The h Bar

0ßelö7

01:40

@ACuriousMind Suppose I have a matter field $\psi$ in an EM field $A$, and the Lagrangian is $$\mathcal L(\psi, A)=|D_\mu \psi|^2-\frac{m^2}{2}|\psi|^2+\frac{1}{p}|\psi|^p.$$ Is the $p$-nonlinearity anything physical or just a nonlinearity for nonlinearity's sake?

Secret

science.sciencemag.org/content/early/2017/09/06/science.aan4701

I think chemistry will start to become electron shell physics when atom by atom synthesis of molecules become possible

Imagine being able to place atoms precisely in space and then they form a molecule. No more complicated organic synthesis procedure needed

I am starting to wonder, whether mathematics can demonstrate the existence of such direct path of synthesis for any complex molecules...

This is because viewing chemical synthesis in a mathematical perspective, it forms a configuration space with reaction conditions as coordinates where paths can be drawn

dmckee

02:37

@Secret There will be a lengthy time when you 'can' build molecules to order but organic chemists and chemical engineers continue to do what they have always done.

Because a mole is such a big number.

On the other hand, tailored biological sourcing could sweep the industry very fast if it turns out to be easy to grown and maintain large cultures

Anonymous

03:50

@0ßelö7 "I say it all the time"<----that is not a reason for it to be correct in the context you used it. A "grammatical rule" could be any rule which conforms to the rules of grammar. Even "Sum of two natural numbers is always greater than the difference of the same two same natural numbers. " is a grammatical rule as it conforms to the rules of grammar. On the other hand "grammar rule" is a "rule of grammar", and is much more widely used.

Anonymous

1

Q: Which is correct: "grammar rules" or "grammatical rules"?

Which is one is correct English: "grammar rules" or "grammatical rules" ? In my opinion, "grammatical" means "conforming with the rules of grammar". Thus, "grammatical rules" should literally mean "rules which conform to the rules of grammar". That is surely not the definition of "grammar...

grammar

Anonymous

1

A: Which is correct: "grammar rules" or "grammatical rules"?

Yes, I think you are correct! Grammar rules means rules about grammar. Grammatical rules are rules that are well-written and grammatical. I think this means you can have grammatical grammar rules. In fact, grammar rules should always be grammatical!

0ßelö7

eye roll

Anonymous

Anyhow, you need to get rid of the notion that your customs are the laws of nature.

0ßelö7

nope

that's a good way of putting it

"law of nature" -- nice!

Anonymous

03:55

I don't see any counter-argument there, so I suppose we're done.

Balarka Sen

04:12

grammar rules is too damn cringy though

@0ßelö7 Well, that works only if $\Omega$ is convex.

But yeah there's some technicalities about smoothing as you approach limit; it needs to be done carefully

@Kaumudi.H Ted has a series of lectures on multivariable calculus, part of a course out of his book (from which I learnt multivariable calculus). Might check it out at some point.

0ßelö7

@BalarkaSen close enough to the boundary we can assume it's convex

I clearly only care about the construction for large $n$...so just cut off the sequence

Balarka Sen

Ok, yes, true.

0ßelö7

@BalarkaSen Hmm. I don't need it to be $C^1$ up to the boundary, just $C^1$ on $(0,1)$, say

user228700

@BalarkaSen Ohh, where (if) are they available to watch?

0ßelö7

And the curve needn't be embedded, self-intersections are allowed

Balarka Sen

04:19

@Kaumudi Look in his profile description; they are all uploaded to youtube

user228700

Ah, oops. Thanks!

0ßelö7

@BalarkaSen Basically I want to estimate $\lim_n f(x_n)/g(x_n)\to 0/0$ and L'Hopital works very well if I can instead do the limit along a $C^1$ curve

Hmm. I might need $C^1$ up to the boundary for L'Hopital...

Amazing

You don't need the functions to be differentiable at that point

Balarka Sen

I have seen something like a C^0 L'Hoptial before in a proof by Thurston

0ßelö7

@BalarkaSen So I think what I described is a legitimate inductive construction. I don't think smoothness could go wrong there.

