The h Bar

David Z

15:02

@Sofia asking people to reveal how (or if) they voted on a particular post is somewhat discouraged. Not really a big deal, but don't make a habit of it.

and for the record, claiming or implying that you know how someone else voted on a particular post is very strongly discouraged.

(just in case anyone was wondering)

Sofia

@DavidZ I am very fond of the Curious Mind. And, somehow, I believe that I know his hand, or, more exactly what he likes. And infi is also very lovely. I didn't ask random people, I asked people that I feel that I can permit myself. It's a very few of them.

@DavidZ I know the rule and I have no objection about it.

David Z

15:20

That last part was meant mostly for other readers who may stumble upon this later

Sofia

@DavidZ what is the word "stumble". Rely on that, i.e. make a custom?

David Z

For instance if someone is reading the chat transcript two days from now and they happen to read this part, they stumbled upon it. The alternative is that they were specifically looking for the statement I made.

Sofia

@DavidZ O.K. I'll take it to my attention. Thanks for telling me.

@DavidZ Aaaaa! But can't you delete my comments if they are not O.K.? Please delete them.

David Z

What comments?

(actually, you can delete your own comments, so if you want them deleted, you should do so yourself)

Sofia

@DavidZ whatever you think that it's not O.K.

David Z

15:27

then if there are other comments that no longer make sense in context, flag for moderator attention so we can delete those

ACuriousMind

@DavidZ She means the chat messages, I think, which users cannot delete themselves after the edit period has passed.

David Z

oh, huh, I didn't know that

well we don't delete chat messages anyway, unless there's some pressing reason to

e.g. if they are egregiously offensive or contain sensitive personal information

Sofia

@DavidZ I'll be more cautious, I promise you.

Sofia

15:48

@ACuriousMind I don't understand this. How $E \to 0$ for $r \to \infty$ doesn't emerge from Maxwell's laws? It's Gauss' law, $Q/\epsilon = \int_S F\cdot dS$. For a spherical symmetry and spherical surface one gets F ~ Q\R^2. So, how the $F \to 0$ for $r \to \infty$ doesn't come from Maxwell's laws?

ACuriousMind

15:58

@Sofia Maxwell's equation do not tell you that the field around a point charge is spherically symmetric - you are putting that assumption in. It is a matter of taste whether you require vanishing at infinity or spherical symmetry as your "physical condition"

Sean

-2

Q: What happens if the Russians decide to nuke the Kola borehole?

I recently came across research on the Kola Superdeep Borehole in Russia. It's one of the world's deepest at 12 km (7.5 mi) down. If someone decided to hoist a nuclear weapon all the way down and set it off, what is the effects and can it be felt worldwide? Is there any simulation package that ca...

Clearly someone is making productive use of their time...

Sofia

@ACuriousMind so, vanishing at infinity is rather a convention - eventually coming from requiring that potential be 0 at infinity, i.e. a reference value.

ACuriousMind

@Sofia No. The field vanishing at infinity is a physical requirement, the potential vanishing at infinity is a gauge choice - a potential that had a finite value at infinity would produce the same physics, a field with finite value would not.

Sofia

@ACuriousMind do you want to say the E \to 0 at r \to infinity just come from the physical observation? I.e. so we see that is happens?

ACuriousMind

@Sofia Yes

noahnu

16:08

There's a buoy in a body of water. It starts out at equilibrium and then an unknown force is applied, submerging part of it. The buoy is weighted so that it remains floating. I need to prove that the motion of the buoy (as it bobs) is SHM and I need to find the period. (density of water, mass of buoy, radius of cylindrical buoy are given) I've determined the acceleration of the buoy as a function of its displacement (x=0;half buoy is submerged/equilibrium). How do I turn a(x) into a(t)?

ACuriousMind

Differential equations need boundary or "initial" conditions, there is nothing surprising about that

Sofia

@Sean I am glad to see you. Why we have a problem in \Delta E \Delta t with the fact that t is not an operator? I really don't understand. This uncertainty relation comes trivially from classical electro-magnetism. The uncertainty in a signal time-duration and the frequency obey \Delta \omega \Delta t \g.e. 1/2.

@Sean now, multiplying by \hbar we get \Delta E \Delta t \ge \hbar/2.

ACuriousMind

@Sofia We "have a problem" because this means we cannot derive this uncertainty in the same way we derive all other quantum uncertainties, because the quantum uncertainty is about standard deviations of operators. Since $t$ is not an operator, it is not clear what the $\Delta t$ is (unless we define it as, e.g., Griffith does)

2

Sofia

@ACuriousMind do you really want to get some heart attack? It may happen even at your age. You work too hard. Would you work a bit less hard? I addressed the question to @Sean exactly for avoiding putting again and again pressure on you. When do you time to sleep, to eat, to relax?

