The h Bar

Sha Vuklia

17:00

no hints please

John Rennie

@Kaumudi.H I suppose you could Google to try and find out what you're suffering from, although Googling illnesses isn't for the faint hearted.

Sha Vuklia

:P

AccidentalFourierTransform

it's not easy to solve this equation the first time you see it

once you know the solution, it becomes much easier

you can set $k/m=1$ for now

and try to tackle the general case after you solve the easier one first

Sha Vuklia

ohhh

but that's easy then

user228700

@JohnRennie I did do some Googling but to no avail :-/ I think it was because I ate bread that had some fungus on it but the doctor doesn't agree. Now, he didn't tell my how I got the fever in the first place but he kept saying "Body pain due to fever" -.-

Sha Vuklia

17:01

$e^{-t}$, no?

no wait

i think i need an $i$

AccidentalFourierTransform

almost, but not quite :-)

Sha Vuklia

yea

$e^{-it}$

AccidentalFourierTransform

correct, that is one solution

can you think of a second one?

John Rennie

@Kaumudi.H it doesn't seem likely that eating mouldy bread is to blame. Not unless you have some remarkably aggressive moulds in Chennai.

Sha Vuklia

I'm guessing $e^{-it}+C_1t+C_0$?

AccidentalFourierTransform

17:03

nope, that doesnt work

user228700

@JohnRennie Yeah, that's exactly what the doctor said as well :-/

Sha Vuklia

oh

AccidentalFourierTransform

try to plug that into the equation, and check if it works

Sha Vuklia

okay

oh haha yea I see

AccidentalFourierTransform

hint: the second solution looks very similar to the first one

user228700

17:04

Gah, I have absolutely no idea what brought this on! It's not even Tonsillitis this time.

Sha Vuklia

we don't differentiate one term

probably $e^{-it+C}$ then

or $e^{-i(t+c)}$

AccidentalFourierTransform

Ill spoil it: the second solution is $\mathrm e^{+it}$

ACuriousMind

@Kaumudi.H Infections are not always caused by something specific. Often, we just have bad luck when we get sick.

AccidentalFourierTransform

so you have two solutions, $\mathrm e^{+it}$ and $\mathrm e^{-it}$

Sha Vuklia

oh yea, of course

AccidentalFourierTransform

17:05

so the general solution is...?

John Rennie

@Kaumudi.H according to Google you have typhoid - oh well, it was nice knowing you :-)

Sha Vuklia

$C_1e^{it}+C_2e^{-it}$?

user228700

@ACuriousMind Ah, well :-/

Sha Vuklia

wait

AccidentalFourierTransform

according to google, you may have "network connectivity problems"

user228700

17:06

@JohnRennie -_-

AccidentalFourierTransform

@ShaVuklia yes

Sha Vuklia

you want the solution in terms of sines and cosies?

oh never mind

AccidentalFourierTransform

well yes, the solution is usually expressed in terms of sins and cos

Sha Vuklia

wow cool, this is literally the first time I did a differential equation :P as you probably know

in fact, it's the first time I've found a function

or, solved for a function

AccidentalFourierTransform

because the solution looks prettier when expressed in terms of real functions

@ShaVuklia nice ^_^

well, in terms of $sin$ and $cos$, the general solution reads $x(t)=a\cos(t)+b\sin(t)$

John Rennie

17:08

@Kaumudi.H fever does cause pain. If you're already doped up then I don't think there's much to do besides wait for it to subside. How long does it normally take?

AccidentalFourierTransform

what if we now bring back the factor $k/m$?

user228700

::Googles "Doped up"::

AccidentalFourierTransform

the solution looks very similar to the previous one

Sha Vuklia

oh god, i quite conveniently forgot about the factor

okay let me see

user228700

@JohnRennie Right, well, yes. I might go to bed soon but like I said, I slept for 5 hours in the afternoon!

user228700

17:10

It usually takes a day and a half.

John Rennie

@Kaumudi.H watch a film? Surf YouTube? Anything to distract you really.

