The h Bar

3:05 PM

@EmilioPisanty Why not Strong Nuclear Bond?

Well, because the energy well of any strong force related phenomenon is so deep that it takes huge amount of energy to break it. There are no materials that binds as strongly as the strong force yet inherently highly unstable and spontanenous to release all that stored energy in a rapid moment. Even if proton decay is true, it is not explosive

Meanwhile, if you somehow cramp a lot of protons in a space and then suddenly remove the confines, then...

ACuriousMind

3:43 PM

@heather @mod doesn't actually do anything, I've seen your post by accident. If you think there's something we should look into, a custom mod flag is the way to go. If it's related to a user and not to a specific post, raise a flag either on an instance of the behaviour or on any post of the user.

0celo7

@dmckee Are you around?

My advisor did something strange

AccidentalFourierTransform

4:04 PM

DHMO

3 hours ago, by DHMO

Is this true? "0 V doesn't mean no charge; no charge doesn't mean 0 V"

@JohnRennie ^

ACuriousMind

@AccidentalFourierTransform The intended meaning is funny, but do you think that "retarded" is an appropriate word to use here?

Slereah

@AccidentalFourierTransform ARE YOU LOOKING FOR A FIGHT

John Rennie

@DHMO Let me have a look ...

Slereah

-+++ is best signature

DHMO

4:06 PM

@AccidentalFourierTransform I don't understand...

Slereah

@ACuriousMind I use it all the time!

It's pretty important for AQFT!

Existence of retarded and advanced solution

John Rennie

@DHMO the voltage is a potential energy. Potential energy has what we call a global gauge symmetry i.e. we can choose its zero point to be anywhere we want.

DHMO

@JohnRennie so is that true?

and is the earth neutral?

AccidentalFourierTransform

@ACuriousMind can I post the same image, but with the dog saying "The intended meaning is funny, but..."?

John Rennie

@DHMO the statement is basically meaningless. It's like asking if apples means no oranges or oranges mean no apples.

DHMO

4:08 PM

@JohnRennie thanks

John Rennie

If charges are present they generally create a potential difference i.e. the voltage changes with distance from them.

But the statement that the potential is zero is just a choice of zero point.

skullpetrol

chat.stackexchange.com/users/-534/xkcd

What is that^?

0celo7

@AccidentalFourierTransform Do you want a fight?

Name one reason why $+---$ is better

Loong

@skullpetrol A feed bot.

skullpetrol

Thanks @Loong

Yashas Samaga

4:15 PM

Isn't this a personal theory question?

-4

Q: Does $ \Delta p \Delta x = \Delta E\Delta t$?

Two formulations of the Uncertainty Principle are $$ \Delta p \Delta x \geq \hbar/2$$ and $$ \Delta E \Delta t\geq \hbar/2$$ Does $ \Delta p \Delta x = \Delta E\Delta t$? I know the Uncertainty Principle is an inequality and not an identity. But I think it is possible that they are equal or onl...

uncertainty-principle

Emilio Pisanty

@YashasSamaga nah, it's an "I don't understand QM" question

Yashas Samaga

lol

0celo7

@ACuriousMind

What is not clear? It does not get much clearer than this. — hooch 22 mins ago

lol

AccidentalFourierTransform

@EmilioPisanty more like "I dont understand basic math"

$a\ge b$ and $b\ge c$ implies $a=b$?

Emilio Pisanty

@AccidentalFourierTransform that too

0celo7

4:18 PM

The two 2H and the one 0.5H inductors are shorted, right?

rob

@0celo7 Half right.

0celo7

Half right?

Seems like the diagonal line is connecting the two nodes directly

rob

@0celo7 Why is the 0.5H inductor shorted?

ghosts_in_the_code

There's some error in @MarkFischler 's answer at physics.stackexchange.com/q/… but I'm not able to rectify it.
If we take the general method of assuming a and b, I'm getting a=b=0 which is weird

Can someone please help?

0celo7

@rob The potential drop across the inductor is the same as across the wire

Because they begin and end at the same node

Emilio Pisanty

4:22 PM

@rob why isn't it?

