The h Bar

Anonymous

17:00

@0celo7 Alright, thanks. I'm reading a bit more about this thing :)

Semiclassical

so the change that comes in can do so in a lot of different ways, but once it goes through a coin machine one just has some number of each kind of coin

Curio

So it emits it in a different way

Semiclassical

it's not that it emits it in a different way. it's still EM radiation

it's that, once the energy has been stored in the system, it only comes out in specific packets

(if anyone wants to chime in on me not explaining this well, please do)

Curio

And this packets are the distribution of the frequencies

Semiclassical

eh, these packets are the photons themselves

0celo7

17:02

@Blue study the differentiability of x^2 sin(1/x). Compute the derivative. Is it continuous?

What is the “tangent line” at zero?

Semiclassical

the distribution is how often photons of one quantized energy are emitted versus photons of a different quantized energy

0celo7

Do this and you’ll understand what I’m saying.

Curio

Yeah I meant them. The packets and the photons and each one has a frequency and the graph changes based on them

Semiclassical

something like that

what gets a bit tricky here is that the blackbody distribution requires a few different parts of physics

0celo7

I think I have to wash my water bottle. I feel slightly nauseous after drinking from it.

Semiclassical

17:04

on the one hand it involves EM radiation, so photons of a given frequency carrying energy

Curio

So the black body absorbs all the energy but it emits it in packets

Which are the initial energy if I sum them

Semiclassical

I'd say that's closer to it, yeah. (I hesitate to call it completely right, because I'm not remembering the details quite right)

@Curio in the sense that the amount of energy it emits via radiation in 1 second is the same as the amount of energy it absorbs in that same second

it's about the rates of energy absorption and energy emission being balanced

getting back to a moment to what I was saying: On one hand, blackbody radiation clearly involves EM waves and therefore the speed of light

0celo7

@Blue @BalarkaSen India is on the top of r/WTF!!

Semiclassical

but it also invokes some thermodynamics/statistical mechanics assumptions

I don't really want to get into that. But it's why Boltzmann's constant k showed up in an earlier statement.

Curio

Is the spectrum of the black body complete?

Emilio Pisanty

17:09

@Curio what do you mean by complete?

Semiclassical

And finally it does involve some assumptions vis a vis quantum theory, namely that the blackbody radiator can be modeled as a set of oscillators with quantized energy levels

Emilio Pisanty

as in, does it emit at every real frequency?

Anonymous

@0celo7 What ?

Curio

All the visible radiations

Semiclassical

I'm not really understanding the question either.

0celo7

17:10

@Blue A bunch of Indians going 70+ in thick smog and creating a huge traffic accident

And then 15 people get out of a 7 person car

It’s amazing

Anonymous

LOL

Anonymous

That's typical

Anonymous

You should see the trains here

Curio

I mean, like the rainbow

Semiclassical

I feel like your question is more about EM radiation than the blackbody

Emilio Pisanty

17:11

@Curio then yes

0celo7

It’s hilarious. Visibility is 10 feet and people don’t slow down.

Semiclassical

If you're asking "does every color we see in nature correspond to light of some frequency?" then yes

Anonymous

@0celo7 Okay, so that is differentiable at $0$ but the derivative is not continuous

Curio

I wasn't asking this, but, does a blackbody emit a continue spectrum

0celo7

Yah. So the tangent line at 0 doesn’t make much sense.

Semiclassical

17:13

That's a bit more subtle, I think

on the one hand, the individual photons emitted from a blackbody should indeed be quantized

however, when you're looking at radiation intensity

Curio

The black body absorbs all the radiations but it emits the total energy in packets. Are these ones correspond to all the frequencies?

