hypothesis and ends up with a differential equation which is identical to TISE.

$E = \frac{p^2}{2m} + V$

@Yashas Ah, yes, it's true for plane wave solutions of the Schrödinger equation, but that you can derive an equation by starting from its solution is not exactly a great insight :P

It just feels like some big coincidence. Because the person assumed plane wave (which is only valid for a free particle) and then used photon related equations and they somehow lead to TISE.

also, once you have non-zero $V$ there are not necessarily any plane wave solutions anymore, and it's unclear what "$E$" and "$\omega$" should be for a general state, anyway

Maybe I should try with a linear combination of plane waves.

Already in trouble. How do I define $\omega$ and $\k$ for a linear combination of plane waves? I think it's not fair to assume separable solutions at the first step.

just take it as another heuristic motivation of the SE and move on

How does $E = \bar{h} \omega$ show/account quantization of light?

Wikipedia says "The relation accounts for quantized nature of light" but I don't see how.

@Yashas it tells you the energy a single photon of that frequency has

@ACuriousMind but how is it related to quantization?

meaning that you "can't" have light of that frequency with continuous energy, but only in multiples of that energy.

frequency can be fractions too?

(this is not strictly speaking true because it is possible to have states that do not have a definite number of photons, but that's what we mean by quantization - the energy comes in discrete packages, not continuously like you can continuously turn up the amplitude of a classical EM wave)

the motivating experiment here is the photoelectric effect - discrete bundles of energy $\hbar \omega$ explain it, continuous energy in a wave does not.

Building on what @ACuriousMind said. Assume that I have a photon of some frequency w and energy E. What quantisation here means is that ALL photons of frequency w will necessarily have energy E. I can't find a photon of frequency w and energy 0.59E or something

I tried to interpret it as you cannot have a photon for an arbitrary value of $E$. Quantization in the sense of energy levels of electrons in atoms. Now I wonder if there are two definitions of quantization in use.

Well, you can. The correct statement would be "you cannot have a photon for an arbitrary value of $E$ and arbitrary value of $\omega$."

yes - what's quantized is the energy levels for a fixed value of $\omega$

So for any given $E$, I can find a photon with some $\omega$?

yes (well... beyond a certain energy scale things can get complicated but in principle, yes)

Not sure why I thought that photons can exist in certain energy states only. Probably a misconception from high school.

(Based on your name, I assume you're from India) Yeah the stuff that we're taught in high school can often be a hurdle to actually understand things

@xcodeking I am from India. I second that. In fact, sometimes I was given wrong information. I think high school teachers focus more on rote learning than conceptual understanding.

@ACuriousMind a question for you, if you don't mind. How is the strings community in EU? Are collaborations encouraged? I am planning to apply for PhD there and would like a better idea. Things in the US seem awfully saturated for now

@xcodeking I don't really know since I'm not in academia, sorry

@Yashas Yup, I also believe it.

@ACuriousMind Ah okay, I got the impression that you were. Thanks

I was told that the uncertainty principle arises due to the fact that the act of measurement disrupts the particle in observation. I wasn't until uni that I learned that that wasn't even remotely close to the actual reason.

I anyway don't understand the need for high school students to learn basic QM. What is its use? I think students in India are learning uni stuff in high school (and this is compulsory for all students).

@Yashas that's a popular misconception historically probably arising from the idea of Heisenberg's microscope

I honestly wouldn't call it basic QM. It's just facts thrown at you, and a lot of them are wrong

@xcodeking Yeah, that's a more accurate description.

the bit of QM we learned here (Germany) in high school was pretty similar, fwiw

so it's not a unique Indian failing - it's just what you get when you want to teach people QM but don't want to/can't do the math :P

Or at least if its being taught, they should teach it well and ensure that the student has a good conceptual understanding. I feel like our textbooks are designed for rote learning. We have to mug up the Schrodinger wave equation for exams. We don't even know what Hamiltonian means (we are just told it's energy; obviously that's a very poor description). We don't even know linear algebra. We just have several dozen graphs, shapes of atomic orbitals, mugging up solutions to the orbitals, etc.

I think a syllabus that would talk about the history of QM would do more justice than this rote learning nonsense.

It REALLY depends on the system that you are talking about

I cannot seemed to think of anything that is uniquely in the meter scale or the centimeter scale

Even nanoscale has its own quirks such as friction and viscosity behaving in some different way such as more sensitive to surface features

well, if your system is a bunch of pellets about a centimeter large, you'll see pretty different things at these scales :P

hmm...

