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Physics Meta

01:19

0

Q: What is the function in the Physics StackExchange website picture?

I'm sorry if this isn't a question fit for this site, but what exactly is the function that's plotted in the website logo/picture? It looks kind of like a simple decaying exponential multiplied by a sinusoid, but I feel like it may be something really interesting that the creator put for a good s...

Physics Meta

05:09

0

Q: What is , "You haven't cast a vote to undo"

I want to upvote an answer, I previously down voted. It shows "You haven't cast a vote to undo". . How do I cast one?

support

John Rennie

05:28

@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.

yuggib

07:35

@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.

Slereah

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..

John Rennie

08:24

@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.

David Z

08:38

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

4

Q: what is the Favicon supposed to be?

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?

Danu

@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

David Z

Is it an HTTPS problem?

Danu

Yes

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

David Z

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

You don't know what HTTPS means?

Danu

Not really, no.

I know vaguely

It's some kind of encryption thing

David Z

08:43

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

Danu

For reference:

0

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

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

David Z

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

Danu

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

David Z

Maybe you weren't accessing meta sites by HTTPS

John Rennie

@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.

David Z

08:45

@Danu let me just post an answer

John Rennie

@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.

David Z

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

Danu

@JohnRennie LOL

John Rennie

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

David Z

@JohnRennie in my onebox?

Danu

08:47

I love the tag "site enhancement" too :D

John Rennie

"My onebox" ?

David Z

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

John Rennie

@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

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

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

David Z

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

Danu

09:06

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.

Danu

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

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

Slereah

11:01

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

ACuriousMind

12:46

@Danu lol

Hitesh Pathak

13:35

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

David Z

@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

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

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.

Hitesh Pathak

@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.

David Z

That was heat

Jiminion

13:50

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?

David Z

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

Hitesh Pathak

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

David Z

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

dmckee

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.

Hitesh Pathak

14:03

5

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

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

David Z

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

dmckee

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

Hitesh Pathak

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

dmckee

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".

Hitesh Pathak

14:23

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

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

dmckee

@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.

dmckee

14:38

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.

John Rennie

Is this a duplicate:

0

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

(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!

ACuriousMind

@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.

dmckee

@ACuriousMind I'm on the case.

John Rennie

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

Jiminion

14:51

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

ACuriousMind

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

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 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?

Jiminion

Is there any proof for micro black holes?

ACuriousMind

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

dmckee

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

David Z

14:56

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

ACuriousMind

@Jiminion What do you mean by "proof"?

John Rennie

@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

Jiminion

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

John Rennie

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 :-)

dmckee

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

ACuriousMind

14:59

@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

dmckee

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.

John Rennie

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

ACuriousMind

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

Slereah

15:03

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?

Jiminion

@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.)

Slereah

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

dmckee

@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.

John Rennie

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

15:06

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

Jiminion

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

ACuriousMind

@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.

John Rennie

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?

ACuriousMind

@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

John Rennie

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).

David Z

15:25

@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.)

John Rennie

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

David Z

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 ;-)

Noah P

@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.

John Rennie

@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 :-)

Noah P

Brill! Thanks :) Would that be Acuriousmind?

John Rennie

15:31

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

Noah P

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

Slereah

Hawking radiation is a complicated thing

ACuriousMind

@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

John Rennie

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.

ACuriousMind

Thanks for that, I guess

Noah P

15:44

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

John Rennie

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

Slereah

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)

Noah P

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

ACuriousMind

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

Slereah

Hawking gotta get paid

ACuriousMind

15:47

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.

Noah P

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

Slereah

Welcome to the internet

ACuriousMind

@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)

Slereah

what paper are you reading btw

John Rennie

@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

ACuriousMind

15:55

@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 ;)

Slereah

thx

ACuriousMind

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

John Rennie

@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.

Slereah

@JohnRennie never forget

People hate rewriting proofs

John Rennie

@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.

ACuriousMind

16:01

@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

Slereah

Depends

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

but the cutting edge people generally use AQFT stuff

yuggib

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

^

yuggib

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

Slereah

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!

yuggib

16:05

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

ACuriousMind

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

Slereah

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

John Duffield

@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

16:56

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

John Rennie

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

Right, I have published my masterpiece on Hawking radiation:

0

A: An explanation of Hawking Radiation

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.

Noah P

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

John Rennie

@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!

Noah P

17:11

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

ACuriousMind

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

Slereah

So reading Hawking's original paper

John Rennie

@ACuriousMind Thanks :-)

Slereah

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

ACuriousMind

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?

Slereah

17:15

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

ACuriousMind

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?

Noah P

yesterday, by Slereah

Hulk smash!

yesterday, by Slereah

Rawr using particles in QFT is wroooong

(Reference for 18:11)

Slereah

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

Moses

Hi all

Slereah

Hello Moses

Do you want us to let your people go

2

Moses

17:23

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

Slereah

Hopefully it's Hawking radiation

It's all we talk about today

Moses

No nothing as advanced.

It's a quantum mechanics query.

Noah P

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

Slereah

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

Moses

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

ACuriousMind

17:30

@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...

ACuriousMind

@ChrisWhite Cautious yes.

Moses

17:33

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

ACuriousMind

@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)

Slereah

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

so called "gravitational squeezing"

ACuriousMind

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

Slereah

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.

ACuriousMind

17:36

@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.

John Rennie

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

ACuriousMind

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

Moses

@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?

Slereah

Eventually a million, I suppose

ACuriousMind

17:42

@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

barrycarter

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

John Rennie

@barrycarter :-)

barrycarter

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

John Rennie

Well Lorentz contraction is really a rotation not a contraction.

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

barrycarter

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

John Rennie

17:53

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.

Slereah

Well it's not really a rotation

It's a hyperbolic rotation

John Rennie

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

barrycarter

@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.

John Rennie

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

Slereah

You can draw hyperbolic rotations

Kinda

barrycarter

17:54

Slereah

Basically

Instead of a unit circle

You have a unit hyperbola

barrycarter

@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.

Moses

@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?

barrycarter

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

ACuriousMind

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

John Duffield

18:01

@NoahP : my pleasure Noah.

barrycarter

Greetings @JohnDuffield

John Duffield

Hi barry.

barrycarter

@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.

John Duffield

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

barrycarter

@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.

Noah P

18:15

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

0

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

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^

John Rennie

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

Noah P

@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...

Moses

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

ACuriousMind

@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

18:31

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

Slereah

Errrr

Not sure

Can't think of a book specifically about that

3075

it can be inside a larger book.

Slereah

Then most QFT books, I guess?

Danu

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

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

ACuriousMind

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

Danu

18:42

...interpreted as...? :P

ACuriousMind

@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?

Danu

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

ACuriousMind

Ohhhh

Slereah

Aren't they gonna be like

Spherical harmonics

Moses

@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

18:51

howdy folks

ACuriousMind

@Moses I'm not sure what you are saying or what point there is to make. You can't start with $\lvert S(t)\rangle = \lvert n \rangle$ to begin with because the l.h.s. depends on $t$ while the r.h.s doesn't. What you say after that also is not really clear to me.

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