Semiclassical

you're not looking at just one photon

you're looking at a whole bunch of photons which are striking your detector at nearly the same time

going back to the change analogy: if someone dumps a whole bunch of change in my lap, there's a lot of possible totals for that

and when you're talking radiation intensity, you're talking about a looooot of photons

Emilio Pisanty

@Semiclassical photons don't have to have a well-defined frequency

Semiclassical

you're right, and I wondered if I wasn't saying that carefully enough

Curio

x photons of a frequency, y photons of b frequency

Emilio Pisanty

17:20

@Curio no, that's not quite right, though

Semiclassical

the quantization is really for the 'oscillators' inside the blackbody

John Rennie

@EmilioPisanty hmm, this takes us back to the what is a particle question. If we take the photon to be the limiting case of a Fock state then by definition it has a precise frequency.

Semiclassical

lol

prooobably should avoid invoking second quantization language in this context :P

Emilio Pisanty

@JohnRennie that's an extremely narrow view of what single-particle states are

Curio

Oscillator?

Emilio Pisanty

17:21

In that view, the superposition of "one particle over here" plus "one particle over there" no longer has single-particle character

John Rennie

@EmilioPisanty it's one end of a spectrum, and the existence of the issue is why I think a canonical question on the topic would be very interesting.

Emilio Pisanty

@JohnRennie maybe

but I would argue that that view is out-and-out wrong

Curio

@EmilioPisanty but we've said that these packets contain different energy so the photons should have different frequencies.

Semiclassical

@Curio fyi, probably should not get too distracted by the other conversation going on here. it's relevant, but higher level than what you're talking about and more likely to be confusing

Anonymous

@0celo7 Hmm. Take $f(x,y)=x^2\sin(1/x)$ and $f(0,0)=0$. Even though the derivative is not continuous the tangent plane $z=0$ is still the best linear approximation as the limit $\lim_{(h,k)\to(0,0)}\frac{f(h,k)-f(0,0)-\mathbf{A}(h,k)}{\sqrt{h^2+k^2}}$ evaluates to $0$. Also, similarly it can be shown that the line $y=0$ is the best linear approximation for $x^2\sin(1/x)$. Could you give me an example where the limit turns out to be non-zero but still the function is differentiable?

Emilio Pisanty

17:23

a superposition of two Fock states with single-particle character (but different frequencies) should still be a single-particle state

otherwise, there are no physically-available states with well-defined particle number

Curio

@Semiclassical we've said that these packets contain different energy so the photons should have different frequencies. Is it right?

Semiclassical

@curio in particular, anything regarding "Fock states" should be taken as outside of what you're going to learn about any time soon

Emilio Pisanty

@Curio continuous spectra are just more tricky to handle

even without quantum mechanics

Semiclassical

here's the picture I'd give, though it's probably not entirely right either

Emilio Pisanty

you can't just say "oh, I have five joules at 500nm"

Semiclassical

17:25

the blackbody absorbs radiation at any frequency and stores it internally

Emilio Pisanty

@Semiclassical @curio I'll second this statement

Semiclassical

however, it stores it in the following way: n1 number of photons at energy level E1, n2 number of photons at energy level E2, and so forth

Curio

And then it emits E1, E2,...

Semiclassical

I should probably not insist on saying that it's n1 photons stored at a given energy level; that's a bit too narrow

but the blackbody has some way of storing the energy in such a way that there's only integer numbers of various energies allowed

so it contains some number of 'energy quanta'

(think of it like a bank choosing to store what it knows about its revenue electronically---it's the same info, just transcribed in a certain way)

the blackbody can then give up one unit of that stored energy as radiation

Emilio Pisanty

@Semiclassical that's true; the mode at frequency $f$ can only have energy $E=nhf$ equal to an integer multiple of $hf$. But it is allowed to have a probability distribution over $n$

Semiclassical

17:29

Right. That's where the statistical mechanics comes in, and more precisely the role of temperature

buuuut that's a harder conversation

Emilio Pisanty

@Semiclassical I'm not quite sure how one can talk about blackbody spectra without talking about thermodynamics, but OK.

Curio

Is all the stored energy emitted?