ACuriousMind

@Sofia I assure you that I'm not risking my health or sanity for answering questions on the internet. I answered because I know that @Sean has repeatedly said that quantum mechanics is not really his expertise.

Sofia

16:18

@ACuriousMind do take care of yourself, it all that I ask. Pleaaase!

alarge

@noahnu In case of a harmonic spring you also have F(x), right? Do you know how to solve that?

Sean

1

Q: Why does the electric field vanish at infinity?

When $r \rightarrow \infty$, $E \rightarrow 0$ for a point charge or set of charges or a finite charge distribution. While this seems obvious, I cannot find a reason why this is true when inspecting Maxwell's equations and the Lorentz force law. I thought however that all of electrodynamics was c...

electromagnetism electricity maxwell-equations

ACuriousMind

@Sofia I appreciate your concern, but you really need not worry about me.

Sean

Am I gonna be wrong and get a million downvotes if I say it's due to Coulomb's Law?

Stan Shunpike

@ACuriousMind I dunno. You live in a country with pole dancing robots.....

Sean

16:21

I'm just surprised no one said that; so I'm afraid it's too obvious and must be wrong

Gowtham

@ACuriousMind yes its very normal for 20+ students to be nightowls ... i myself was one

ACuriousMind

@Sean Look at the question - it explicitly says that we want to go from Maxwell's equations and Lorentz force law. To get Coulomb, you have to make a spherical symmetry assumption that is not inherent in them.

Sean

So it was too easy to be true

darn

ACuriousMind

@StanShunpike Heh, touchè

dmckee

This ...

27

Q: Why do schools teach arrays over List?

Most of the assignments in my school for the initial programming classes required me to use arrays. I work full time now, and I never used an array for any project that I have worked on. Even in the exisiting projects I never saw the use of arrays anywhere. In my opinion, List is easier to use an...

java programming-languages programming-practices

reminds me of why I don't like Programmers. The OP has been heavily rewarded for a provincial, ignorant rant.

Sean

16:24

@Sofia my answer would have been very similar to what @ACuriousMind mind told you. Have you read the deriviation in Griffith's?

dmckee

On the other hand, some of the answers remind me of why I do like programmers.

ACuriousMind

@dmckee Well, the idea of programmers kinda was "Stuff that we don't want on SO", wasn't it?

Stan Shunpike

@Sean I love Griffiths. I will always remember his QM book fondly because prior to him, I felt like everyone who explained QM was speaking gibberish. But I am glad to be beyond him because he doesn't really address Hilbert spaces.

dmckee

@ACuriousMind That's a long story, where it was, then it wasn't.

noahnu

@alarge there's no actual spring. I know the equations for SHM, however I don't know angular frequency or anything with time. My a(x) comes from $\Sigma B - F_g$.

Sean

16:31

@StanShunpike yeah he pretty much says "hilbert space is important" and then moves on

I still really don't "get" Hilbert space as much as I would like

Stan Shunpike

Yeah! Exactly! That drove me nuts. As soon as I saw that, I was like "yeah okay, I clearly need a supplement because that's all @ACuriousMind talks about" ;p

alarge

@noahnu Right, but the differential equation just has the form d^2 x / dt^2 = -k(x-x_0), no matter if it's an actual spring or not.

@dmckee Well I suppose the core of the question is useful (when to use list vs. array), although it is, unfortunately, in the form of a rant.

Sean

@StanShunpike so what did you read to help you with that?

Stan Shunpike

@bolbteppa Told me to read Elements of the Theory of Functions and Functional Analysis. This is just a math book but this is giving me the tools to move beyond just vector spaces to more interesting vector spaces like Hilbert spaces. But apparently you need analysis for this (according to math chat), so I am supplementing with Principles of Mathematical Analysis by Rudin.

@Sean ^

@Sean From what I understand, once you understand the math for Hilbert Spaces and related stuff in those books, you then have the math to understand why exactly they are useful physically. I think they are subtle points, but I don't really know yet. I just sort of ask on SE to get pointed in the right direction and then blindly read with the faith that they have told me the right thing :)

David Z

0

Q: Who created the energy conditions?

The earliest text I've been able to find that explain the GR energy conditions is "The large scale structure of space-time" (1973) by Hawking and Ellis. However in Barcelo and Visser's paper "Twilight for the energy conditions?" they speak of the trace energy condition which was popular in the ...

general-relativity resource-recommendations

off topic?

Stan Shunpike

16:41

@DavidZ What energy conditions does s/he mean?

David Z

I couldn't say

You could post that as a comment

Stan Shunpike

Okay I will

ACuriousMind

Energy condition

In relativistic classical field theories of gravitation, particularly general relativity, an energy condition is one of various alternative conditions which can be applied to the matter content of the theory, when it is either not possible or desirable to specify this content explicitly. The hope is then that any reasonable matter theory will satisfy this condition or at least will preserve the condition if it is satisfied by the starting conditions. In general relativity, energy conditions are often used (and required) in proofs of various important theorems about black holes, such as the no hair...