@Kaumudi.H A day and a half! Ooooooooooouuuch :-(

Sha Vuklia

$$
e^{i\sqrt{k/m}\ t}?
$$

user228700

@JohnRennie Exactly! :'-(

user228700

@JohnRennie I don't have the energy to do any of those things :-/

AccidentalFourierTransform

@ShaVuklia yes :-)

Sha Vuklia

17:11

niiiiiiiiiice

AccidentalFourierTransform

but, as before, its prettier if you write it in terms of sins and cos :-P

Sha Vuklia

i wonder if i will continue enjoying differential equations :P I had no idea they were so much fun

AccidentalFourierTransform

so in this case you define $\omega=\sqrt{k/m}$

and write thee quation of motion as $\ddot x+\omega^2x=0$

ACuriousMind

@Kaumudi.H I didn't know it took energy to watch silly YouTube videos :P

AccidentalFourierTransform

whose solution is $x(t)=a\cos(\omega t)+b\sin(\omega t)$

i.e., the equation for a harmonic oscillator

user228700

17:13

@ACuriousMind :-) Looking at the screen is draining out whatever energy I have left.

Sha Vuklia

ah right, that makes sense, since $F=-kx$

user228700

Silly YouTube videos huh? Any recommendations? :-)

AccidentalFourierTransform

exactly

Sha Vuklia

I will only have to check why the solution in terms of $\cos$ and $\sin$ is as it is

AccidentalFourierTransform

so now you understand why sins and cos are so important in physics

Sha Vuklia

17:14

haha yes

AccidentalFourierTransform

well, you can write $\mathrm e^{ix}=\cos x+i\sin x$

Sha Vuklia

yea yea I know:P

i just have to verify it

AccidentalFourierTransform

ok then

Sha Vuklia

alright, so in our case we'd get $(C_1+C_2)\cos t+(C_1-C_2)\sin t$

ACuriousMind

@Kaumudi.H I suddenly realize I almost never watch YouTube videos :D

AccidentalFourierTransform

17:16

yep, and now you just relabel the arbitrary constants and youre done

(there might be a factor of $i$ missing somewhere, but it's not really important)

Sha Vuklia

okay cool. you know that i entirely skipped the course Vibrations and Waves because in the first of second lecture, our teacher told us we would be "guessing solutions" (when he was talking about differential equations)

i just immediately quit that course

user228700

@ACuriousMind :-P OK...

AccidentalFourierTransform

o.O

Sha Vuklia

it scared me so much, I thought it would make no sense at all

user228700

Anyhoo, I think I'll just go to bed. Bye, folks!

AccidentalFourierTransform

17:18

@Kaumudi.H bye and get well

Sha Vuklia

how can we "guess" things :P I thought no way am I going to torture myself like that. but it's pretty alright

user228700

@AccidentalFourierTransform Thanks :-)

Sha Vuklia

bye @Kaumudi!

AccidentalFourierTransform

@ShaVuklia if you already know about Taylor's theorem, next time we'll have fun solving the equations of motion by a different method

Sha Vuklia

ohhhhhh!!!!!

AccidentalFourierTransform

17:19

a very neat method

Sha Vuklia

Taylor is my favourite

John Rennie

@Kaumudi.H good night. Sleep well!

Sha Vuklia

I know everything about Taylor

I even know it it more dimensions

AccidentalFourierTransform

noice

Sha Vuklia

I literally know so much about Taylor :P

sorry, Taylor excites me :P

AccidentalFourierTransform

17:19

well, you seem to be ready to learn advanced QFT then

Sha Vuklia

hahahaha, I'll first finish my review of classical mechancis XD

AccidentalFourierTransform

we'll discuss that some other day then :-)

but yeah, Taylor is very cool

Sha Vuklia

I literally learned about scalar/vector line integrals today.

John Rennie

@ShaVuklia you have to call it an ansatz!

Sha Vuklia

because our uni hasn't taught us that yet.

so i have to do this basic vector calculus on my own

AccidentalFourierTransform

17:21

you're gonna have so much fun when you learn complex analysis :-P

Sha Vuklia

LOL

@John

I remember my teacher calling "something" an ansatz

I had no idea what he was talking about

Lol I think too then! @Accidental

Bernardo Meurer

@JaimeGallego Yo

I'm here now

Jaime Gallego

I'll set up a room

John Rennie

@BernardoMeurer: hey I bought that sound bar!

Bernardo Meurer

@JohnRennie Hey, nice!!

I want to know how it sounds :)

John Rennie

17:25

I'm hoping it will be real hi-fi quality. The reviews suggest that it is a big step up from the usual computer audio systems.

I'll know in two days :-)

Bernardo Meurer

I recently got a pair of IEMs just to be disappointed with their fitting

My ears really don't like IEMs

Slereah

I Can Time Travel!?

►

John Rennie

@BernardoMeurer I hate earplugs. I'll only ever use over the ear headphones.

