The h Bar

5:28 AM

@lucas: Hi Lucas. re your comment about my answer to the bow and arrow question - yes I think my answer is very close to being a homework answer and might well be deleted. It is right on the edge of what is acceptable.

7:35 AM

@JohnRennie @NoahP @Jiminion For those interested in a detailed (and rigorous) discussion of Hawking's radiation, you should take a look at this

lucas

@JohnRennie Thank you for your attention. But it isn't deleted yet and I guess that it won't be deleted for ever, because it is your answer and you aren't me. You are admirable one and I am nothing.

i don't have a very good track record for getting answers to my questions :p

user3183724

I have a quick question that is probably more suitable for the chat than a standalone question. I'm simulating the mollow triplet for a single qubit with a classical drive (using python package QuTiP), the Hamiltonian for which would be $H = \omega_r a^\dagger a + A \sin(\omega_d t) (a + a^\dagger)$ which can be written as $H = (\omega_r-\omega_d) a^\dagger a + \frac{1}{2} A (a + a^\dagger)$ in the RWA. However, normally you set your drive frequency equal to the transition frequency

To look at the triplet, that is. But doing that in the RWA makes your qubit term drop out, and not applying the RWA gives a time dependent Hamiltonian which complicates things I'd say..

8:24 AM

@lucas I don't get any preferential treatment, believe me! :-)

The reason I wrote that answer is because the OP had already solved the problem but using a very complicated method. I wanted to point out that there is a simpler method available without actually doing the calculation. Answers that discuss ways a calculation could be done, without actually doing it, are kind of borderline.

Q: what is the Favicon supposed to be?

8:38 AM

The question about the favicon gets asked frequently and always gets closed as a duplicate of this:

4

What is the favicon of the site supposed to represent? My initial impressions are that it looks like a partially unrolled condom. I'm not trying to be a troll, but I just don't see it making a good first impression. Am I missing something obvious?

discussion site-enhancement

so I wonder if we shouldn't edit that into better shape and mark it faq?

@DavidZ There's something wrong with following links from h bar to meta.PSE for me...

I'm about to post a meta question about it

Is it an HTTPS problem?

Yes

But I don't know what exactly that means/what I should do about it

Yeah, that's well known. Meta sites don't work with HTTPS.

You don't know what HTTPS means?

Not really, no.

I know vaguely

It's some kind of encryption thing

8:43 AM

Yeah, HTTPS means that your browser encrypts the data that it sends to the web server, and vice versa, so that nobody(*) can intercept the data being sent back and forth

(*) indicates that there are some technical caveats

Q: I can't follow links from the h bar chatroom to meta.PSE. What's going wrong?

For reference:

0

Whenever I click a hyperlink to meta.PSE (as posted by e.g. the "Physics Meta" bot) in the h bar, I get the following screen: When I choose to proceed anyways (bottom hyperlink), I get this: Note that the URL looks like this: Some other times, I just immediately get this screen: I do...

bug

Yeah. I was about to say that this error is well known and has been known for years.

I wonder why it never caused any problems for me before.

Maybe you weren't accessing meta sites by HTTPS

@DavidZ a partially unrolled condom? That had never occurred to me, and even with the thought in mind I still don't see it. Gallamine must have a very strangely shaped tadger.

8:45 AM

@Danu let me just post an answer

@DavidZ I think dmckee's answer is just fine. After all, this is a physics site and we should assume users are capable of following links and Googling for extra information.

Yeah, I meant more like cleaning up the question. Not sure we want everyone who wonders about the favicon having to picture it as a condom :-P

@JohnRennie LOL

@DavidZ the link I see is http not https. I wonder why you see it as https ...

@JohnRennie in my onebox?

8:47 AM

I love the tag "site enhancement" too :D

"My onebox" ?

The link I posted to to the meta question about the favicon

Q: I can't follow links from the h bar chatroom to meta.PSE. What's going wrong?