@DavidZ Well, I think it is the kind of question that could be asked under specific-reference, isn't it?

Gowtham

@DavidZ HSM seems good for it.

Sean

Migrate to HSM?

Sofia

16:45

@ACuriousMind and @Sean : Griffith's proof is for open systems, s.t. it is not general. How can the average of the energy vary in time? Where from the supplement of energy of the system, to the more, or to the less? And, if it is for open systems, these ones don't have a well-defined Hamiltonian, as it is in exchange of energy/particles with the environment.

@ACuriousMind and @Sean To the contrary, every wave-function limited in time obeys the time-energy uncertainty. No need for open system, and the average energy can be constant in time.

David Z

@ACuriousMind yeah, I believe so

alarge

@StanShunpike My intro to functional analysis course was probably based on the book by Kreyszig (well, the contents was very similar, but we only had lecture notes so I don't know for sure). For that at least I don't think you really needed much background in anything too mathematical. Hell, even my intro to digital signal processing courses did a bit of engineering math Hilbert spaces.

David Z

@Gowtham @Sean I'm not asking whether it fits on HSM, I'm asking whether it's off topic here

Stan Shunpike

@alarge really? What are Hilbert spaces used for in signal processing?

ACuriousMind

@Sofia No, it is not assumed that the average of the energy varies in time, and I see no mention of an open system.

Gowtham

16:50

yes +1 on off topic, i flagged it

ACuriousMind

At least, joshphysics' answer doesn't mention that with a single word, and it looks correct to me

Sofia

@ACuriousMind I jump to eat, and I return.

Sean

@DavidZ well don't the two go hand in hand? Its not on topic, and is about the history of science

ACuriousMind

@Sean That something fits on HSM doesn't make it off-topic here.

alarge

@StanShunpike You deal with vector spaces, Fourier transforms and all that jazz. Were you the econ major, by the way? You use a lot of Hilbert spaces in finance as well (take eg. Hansen's work).

Stan Shunpike

16:54

Yup, I'm the budding econ major. I'm taking my first econ class this quarter. I'm pretty stoked. I'm actually glad I learned most of my math through physics though. It gives me nice ways to think about it outside of econ that, for me, are more natural. I really want to take some signal processing courses.

David Z

@Sean no, they don't. A question may be on topic here and on topic at HSM, or on topic here and off topic at HM, or off topic here and on topic at HSM, or off topic on both sites.

alarge

@StanShunpike Here's one that does seem to have some sort of an intro to Hilbert stuff as well (I'm guessing nothing very in depth, i.e. probably just the concept, but anyway).

David Z

I don't care (yet) whether it fits on HSM; I care whether it's off topic here.

Stan Shunpike

@alarge Nice, a decent start. Do you have a favorite textbook(s)? I haven't found any good book I like on signal processing, which is unfortunate as I study neuro for fun. I remember taking IT and that was like the coolest course ever. I thought Boolean algebras and stuff like that were so amazing.

alarge

@StanShunpike I don't actually. For the basic signal processing I only have my old lecture notes somewhere. I think there are a couple of canonical books out there, though. I have Haykin for neural networks, but there may be better books out there.

Stan Shunpike

17:02

@DavidZ For me, the only gripes I have are (1) is it mainstream physics and (2) is it a resource requests. he's trying to get seemingly information about a topic that he's having trouble finding basic factual information on, so I think that's fine. But even so, if (1) or (2) is a problem, then it might be off topic.

alarge

Oh, here is one use for Hilbert spaces in finance (by Hansen). In case you're interested.

noahnu

@alarge I have the period now, but that is based off the assumption of SHM which the question says to first "show that the buoy will execute SHM".

Stan Shunpike

@alarge Nice! Hansen was the Nobel winner right?

alarge

@StanShunpike Yes. For GMM from what I remember.

@noahnu Well I'm not sure what definition of SHM you have is. In principle if you have the force as a linear function of the displacement (of the form F = -k(x-x_0)), then SHM is what you'll get.

noahnu

@alarge oh okay, thanks!

Stan Shunpike

17:09

@DavidZ Do you think it's off topic?

user54412

17:25

@Sean honestly, all you get from the promotion vector space -> Hilbert space is well-defined infinite sums

user54412

if you really want to understand completeness (and I don't think there's much to understand), you can do it with just a metric space without all the clutter of a vector space

alarge

17:37

@ChrisWhite While this is true, studying Hilbert spaces (or rather functional analysis) might give you some insight into certain things you have not previously thought about in your courses on linear algebra like the boundedness of operators, spectral theory etc.