Bernardo Meurer

@JohnRennie I have good over the ear headphones, but I wanted something to use when going around the city or walking to class

Slereah

that guy is a tool and I hope he gets ebola

AccidentalFourierTransform

17:33

american wine is better than french wine

ACuriousMind

@Slereah Be nice even to such people, please

Slereah

what disease should I hope he gets instead

AccidentalFourierTransform

let us hope someone kills his grandma in the past

Slereah

I'm gonna sleep with his grandmother and never call back

ACuriousMind

@Slereah I don't know or care, just keep that hope to yourself ;P

Slereah

17:35

I'm not even gonna time travel to do it

ACuriousMind

@Slereah That reminds me of a Misfits episode...

dmckee

@Phase That's all wrong. Everyone knows the supports are elephants and the stand on the back of a turtle.

AccidentalFourierTransform

17:50

so given the Einstein equations $G_{\mu\nu}=T_{\mu\nu}$, the non-relativistic limit is $-\nabla^2\phi=T_{00}$, right?

or is it $=T^{00}$? or $=T^0_0$?

or is the index position relevant at all in the non-relativistic limit?

no it isnt

thanks!

Slereah

In the linear limit you use $\eta^{\mu\nu}$ to raise indices, yes

so it doens't matter too much

The inverse metric tensor is fairly annoying to deal with since it's an infinite sum but you can show it's $g^{ab} = \eta^{ab} - \lambda h^{ab} + \mathcal O(\lambda^2)$

Where $h^{ab} = h_{cd} \eta^{ac} \eta^{bd}$

Moses

Hi all

In Sakurai he states that $\sigma = \text{tr}(\rho \text{ln} \rho)$ where $S = k\sigma$ (where $S$ is entropy). And then he states "let us maximize $\sigma$ by requiring that $\delta \sigma = 0$." Does anyone know what $\delta \sigma = 0$ means?

0celouvskyopoulo7

@AccidentalFourierTransform it should, you'll pick up a sign in -+++ from the Minkowski metric

In +--- there might not be a sign change

AccidentalFourierTransform

18:06

@0celouvskyopoulo7 you know I always use the correct sign convention ;-)

Slereah

@Moses functional derivative?

you can show the inverse metric via the binomial inverse theorem, btw

0celouvskyopoulo7

@Slereah what?

Slereah

$$ \left({\mathbf {A}}+{\mathbf {B}}\right)^{{-1}}={\mathbf {A}}^{{-1}}-{\mathbf {A}}^{{-1}}\left({\mathbf {I}}+{\mathbf {B}}{\mathbf {A}}^{{-1}}\right)^{{-1}}{\mathbf {B}}{\mathbf {A}}^{{-1}}$$

Just replace $\mathbf A$ with the Minkowski metric

Moses

@Slereah mmmmmmm that seems like a promising suggestion. Have you read Sakurai at all?

Slereah

i have not

Feynman is the place where I learned how to do the linearized GR properly :p

He doesn't just go "Eeeeh it's small anyway"

Moses

18:16

Oh okay. I'm going through Sakurai stuck a bit on this one thing where he tries to maximize $\sigma$ using Lagrange multipliers and functional derivatives apparently. I don't know anything about functional derivatives but at least the notation is consistent on wiki so it probably is that.

0celouvskyopoulo7

@Moses I believe you two are having very different conversations.

Moses

@Slereah So $sigma$ would be the functional, where it takes $\rho$ to a field of scalars.

Slereah

Oh right, I confused him with @AccidentalFourierTransform for a moment

Yes, that is what the trace does

$\textrm{Tr} : \mathfrak{B} \to \Bbb R$

With $\mathfrak B$ operators on the Hilbert space

I don't know how rigorous it would be to do a functional derivative on it but that's probably okay

ACuriousMind

@Moses He means that for $\rho + \delta \rho$ where $\delta \rho$ is an infinitesimal variation, the change in $\sigma$ will be zero. It's the same as imposing $\delta S = 0$ in the principle of least action.

Slereah

The more rigorous way would be $$\frac{\delta S}{\delta \rho} = 0$$

AccidentalFourierTransform

18:20

$\rho\log \rho$ is trace-class so its fine, right?

Slereah

Probably yeah

AccidentalFourierTransform

$\rho^2=\rho$

?