@DavidZ Yes. If I right click and choose "copy link address" then paste into notepad the link is http: not https:

That's with Chrome on Windows

Physics Meta

0

Whenever I click a hyperlink to meta.PSE (as posted by e.g. the "Physics Meta" bot) in the h bar, I get the following screen: When I choose to proceed anyways (bottom hyperlink), I get this: Note that the URL looks like this: Some other times, I just immediately get this screen: I do...

bug

Perhaps I see it as HTTPS because I use the HTTPS Everywhere extension. That might be rewriting links to use HTTPS.

9:06 AM

Thanks for the answer @DavidZ

That seems to work.

lucas

@JohnRennie But before you, H.Tofaili had been mentioned your answer. And it isn't the matter. I say They don't delete your answer. They get preferential treatment and I think you know this.

I think the TeX.SE guys are starting to get annoyed by me... :D

in TeX, LaTeX and Friends, 3 mins ago, by Christian Hupfer

Shing

10:01 AM

In quantum mechanics, how do we measure rotational angular momentum?

I have read this: physics.stackexchange.com/questions/32498/how-is-angular-momentum-measured-in-ex‌periments-in-practice

but I don't quite understand what meant by angular distributions

11:01 AM

the angle distribution at which a particle is scattered in a collision

12:46 PM

@Danu lol

1:35 PM

In classical thermodynamics do we define the work done (on or by the system) in tge same ways as we do in classical mechanics?

Q: Using (only) scan/photograph/screenshot of text as an answer?

@HiteshPathak it's usually pressure times volume change in thermo, rather than force times position change in classical mechanics, but yeah, basically the same thing AFAIK

Other news: this "rule" we have about not including images with text. Is that written down anywhere? There is this:

27

This answer* is a picture of some text. Now I thought about this for a moment. This text could be considered an answer. After all, it attempts to provide a solution. I didn't read the question to see if it's a right solution, but nonetheless a solution is provided. In reference to all the questi...

discussion answers images copyright

but that's about posts which only include an image of text, it doesn't focus on the mere fact that something which could have been typed out was not.

@DavidZ i read about joules paddle experiment in which the mass was lowered and the temperature of the system increased ..in this case the work was done on system evrn when its volume was constant

@DavidZ therefore i dont think that this definition of work is valid in general.

That was heat

1:50 PM

So, does the particle/anti-particle notion mean that a 3rd particle can translate or 'jump', by merging with the anti-particle and the remaining particle taking it's place? Is HR then somewhat akin to quantum tunneling?

Not really. Even after the third particle and the antiparticle merge, they turn into something.

@DavidZ i have Classical Thermodynamics by AB Pippard and he says that it should not be legitimately called heat

I say he's being pedantic :-P

I mean, yes, I see how you could call it work. It's the result of a dissipative force (viscosity) acting over a distance. But that's a mechanical view, not a thermodynamic one.

Of course most methods of energy transfer in thermodynamics are just mechanical work under the hood.

So I guess if you were asking whether anything that is considered work in mechanics would be considered work in thermodynamics, the answer is no.

But that's not what I thought your question was about.

Arnold Neumaier

@NoahP - Your paragraph from yesterday makes a good fairy tale but is still no good science. See my answer to your question on the main site, physics.stackexchange.com/a/252183/7924

As David says the division of energy inputs into "work" and "heat" is a matter of convention.

For the purposes of textbooks we often define "work" to mean processes which change the extensive state parameter (i.e. volume in a confined fluid system) and let everything else be lumped into "heat".

Even though the falling weight in Joule's paddle rig is clearly doing mechanical work.

But that energy is getting promptly randomized, so it appears in the confined system as heat would.

Q: How can this result in Thermodynamics be rigorously proved?

2:03 PM

5

In Fermi's "Thermodynamics" there's a proof of the formula: $$W=\int _{V_1} ^{V_2} p\,\text dV,$$that is, the work done by the pressure of a gas that expands from a volume $V_1$ to a volume $V_2$ on the surface that contains it is equal to the integral above. The proof goes like this: Conside...

physics fake-proofs infinitesimals

The top answer to this question defines work as something whose sole result is to lift up or lower down a mass in surroundings

@HiteshPathak I took the liberty of removing the period so we would see the question

Notice that he uses the word "sole" in there. That is important.

