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17:01
@manshu it can if it starts with a non-zero component of velocity in the direction up the plane ...
lol...I never thought like that :p
and another thing.
does friction acts in the same direction if it is coming down the plane and if it is going up the plane?
@manshu friction always acts in the opposite direction to velocity
ohhkk,
thanks again Sir.
17:50
0
Q: Why would a black hole evaporate?

JiminionHawking radiation doesn't make sense to me, with respect to black holes getting smaller. It would seem that any particle (or anti-particle) leaving the Schwartzchild radius would have a similar anti-particle (or particle) similarly entering the black hole as well. So, over time, I don't see how...

Here we go again... (I didn't post it...) @JohnRennie @Slereah @ACuriousMind
Sometimes I feel like half the people here are to learn physics and the half are here to prove physics wrong :p
It's a common thing where people prefer common horse sense to science
There are some things where common people have Opinions
common horse sense? @Slereah
common sense
Some science rubs people the wrong way and they would prefer it being false
Ah. I'm not even sure what kind of typo or autocorrect could have lead to hr=orse...
Jesus... Horse*!
17:57
you mean to say unicorn?
Not a typo
"Horse sense" is used as a synonym for "common sense"
(idiomatic) Common sense, especially with a connotation of folk wisdom independent from, and trumping, formal education.
I'm going to have to flag that as off-topic...
the chat has no rule for that
Hmmph
@NoahP Hi Noah. See the comments I've made to that question.i would have closed it as a duplicate but I've used all my close votes.
18:06
@JohnRennie I just cast my last one one it.
I think it's actually quite a perceptive question for someone who has only seen the pop science description of particle/antiparticle pairs. However it is a duplicate.
I really, really must get around to answering Noah's question so I can use it as a reference for this sort of duplicate.
18:33
@JohnRennie : I look forward to reading your answer and commenting upon it. Meanwhile, don't forget this.
@NoahP Ha. That was me....
@@Jiminion Huh? Sorry, lost track!
@ACuriousMind I'll delete if you want. (I might be a simpleton, but I can be reasoned with..)
@NoahP Black hole evaporating... yada yada
@Jiminion No need to delete duplicates, you should be seeing a button to agree with the duplicate vote, though (which you should click if you agree it's a duplicate)
@Jiminion Ahhh okay!
18:44
(to avoid duplicate questions - SO needs a better searching mechanism. I've said this like a billion times...)
@ACuriousMind Yep, I'm working on it. I still don't know if John Duffield thinks Hawking radiation is a farce.
And they've said a million times that implementing good search algorithms is hard!
@JohnDuffield So, are you saying Hawking radiation is not real?
The best current way to search is to just use Google with site:physics.stackexchange.com
@ACuriousMind The site should have its own search.... (but I've heard that b4)
@Jiminion : I'm afraid I'm not allowed to talk physics in this chat room.
18:46
@JohnRennie Can the referee please step in here?
@JohnDuffield Hmm. That's not a particularly good or encouraging sign w.r.t. your comment on the question....
@JohnRennie On a different note, do you have a minute to review my edited paragraph on HR?
@NoahP JohnRennie mentioned particles with 'negative energy'. I've never ever heard of that before. Ever.
@Jiminion : sorry. Follow my link to what John Rennie said in his answer.
@Jiminion Well...it's kinda what you have to introduce when you want to rescue the idea of pairs of particles being generated at the event horizon and one falling in, and as with the "negative energy" solutions back in the day of the Dirac equation, it is an indication that something is wrong. In this case what is wrong is the idea that Hawking radiation actually has something to do with these pairs of particles.
It's a classic "garbage in, garbage out" situation - if you start with that botched analogy, you won't get any reliably correct statements out.
18:52
@ACuriousMind Hmm, so sort of like two wrongs making a right? ;)
@Jiminion Don't let Slereah here you say that...
@Jiminion Hmmm...in this case they just combine into a greater wrong, I think ;P
Every substantive physics question (to a neophyte) turns into Feymann bitching at a reporter about magnets....
I can explain a magnet. But not here!
18:59
I heard my name being taken in vain
@JohnRennie Sorry bout that
No problem. Did you want anything or has the debate sorted itself out?
@JohnRennie We were actually saying in a prayer, hoping you would appear....
@JohnRennie My request is totally unrelated to the now sorted issue
(now sorted possibly)
@NoahP You wanted me to read your updated article on Hawking radiation? Do you have a link for it?
19:02
It's a small paragraph - I can paste it here, or would a separate chat be more appropriate?
Paste here is fine
A correct and accurate description of the process behind Hawking radiation requires a deep knowledge of Quantum Field Theory (QFT). Therefore, an analogy given by Hawking himself will be used here, with its flaws pointed out.
Within the ergosphere of the black hole, virtual pairs of particles and anti-particles are constantly present due to vacuum fluctuations. Typically, the pair would then annihilate before this could be of any consequence, by the particles becoming real. However, as one of these two particles will be closer to the hole than the other, it will experience a greater gravita
As you wish
Thanks! :
:)*
I really don't think that invoking virtual pairs is at all necessary for Hawking radiation because it happens even without considering disconnected loops
I would delete the last paragraph. People will ask you what it means, and you'll be forced to admit you don't really know. Other than it's as good a description as I've read of the process.
@NoahP This helps, in the sense it tells me what hawking radiation is NOT. At to what it is, I might need 15 minutes or so to grok QFT.....
19:07
@JohnRennie I need to then continue into using the equations though, such as calculating luminosity through HR. How would you suggest doing this?
@NoahP : delete the whole lot until you get a satisfactory answer.