AccidentalFourierTransform

@ghosts_in_the_code broken link

ghosts_in_the_code

1

Q: Does this system perform simple harmonic motion?

Does the above system perform SHM when displaced slightly? Mathematically I couldn't prove that it will, though generally such spring systems do perform SHM. Is this just an exception? P.S. You can ignore gravity. P.S.2 This is not a homework question; I was just solving some problems on SH...

homework-and-exercises newtonian-mechanics harmonic-oscillator spring

Emilio Pisanty

@ghosts_in_the_code actually, not a link at all

ghosts_in_the_code

Sorry

rob

@EmilioPisanty Oh, you're right, and I'm a dope. Sorry.

0celo7

4:23 PM

He's right :(

I said it first

AccidentalFourierTransform

you can find the answer in C:\Users\AFT\Desktop\random_stuff

Emilio Pisanty

@0celo7 your textbook is trolling you

serves you right, probably

0celo7

@EmilioPisanty what?

ghosts_in_the_code

So no help then?

AccidentalFourierTransform

I dont know anything about Newtonian mechanics, sorry

ghosts_in_the_code

4:26 PM

Oh ok no problem

Emilio Pisanty

@0celo7 you kinda troll people a bit on occasion =P

0celo7

@AccidentalFourierTransform you still haven't explained why liking $-+++$ makes one mentally retarded

Emilio Pisanty

@ghosts_in_the_code "The above system" isn't particularly well defined.

ghosts_in_the_code

Why?

0celo7

@EmilioPisanty you'll be hearing from my lawyers

ghosts_in_the_code

4:30 PM

Only the initial phase is not given

Emilio Pisanty

If that is two equal masses linked to each other and to the wall by two equal-constant springs, you need to say so.

the diagram is awful

as is your handwriting

ghosts_in_the_code

Oh ok

Emilio Pisanty

no offense meant

AccidentalFourierTransform

@0celo7 bc -+++ is stupid

it just is

ghosts_in_the_code

Should I edit?

Or will it not matter?

Emilio Pisanty

4:31 PM

@ghosts_in_the_code not sure

ghosts_in_the_code

And btw do u know the answer?

Emilio Pisanty

Also, what exactly do you mean by "perform SHM"?

there's two degrees of freedom

ghosts_in_the_code

The given answer seems a bit wrong

I haven't learnt about degrees of freedom

Emilio Pisanty

@ghosts_in_the_code you've got two independent variables

Yashas Samaga

0

Q: Difference between single and double slit diffraction pattern

I'm trying to understand the diffraction pattern for a single slit compared to a double slit. I understand that the wavelets passing through the slit diffract and interfere with each other producing a pattern. Why is the central maximum double the width, and why do the intensity of the maximum's ...

visible-light

Emilio Pisanty

4:32 PM

i.e. the positions of the two blocks

Yashas Samaga

can the answer in that question be called as an answer?

ghosts_in_the_code

Ok if we have to just write the position time relation for each body

Nvm if it is called SHM or not

Emilio Pisanty

@YashasSamaga not really. Flag as NAA and move on.

0celo7

@AccidentalFourierTransform explain

Emilio Pisanty

@ghosts_in_the_code what?

How is that not the core of the question?

0celo7

4:34 PM

how (redacted) do you have to be to make spatial directions negative??

ghosts_in_the_code

Basically I want to just know if there is a simple calculus way to understand the situation

Don't actually need the definition of shm

Btw what is the definition of SHM then?

AccidentalFourierTransform

@0celo7 in -+++ all formulas look silly

Yashas Samaga

1

Q: How well do pressure measurements translate to temperatures?

At low temperatures we measured the changing pressure in a cryostat and converted these to temperatures using the function given by Donnelly (p1267). But it seems that this is only accurate if cooling is consistent. As in, if you're cooling helium by pumping on it, and you use the pressure to fin...

experimental-physics low-temperature-physics

Whoever has to answer that will have to read the research paper, so is that Q allowed?