According to this definition the system in joules experiment is being worked on

If we run the Joule process backwards, you get the weight rising, which is fine, but all that work comes from a conversion of thermal energy into work which violates the 2nd law, so for the process to actually raise a weight, the system would have to be in contact with a thermal bath as well and the process would do things other than just raising the weight.

@HiteshPathak yeah, and the author of that answer ducked on getting a definition for work done on the system. I can understand why, it's going to be hard to frame in similar language.

The point here is that to reason about thermodynamics it is helpful to have simple definitions of the words "work" and "heat" that apply to a nice imaginary system of some kind.

But these simple definition don't cover all the ground that is possible with real system (and don't correspond to the most general defintions). So you have to generalize by understanding the roles that work and heat played in the simple system.

You say "In this system the energy supplied by the paddle wheel ends up in thermal motion without changing the macroscopic configuration of the system, so we treat it as heat".

2:23 PM

@dmckee are you saying that the work must change the macroscopic coordinates?

I am actually confused with the distinction between "work" and "heat"

@HiteshPathak Depends on the system. The thing to keep in mind is that both work and heat are transfers of energy. There is going to be a level at which they do the same thing. So the distinction is in some sense arbitrary.

For the purposes of making learning the subject tractable it is usual to make a very strong arbitrary distinction. Something like "for a confined fluid system 'work' means $P\,\mathrm{d}V$.'

And yeah, that leaves you with system where your mechanics text would describe something as "work" and your thermo text describes it as "heat".

The real distinction to make about the Joule paddle wheel is that it is highly non-reversible.

user116211

@HiteshPathak Heat is the process of increase of mean energy of the system keeping external parameters fixed. Work is the process of increase of thermal energy which is adiabatically isolated.

2:38 PM

The fun is that he's been asking about Joule's paddle-wheel experiment. The falling weight is "work" in a mechanics sense, but we usually describe the energy input to the fluid system as "heat".

user116211

$$\delta Q= \sum E_i\mathrm dP_i\\ \delta W= \sum P_i\mathrm dE_i$$; note that the in heating, only the probability changes infinitesimally.

Q: The derivation of the Mass-Energy(including mechanical energy) equivalence principle

Is this a duplicate:

0

(Not duplication! I couldn't find the answer for the general case, instead of the special case that I've already seen in Youtube I hyperlinked.) I will fully satisfy with not only a mathematically rigorous proof, but also just any argument. For example, I was impressed by the thought experiment...

general-relativity special-relativity mass-energy

Iclosed it as a duplicate of:

8

Q: How to derive $E=mc^{2}$?

Is there some way to derive $E=mc^{2}$? I can understand that energy in something is proportional to its mass, but the $c^{2}$ part. I have no idea. It seems like the way the units are going it would end up as $kg*(m/s)^{2}$, which I think is the unit for Newtons, not energy. And why is there no ...

special-relativity units mass-energy

but the OP deleted the closed question then reposted it.

I'm reluctant to VTC again as obviously the OP disagrees with me!

@JohnRennie Flag for mod attention, so they can undelete the old question and close as duplicate.

Regardless of whether OP agrees with a duplicate vote, just deleting and reposting a question is not okay

And it is an obvious duplicate - both questions ask "How to derive $E=mc^2$?". If OP says "not duplication", they need to make clearer what specifically is dissatisfactory about the answers to the other question, and how their question is supposed to be different.

@ACuriousMind I'm on the case.

@ACuriousMind: I'm stuck on an aspect of this Hawking radiation thing. Is it something you could comment on?

2:51 PM

@DavidZ Well, they give back the energy "borrowed" to produce the part./anti-part. pair.

@JohnRennie Perhaps. But the interplay between GR and QFT is by far not my best field, I like my relativity special ;)

Thanks :-) It appears that demonstrating there's a difference between the vacua is just the first step. You still don't get a net flux of radiation away from the black hole unless a horizon is present.

I think this is effectively a red shift i.e. near the event horizon the vacuum may appear to have more energy but it loses some of it in escaping to $\scri^+$.

For a massive body without a horizon, i.e. a neutron star, all the energy is lost so there is no Hawking radiation. With a horizon not all the energy is lost and some escapes thus taking energy away from the black hole.