Slereah

Well it's a density matrix so one would hope

otherwise something might be wrong with the universe

although that's only for pure states

ACuriousMind

@AccidentalFourierTransform Only if it's pure

AccidentalFourierTransform

whatev

ACuriousMind

18:22

The entropy of a pure state is zero precisely because of that property

Slereah

I don't have to work with experiments so I just always pretend states are pure

0celouvskyopoulo7

sick

you pretend a lie!

AccidentalFourierTransform

how about we just kill non-pure states

Slereah

Or do I pretend

THE TRUTH

Moses

@ACuriousMind What is $\delta \rho$? $\rho$ is the density operator so is $\delta$ some infinitesimal scalar?

Slereah

18:22

No quantum state is mixed in real life

ACuriousMind

@AccidentalFourierTransform Careful.

Sha Vuklia

guys! maybe a silly question, but does anyone know the name of the piece that starts playing at 1:03? youtube.com/watch?v=IJcvRLKO_pU

Slereah

We just describe them as such because we don't know the real state

AccidentalFourierTransform

it was a Harry Potter reference!

0celouvskyopoulo7

I thought it was a Nazi reference.

ACuriousMind

18:23

@Moses No, $\delta \rho$ is one symbol. It's not $\delta$ times $\rho$ it's just $\delta \rho$.

Precisely like the variations $x\mapsto x + \delta x$ in classical mechanics, where $\delta x$ is an infinitesimal variation of $x$.

0celouvskyopoulo7

pls

at least use the jet functor formalism

Slereah

there's a real definition

It's basically like $(S[\rho + \epsilon f] - S[\rho] )/ \epsilon$

for some function $f$

ACuriousMind

Of the many things I find lacking in the physicists' description, the treatment of variations actually does not belong among them.

Slereah

Usually $\delta$ is picked as the "function"

Moses

@ACuriousMind I would consider $\delta x$ as just being an infinitesimal displacement in the direction of $x$?

ACuriousMind

18:25

@Moses Oh...you haven't seen Lagrangian mechanics/the principle of least action before?

AccidentalFourierTransform

no, its a displacement in an arbitrary direction

Moses

@ACuriousMind No I haven't unfortunately? Is that game over?

ACuriousMind

Sorry, I tend to assume that people doing QM have seen both Lagrangian and Hamiltonian formalisms in classical theoretical mechanics since that's how it works here

0celouvskyopoulo7

Slereah

Is that the famed jet bundle

0celouvskyopoulo7

18:27

That's how you do the variation.

ACuriousMind

Then I'm not actually sure how to communicate to you $\delta \rho$ to you without explaining calculus of variations.

Bernardo Meurer

Today I learned about Lagrange multipliers

I am lost as fuck

Slereah

I don't see any jets at all

ACuriousMind

@BernardoMeurer Ask @DanielSank about them, he loves to explain those

Slereah

West Side Story-Jet Song

►

0celouvskyopoulo7

18:28

@Slereah there are jets hidden in there

I have no clue what it means though

Category geometry is too damn hard

Moses

@ACuriousMind Okay no prob. Is calculus of variations covered in Lagrangian and Hamiltonian mechanics?

ACuriousMind

It's necessary for them, yes

0celouvskyopoulo7

that's kind of the whole point @Moses

ACuriousMind

And I really think it's kinda bad to have people learn QM before they have learned theoretical classical mechanics

Noether's theorem, man, Noether's theorem...

0celouvskyopoulo7

When have you actually used it?

I never see it

And why would you use it in QM?

How would you use it?

AccidentalFourierTransform

18:30

oh shut up

0celouvskyopoulo7

Please apologize

Moses

@ACuriousMind What's a good introduction to Lagrangian and Hamiltonian mechanics?

Quite funny "shut up" -> "please apologize"...

ACuriousMind

@0celouvskyopoulo7 I just think it's a very profund theorem that one should know to understand the relation between symmetries and conserved quantities. Knowing its Hamiltonian (almost trivial) version elucidates why QM symmetries must commute with the Hamiltonian, and generally Hamiltonian classical mechanics is useful to know to see where QM comes from, imo

@Moses I'm terrible at recommending resources, sorry.

0celouvskyopoulo7

Almost trivial?

It requires the concept of cohomology

Anyway

I just think it's strange you mentioned Noether for basic QM

Slereah

Yeah I dunno if it's terribly important

QFT, sure

Basic QM, i dunno

ACuriousMind

18:37

I just find it strange to do QM without having that classical mechanics background. It's not "important", sure.