Stan Shunpike

18:00

@alarge yeah I feel like I'm shaky on the spectral stuff. Know of anything that goes into the in detail? A bunch of posts qmechenic linked me to said that was one of the main uses for Hilbert spaces, ie discussing spectral stuff in QM

David Z

@StanShunpike I'm not so sure. This sort of thing does seem like one of the use cases our specific-reference tag was designed for.

alarge

@StanShunpike If QM is what you want to learn (or rather you want to learn functional analysis for QM), you should probably pick up a book on QM. That said. the book intro to functional analysis by Kreyszig (I mentioned it earlier) also does seem to have a chapter on QM (not sure if it is any good for your purposes, though).

Pi Day

18:28

Hello ! I'm having some trouble with my maths. I'm trying to get to the result (15a) in this paper. I already got the LHS and the first part of the RHS, but I can't find the second part (i.imgur.com/kFvsYQ8.png). So far here's what I've done : i.imgur.com/p9MFJmX.png, i.imgur.com/yHCFbOu.png .

I get to a formula that's rather close to the one given; however the $\alpha^2$ and the $\epsilon^2h'^2$ both seem to be on the wrong side (left part of the numerator / right part of the numerator), and I'm off by a minus sign in the right part. Could someone help me ? Thanks

Stan Shunpike

@PiDay is that sonic the hedgehog?

Pi Day

@StanShunpike s A nic

Sean

@DavidZ I didn't realize that. I assumed on-topicness was mutually exclusive between sites

Stan Shunpike

@Sean thx to mathmaticians, every time I hear the phrase mutually exclusive I think of the empty set

@PiDay ?

Huh?

Pi Day

knowyourmeme.com/memes/sanic-hegehog @StanShunpike

Stan Shunpike

18:38

Ah, lol. So it is...sanic

@DavidZ If there's a tag for it, I would just leave it. Its not the best question ive seen, but its not the worst either and seems to be substantive / not a crank. It just is more of asking for a reference but I do think it fits that tag well.

@alarge Yeah, I have several books on QM now. I try to read across areas both in math and physics so I don't just get a physicists view point. This also has the added benefit of seeing what other areas I can apply the math too.

Pi Day

@StanShunpike Any idea on my question above ?

Stan Shunpike

@PiDay I'm not knowledgeable enough to answer. But I think if you post that as a question you will receive a nice response. People tend to be more willing to write in Latex if you post as a question than in chat.

ACuriousMind

@StanShunpike @PiDay: That's a homework-like question, please don't post it. It will be closed in that form, anyway

Pi Day

@StanShunpike Is it in the scope of what can be asked ? I heard that PSE didn't like 'homework-like' questions

Oh well @ACuriousMind answered that

alarge

@PiDay Have you checked that with Mathematica (or another CAS)?

Stan Shunpike

18:47

Ok there ya go

@ACuriousMind thank you for correcting me. My bad

Pi Day

@alarge I unfortunately don't have those. The result in the paper should be good, since it matches their experiments really well though

alarge

@PiDay wxMaxima is free.

Kyle Kanos

^just about to recommend that

alarge

Python also has symbolic libraries. And if the problem is small enough, wolframalpha.com might be able to handle it.

Pi Day

It's quite a bit computation

I've checked my maths three times but can't find my error

alarge

18:50

@PiDay Not for a computer it's not. I'm not entirely clear on if wolfram alpha kind of restricts by length as well.

Pi Day

downloading maxima

alarge

If all else fails, just write it in any programming language and throw random numbers in place of all the symbols. If the equation holds, it's probably right.

Pi Day

It won't tell me where my error is though if I'm wrong

Kyle Kanos

If you're smart about it, you can figure it out

Take each derivative/integral/computation one step at a time

Pi Day

The thing is I tried that three times :/

Kyle Kanos

18:52

If the CAS gives you a different answer than your handiwork at one particular step, you'll know where to look

Pi Day

I started again from scratch

and I'm 99% sure my result is different from theirs. The factors before the h'^2 don't even match

Stan Shunpike

@ACuriousMind CM has to do with predicting the path of a particle/object right? What does QM claim to predict? The evolution of the probability amplitude?

ACuriousMind

@StanShunpike Uh. QM predicts what you measure. I try to not impose any ontology on it.

Kyle Kanos

^ = "shut up and calculate" :D

Stan Shunpike

LOL

Okay, Ballentine suggests on page 99 that we shouldn't interpret the wave function as a wave. Why do we call it a wave function then? Why not just call it a state function?

bolbteppa

19:09

That's so funny lol knowyourmeme.com/memes/sanic-hegehog

ACuriousMind

@StanShunpike It's a historical accident because we "discovered" QM, among other things, through the double slit, and so it seemed that the "particles" we shot through the slits were "really waves"

I hate the term wavefunction and wish it would indeed be replaced by something like state function.