Emilio Pisanty

@ghosts_in_the_code there isn't a question yet

@ghosts_in_the_code there's multiple possible definitions

the fact that you haven't specified which one you are working with is one of the key flaws in the question.

ghosts_in_the_code

Oh ok thanks

Btw the method he used to find a position time relation was interesting. But does it really work for that particular situation?

AccidentalFourierTransform

4:39 PM

but yeah, all systems perform harmonic motion if there is a stable configuration and the matrix of second derivatives is non-negative

Emilio Pisanty

@ghosts_in_the_code who is "he"?

@YashasSamaga looks like a reasonable question to me.

ghosts_in_the_code

@MarkFischler

The answer given

Yashas Samaga

@EmilioPisanty he just added the page number

Emilio Pisanty

@ghosts_in_the_code looks correct to me. Why?

ghosts_in_the_code

Is u = x2-x1 or x1-x2?

Emilio Pisanty

4:42 PM

@ghosts_in_the_code don't have time to do the details on that math

ghosts_in_the_code

Oh ok

I'm getting a=b=0

Which makes the method fail

0celo7

@AccidentalFourierTransform What?

ghosts_in_the_code

But idk if I'm correct

0celo7

what formulae are you looking at

Slereah

@AccidentalFourierTransform how are you supposed to do a Wick rotation in (+---)

user123733

4:44 PM

how can we find maximum KE in a standing wave on a string

Physics Meta

0

Q: What happened to this question's commutative diagram?

An old answer of mine was written, in March 2014, with a commutative diagram, $$ \require{AMScd} \begin{CD} V @>{T}>> W\\ @V{\sigma_V}VV @VV{\sigma_W}V \\ V^* @<<{T^*}< W^* \end{CD} $$ which looks like this and which on meta renders correctly (at present) as $$ \require{AMScd} \begin{CD} V @>

bug mathjax

AccidentalFourierTransform

@0celo7 one of the simplest formulas, $p^2=m^2$, in -+++ becomes $p^2=-m^2$. That looks utterly ugly to anyone with good taste

@Slereah how is the signature relevant to the rotation?

0celo7

@AccidentalFourierTransform you mean $p^2+m^2=0$?

AccidentalFourierTransform

ha

Slereah

the point of a Wick rotation in path integrals is to make the term negative

AccidentalFourierTransform

4:46 PM

no, $p^2=m^2$

Slereah

if you do a wick rotation of time in a path integral, you end up with a positive term

Which is divergent

0celo7

@EmilioPisanty I read the post of yours linked in that meta post

AccidentalFourierTransform

@Slereah can I say that Wick rotations, at the level of the PI, are idiotic too? Or you people will hate me for that?

0celo7

@EmilioPisanty It's interesting to note that we analysts call the contravariant form of $df$ the gradient but for some reason physicists insist on calling $df$ the gradient.

Slereah

Well how else do you define a path integral, then

There's no measure for Lorentzian path integrals

AccidentalFourierTransform

4:49 PM

@Slereah I dont. Path integrals are idiotic too.

0celo7

@AccidentalFourierTransform ah, so you're trolling

Slereah

Ban @AccidentalFourierTransform

3

AccidentalFourierTransform

No I'm not!

0celo7

Do you insist on doing canonical quantization, i.e. without path integrals?

Emilio Pisanty

@0celo7 I don't see the difference there

oh, wait

"of" $df$.

what are you defining as "the contravariant form of $df$"?

0celo7

4:50 PM

@EmilioPisanty $df$ is just the "differential" or "derivative"

AccidentalFourierTransform

@0celo7 are you implying that CQ is worse than PI?

0celo7

@EmilioPisanty $g^{ij}\partial_j f$

ACuriousMind

@AccidentalFourierTransform The Osterwalder-Schrader reconstruction theorem ensures that you can analytically continue the Euclidean results to the Minkowski results, so Wick rotation is not idiotic...it just needs considerably more work to be justified than is usually done.

Emilio Pisanty

@0celo7 well, physicists kinda do that too

0celo7

"To make things easier to understand to undergraduates who are fresh out of 1D calculus, this linear map is most often 'dressed up' as the corresponding vector, which is uniquely obtainable through the Euclidean structure, and whose action must therefore go back through that Euclidean structure to get to the original $df$."