Balarka Sen

I'll believe you!

user228700

04:31

Sigh, it's 10 o'clock on a lovely Saturday morning and this, this is what I'm doing:

user228700

John Rennie

04:46

@Kaumudi.H My sympathy is limited because at your age I was absolutely fascinated by computers and would happily have spend every waking moment playing with them.

0ßelö7

^sad

John Rennie

@0ßelö7 no, I was very happy as a teenage computer nerd. It was teenage computer nerds like me (but with more ambition) who built todays computer industry.

And to be honest in the UK in the late 70s there was frack all else to do :-)

0ßelö7

@JohnRennie golden age of global calculus

you could have done that

user228700

I have a quick question! Do two waves need to be coherent to interfere? Not to interfere completely constructively/destructively, merely to interfere.

Anonymous

@Kaumudi.H Define coherence and you'll get the answer

user228700

04:59

Huh, well...

Anonymous

Mathematically, the answer is "yes"

Anonymous

But practically speaking there's something called "degree of coherence"

user228700

Clearly, this is not a situation in which having little information about the definition of coherence is helpful :-/

Anonymous

@Kaumudi.H Did you get your answer?

user228700

Only sort of.

John Rennie

05:02

@Kaumudi.H any waves will intefere, coherent or not, but to get a stable interference pattern you need coherent waves.

Interference is just vector addition of the waves.

user228700

^ This was my initial guess as well :-/

user228700

I was confused for a moment only because my textbook has defined it differently, to suit the context.

Anonymous

For sustained interference you need them to be coherent

Anonymous

But, yeah

John Rennie

@Blue for a stable interference pattern you need them to be coherent. I'm not sure what sustained interference means in this context.

Anonymous

05:05

@JohnRennie physics-assignment.com/condition-for-sustained-interference

user228700

Heh, I have never liked this topic :-/

Anonymous

It's same as stable interference

Anonymous

:P

user228700

@JohnRennie :-) Well, not me!

Anonymous

@Kaumudi.H This topic forms the basics of QM

Anonymous

05:07

It's actually very interesting

John Rennie

@Kaumudi.H we can't all be perfect :-)

user228700

Well, I've never been able to form an intuitive understanding of it, hence the dislike.

user228700

@JohnRennie :-)

user228700

Anonymous

@Kaumudi.H Don't worry. Even the elite physicists of the 19th and 20th century were boggled by this. :P :)

user228700

05:08

Itza today!!

Anonymous

Wow, cool

user228700

@Blue Yeah :-)

John Rennie

@Kaumudi.H Have fun! It looks like it will be a great time.

user228700

@JohnRennie Well, I hope! :-) Thanks!

user228700

*INCLUSIVE OF DINNER (Huzzah!!)

Anonymous

05:14

@Kaumudi.H It's not free. You paid for it (indirectly). :'Dd

Anonymous

Do enjoy :)

user228700

No, it's not, haha :-) STILL! DINNER! At a TED event! :-P

user228700

Thank you :-) What are your plans for the weekend?

Anonymous

@Kaumudi.H I'm going out for lunch with family and relatives (uncle, aunt and cousin) today. :D

Anonymous

Then I'll go for Durga puja shopping

user228700

05:18

Oh, nice! :-)

user228700

@Blue Ah, when is it, again?

Anonymous

@Kaumudi.H Starts from 23rd I think

Anonymous

I'm not interested in the "puja" part though :P

user228700

Cool!

Anonymous

Only the food

Anonymous

05:18

lol

user228700

Hehe :-)

user228700

Oh, BTW, this is the theme of the event:

user228700

It's a bit vague, but we'll see!

user228700

05:41

Can anybody please help me with the following equation:

user228700

$l_o={{\lambda}_o}^2/\delta\lambda$

user228700

Where $l_o$ is the coherence length, $\lambda_o$ is the wavelength of the light in question and $\delta \lambda$ is the wavelength spread.

user228700

How does this work?