2

Slereah

Classical mechanics is important, yes

You don't 100% need it, but it's simpler

Else you'll wonder "what is a hamiltonian and why are we bringing it up"

Really lagrangian + hamiltonian + wave mechanics I'd say is best before you learn any QM

Moses

@sleerah These things have been...wondered...But accepted.

AccidentalFourierTransform

I mean, canonical transformations are important

Moses

I will learn it at some stage just don't have time now.

Slereah

My 70's QM book is pretty nice for that because it brings up QM from a wave mechanics perspective

While wrong, it's a nice way to motivate things and making them a bit more intuitive

Moses

18:41

Is there Lagrange multipliers in Lagrangian and Hamiltonian mechanics?

I mean is it used often?

AccidentalFourierTransform

for basic problems, no

they are not used very often

Moses

This is the material from Sakurai I referred to

AccidentalFourierTransform

my advice is that you should move on

if you dont understand the details of a particular derivation

try to understand the result at least

what it means, and why it is important

and come back later on, or pick a different book if you really want to understand the actual proof

Moses

@AccidentalFourierTransform That's good advice. Yeah will do that instead of wasting time where I would not really understand it.

Slereah

Hm

Quantum spaghettification

18:56

Can I have someones opinion on whether this answer: physics.stackexchange.com/a/306672/70392 is right specifically the last paragraph - I am dubious.

Slereah

Since all Hilbert spaces are isometric, I'm guessing that the Fock space description will work for any theory, but that the action of field operators on it will not be the same, or even meaningful

So for an interacting theory, the Fock space won't really have anything to do with the notion of modes of the field

is that correct

AccidentalFourierTransform

@Quantumspaghettification I don't know much about this topic, sorry

ACuriousMind

@Slereah I think so, yes

Slereah

Ah good

I am slowly getting a grip on it

Quantum spaghettification

@AccidentalFourierTransform No worries, there doesn't seem to be much in the literature about this topic - its driving me crazy trying to search for a reliable explanation :) .

Slereah

19:03

It would be nice if some theory actually had a well defined Hilbert space!

sine gordon is almost well defined but I don't think I ever saw any Hilbert space explicitely built for it

sine gordon is fairly "simple" classically

it has a bunch of "particle" solutions, and "bound" solutions

And they have some kind of superposition principle

it's fairly nice

Too bad it is 100% useless

Emilio Pisanty

19:22

@Quantumspaghettification It's about right, I should think.

It's obviously not 100% accurate

Abcd

Emilio Pisanty

but then again if you're writing $E\propto 1/n^2$ you're doomed to a pretty low level of accuracy

Abcd

How to do this question?

Please ignore the scribbling there.

AccidentalFourierTransform

19:36

inb4 Kirchhoff

Moses

19:46

@AccidentalFourierTransform If you have a chance, please have a look at my post.

no_choice99

20:15

hey guys is it just me or did the latex break on PSE since a few weeks?

ACuriousMind

@no_choice99 It's just you. You can still post a bug report on meta, though

DanielSank

@BernardoMeurer Lagrange multipliers are awesome.

You have to draw the right diagram.

We've actually talked about this before.

Sanya

@ShaVuklia sorry, had to leave yesterday - but even with your thought experiment, the definition seems to be, at its core, circular to me, because the division into interactions is usually done via their forces ...

@0celouvskyopoulo7 books that you have to read as in "are forced to" or "feel compelled to"? :D

@all physics.stackexchange.com/questions/330744/… can someone explain to me in which way this is better than your usual "help me with my basic problem about newtonian mechanics"-question? (and yes, this discussion has probably taken place before, but maybe someone might offer me a new perspective ... :) )

no_choice99

thanks @ACuriousMind , strange, I can;t see any LAtex with neither my desktop nor laptop pc

@DanielSank can you use Lagrange multipliers to find the shape of a rod that transmit most heat if we restrain the surface of say its top and bottom sides to have a fixed area?

ACuriousMind

20:31

@Sanya I would not be caught arguing it is better :P

Sanya

@ACuriousMind that is reassuring (which leads me to the point of asking why one gets closed and the other is allowed :P)

ACuriousMind

@Sanya Could you propose an (at least reasonably) objective criterion by which this question should be closed?

Sanya

@ACuriousMind low effort, but no one likes that - and of course, its level of objectivity scales with the level of physics knowledge

you could also call it a "do my work" question

where "work" now includes both simple research and simple calculations/problems

0celouvskyopoulo7

@Sanya I'm compelled to read a lot of books, but my summer project is about conformal methods in GR

So I have to learn spinorial PDE

So basically forced

I wanted to read the book anyway but now that I'm expected to...well

Sanya

@0celouvskyopoulo7 takes away a bit of the excitement and sense of "free time"? :D

0celouvskyopoulo7

20:40

exactly

DanielSank

@no_choice99 sort of.