Sir Cumference

4:52 PM

So many people in the chat

skullpetrol

Welcome.

Emilio Pisanty

or rather, the standard is to call $\partial_j f$ the gradient and pretend that that is equal to $g^{ij}\partial_j f$.

which of course works until it doesn't

0celo7

That's purely a physics phenomenon

I take that back

Emilio Pisanty

@0celo7 good

0celo7

People get tired of writing $\mathrm{grad}f$ in analysis

AccidentalFourierTransform

4:53 PM

@ACuriousMind I know that it can be analytically continued to the Minkowski result (otherwise why would people even do the W. rotation?). But that doesnt make it any less idiotic.

Slereah

A lot of theorems of QFT are hard to show in any other way than with path integrals or canonical quantizations

0celo7

So they write $\nabla f$

Slereah

You kinda need both

ACuriousMind

@AccidentalFourierTransform Uh...why would it be "idiotic" if it works?

0celo7

But if you pressed a mathematician to define the gradient, they would say $df^\sharp$

ACuriousMind

4:54 PM

If you're trolling me you should say so now :P

Slereah

heh

skullpetrol

Do not ban @AccidentalFourierTransform

3

Slereah

When you see musical isomorphisms, you know it's a math guy

AccidentalFourierTransform

@ACuriousMind what do you mean by "it works"? just do everything in M. spacetime, that works too!

0celo7

it does?

AccidentalFourierTransform

4:55 PM

nothing ever works in QFT

sometimes we pretend that some things do

BUT THATS A LIE!

ACuriousMind

@AccidentalFourierTransform The Euclidean path integral at least can be well-defined in 2d and 3d, unlike its Minkowski version.

Emilio Pisanty

@0celo7 what's that sharp thing?

Slereah

I think you can kinda do Minkowski path integrals, but like

Emilio Pisanty

index raising?

Slereah

It's hard to define for anything but free fields

@EmilioPisanty : yes

Those are called the musical isomorphisms

ACuriousMind

4:57 PM

@skullpetrol It's a "pseudo-user", which shows up as the author of the feeds that are associated to it, but it's not actually an account.

Slereah

$a^\sharp$ maps dual vectors to vectors

AccidentalFourierTransform

@ACuriousMind if the M. version agrees, in some sense, with the E. version, then the M. version can be well-defined too. People just dont know how to yet

0celo7

@EmilioPisanty yes

skullpetrol

Thanks @ACuriousMind

ACuriousMind

@AccidentalFourierTransform Nonono, that you can continue the results does not mean that the path integral in both versions is well-defined.

Slereah

4:58 PM

Wait

0celo7

Implying path integrals are well-defined at all in either case?

Emilio Pisanty

@0celo7 Yeah, I gotcha. I think yeah, the standard physicist practice is to define $df$ as the gradient and pretend it lives in the tangent space.

ACuriousMind

@0celo7 In 2d and 3d they are for the Euclidean case, cf. the work of Glimm and Jaffe.

Slereah

@0celo7 : It's well defined for polynomial lagrangians in 2D

AccidentalFourierTransform

duh, $$\int\mathrm d\phi \mathrm e^{iS_M}=\mathrm{A.C}\left[\int\mathrm d\phi \mathrm e^{-S_E}\right]$$

see, well defined

0celo7

4:58 PM

@ACuriousMind I knew that.

But I thought we lived in 4/10/11 dimensions.

Slereah

This is all an illusion

We are all lines

Living in a line

AccidentalFourierTransform

@ACuriousMind if initial M. theory is not defined, then there is no result to continue

because, result of what?

Slereah

^the universe

more importantly

I just bought a waffle iron

skullpetrol

Where do you get this stuff?

John Rennie

@Slereah so you can flatten out your crumpled waffles?