Does this sound plausible?

Is there any proof for micro black holes?

@JohnRennie What do you mean by the energy being "lost"?

@Jiminion Proof for what about them? That they exist?

2:56 PM

@Jiminion if it's part of a larger process, sure

@Jiminion What do you mean by "proof"?

@ACuriousMind Lost means some energy/particles/whatever escape to infinity and the mass of the black hole decreases as a result. One of the papers I've been looking at, which seems to give a simple explanation is at:

theory.tifr.res.in/~padmanath/Research/GTR/HRGRR2009.pdf

@dmckee Yes, proof that they exist. Has anyone ever detected one?

Could you glance at and see if my interperation of what it says is correct?

But if you're busy it doesn't matter. I will simply read on and hopefully learn something :-)

@Jiminion I'm sure it would have been big news.

2:59 PM

@JohnRennie On it

Secret

http://arxiv.org/pdf/1603.04671v2.pdf
This preprint had became one of the latest buzz by some 'media' including dailymail etc.

Best to be treat this as highly speculative until we have a bag of axions to test

Their creation and decay in an accelerator context has been modeled, but as I understand it the LHC results are negative, which only leaves astrophysical observation of presumed primordial holes.

@SevenSidedDie: hello, what have we done now? :-)

It seems to me the original article by Hawking doesn't say that a horizon is at all necessary to have radiation, he just says horizons may be the only regions in the universe where the curvature is strong enough such that this effect becomes non-neglegible.

I'll now read what you linked

Amith KK

Hello

3:03 PM

A horizon isn't necessary.

It's a generic effect of gravitational field

It can happen at any distance from the horizon.

Amith KK

In an AC circuit consisting of an inductor and a capacitor, is current and EMF never out of phase?

@dmckee I don't think we'd be able to see a micro-blackhole in space. (I'm referring to the really tiny ones - electron sized or whatever.)

I mean it even happens with no gravity at all in a Rindler space

@Jiminion The way you would detect primordial black holes astronomically would be by observing the explosive final evaporation of one. As I understand it that'd be visible over a non-trivial distance if you happened to have the right instrument looking that way.

On page 8 the paper specifically says:
The presence of the Killing horizon is crucial for this effect. In fact a neutron star in the
Schwarzschild spacetime doesn’t emit Hawking radiation.

Amith KK

3:06 PM

I would ask a question on the main site, but I suppose it's a basic thing

@AmithKK It shouldn't be hard to work out the diff. equation.

@JohnRennie Ah, yes. The point is that the observers sit at different points! For Hawking radiation, the observer sits at spatial infinity, and if there is no horizon, all radiation just gets infinitely redshifted. Compare with the preceding section, where the Rindler observer at spatial infinity won't see any radiation. The presence of the horizon is necessary for the presence of radiation for inertial observers, but not necessary for the idea of radiation as such.

Thanks, that sort of fits with what I had guessed. What I'm unclear about is the mechanism i.e. what is it about the horizon that causes a finite, as opposed to infinite, red shift at spatial infinity?

@JohnRennie The timelike Killing field - the thing that conserves energy - "dies" at the horizon (it gets zero there, and is spacelike thereafter, since radial coordinate and time coordinate switch roles after the horizon). Therefore, all energy transport happens between the horizon and infinity, while in the absence of a horizon, it happens from the middle of the star to infinity.

And that additional distance is what is enough to redshift all radiation possibly emitted to zero for an observer at infinity.

At least, that is my interpretation of section 5 in what you linked

That fits with what I thought, but I don't understand the calculation. If we have a neutron star just too big to be a black hole then we calculate the energy flux by integrating something from $r=0$ to $r=\infty"? Is that the difference? If so what is the calculation?

And how does cutting it off at $r_s$ make a difference?

Or is that not in the paper (I couldn't see it).

3:25 PM

@JohnRennie BTW for future reference, the mere fact that the OP disagrees with you in a case like this shouldn't be enough to change your opinion. They should be able to explain why they disagree, in a way that can satisfy everyone else too, like ACuriousMind said. (In case that wasn't said already.)

