Ah, ok, this is not 3-sorted :P

I was coming up with a paradox I observed over lunch that if you 2-sort it you'd get {10, 11, 12, 2, 14, 13, 15} which is not 3-sorted.

But yeah, that's not it.

@Blue In any case, your argument doesn't work. It doesn't deal with the middle entry at all.

These are the kind of examples where your argument would fail.

The point is this.

$x_0 < x_n < x_{2n}$ is a necessary but not sufficient condition for saying $\{x_0, \cdots, x_{2n}\}$ is $n$-sorted.

Whereas $x_0 < x_n$ is both a necessary and a sufficient condition for saying $\{x_0, \cdots, x_n\}$ is $n$-sorted.

Size matters :P

lol

In any case I'll think about it a little more.

@JohnRennie why stopping potential depends on frequency?

@BAYMAX one photon ejects one electron by transferring the energy of the photon to the kinetic energy of the electron.

The photon energy is $E = hf$, so the (maximum) kinetic energy of the ejected electrons is equal to $hf$.

The stopping potential measures the kinetic energy of the ejected electrons.

@Blue K, I think I got it.

@JohnRennie hmm,I find this twisting

The stopping potential, $V$, is the potential at which $eV = \tfrac{1}{2}mv^2 = hf$, where $e$ is the electron charge.

That is, to cross a potential $V$ an electron requires an initial energy of $eV$.

at stopping potential velocity of photoelectrons is 0

So we end up with the stopping potential proportional to the frequency of the light.

@BAYMAX at the stopping potential the final speed of the electrons is zero.

final speed?

@BAYMAX That is they start, i.e. leave the metal surface, with a speed given by $\tfrac{1}{2}mv^2 = hf$ and the stopping potential slows them to a halt.

@Blue Say $A = \{x_0, \cdots, x_n, \cdots, x_{2n}\}$ is $n$-sorted. We can assume the subarray $\{x_0, \cdots, x_n\}$ is $m$-sorted WLOG and because of what we proved earlier. Suppose now that while $m$-sorting, $x_n$ and $x_{n+m}$ are switched; then $x_{n+m} < x_n$. And then on the last entry, $x_k$ (in the $2n-m$ th position) and $x_{2n}$ are switched. Then $x_{2n} < x_k$.

Here's the trick.

After the $m$-sorting is over, the $n$-gap subarray starting at $x_0$ is $\{x_0, x_{n+m}, x_k\}$. We have $x_0 < x_m < x_{n+m} < x_n < x_{2n} < x_k$, where $x_m < x_{n+m}$ is because $A$ is $n$-sorted. :P

after leaving the metal surface with speed $v$ who controls the photoelectrons?

@JohnRennie

after they leave,the photelectric equipment and phototelectron are two different systems now!

@BAYMAX That's what the stopping potential does. Basically we have a negative charged plate above our metal surface and the negative charge of the plate repels the electrons that come flying out of the metal surface.

If the negative charge is small the electrons are not completely repelled and they still reach the charged plate. If you make the charge high enough it repels the electons so strongly they can't reach the plate.

The stopping potential is the potential necessary to produce that charge.

@Blue No, because of $n$-sortedness! $x_0 < x_m < x_{n+m}$. $x_m < x_{n+m}$ because they are $n$-apart in $A$.

ha ha ,i thought they were flying,so silly a bit,nice1 @JohnRennie

$A$ being $n$-sorted means precisely that any two elements of $A$ which are $n$-elements apart are in ascending order of appearance. $x_m$ and $x_{n+m}$ do happen to be positioned $n$ elements apart.

hi

so I now see the reasoning that why in the above graph the curve touches the potential axis at different points@JohnRennie

so we now have that stopping potential depends on frequency rather 'stopping potential $V$ is directly proportional to frequency $f$'

@JohnRennie

@Blue, now, there's a little more trouble :P

What about the subarray $\{x_i, x_{i+n}\}$?

is probability extremely present in physics? or under which sub-field of physics?

That's also an "$n$-gap subarray" We know $x_i < x_{i+n}$. Does $m$-sorting change that order? If it does we're frigged

@BAYMAX: oops, the photoelectron energy is $hf - \phi$, where $\phi$ is the work function. So yes, I was wrong to say $V \propto f$.

It's actually: $$ eV = hf - \phi $$

><

@Blue ah, looks cool

plot b/w $V$ and $f$ will still be a straight line with positive slope?@JohnRennie

Yes, and the intercept will give you the work function.

so they are directly proportional right

@JohnRennie

@Blue No I mean we checked (0-th position) < (n-th position) < (2n-th position) after $m$-sorting. But, we discussed that doesn't suffice to prove our new array is $n$-sorted.