Slereah

5:01 PM

@skullpetrol the supermarket

skullpetrol

:D

ACuriousMind

@AccidentalFourierTransform A Wightman QFT is well-defined and has well-defined correlation functions. It does not, generically, have a well-defined path integral. The OS theorem says that there is a 1:1 correspondence between Wightman (Minkowksi) and OS (Euclidean) QFTs, and that the correlation functions are related by analytic continuation. So, if you manage to compute the c.f. of the OS theory by a Euclidean path integral, this does not imply that there's a well-defined Minkowskian path integral.

Now, you can raise several objections (the most serious, imo, "Our QFTs aren't Wightman QFTs"), but there is really no need for the Minkowskian version of the path integral to exist just because the Euclidean one exists.

AccidentalFourierTransform

@ACuriousMind Im not saying that there is a well-defined Minkowskian PI. Actually, I think I have already expressed very clearly my loathing of PI.

Slereah

@ACuriousMind What about the fact that you can't do Wick rotation in a curved spacetime

ACuriousMind

@Slereah Another data point that QFT in curved space is a fundamentally misguided idea :P

AccidentalFourierTransform

5:05 PM

what I'm saying is that, if we pretend that the E. theory (either through PI or any other quantisation method) is well defined, and that it agrees with the M. theory upon analitical continuation, then we must admit that the M. theory is well defined too, in some sense that may not be clear yet

ACuriousMind

@AccidentalFourierTransform That the well-definedness of the Euclidean theory implies well-definedness of the Minkowskian theory is something I agree with; that's somewhat the essence of the OS theorem.

I've lost track of how all this is supposed to support "Wick rotation is idiotic".

AccidentalFourierTransform

I guess that I've always been told that "Minkowski QFT is meaningless, but the Wick rotated theory is well-defined, so we shall work in the E. version"

which never made sense to me: if the M. theory is meaningless, then no matter how good the E. theory is

because it is not related to a physically useful theory, and as such, we dont really care for its well-definedness

so using the W. rotation is idiotic if we stick on that philosophy

ACuriousMind

@AccidentalFourierTransform Yeah, that's nonsense. The "proper" statement would be that if we can show the Euclidean version is well-defined, then this shows the Minkowskian theory is also well-defined, and it's - in some very select cases - actually tractable to show well-definedness of the Euclidean version.

Slereah

IIRC you can show that the path integral for the imaginary Gaussian measure thing converges alright

ACuriousMind

Alas, the continuation of this programme to gauge theories and 4d theories seems to have produced no real results so far

Slereah

5:10 PM

But 'course it's harder to show for generic lagrangians

ACuriousMind

And, make no mistake, none of the QFTs we usually work with is rigorously well-defined in either incarnation

Slereah

Is there any theory that actually work alright, as far as we know

Like one of the many quantum gravity theories

AccidentalFourierTransform

@ACuriousMind Ok. I might concede that the W. rotation is not as idiotic as they've led my to believe.

but only when properly applied!

ACuriousMind

@Slereah (s)CFTs and other two-dimensional theories are pretty good on that front, also many topological QFTs since the path integral often degenerates into an integral over a finite-dimensional space.

skullpetrol

Didn't Dirac say if you wait for everything to be rigorously defined you won't get anything done?

AccidentalFourierTransform

5:13 PM

@skullpetrol Dirac banged his cousin

Slereah

It's all his fault if nothing is well defined!

skullpetrol

@AccidentalFourierTransform eww

ACuriousMind

As for quantum gravity: I think non-perturbative renormalizability ("asymptotic safety") of the naive "GR as a QFT" is still an open question, no one knows the proper non-perturbative formulation of string theory/M-theory and I have no idea what the hell the LQG people do.

Slereah

From what I hear LQG has troubles working in fermions

Also LQG has violations of Lorentz invariance which is maybe not shown in experimental data?

AccidentalFourierTransform

From what I hear LQG has troubles working.

Slereah

5:15 PM

Also that, yes

But then again that's true of most QG theories

ACuriousMind

@Slereah Doesn't surprise me, putting fermions on lattices is horribly difficult.

Slereah

Basically the only QG theories that work are the ones that don't work

Covariant quantum gravity and semiclassical gravity

Also the inbetween one

What is it called

Stochastic gravity?

Those are theories you can actually get results from

but of course they're all bullshit

AccidentalFourierTransform

@Slereah engineering?