@DavidZ the gold tag badeg weighs heavily on me in these circumstances :-)

Oh right, there's that. Yeah, I guess when you have a unilateral close vote, it's worth being a little more careful, but still, don't automatically defer to the OP whenever there is a disagreement.

If a poster thinks their question is not a duplicate, they do need to explain themselves.

Not to mention, we don't really want to reward people for going against the rules ;-)

@JohnRennie There's a 200 rep bounty being dangled like a carrot now...

@JohnDuffield Thanks for putting the bounty on - I haven't got enough reputation yet to be able to do so.

@NoahP see:

37 mins ago, by John Rennie

Thanks :-) It appears that demonstrating there's a difference between the vacua is just the first step. You still don't get a net flux of radiation away from the black hole unless a horizon is present.

I now have our resident QFT expert trying to find a way to describe the calculation in layman's terms :-)

Brill! Thanks :) Would that be Acuriousmind?

3:31 PM

It would indeed. I suspect he is now wishing he had never agreed to help :-)

This question seems to have become much bigger than I anticipated!

Hawking radiation is a complicated thing

@JohnRennie I think the calculation is eq. (37) - it calculates the redshift factors for the horizon and infinity. The point is that, for the star, the expression for $V$ is only valid until the surface of the star, and will look differently inside the star (My guess: linear in $1-r$ for homogeneous, spherical star). Then, when you compute the asymptotic redshift factors, you get 1 for the redshift factors in both cases, meaning they agree on the radiation coming from the star - none.

I find it funny that Hawking's original article contains the "virtual particle analogy" with an explicit warning that that is not the reason! It says: "One might picture this negative energy flux in the following way. [virtual particles, blah, blah]. It should be emphasized that these pictures of the mechanism responsible for the thermal emission and area decrease are heuristic only and should not be taken too literally."

2

For shame, everyone who ever claimed that Hawking himself "used that explanation"!

First time I've actually bothered to look that up

Thanks. I'm still not sure I get it, but you've given me enough to reread the paper, or look for other descriptions, with a bit more insight.

Thanks for that, I guess

3:44 PM

@ACuriousMind This warning isn't given in any of his books - He just writes as if it is totally correct.

@NoahP in his popular science books, yes. Not in his scientific paper!

Yeah I don't think he would say that in his serious books :p

He would probably say THE FOCK SPACE DECOMPOSITION ON THE CONVEX NORMAL NEIGHBOURHOOD IS blablabla

(Hawking science books are pretty dry)

Nevertheless, it's disappointing that he doesn't even mention that it is only heuristic.

@NoahP I think he considers that warning implicit when writing pop-sci books.

Hawking gotta get paid

3:47 PM

Almost nothing in those books can be taken as the rigorous explanation because there is no math! It's just-so stories for people who don't want to spend years learning to decipher arcane symbols.

@ACuriousMind Yet people who have read it are under the impression that they themselves are theoretical physicists...

Welcome to the internet

@JohnRennie Well, I think the real reason the horizon is needed is that the horizon is something that is relevant in scattering - if I look at how a wavepacket (which a free scalar field essentially is) behaves when moving from the asymptotic past to the asymptotic future when you fire it that the Schwarschild object, it doesn't do anything interesting when it hits a star, but it does split into two parts with changed frequencies when it hits a horizon

(due to the nature of that horizon changing the time-like Killing field and hence the nature of propagation in time)

what paper are you reading btw

@ACuriousMind Aha, now that sounds like the sort of explanation I'm looking for. We're calculating how the extra energy in the vacuum near the star/horizon propagates and the horizon makes a difference to this. I have heard something along those lines before.

@Slereah theory.tifr.res.in/~padmanath/Research/GTR/HRGRR2009.pdf

3:55 PM

@JohnRennie That is the exact explanation in Hawking's original paper, although it's of course interpersed with all the actual math he is doing ;)

thx

He computes how the scalar field scatters off a collapsing black hole, then compares the frequencies, since the "most obvious" mode/frequency decomposition is what defines the vacuum in this case

@ACuriousMind I hadn't thought to look at Hawking's original paper. I assumed that 30 years on we'd have easier to understand explanations. But now you've provided the link I have a read and see what I can get out of it.