Kyle Kanos

@ACuriousMind Write an Intro to QM book and use state function instead.

Stan Shunpike

Okay, I'm going to be obstinate and use state function from now on. I hate the term wave function

Kyle Kanos

Probably because I'm not really a QM guy, but the name doesn't bother me.

bolbteppa

But the wave function acts like a wave in the sense that the quasi-classical approximation is just the geometric optics wave function, wave function makes perfect sense

ACuriousMind

19:15

@bolbteppa Yeah, but the problem is that it isn't what you'd intuitively call a wave in the full quantum theory. I think naming stuff after what it behaves like in some approximation is not a good idea.

Stan Shunpike

Question: Ballentine defines $|\psi\rangle = \sum_i c_i |u_i\rangle$ where $\lbrace |u_i\rangle \rbrace$ is a set of basis vectors. He then writes $c_i = \langle u_i | \psi \rangle$. Can someone explain why multiplying $|\psi \rangle$ by the bra vector yields a constant? I thought he just defined it as a sum. Wouldn't only $\langle u_i | u_i \rangle$ yield $c_i$?

Kyle Kanos

Holy crap...so the guy at The Bank who is going to interview me in 10 days got a PhD under Lawrence Krauss.

bolbteppa

@StanShunpike I don't think I actually told you to read Kolmogorov/Fomin, I referred to a quote in the amazon reviews to that the book amazon.com/Elements-Functions-Functional-Analysis-Mathematics/… and the l2 L2 isomorphism you were asking about the wave function vs. Hilbert space ways of viewing QM.

If I'd specifically told you to read it I would have said you should read the "Introductory Real Analysis" by Kol/Fomin to prepare for it tbh e.g. the first 5 chapters are a more expanded/developed exposition of what's in the functional book, but tbh Kolmogorov is hard, I would have said to read Shilov to prepare for that lol and Rudin is far too brief but you gotta refer to Rudin no matter what you read.

Danu

@ACuriousMind State function? Yuck

@KyleKanos Am I an ass for not knowing Krauss?

Stan Shunpike

@bolbteppa I'll be more careful when I quote you lol. But I liked the references and Kolmogorov is great.

Kyle Kanos

19:22

@Danu Yes. Absolutely yes. (okay not really)

Danu

Ah, he's a science promotor (and scientist)

bolbteppa

@StanShunpike if you like that kind of Russian analysis then you could view Shilov's 3 volumes
http://www.amazon.com/Linear-Algebra-Dover-Books-Mathematics/dp/048663518X/
http://www.amazon.com/Elementary-Complex-Analysis-Dover-Mathematics/dp/0486689220/
www.amazon.com/Elementary-Functional-Analysis-Dover-Mathematics/dp/0486689239/
as the easy version (his analysis book is like a better Rudin in lots of places) and then these three books
http://www.amazon.com/Lectures-Linear-Algebra-Dover-Mathematics/dp/0486660826/

Cool, yeah I remember the proof of the uncountability of the reals in Kolmogorov's real analysis book as being the best one I've read because it was so intuitive/nice

Stan Shunpike

I got Shilov! Georgi boy! Wait, 3 volumes?!? I only have one.

That's so weird how different books spell Gelfand...why?

bolbteppa

@StanShunpike he took the inner product of $|\psi>$ with the dual $<u_i|$ of $|u_i>$ where the basis is normalized

Danu

@Sofia I really wonder: Have you never thought about this? Have you not read any of the standard textbooks on QM?

ACuriousMind

19:37

@StanShunpike Because transliteration of Hebrew is not uniform

Stan Shunpike

@bolbteppa so $\langle u_i | u_j \rangle = \delta_{ij}$?

user54412

@ACuriousMind Is transliteration of any language uniform? Even if they largely share the same alphabet?

Stan Shunpike

@ACuriousMind if the Hamiltonian isn't independent of time, does $U(t,t_0) = e^{-i(t-t_0)H}$? No right? Because you couldn't do that operation iteratively. But then what form does the unitary operator take when the Hamiltonian isn't independent of time?

ACuriousMind

@ChrisWhite No, transliateration between two different alphabets is never quite uniform. Greek to Latin comes close though, I think, there's not much ambiguity there

Dyson series

In scattering theory, a part of mathematical physics, the Dyson series, formulated by Freeman Dyson, is a perturbative series, and each term is represented by Feynman diagrams. This series diverges asymptotically, but in quantum electrodynamics (QED) at the second order the difference from experimental data is in the order of 10−10. This close agreement holds because the coupling constant (also known as the fine structure constant) of QED is much less than 1. Notice that in this article Planck units are used, so that ħ = 1 (where ħ is the reduced Planck constant). == The Dyson operator == Suppose...

Danu

Is learning about the universal property of the quotient topology a totally standard part of any undergrad course on topology?

ACuriousMind

19:48

@Danu We don't have any normal "topology" course here, so no idea.