ACuriousMind

@Sanya low effort has been proposed under the guise of "insufficient effort". I personally think that's what downvoting, not closevoting, is for, but I'd have also no problem if we obtained consensus to close for insufficient effort.

@Sanya All questions are "do my work", if you look at them from the right (wrong?) angle, I don't think that's a useful criterion

DanielSank

@ACuriousMind I think we should comment, edit, and maybe down-vote in that case.

Sanya

@ACuriousMind I did have that feeling as well and I often do think that most questions do not follow the meta-rule of "do your own research first", so it might indeed not work; I think that in the end the feeling of whether a question is justified here comes down to whether we feel appropriate effort went into the asking of the question - but as you said, insufficient effort has been proposed and never gotten enough support :<

0celouvskyopoulo7

@Sanya And now I have the urge to read books I don't have to read!

Sanya

20:53

@0celouvskyopoulo7 :D :D :D well, you will just need to spend more time reading ;D

Slereah

Does the basic example of quantum measurement as entanglement work if we use some bosonic or fermionic particle of the same type?

The system pre-measurement will be $\psi^O \otimes \psi^S + \psi^S \otimes \psi^O$

Will it evolve "properly" into $$\sum_i a_i (\psi^O_i \otimes \phi_i + \phi_i \otimes \psi^O_i)$$

0

Q: What is an observer in QFT?

In non-relativistic quantum mechanics, an observer can be roughly describe as a system with wavefunction $\vert \psi^O$ which, upon interaction with another system $\vert \psi^S\rangle$ (in some way that measures the observable $\hat A$) evolves into the following system $$\vert \psi^O \rangle \...

plz halp

Or maybe I'm overthinking this and we don't actually need to decompose it into a product for things to work

I dunno

@DanielSankTheTitanic what say you

AccidentalFourierTransform

22:02

@Moses I think that the problem is that you assumed that $[H(t),H(t')]=0$, which Griffiths does not. Therefore, the expression in the book is more general than yours

$\log\log\log\log{(\text{WFT})}$

Slereah

He proved it, but

why

0celouvskyopoulo7

22:29

what even is that

macco

22:52

With the Fresnel equations I can calculate the amount of light that is reflected and the amount of light that is refracted. But is the reflected light part always specular reflection?

dmckee

@macco I think that is a fair characterization if you examine a single incoming ray.

But as soon as you allow for the interference effects of light reflecting off of a finite area the effect can be non-specular. Consider, for instance, a reflection mode diffraction grating.

macco

So where would the diffuse reflection come from? Would that be the refracted part that is scattered after penetrating the medium?

dmckee

@macco When you talk about 'specular' reflection in the context of, say, ray tracing there is a unstated assumption that you are dealing with a large enough area to be visible.

Real surfaces may be course on the scale of those patches, and the diffuse reflection is made up of the light reflecting off of microscopic patches whose orientation differs from that measured for the surface at a larger scale.

macco

Not only, most diffuse materials are diffuse because of sub-surface scattering

dmckee

For very smooth surfaces, the reflection is almost wholly specular. We call those surfaces 'mirrors'.

@macco Yeah. This is another effect.

macco

23:00

But if I have the index of refraction of a medium (light enters it from air), I can then calculate the amount of specular reflection using Fresnel and the amount of diffuse reflection would be the refracted part minus the absorption?

Smooth surface

dmckee

I don't think that the Fresnel formalism knows anything about diffuse reflection.

macco

Yes, but you calculate the amount of refracted (i.e. Transmitted) light with Fresnel, and from that amount you should be able to subtract the amount of light that is absorbed and get the diffuse reflection?

rob

23:59

Fyi, this is now: meta.stackexchange.com/questions/295285/…

191

Q: Brief outage planned for Wed, May 3, 2017 at 8pm US/Eastern (00:00 UTC) (like a fire drill for computers)

MicroVersion: Planned service degradation: All Stack Overflow/Stack Exchange sites read-only for 20 minutes on Wed, May 3, 2017 shortly after 8PM US/Eastern (midnight UTC). If you blink, you'll miss it. Short version: There will be a service degradation for up to 20 minutes shortly after 8PM US...

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