@JohnRennie never forget

People hate rewriting proofs

@Slereah I think it's generally true, even in maths, that the first derivation/proof of something is rarely the simplest and most efficient derivation/proof.

As it progresses from leading edge to undergraduate textbooks it gets simplified and easier to understand.

4:01 PM

@JohnRennie From reading that paper, it seems QFT in curved spacetime has not really changed in that time span. I looks like something we could have talked about in our seminar on that topic last year

Depends

Most people still use the same methods in QFT in CST for calculations

but the cutting edge people generally use AQFT stuff

8 hours ago, by yuggib

@JohnRennie @NoahP @Jiminion For those interested in a detailed (and rigorous) discussion of Hawking's radiation, you should take a look at this

it's only 25 years old ;-P

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

^

and judging by the authors, there should be a lot of AQFT

oh you never know

it's always the pastors that end up in drugs and sex scandals

2

Maybe Haag preaches AQFT a lot but may not practice it!

4:05 PM

it doesn't seem to be so AQFT in fact...

It seems the argument is still the wavepacket scattering from Hawking, just elevated to Haag's standards of rigor ;)

Yeah AQFT is fun and all but doing calculations with it smacks of hubris

@JohnRennie : Since I've offered a bounty, I should draw your attention to this: "An atom absorbs or emits light at a frequency which is dependent on the potential of the gravitational field in which it is situated". The photon is emitted at a lower frequency at a lower elevation. It doesn't reduce in frequency as it ascends. Instead you and your clocks run faster as you ascend, so you measure the frequency as reduced.

@JohnRennie : good luck with that.

SevenSidedDie

4:56 PM

@JohnRennie Nothing! I've just been taking random samples. :) (Interim results: looking pretty healthy in here.)

A: An explanation of Hawking Radiation

@SevenSidedDie We're OK chaps, SevenSidedDie hasn't seen anything :-)

Right, I have published my masterpiece on Hawking radiation:

0

To answer this we need to talk a bit about how particles are described in quantum field theory. For every type of particle there is an associated quantum field. So for the electron there is an electron field, for the photon there is a photon field, and so on. These quantum fields occupy all of s...

I rushed it out because that Neumaier chap is trying to steal my thunder :-) I still have to add all the links and possibly tidy up some of the text. Please feel free to criticise, but ideally do it here to avoid a massive comment trail under the answer.

Impressive! What has been explained makes sense - though I am taking it on complete faith that the red shift is finite!

Thanks very much @JohnRennie

@NoahP Between you and me I have no idea why the red shift is different with and without the horizon.

The simplified versions don't say, and Hawking's maths in his paper is way beyond me!

5:11 PM

@JohnRennie Shhh don't say that, you'll awaken the hulk! (Acurious...)

@JohnRennie Fine job! (Except for what I perceived to be a misuse of the term of art "scattering amplitude" which I have corrected)

So reading Hawking's original paper

@ACuriousMind Thanks :-)

there is indeed an inward flux of negative energy!

Phew

For a moment I thought I imagined it

Also he uses BOTH EDDINGTON COORDINATES

I didn't know you could do that

How, exactly, is an "inward flux of negative energy" different from an "outward flux of positive energy"? Seems to be a simple switch in interpreting the signs, no?

5:15 PM

Well there is both!

Also at some points the negative energy is higher than the positive

You might recall that there's a theorem saying a black hole's horizon always increases in surface

One of the assumption for this is that the WEC holds

No, I mean, given a current of 1A, without further information, I can't tell whether there is 1C/s of +-charged objects moving to the right or 1C/s of --charged objects moving to the left, right?

So what make us decide a certain energy flux is inward and negative instead of outward and positive?

yesterday, by Slereah

Hulk smash!

yesterday, by Slereah

Rawr using particles in QFT is wroooong

(Reference for 18:11)

It is probably somewhat arbitrary, in the end

Might depend on the basis you choose for the field

But the energy conditions are Lorentz invariant, though

Hi all

Hello Moses

Do you want us to let your people go

2

5:23 PM

Keep them for now...Have some stuff to figure out first.