Danu

I'm asking because it is used in my diffgeo notes as if it should be known

...but then again, I don't trust my prof haha

ACuriousMind

@StanShunpike Essentially, you have to integrate over the time dependent Hamiltonian in the exponent instead of simply multiplying it by the time. It's the same as going from $W = F\cdot s$ to $W = \int F\mathrm{d}s$ in the definiton of work.

Danu

I've always found the Dyson series a lame trick

Time-ordered exponentials, lame

ACuriousMind

@Danu Heh, tell that to the Wilson loops :P

Danu

@ACuriousMind Blissfully ignorant of those for now

user54412

19:59

@Danu we mixed category theory and algebra liberally, but topology was rather separate -- just plain point-set, differential, algebraic

Danu

@ChrisWhite So, did you learn about the universal property or not?

user54412

doubt it

bolbteppa

@StanShunpike if the basis is normalized yes

Danu

@ChrisWhite Damnit, professor Leeb! :P

Stan Shunpike

If I want to define a quantum state in a position representation, I have to define the vector using an orthonormal basis of position eigenfunctions. And when I do this, if I multiply the state vector by a position basis bra vector, I get the probability amplitude. Am I understanding all that correctly?

user54412

20:15

2

Q: In theory, could gravity waves be used to make a "gravity laser"

The sources I've read compare gravity waves to electromagnetic waves, I'm curious to what extent this is. In theory, could gravity be harnessed in similar ways to how we've used electromagnetic radiation such as in lasers? If so: what differences would this have to a regular laser? If not: What...

general-relativity gravity gravitational-waves

user54412

so all the answers seem really pessimistic

user54412

but they're all based on "I can't think of a source"

user54412

am I the only one not satisfied by this?

Kyle Kanos

Would you rather an answer that says, "Well if we can generate single gravitons, then ..."

ACuriousMind

@ChrisWhite I think the correct answer to that question is: "Lasers are devices relying on quantum processes. Since we have no quantum theory of gravity, this question is unanswerable".

user54412

20:17

like, in general, when gravity doesn't act like EM, I expect the reason to be traceable back to the positivity of mass

user54412

@ACuriousMind Quantum explains how to get stimulated emission, but isn't the propagation of in-phase EM waves treatable purely classically?

ACuriousMind

@ChrisWhite Oh, you think the question is about whether or not there could be a "beam of monochromatic gravity radiation"?

user54412

maybe that's what I'm saying

ACuriousMind

Do/can gravity waves even have a single wavelength? (I don't know much about them)

user54412

sure -- just generate them with a single frequency, since their speed is c

ACuriousMind

20:22

@ChrisWhite Are they waves in that sense? They are excitations of a tensor field, not a vector field, after all...

user54412

i.e. a slowly decaying binary orbit (the closer to a pure quadrupole the better)

ACuriousMind

But yeah, it probably works

I think the question should be rephrased then, because the answers are all attacking the "laser" bit. If it is about whether or not just a "coherent beam" of gravity radiation is possible, that's a slightly different question.

user54412

well I don't know what the OP really wants

ACuriousMind

I'm not sure OP does :P

user54412

also, is there no such thing as a classical laser (i.e. phase-correlated stimulated emission)?

user54412

20:26

like, out of all the wave phenomena in the universe, can we not think of something that does this?

ACuriousMind

@ChrisWhite What does "stimulated emission" even mean classically?

Oh, lookee here

user54412

@ACuriousMind I don't really know what I'm saying. But couldn't you have a (powered) system that amplified whatever waves passed through it?

ACuriousMind

I can't access the article, but it seems there is "classical stimulated emission"

@ChrisWhite I only see laser and semi-conductor amplifiers as possible optical amplifiers

Buth aren't really classical systems

user54412

"Because stimulated emission is crucial to the derivation of the Planck spectrum and because it is usually described in terms of a transition rate between quantum states, it is sometimes re- garded as a distinctly quantum-mechanical effect. One purpose of this paper is to show that stimulated emission arises naturally also in purely classical systems. This point was made some time ago by Gaponov et al. and one of the present authors formulated the problem in semiclassical terms."

user54412

looks like a good paper

ACuriousMind

20:32

It certainly sounds interesting

Damn you, paywall!

Stan Shunpike

Paywalls are annoying.

What kind of number is the eigenvalue of an observable?

Is it {0,1}? $\Bbb{R}$? $\Bbb{C}$?

ACuriousMind

@StanShunpike $\mathbb{R}$

Observables are self-adjoint operators, whose spectrum is real.

Stan Shunpike

20:48

@ACuriousMind why do observables need to have eigenvalues? What does multiplying a state vector by a number achieve?

ACuriousMind

@StanShunpike What kind of question is that? We don't want to achieve anything when we multiply by a number.