Slereah

Well experimental results, anyway

AccidentalFourierTransform

physics is not about results you silly boy

Slereah

5:17 PM

@AccidentalFourierTransform has gone too far on the abstract scale

He has become a mathematician

John Rennie

Cor, did you know that _ is a valid variable name in C++?

Slereah

One thing that bothers me with quantum gravity is that passions are very high in it

John Rennie

That language!

Slereah

If you get Lubos Motl and Carl Rovelli in the same room someone might get knifed

AccidentalFourierTransform

Slereah

5:20 PM

Sounds like category theory alright

skullpetrol

Wow @JohnRennie that is interesting.

Slereah

Category theory is the theory of relations between things

AccidentalFourierTransform

arxiv.org/abs/1203.6902

wtf

Slereah

(Submitted on 1 Apr 2012)

AccidentalFourierTransform

ok that makes sense

ACuriousMind

5:22 PM

@Slereah Category theory is the theory of relations ~~between things~~. Do you not know that to speak of objects is evil, cf. ncatlab.org/nlab/show/evil?

0celo7

"Axiom 1. The number of gods in a 3-cell is one."

Slereah

sounds reasonable

I've never seen more than one god in a 3-cell

AccidentalFourierTransform

@Slereah fermionic gods or bosonic gods?

ACuriousMind

Can two-dimensional gods be anyonic?

Majorana gods

Slereah

I don't think gods are fermionic or bosonic

I think gods are distinguishable

Hence unrelated to statistics of such types

AccidentalFourierTransform

5:24 PM

Trinity

The Christian doctrine of the Trinity (Latin: Trinitas, lit. 'triad', from trinus, "threefold") holds that God is three consubstantial persons or hypostases—the Father, the Son (Jesus Christ), and the Holy Spirit—as "one God in three Divine Persons". The three persons are distinct, yet are one "substance, essence or nature" (homoousios). In this context, a "nature" is what one is, whereas a "person" is who one is. According to this central mystery of most Christian faiths, there is only one God in three persons: while distinct from one another in their relations of origin (as the Fourth Lateran...

Slereah

The Holy Trinity | Music Videos | Axis of Awesome

►

AccidentalFourierTransform

that was... something... I guess

Yashas Samaga

0

Q: Calculating saturation current and energy associated with a band of wavelengths

Consider white light whose wavelength spread is from 400nm to 700nm. Its energy is uniformly distributed in this spectrum. The light is incident on metal A of work function 1.55eV. Saturation photocurrent is 6mA. Now the same light is incident on metal B, work function 2.48eV. Calculate the total...

photons photoelectric-effect

I'm still stuck.

$\lambda dn + n d\lambda$ = 0

AccidentalFourierTransform

start a bounty

Yashas Samaga

I just want a hint lol

Let k be the no_of_photons/mA_of_current

AccidentalFourierTransform

5:34 PM

start a teeny tiny boutny

Yashas Samaga

if I integrate dn, I should get k (6ma)

I am going to try this till 6AM in the morning. I'll start a bounty if I fail to solve it.

That is a high school question and I am feeling ashamed.

dmckee

@EmilioPisanty I finally got around to watching those slide-rule videos. I'm pretty sure that my father's rule (which I learned on) was the same model as Neil Armstrong's. It was about as swanky as a general purpose linear rule got.

DHMO

This is amazing

Emilio Pisanty

@dmckee I'm really tempted to buy one

probably will do pretty soon

a Neil Armstrong model sounds like a good place to start ;-)

ACuriousMind

@AccidentalFourierTransform @JohnRennie Pretty simultaneous commenting on that PhD question ;P

AccidentalFourierTransform

5:40 PM

dmckee

@EmilioPisanty I own a couple. One is a EE's rule with lots of goofy scales for impedences and whatnot. The other is a Korean War artillery spotter's calculator for 155mm howitzers.

John Rennie

:-) Within 4 seconds of each other!

dmckee

You can often find special purpose ones at antiques shops for quite reasonable prices.

Yashas Samaga

@DHMO You could have fused hydrogen directly into Helium instead.