Hopefully it's Hawking radiation

It's all we talk about today

No nothing as advanced.

It's a quantum mechanics query.

I'm actually quite proud of the fact that I've only been on Physics SE about a week and my question has led to most of the top end of the community being involved, and a 200 rep bounty

Meanwhile nobody seems to move much to answer my questions :V

Could anyone confirm that my comment is correct in this post. I commented after Emilio's answer.

5:30 PM

@Moses No. You have $\psi_n(x) = \langle x\vert n \rangle$ and $\lvert n \rangle = \int \psi_n(x)\lvert x\rangle$.

I'm not sure how you came up with what you wrote there

user54412

@ACuriousMind Since you've jumped into this Hawking radiation thing, I'm curious what your responses are to my earlier comments:

user54412

2 days ago, by Chris White

also, do you mean to say that if I have a 1 g neutron star at 1.0001 Schwarzschild radii, it will sit there all day and not radiate, but if I collapse it to a black hole it will all of a sudden emit a burst of radiation?

user54412

2 days ago, by Chris White

moreover, that any real black hole will have no Hawking radiation because real black holes never quite get to having a true horizon in finite time?

user54412

2 days ago, by Chris White

This would seem to indicate a way of testing for true horizons...

@ChrisWhite Cautious yes.

5:33 PM

@ACuriousMind I'm really just starting to learn about qm. How did you come up with $\langle x | n \rangle$ equal to that?

@ChrisWhite Cautious no. Hawking does his original calculation actually for a collapsing black hole, not an eternal Schwarzschild BH. What exactly do you mean by "true horizon"? (I think it is the presence of a Killing horizon that is relevant, but I could be wrong)

I seem to recall that the effect of WEC violation for the vacuum is a generic effect of massive objects

so called "gravitational squeezing"

@Moses That's...kinda the definition of the wavefunction.

I'm not sure if this translates into hawking radiation for everything

user54412

@ACuriousMind I mean not just an apparent horizon/trapped surface, but actually the boundary of the past of null infinity or whatever it is.

5:36 PM

@ChrisWhite Ah, yes. I think having a Killing horizon (one where a time-like Killing vector vanishes) is the type of horizon required (for the scalar wavepacket to split into two when hitting it), and that seems to have nothing to do with your true horizon.

@ACuriousMind well the two are the same in the limit of infinite time.

I could be horribly wrong though because I have no idea about all this "past null infinity" business

@ACuriousMind I'm using the book Introduction to Quantum Mechanics by Griffiths. He doesn't mention the inner product of the position eigenfunction and hamiltonian eigeinfunction being equal to the stationary states, he mentions each of these things separely but never states that. This is why I'm asking.

user54412

@ACuriousMind What about an accreting black hole? Something that starts out at like a 1000 solar mass Schwarzschild and then slowly gains another million solar masses. Will its Hawking temperature be that of a $1000\ M_\odot$ or a $10^6\ M_\odot$ black hole?

Eventually a million, I suppose

5:42 PM

@ChrisWhite How is an accreting black hole different from the collapsing one Hawking considered?

Or is your question how it will radiate while it is still accreting?

For that, I don't have the slightest clue, as you can't go to the asymptotic future there

@JohnRennie I'm beginning to think relativity might be a geometric theory. Crazy, huh?

@barrycarter :-)

@JohnRennie And the Lorentz contraction and simultaneity don't really "exist", and, in fact, cancel each other out.

Well Lorentz contraction is really a rotation not a contraction.

The rotation just means the spatial component of the four vector changes.

@JohnRennie Sort of... what confused me about the word "rotation" is that it's not a rigid rotation. IE, angles and lengths change.

5:53 PM

I guess you could regard the failure of simultaneity in the same way. Put the two simultaneous events at either end of a space-like four vector and that vector will rotate under a Lorentz boost meaning the events are at different times.

Well it's not really a rotation

It's a hyperbolic rotation

@barrycarter the proper length of the four-vector doesn't change.