The observable has to have an eigenbasis so that there are states into which its measurement can project - after measuremnt, we find the system in an eigenstate, and the eigenvalue is the value we measured. This is what is commonly called collapse.

Stan Shunpike

Oh, I get it. Okay. That's what I thought but somehow I talked myself out of it.

I'm still trying to get used to using these vectors differently. As states.

Sean

Geez, reading this conversation makes me realize how badly I need to re-read griffiths (and then maybe something else)

Physics Meta

21:29

0

Q: off-topic question

I posted a question on Stack Exchange Physics and the question was closed as off-topic. As a matter of fact I did a mistake when selecting the tags, since I used the homework tag. When I realized that I did a mistake I modified the tag, however the question was closed as off-topic. I really would...

support

0celo7

@ACuriousMind You know anything about complex diffgeo?

ACuriousMind

@0celo7 Heh, Danu asked me the same thing a few days ago. No, I don't know it.

0celo7

@ACuriousMind So if I asked you why if $\operatorname{dim}_\mathbb{C}\mathcal{M}=m$, then why does $\operatorname{dim}_\mathbb{C}(T_p\mathcal{M}\otimes\mathbb{C})=2m$, you wouldn't know?

This isn't a question appropriate for this site and will get lost on Math.SE.

ACuriousMind

@0celo7 No, but I would ask you what base field the tensor product is taken over, and as what kind of object we consider $T_p\mathcal{M}$ here. (Since $\otimes_\mathbb{C} \mathbb{C}$ is a do-nothing operation).

0celo7

@ACuriousMind My hand wave argument is that since $\operatorname{dim}_\mathbb{R}\mathcal{M}=2m$, when we complexify the tangent space, instead of being isomorphic to $\mathbb{R}^{2m}$, it becomes isomorphic to $\mathbb{C}^{2m}$, thus having complex dimension $2m$.

ACuriousMind

21:46

@0celo7 I also suspect that that is meant, but, really, writing $T_p M\otimes\mathbb{C}$ isn't well defined without saying that we are considering the tangent space as a real vector space and taking the tensor product over $\mathbb{R}$ with $\mathbb{C}$.

0celo7

@ACuriousMind Is this formally what "complexify" means?

Sofia

@Danu I don't know that Griffiths' textbook, and I don't care of it. Nobody is a supreme authority in physics. If I find that a proof is not satisfactory I don't care who stands behind it.

ACuriousMind

@0celo7 Yes, to complexify something means to take the tensor product with $\mathbb{C}$ over the field/ring it is originally defined over.

In all cases I know, this is the tensor product of modules (of which the tensor product of vector spaces is a special cases) with $\mathbb{C}$ over $\mathbb{R}$.

0celo7

@ACuriousMind One more question, this quite general. If a vector space $V$ splits like $V=V_1\oplus V_2$, does $\operatorname{dim}V=\operatorname{dim}V_1+\operatorname{dim}V_2$?

ACuriousMind

@0celo7 Yes

Similarily $\dim(V\otimes W) = \dim(V)\cdot\dim(W)$.

0celo7

21:50

@ACuriousMind Cool. Now I just have to work out the splitting of complex exterior bundles.

Or whatever they're called.

ACuriousMind

I don't know what you are talking about :)

0celo7

What is the exterior algebra called when viewed as a bundle?

ACuriousMind

Or perhaps I do, but not under that name

@0celo7 Uhhhh. Do you mean the space of $p$-forms?

0celo7

@ACuriousMind Forms are sections of ___?

ACuriousMind

Ah

0celo7

21:52

Something bundle?

ACuriousMind

I'd guess it's the exterior bundle, but I cannot recall if I've ever really heard a name of that thing

You more commonly talkabout the "space of $p$-forms" instead of the bundle, I guess

Yay, @Jimnosperm! How are your cones today?

0celo7

@ACuriousMind $$\Lambda^kT^*_\mathbb{C}\mathcal{M}=\bigoplus_{j=0}^k\Lambda^{j,k-j}\mathcal{M}‌$$

This stuff is horrible.

ACuriousMind

lol, I agree, that looks horrible

0celo7

@ACuriousMind This noob (author) doesn't even \mathrm his differentials.

Why are mathematicians inconsistent about that?

I don't know what you're talking about ;)

ACuriousMind

Me neither :P

@0celo7 Because they're lazy and/or don't all share the same sense of aesthetic?

0celo7

21:58

@ACuriousMind There are packages for that.

I should DL the physics package.

Or maybe it came with my distro.

ACuriousMind

@0celo7 You would be horrified how little some mathematicians and physicists care about writing proper TeX, or even readable articles and books.

0celo7

Lol 1057 package updates. Looks like I haven't had the utility open in a while.

Oh god it wants me to do stuff. ::presses random buttons::

@ACuriousMind $\LaTeX$ support?