We need to check eg (1-th position) < (n+1-th position)

@BAYMAX No, because directly proportional would mean $$ V = kf $$ for some constant $k$ i.e. when $f = 0$ we'd get $V = 0$ and vice versa.

ohk

@BAYMAX In fact $$ V = \frac{h}{e} f - \frac{\phi}{e} $$

$f$ increases implies $V$ increases?

@JohnRennie

@BAYMAX Yes

@Blue Turns out that's easy to deal with. If we are to prove ($i$-th position) < ($n+i$-th position) in the new array, we just need to check it for $i \leq n$, right? But that means the element in the $i$-th position is $x_i$ because $\{x_0, \cdots, x_n\}$ is $m$-sorted by assumption.

@Blue On the $\{x_0, \cdots, x_n\}$-th half of the array, yeah!

Because we assumed it's $m$-sorted.

when $V =0$ photocurrent is independent of frequency ?

@JohnRennie

@BAYMAX the current is proportional to the number of electrons per second, and that is proportional to the number of photons per second hitting the metal surface.

so current is proportional to the frequency of the incident light ?@JohnRennie

When you increase the frequency the energy of each photon increases. So if you increase the frequency but keep the light intensity constant the number of photons per second goes down.

And that means the number of photoelectrons pere second goes down.

So if you increase the frequency but keep the lght intensity constant you actually get fewer electrons per second - the photocurrent decreases.

then why the curves land at the same point on the y axis @JohnRennie

Well, yeah, because you only have to look at what happens if you $m$-sort the subarray $\{x_i, \cdots, x_{i+n}\}$.

But that's an array of length $n+1$, so we're done.

@BAYMAX Hmm, I have to say I think your book is wrong.

ohh,it's Arthur Beiser

I suspect you should be able to generalize this to $N = kn + 1$ length subarrays by induction on $k$. This is a rather elaborate bitchery.

For $N$ such that $(k-1)n+1 < N < kn + 1$ the argument is the same as $N = (k-1)n+1$ except for taking care of the extra elements.

I'd like to see a slick proof of this

@BAYMAX Hmm, let me think about this ...

okay@JohnRennie

h

pretentious minimality

@Blue I could try asking someone on the math chat when he comes.

He's the sort of demigod on these types of arguments

Akiva

nah he's a high schooler and does math

with probability 0.9 you'd get a slick proof out of him

true there are way too many high schoolers frigging around

them frigging frigs

i do not identify as a highschooler

meh university students are nerds. i do not identify as a nerd

@Blue is a flipping nerd

Those who are confident with undergraduate linear algebra (i.e. intuition and clear with nearly everything in it) say ***!

(just curious)

@BAYMAX I've been looking through my physics books to see if I could find a figure to show you, but none of them give a graph of photocurrent vs frequency at constant intensity. However I'm sure I am right and the current would fall with increasing frequency.

@Kaumudi.H: you chilling now? :-)

@Kaumudi.H Quiet

JU students are a bunch of a... um... nevermind.

I can't say that stuff here

lift yr skinny fists like antennas to heaven

Students are always revolting

Okaybut we should remember that it is the case when $V = 0$@JohnRennie

(only true music nerds will understand that reference)

@JohnRennie Hah!

I see what you did there

@JohnRennie :P

@BalarkaSen fortunately Google is an effective alternative to being cool :-)

I've made another (probably) poorly formed question.

but google wouldn't tell you how much of a meme that album is

so you may be cool, but you need to do better to be a memelord

it's a mix of the two

like speedballs

maymay is how it should be pronounced. meme is how it is pronounced. but, as it turns out, the spelling has also been memefied

People spell good memes as "god maymays"

So that's how the terminology goes

ahaha, so relevant.

not sure if the issue with attack being on the title is fduckery

but that's hilarious as hell

I think the point is not position. The notion of position is not well defined in QFT afaik

hi @EmilioPisanty

Probably, but surely there's some function for probability of the interaction based on some measure of 'seperation'

otherwise every electron in the universe surely would be equally as likely to interact with a single positron

@Blue Only that sentence is completely wrong, because fermions also obey "quantum statistics" but can never occupy the same quantum state :P

@Blue I suppose the answerer had bosons in mind and didn't carefully word that sentence; I left a comment

@Blue Which question? The one whose answer you just linked?

Maybe he meant particle in a less specific sense just to mean "small things", which would include molecules being in the same state configuration

that is a bit of mental gymnastics to interpret it though

@Blue Well, I can't answer it because I don't agree with the quote - M-B statistics can perfectly well apply for classical indistinguishable particles.