@JohnRennie So when you say "Earth is 4.8 light years away at time t= something", you're referring to a different Earth than when you viewed in from the stationary frame.

@JohnRennie Yes, but standard norm length does.

@Slereah Yes, the formulas come out to show that (the standard rotation matrix with cosh and sinh), but I'm not sure how to draw it on graph paper, though I'm sure it's possible.

@barrycarter if you will insist on using Euclidean geometry where it doesn't apply :-)

You can draw hyperbolic rotations

Kinda

5:54 PM

Basically

Instead of a unit circle

You have a unit hyperbola

@JohnRennie Well, I'm talking about the graphing for now.

@JohnRennie I actually considered graphing lines where ds^2 = 1, ds^2 = 2, etc, but I'm pretty sure I'd end up with that monstrosity.

@ACuriousMind
Would it then be that $c_n e^{\frac{-i E_nt}{\hbar}}$ correspond to the finite dimensional $e_j \cdot A$ as is stated in the answer?

I'm still trying to give a better answer to those special relativity questions. Next: I hope to draw the grid for constant acceleration.

@Moses They are $\langle n \vert \psi(t) \rangle$, and yes, they correspond to $e_j\cdot A$ for the basis made of the $\lvert n \rangle$:

6:01 PM

@NoahP : my pleasure Noah.

Greetings @JohnDuffield

Hi barry.

@JohnDuffield I've come up with a shiny new understanding of SR that shows LC and simultaneity are basically two wrongs that cancel each other out.

@barrycarter : sounds promising. Like I said there's less to this than meets the eye.

@JohnDuffield Yes. I'm still not convinced the CMBRF is absolute, but you can certainly use it to draw Minkowski diagrams, even if neither object is in that frame. That way, you still get traditional SR answers, but with a fundamental fixed frame.

Q: is the energy of super nova greater than that of it existed star

6:15 PM

physics.stackexchange.com/review/suggested-edits/128135 The OP looked like it was almost certain to be closed. Does this edit help? Thanls

0

how much energy is produced during a supernova? and where does the energy comes from if it previously existed star energy has an amount[if the energy of the supernova is > energy of it previous existed star]

solar-system

OP^

@NoahP a valiant effort, but it's basically a crap question and shows absolutely no effort to research it before posting it here.

@JohnRennie I thought this may be the case, I saw the rep of 1 and 2 questions asked straight after each other, just thought I might as well try rescuing it...

@ACuriousMind Can we see that $\langle x| n \rangle = \psi_n(x)$ by noting that $\int \delta(x-y)\psi_n(y)dy = \psi_n(x)$?

@Moses How would that show that? From what do you want to derive that? As I said, that the wavefunction of the state $\lvert n \rangle$ is $\langle x \vert n \rangle$ is how you define wavefunction in the formalism of abstract states, it is not a derived identity.

3075

6:31 PM

what is a good book to learn about maxwells electrodynamics as U(1) gauge theory? (introduction)

Errrr

Not sure

Can't think of a book specifically about that

3075

it can be inside a larger book.

Then most QFT books, I guess?

Is there any obvious interpretation of the complex eigenvalues of a rotation matrix?

...I should probably ask in the math chat room.

@Danu Isn't it just $\mathrm{e}^{\pm\mathrm{i}\phi}$ where $\phi$ is the rotation angle?

6:42 PM

...interpreted as...? :P

@Danu What? What is there to interpret, the angle the complex number makes on the unit circle with the real axis is exactly the angle of rotation?

@ACuriousMind I'm talking about general dimensions (not just 2)

Ohhhh

Aren't they gonna be like

Spherical harmonics

@ACuriousMind I understand what you are saying. Given the state $|\mathcal{S}(t)\rangle = |n \rangle$ and the wave function $\Psi(x,t) = \langle x | \mathcal{S(t)} \rangle$, where as we have shown in the basis made up of $| n \rangle$ we have $\Psi(x,t) = \sum_n c_n e^{-\frac{i E_n t}{\hbar}}\psi_n(x)$ it follows that $\langle x | n \rangle = \psi_n(x)$. Is this closer to the point?

user507974

6:51 PM

howdy folks