Well that's true, but there does not seemed to be a physics law or behaviour that is most significant to that scale. Well what I mean is that, say if I want to describe what is unique about the quantum scale, is that things obey superposition and hence locations are not as clear cut. But if I want to do a similar pitch for centimeter vs meter and kilometer scale, there does not seemed to be a unique thing that distiniguish them each from the other length scales

quantum behavior is not a matter of scale

think Schrödinger's cat, where the microscopic decay "infects" the macroscopic system (the cat) with "quantumness"

what's true is that quantum behaviour is often more relevant at small scales, but it is an error to think there is something magical about the scale

$E = \frac{p^2}{2m}$ vs $E = \frac{1}{2} mv^2$ Which is more fundamental? I think the former is but is there an explanation for that?

define "fundamental"

$p$ seems to be more natural than $m$. We can talk about momentum of photons but not the mass of it.

...but both your equations contain $m$

neither is appropriate for the photon

Ah, I didn't realize that both had $m$. Silly me.

does $p=mv$ hold always in the non-relativstic limit?

what's your definition of non-relativistic $p$?

in many contexts, what you've just written down is its definition, so it is meaningless to ask whether it holds

Good question. I have no idea.

I definitely have some serious conceptual misunderstandings. I probably need to start from scratch.

What does $p$ for a photon mean? Is it a bad question in the non-relativistic limit?

Is the momentum of a photon a consequence of special relativity?

i.e. there is no classical explanation

you can't take the non-relativistic limit of a photon, because its velocity is constant at $c$, but the non-rel. limit is that $v$ is much smaller than $c$!

at least p = mv for a photon doesn't make a lot of sense in the classical world

@ACuriousMind is it meaningful to apply the free particle solution derived from non-relativistic SE to photons?

Not really. Photons don't have wave functions because there's no proper relativistic position operator, but that's another can of worms and you can often get away with pretending they do

Is de Broglie's wave equation $p = \frac{h}{\lambda}$ universal?

If I were to start learning physics from scratch, where do you recommend I read from? I was thinking of reading Feynmann's lectures on physics. I will probably start from volume I itself. I think I have lost the whole plot and chronology of stuff in physics.

My fundamentals seem irreparably broken. It's probably a good idea to start from scratch than fix here and there.

I don't think there's such a thing as a unique way of "learning it from scratch", really

learning physics is a constant process of revising what you thought, that's normal

first you must learn the way of the ball

How does a ball fall down

when we first learn Newtonian mechanics, we learn all sorts of concepts that no longer universally apply when we move beyond it. That doesn't mean you should forget all about it, just get used to it

@xcodeking I don't know anything in particular about the string theory community or the EU but the job market is always tightening in the US and the next two years will be very, very rough for people on the market. If you have the opportunity to apply in the EU, and your desire is to stay in academia, it would probably be in your best interest to do so

@MikeMiller thanks. How rampant is publish or perish in theory (math or physics) in the US?

I dunno that I would call it rampant, it's the logic inherent to the academy as opposed to a bizarre beast we didn't expect to see

Only some percentage (say, 50%) of grad students get postdocs; only some percentage (say, 20%) of postdocs get tenure-track jobs; only some percentage (no guess here) of tenure-track faculty get tenure

At each of these levels you're in a competition (the last level one of a different sort)

And you will be compared against other candidates in a number of ways. One of those is the depth and breadth of your research program

If someone else has a more expansive research program that gets cited more, by the usual metrics they are a better candidate

(Of course who you/your advisor know also plays a significant part)

Thus the job market speaks, expelling those who didn't make the cut onto the general labor market. That's all publish or perish is; it's the statement that the academic labor market is expanding and so competition for positions deepens

The average person is a 28-year-old Chinese man.

By almost all important measures, the world is a better place to live today than at any other time in human history.

Stephen Pinker has entered the chat

Take all the known data from the beginning of human history, up to year 2003. Currently, we produce an equivalent amount of data every 2 days

Fishing is the most dangerous occupation in America (128.9 deaths per 100,000 people), compared to logging (115.7), the military (82, the leading cause being "accident"), flying (72.4), structural iron and steel working (46.4), mining (35) and policing (11.1).

The Pew Internet & American Life Project study, published in Aug. 2011, found that 8 percent of Internet users do not use email or search engines. Somehow, this sizable portion of the online population manages to surf the web some other way... perhaps by typing in URLs by hand?

100% of hbar users do not openly share their appreciation for JingleBells' valuable statistical sharings.

Ok I had a loong think about what we were talking about earlier, and I think it is clearer now. We have a representation of, say, the Lorentz group - when we perform a coordinate transformation in the active transformation picture, we take an element of the Lorentz group, $g_1$, identify it with an endomorphism over the representation space, and then "act" on every element of the vector space with this endomorphism.

On the dual space, in order to preserve the inner product, we "act" on each element of the dual with an endomorphism (from the set of endomorphisms of the dual) which corresponds to the group inverse of the original element of the Lorentz group, $g_1^{-1}$.

And so in this sense vectors and covectors transform "oppositely"

@Charlie yes, exactly

@JingleBells if I want random statistics I'll google them myself :P

@JingleBells There's an entire site dedicated to generating such spurious correlations.

@ACuriousMind awesome, ty!