Semiclassical

@EmilioPisanty for the purposes of a derivation of the Planck distribution, I entirely agree.

for an elementary discussion of how the Planck distribution arises, though...eh

There's a lot of physics which converges at the same time in the blackbody problem

quantum, thermo, and EM

@Curio if it's going to remain in thermal equillibrium, the blackbody had better give up energy at the same rate as it absorbs it

talking more specifically about the 'timing' of that process is not simple to do, I think

and it's really not necessary, since the radiation intensity we observe is going to be measured over some particular range of time

Curio

So it doesn't emit all the energy it absorbs

Semiclassical

"if it's going to remain in thermal equillibrium, it had better give up energy at the same rate as it absorbs it"

in that particular sense, it does emit all the energy it absorbs.

Curio

17:36

I meant, not all in one time

Semiclassical

ah

that's pretty much right, then. it emits as much energy in one second as it absorbed in that same second

that's a statistical statement, though. in one second it emits and absorbs a lot of photons

(I'm not actually sure how many photons a blackbody would radiate per second.)

Curio

Can it emit all the frequencies?

Semiclassical

As individual photons, I'd say no.

Curio

I mean, radiations with all the frequencies

Semiclassical

It can emit photons of energy $h\nu$, $2h\nu$, etc.

however, we're again talking about a huge number of photons

0celo7

17:39

@Blue what is $A(h,k)$?

Curio

So not all the frequencies because the energy is quantized

Semiclassical

in which case it's a bit like asking whether, if we've got a trillion pennies, we can represent any amount of money between 0 and 10 billiion

Emilio Pisanty

@Semiclassical that's incorrect

Semiclassical

is it? I was worried I was wrong on that.

Emilio Pisanty

@Curio yes, it can emit radiation at all positive frequencies

Anonymous

17:40

@0celo7 $A:\Bbb R^2\to \Bbb R$, given by $A(h,k) = [\frac{\partial f}{\partial x} \space \frac{\partial f}{\partial y}][h \space k]^T$

Emilio Pisanty

@Semiclassical it can emit energies $h\nu$, $2h\nu$, etc., at each frequency $\nu$.

Semiclassical

durrrr

yes

Curio

So the spectrum is continue

Like stars' spectrum

Semiclassical

hmm

Anonymous

For that choice of $A$ the limit can never be non-zero. It can either be $0$ or not exist at all (think so)

Emilio Pisanty

17:42

@Curio yes

Semiclassical

@EmilioPisanty something doesn't sound right here, come to think of it. emit those energies as what?

Phase

is the dimension of the kernel just the amount of orthonormal basis vectors required to describe all the vectors in the preimage set that map to the zero vector?

i.e. if the Kernel is just zero, then it's zero dimensional?

Emilio Pisanty

@Semiclassical as electromagnetic radiation at frequency $\nu$

Semiclassical

hmm

I guess you're not wanting me to use the word photon for that

Emilio Pisanty

the different outcomes $0h\nu$, $h\nu$, $2h\nu$, etc, occur with different probabilities

you're essentially looking at the counting statistics of that mode

Curio

17:44

So does it absorb radiations of all frequencies?

Emilio Pisanty

(once it's been monochromatized)

@Curio yes

Balarka Sen

@0celo7 OK, I will

Phase

If it didn't absorb radiation of all frequencies it wouldn't be a black body :^)

Semiclassical

^

Balarka Sen

@0celo7 lmaooo

Phase

17:44

It might be a Pea-green body, maybe clementine-orange

Balarka Sen

#proudtobeanIndian

Semiclassical

that it absorbs at all frequencies is a definition

Balarka Sen

@Phase So you figured out dual space schtick?

Curio

@Semiclassical what did you mean so?