How do I get a package that I dl'd from CTAN into LaTeX?

ACuriousMind

@0celo7 Luckily, I've never needed to do that. I have no idea

I had enough trouble making it recognize a newly installed font properly

0celo7

@ACuriousMind I want to get the nice $g$ with the squiggle on the bottom and change the default $\Psi$, but I have no idea how to do that.

I have a .zip.

Uh

ACuriousMind

@0celo7 Me neither, you'll have to ask the guys at TeX.SE

0celo7

22:08

My god why is this so hard?

What is this "repository address" business?

Is anyone who frequents h Bar a TeX god?

Lol, I don't even have the package in my distro, but adding \usepackage{physics} in the preamble worked.

Sean

How godly do I need to be @0celo7

0celo7

::shrugs::

@Sean I figured it out, I think.

I don't have the package, but it works somehow.

Sean

Ok good because I looked back earlier in the conversation and I'm definitely not that godly. Math.SE has some pretty complete posts on tex in meta though, just so you know

Sofia

@DavidZ I was very interested in a question asked a few days ago, about tachyons. I don't work in a suitable domain, but it seemed to me illogical that tachyons can be found. So, I asked experts. Am I allowed to say/quote what I learnt from them?

Stan Shunpike

@0celo7 If my potential for the Hamiltonian is the Coulomb potential, what is the operator? How do I write the Coulomb potential as an operator? Would it be the $\frac{1}{r^2}$ term that's an operator?

I mean 1/r sry my bad

0celo7

22:18

@StanShunpike $1/\hat X$ IIRC.

Don't remember exactly.

If @ACuriousMind has doesn't answer, I'll go check Shankar.

Sofia

@DavidZ I don't work in a suitable domain, so, I asked experts. Am I allowed to say/quote what I learnt from them? Of course the bottom line is no tachyons, but they explain why and how.

0celo7

@StanShunpike Yes, it's just $$\frac{1}{\sqrt{X^2+Y^2+Z^2}}$$

This is probably meant to be interpreted in the spectral theorem way.

@ACuriousMind Do you have $\mathrm{e}$ and $\mathrm{i}$ macros?

ACuriousMind

@0celo7 Usually \ee and \ii, yes

0celo7

@ACuriousMind How do I define operators in the preamble?

ACuriousMind

@0celo7 \DeclareMathOperator

That produces stuff like $\lim$

With proper limits and spacing

0celo7

22:28

@ACuriousMind What's the syntax?

ACuriousMind

I'm not sure that was what you meant with operator...

0celo7

No I meant like \ee

Is that \newcommand?

ACuriousMind

@0celo7 Oh, that's just \newcommand

For writing here on SE, having something like \newcommand{\ket}[1]{\lvert #1 \rangle} is handy, by the way. You can put it into a post and it works on the entire page

0celo7

@ACuriousMind So \newcommand{\ee}{\mathrm{e}}?

ACuriousMind

@0celo7 Yep

0celo7

22:33

@ACuriousMind Thank you based skull.

bolbteppa

@0celo7 doesn't $\mathbb{C}$ have a $\{dz,d\bar{z}\}$ basis so that $T_p(M) \otimes \mathbb{C}$ will have $2m$ possible products?

0celo7

@bolbteppa No, because a complex space of dimension $m$ is essentially a real space of dimension $2m$ with a complex structure.

Or not.

I'm not too sure about this.

bolbteppa

That doesn't affect the argument though

Stan Shunpike

If we aren't dealing with stationary states, does $\hat{H}|\psi\rangle$ yield an eigenvalue? I know for example it does in the stationary state of energy levels in an hydrogen atom. But I am unsure what happens in a nonstationary state.

0celo7

@bolbteppa I don't think you can have $2m$ coordinates on an $m$-dimensional manifold and have everything make sense.

Oh god I'm confused again.

Stan Shunpike

22:43

Nvm I found an answer. Yay!

ACuriousMind

@StanShunpike Stationary states are by definition the eigenstates of $H$. If it isn't stationary, it isn't an eigenstate.

bolbteppa

I'm sorry I'm not gonna make an argument I could be wrong, I'm not sure if the m barred variables are always there implicitly or whether people do tricks to pretend they don't exisst

0celo7

@bolbteppa Yeah I'm not sure either.

ACuriousMind

@bolbteppa Basis over what it the crucial point here. Over $\mathbb{R}$, that's right, but over $\mathbb{C}$, you only need one.

@0celo7: Basically, your text should also be careful with distinguishing $\dim_\mathbb{R}$ and $\dim_\mathbb{C}$.

bolbteppa

I hate this stuff

user54412

22:59

@0celo7 A lot of tex distributions automatically download missing packages from a central repo. It won't work if the package isn't registered there, and for some distributions it won't work if the distribution hasn't been updated since the package was released.

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