Emilio Pisanty

@dmckee whoa

dmckee

5:41 PM

The general purpose ones are a little pricier. Especially the really good ones.

DHMO

@YashasSamaga but that is not a cycle!

Emilio Pisanty

No, I think I want a more standard one

@DHMO your image isn't a cycle either

dmckee

I looked up Dad's model once and there were a couple on EBay for about $300. Which is five time the original sticker price.

DHMO

@EmilioPisanty It is a cycle

Emilio Pisanty

you are describing a catalyst, though

Slereah

5:42 PM

$a^a^a^a^a^a$

Too much!

John Rennie

@DHMO That's just the CNO cycle isn't it?

Emilio Pisanty

@dmckee gulp

DHMO

@JohnRennie yes that is

Slereah

@dmckee : one of my friend owns a curta calculator

it's quite neat

dmckee

@EmilioPisanty The basic student models can be had for around US$50.

John Rennie

5:43 PM

@YashasSamaga you can't fuse hydrogen directly into helium

Yashas Samaga

by hydrogen, I meant a suitable isotope of hydrogen

but well, two protons cannot be fused directly, right?

John Rennie

No

Thre is no bound state of two protons

Yashas Samaga

it is going to be really hard to bring them together

Emilio Pisanty

How is $^{12}\mathrm C+4p\to {}^{12}\mathrm C+{}^{4}{\mathrm{He}^{2+}}+2e^{+}+2\nu_e+3\gamma$ a cycle, but $4p\to{}^{4}{\mathrm{He}^{2+}}+2e^{+}+2\nu_e+n\gamma$ isn't?

Yashas Samaga

ah yea

there has to be a neutron

DHMO

5:45 PM

@EmilioPisanty because the big guy is regenerated!

2

Emilio Pisanty

@DHMO what 'big guy'?

John Rennie

@YashasSamaga You have to hope that one of the protons changes to a neutron by a weak process so you form a deuterium nucleus, but that is very low probability. That's why fusion in the Sun is so slow.

Emilio Pisanty

the carbon?

DHMO

ya, the carbon

@JohnRennie I heard that two protons directly collide with one of them turning into a neutron...

Emilio Pisanty

@DHMO no, I don't see it

it's not 'generated', you started with it

John Rennie

5:46 PM

98

Q: Why does the Sun's (or other stars') nuclear reaction not use up all its "fuel" immediately?

The temperature and pressure everywhere inside the Sun reach the critical point to start nuclear reactions - there is no reason for it to take such a long time to complete the reaction process. Just like a nuclear bomb will complete all the reaction within $10^{-6}$seconds. Why does most of the...

nuclear-physics sun stars

DHMO

@EmilioPisanty I corrected it shortly after!

user123733

Hii , may I ask a question on waves

Emilio Pisanty

@DHMO I still don't see the difference

DHMO

@JohnRennie thanks!

dmckee

@user123733 Sure. Someone might even answer it.

Emilio Pisanty

5:47 PM

you want to call it catalysis, then sure

Yashas Samaga

@JohnRennie Yay for our amazing universe!

user123733

@dmckee thanks

dmckee

My pleasure.

Emilio Pisanty

A cycle gets you back to where you started. Your image doesn't.

user123733

How can we find maximum KE in stationary wave on a string

Yashas Samaga

5:48 PM

Split the string into pieces.

Emilio Pisanty

You might call it a squint-cycle, though: a process that gets you back to where you started if you squint and ignore the the >50% that doesn't have the properties that you want it to.

Yashas Samaga

and sum the kinetic energy up

dmckee

@EmilioPisanty The astro folks call that "the CNO cycle" not withstanding that four protons are built into a helium on each run.

user123733

@YashasSamaga in what way

Emilio Pisanty

@dmckee as long as the astro folks won't claim that it is actually a cycle when presented with the fact, then yeah, it's a catchy name

dmckee

5:49 PM

@user123733 Do you have the mathematical expression for the standing wave handy?

If you do then you can see that this is just a continuous array of harmonic oscillators, and you can sum (i.e. integrate) their energy.