Semiclassical

pretty much what I said.

it can only store the radiation energy it absorbs in certain quantized energy levels

Phase

17:47

@BalarkaSen think so

I came to a conclusion about inner products

that seemed to be right

Semiclassical

@EmilioPisanty I feel like we're using terminology differently (which means I'm probably remembering stuff wrong)

Curio

And the radiations of frequency which it can't store go away immediately

Emilio Pisanty

@Curio no

Curio

Doh

Emilio Pisanty

for it to be a blackbody it needs to be able to interact with all positive frequencies

there are no "frequencies which it can't store"

Curio

17:49

So it stores all

0celo7

@BalarkaSen it seems to have been deleted

Balarka Sen

aw

Phase

@BalarkaSen is the dimension of the kernel just the amount of orthonormal basis vectors required to describe all the vectors in the preimage set that map to the zero vector?
i.e. if the Kernel is just zero, then it's zero dimensional?

Emilio Pisanty

@Curio it absorbs, stores and emits at all positive frequencies

if it is an ideal blackbody

real black bodies are only approximations so they might have gaps

Curio

Are there negative ones?

Semiclassical

17:51

not for these purposes

Phase

No

0celo7

@Blue I don't know why you're computing that thing. If it has to do with linear approximations, I'd just recommend ignoring them. They never show up in practice.

At least that expression never shows up.

Phase

in this case frequency is just a scalar that measures the periodicity of the emitted light

rather than any direction

Balarka Sen

@0celo7 Uh, this is wrong dude.

Phase

So it's fine in the positive numbers

0celo7

17:52

@BalarkaSen ??

the gradient would be zero if the derivative is continuous

at least that's what he said

Semiclassical

too many conversation threads at once

Curio

So it continually absorbs and emits

Anonymous

@0celo7 Which thing?

0celo7

@BalarkaSen youtube.com/watch?v=qjzdolHjGpM

there's a video of the actual crash but I can't find it

Balarka Sen

$f$ is differentiable at $a$ iff there is a linear transformation $A$ such that $\lim_{h \to 0} (f(a + h) - f(a) - Ah)/\|h\|$ is zero.

0celo7

17:53

I thought he was doing a difference quotient.

Balarka Sen

Ah, no, that's not what he's doing

0celo7

You can't do difference quotients for non-C^1 things

Balarka Sen

He's just writing the defn of derivative

aka what i said

0celo7

Yeah but he's doing it with partial derivatives.

Which is wrong in the non-C^1 case

Phase

Do all matrices with Linearly independent columns have nullity zero, and thus are surjective?

Curio

17:55

Let's consider the graph again. If it is at 5000K it stores x energy and it emits y and it absorbs y.

Balarka Sen

@0celo7 I agree.

Anonymous

Uh. I would just use the limit definition of $\partial f_x$

Anonymous

Would that be wrong?

Balarka Sen

Have you checked if your function has continuous partial derivatives?

Curio

If it is at 10000K it stores 2x energy and it emits 2y and absorbs 2y

Balarka Sen

17:57

You need to check that before plugging in $A = [f_x, f_y]$

0celo7

@Blue You can't calculate a Frechet derivative using partials unless they are continuous functions.

You're getting a strange result because you're assuming it's continuous, I think.

Balarka Sen

@0celo7 I'll have a look after I finish dinner

Emilio Pisanty

@Curio no

Curio

!?

Balarka Sen

@Phase Dimension of kernel is equal to the dimension of the null space is equal to the dimension of a basis of the null space, yes.

Anonymous

17:58

@BalarkaSen Oh, and what's the reason for that? I forgot :/

Emilio Pisanty

@Curio why would you think that?

for one, is that per-frequency, or integrated over the spectrum?

Balarka Sen

Doesn't need to be orthonormal (which doesn't make sense if you don't have a inner product beforehand)

@Phase Wait, no.

0celo7

@Blue Because it's not true otherwise?

Phase

I'm confused by this whole notion of a basis without an inner product

0celo7

You need continuity for the estimates to work.

Phase

17:59

How would you even know what basis to choose

Anonymous

@0celo7 Oh, got it

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