Hello all
The best you can do is just write the equations citing Hawking as the source. The derivation of the equations is too hard for the vast majority of (non-theoretical) physicists let alone A level students.
@JohnRennie : Noah is waiting for your answer.
I have a question
Let's take an example the potential difference (PD) across a resistor. if there's a current flowing, the power lost is equal to the current x PD. But in reality, the electrons inside the resistor are losing this potential energy right? If so, what is the relationship between the potential energy (or the quantised energy) of an electron inside the resistor and voltage drop?
19:11
@Tonylb1 Electrons losing energy in a resistor is like a ball bouncing down the stairs.
Yes but what is the relationship with electron energy mentioned in quantum mechanics
If the ball can fall straight down it accelerates at 9.81 m/sec^2 so it converts PE into KE and is moving fast by the time it hits the ground.
If it's falling down the stairs it keeps hitting the next step and stopping, so it converts the PE into KE then loses it as heat
@JohnRennie how do the electron know that how much energy to lose?
@Tonylb1 Ah OK. The conduction electrons in a resistor are essentially free so their energy isn't quantised.
oh I see
user54412
19:15
It's a great shame, but quantum mechanics has as little to do with quantization as any other field of physics. The theory itself makes no reference to quantization; quantization merely arises from particular boundary conditions in certain particular problems.
@JohnRennie Ok, will do. Thanks!
user116211
Emilio is in Berlin.... just saw his profile.
thanks chris for the info
@ChrisWhite You mean discretization and not quantization, right?
Because "quantization" is the technical term for the procedure that produces the corresponding quantum system for a classical system.
19:25
The ol' quantizaroo
Is there a name for a sequence of the form $a_{i+1} = d + \frac{r}{a_i}$
@NoahP : keep seeking a correct and accurate description that 1) you can understand and that 2) makes sense. If you don't get one, you might want to check this out.
OK, gotta go.
19:42
I'm trying to understand the 'big bang'. What I've gathered so far is that initially the sum of all distances in the universe was zero or the distance between any two points in the universe was zero? Is that an accurate description of the current thoughts on the big bang?
How does this look for part one of the answer? I'll complete it tomorrow as I'm off to bed now ...
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 spacetime i.e. they exist everywhere in space and everywhere in time. It’s important to realise that a quantum field is a mathematical object not a physical one - more precisely it is an operator field - however it’s common to talk as if quantum fields are real objects and I’m going to commit this si
2
@JohnRennie Looking good!!!
@NoahP I wouldn't judge until @ACuriousMind has had a chance to rip it to shreds :-)
@JohnRennie it looks correct to me. Perhaps " the vacuum state is the state for which the number operator returns zero" is not 100% accurate, because the number operator is not well defined for interacting theories. The vacuum state is defined as the state for which the Hamiltonian returns the minimum possible eigenvalue, that is, the state with the lowest energy
@JohnRennie Even from my limited experience on here, that seems like good advice...
19:49
BUT WHAT ABOUT THEORIES WITH NO LOWEST HAMILTONIAN EIGENVALUES
Nitpicking war :p
WIGHTMAN SAYS THAT IS NOT A QFT
@JohnRennie but at an introductory level, what you say is perfectly fine IMHO
Oh god. Please keep the shouting down at this kind of time, you might wake the neighbours...
3
@JohnRennie It looks fine to me
@Slereah What kind of theory is that?!
@ACuriousMind C'est une miracle. Thanks guys, I'm off now. Looking forward to an answer tomorrow!
OK, thanks all. Next step is to explain how the presence of a horizon creates a net flux of particles. But that will have to wait for tomorrow as I have to be at work in 9 hours and 7 minutes.
19:54
@john gnight :-)
@ACuriousMind $\varphi^3$
Also commuting fermion fields :p
I dont even know if $\bar \psi\not A\psi$ makes $H_\mathrm{QED}$ unbounded below :-(
@Slereah Hmmmm
@Slereah Well, the only time we have those they are designed to kill another pathology, so that doesn't count
Also I object to "The vacuum state"
You can have more than one vacuum!
@Slereah Perturbatively, THERE CAN ONLY BE ONE! ;)
And I think you need to introduce a linear term in $\phi^3$ to make it tractable perturbatively
20:00
Do you?
You can do perturbation theory around several vacua
$a\phi^3+b\phi^4$ is well defined though
Like perturbation around a domain wall
user54412
@ACuriousMind Yes, but we only retroactively made "quantization" have that meaning later on. The original Latin root certain had nothing to do with "generalizing a commuting operator theory to a noncommuting one."
@Slereah The LSZ formalism needs the presence of a single vacuum sitting at zero energy. You can certainly do something, but you're not doing standard perturbation theory if you have more than one vacuum
I'm too cool for standard perturbation theory
^me
20:03
@ChrisWhite Okay, I agree. Although the Latin root also doesn't have much to do with putting things into discrete chunks.
"From Proto-Indo-European *kuiʰent- ‎(“how much, how many”)."
Let's go back to our Kurgan roots
@JohnRennie I remember hearing and reading "time is a coordinate, not a parameter" to which I would nod but, in fact, it took a relatively long time to come to appreciate all that that simple phrase means.
20:46
@Slereah "Hello Ladies!" - Clancy Brown (The Kurgan)
21:37
I have a hunch that the path integral between two boundaries I have will converge to something like...
$$e^{\alpha N (\sum_i \varphi(t_0, x) + \varphi(t_N, x))^2 }$$
Or in the continuum limit,
$$e^{\alpha (\int d^nx (\varphi(t_a, x) + \varphi(t_b, x)))^2 }$$
Does this sound like it makes sense for a transition amplitude
This is Euclidian so you can probably throw in a i for the real one
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