(You also did that thing again where you give a quote but not the reference it was taken from, please try to be diligent when quoting in questions and answers)

Oh, wait

I think I know the answer

(and will write it as a proper answer)

this is a dumb pointless question but if you have an electron, and measure it incredibly precisely w.r.t position, such that the uncertainty in the momentum is incredibly uncertain, since classically dispersion is a function of wavelength and wavelength is related to momentum, how do you model the dispersion since wavefunctions evolve deterministically there must be some well defined dispersion for it right? And is the calculated dispersion limited by $c$?

Does somebody have good sources on understanding the effective action in the path integral formalism? I am staring at some papers which suggest applying the Legendre Fenchel transform to the cumulant generating functional to get the effective action, but I entirely don't get the motivaation for that.

@Phase would it help if I mention that, at relativistic speeds, the expression $p=mv$ is not remotely valid?

I probably wasnt clear but I only meant that dispersion is a function of wavelength in classical treatment of waves, I'm taking wavelengths relation according to Plancks

@Phase perhaps the best way to get a solution would be to think of what this means in terms of what's happened to omega and the derivatives w.r.t. momentum, then reconcile this with what dispersion actually is

@GPhys No, it's correct.

See page 234 of Zee.

@0ßelö7 can you explain what he means?

What Weinberg means?

I meant a typo in the textual explanation not the equation (since I was unable to get close enough to an equation to comment on it)

ok I'll work it out

i.e. he tells you to plug in one thing and then a different thing to the same number

hmm, yeah

this is some GR index trickery

brb off to the index black hole

I could never quite get anything like this only many terms similar to this when playing around with taking the derivative

ok here's what you need to do

you solve the time equation $\alpha =0$ for $\partial_\mu [(\rho +P)U^\mu]$

the answer is $U_0^{-1}[\partial_0 P-(\rho +P)U^\mu \partial_\mu U^0]$

now, you get that same term in the $\alpha=i$ equation, so substitute

Hey @0ßelö7 I think I just wasted my time

that depends on what you were doing

but it's quite possible

I took the LP transform of the Taylor series and got the sum of terms of the form $\frac{f^{(n)}}{s^{n+1}}$

@Phase what is wrong with you?

Ikr

Honestly i only did it to practise, didn't expect anything useful

For future reference though, what is it about what I did that makes what I got so useless

whoops got a typo

$\frac{f^{(n)}(0)}{s^{n+1}}$

@GPhys hmm, that has a sign error

@0ßelö7 what does? the book?

no, what I wrote

crap, is this a +--- issue

oh that's fine I wasn't paying close attention to it anyway I just was taking the method

so I could work it out myself

did it work?

I am trying now (about to head to class so was in the middle of some things)

yes this seems fine

oh, sneaky Weinberg

that's actually a $\partial^\alpha$

the derivative on the $p$ is a contravariant derivative

yes I have it in my notes the correct way

he has it correct as well

maybe it's just not as obvious because he wrote it as a derivative

it's very poor notation

it was easy for me to see in my notes, because when I derived it I had to flip it with the eta

oh god he has $\partial/\partial x^\alpha$ and $\partial/\partial x_\alpha$

yes

for 99% of people those are the same thing

of all the things I could not figure out, I figured that one out :P

it's actually an abuse of notation because $x^\alpha$ isn't a vector

shame on you, Steve

@GPhys Well, glad to know my method worked

I put some BS argument for the variational problem since the path is along something like sqrt(1-v^2)

I tried to do it but I kept getting nonsense answers

why do you do this last minute :P

I would like to say that he gave us 4 days for this homework and I started over two days ago

what is wrong with him lol

is it a prelim course?

no!

I don't know

we asked him last week to please give entire week but maybe this week he can

"can"

oh don't make him out to be innocent here

he wants you to suffer

first quantum field theory homework is due Friday

haven't been working on it because I was doing this

before that I really need to finish filing my taxes though

Q_Q

...taxes?

in September?

I was deathly ill in April :D

taxes are due due in mid oct

if you filed for the extension

>.<

so wait until mid october

I was dead this week end

Rotting in the ground

a skeleton

cool story bro

Much like Jesus I arose on the third day

To go back to work

even jesus wakes up when he's got to finish his phd

so does $f=o(1)$ just mean that $f\to 0$

@BalarkaSen do you meme ironically?

what does it mean to meme unironically

@BalarkaSen would you meme if you were unable to tell anyone about it

oh lol. well by instincts, i can imagine that happening

Alzheimer's paper with Sobolev spaces arxiv.org/abs/1709.05671

Bogus?

ahahaha

what the heck is a $\Bbb Q$-Fano fibration

The bogus man