John Rennie

Suppose you have light made up from two wavelengths $\lambda$ and $\lambda+\delta\lambda$.

user228700

OK...

John Rennie

05:46

And they start in phase.

user228700

OK.

John Rennie

Because their wavelengths are different they will get out of phase after a certain number of wavelengths.

user228700

Hmm.

John Rennie

Suppose we kind of wave our hands around and argue that they will get out of phase when the difference is one wavelength. The number of cycles this takes is $\lambda/\delta\lambda$.

user228700

Right.

John Rennie

05:49

e.g. if they differ in wavelength by 1% then it takes 100 cycles for them to differ by one wavelength.

user228700

Ah, right.

John Rennie

The distance is takes is the number of cycles, $\lambda/\delta\lambda$ times the wavelngth $\lambda$, hence the formula you cite.

user228700

AHHH.

user228700

I s'pose I must let go of any hopes to understand what that hand waving involved, exactly :-P Though I'm satisfied with this much!

John Rennie

Of course in reality we have a continuous spread of wavelengths and $\delta\lambda$ would be something like the half width of the wavelength distribution.

So it's all a bit hand waving, and the coherence length is somewhat vaguely defined.

user228700

05:51

@JohnRennie Huh?

user228700

I wasn't able to parse that properly. What dyou mean, exactly?

John Rennie

In real life light isn't a single wavelength $\lambda$. If you spectrum analyse it you get a spread of wavelengths centred on $lambda$.

user228700

Ah, right, right, OK.

John Rennie

So what I'm saying is the width of this wavelength distribution would be what $\delta\lambda$ means.

user228700

Yes, it says:

user228700

05:55

> "The emission will not be strictly monochromatic and will have a wavelength spread (spectral broadening) due to a variety of reasons such as the inevitable width of the energy levels and the Doppler broadening due to random motion of the emitting atoms and molecules"

user228700

$=\delta \lambda$

user228700

Alright, thanks so much! :-)

user228700

@JohnR: Are you working ATM?

John Rennie

Kind of, but I have attention to spare for physics if you want to ask anything.

user228700

Yes, I was wondering if we could complete that discussion on DSHO but I'm not too sure what I'm looking to understand, really, because, see, this is what my textbook is like:

John Rennie

06:07

Yes, do you want to do it in the deep end or here?

user228700

AH, the amount of time that took!

user228700

I dunno what to do, really, whether to learn this by heart without understanding a thing or to spend hours upon hours trying to.

John Rennie

I would suggest we take a step back and let me explain how we approach the DSHO. I think the time taken will be worth it in the long run.

user228700

@JohnRennie OK.

John Rennie

06:11

Ok, you remember how I approached the undamped SHO? I guessed the solutionwas $e^{i\omega t}$, fed this into the differential eqaution and showed that (a) it was a solution and (b) we got the value of $\omega$.

user228700

Yes, I remember.

John Rennie

OK, we adopt exactly the same strategy for the DSHO.

user228700

OK...

John Rennie

In the limit of no damping the solution has to be $e^{i\omega t}$, so my guess is that with damping the solution will be $e^{i\omega t}$ times some decay function e.g. an exponential decay $e^{-kt}$. Does this seem sensible?

user228700

Yes, absolutely.

John Rennie

06:16

So the solution is going to be $$x(t) = e^{-kt} e^{i\omega t}$$

user228700

Hmm, yes...

John Rennie

And a bit of rewriting gives $$x(t) = e^{(-k + i\omega)t}$$

user228700

Yes...

John Rennie

Or: $$ x(t) = e^{at}$$ where $a$ is a complex number

user228700

Ah, right.

John Rennie

06:19

So what we do is feed this into the equation for the DSHO: $$\frac{d^2x}{dt^2} + \frac{k}{m}\frac{dx}{dt} + b^2x = 0 $$ (I think that's all correct though it's worth checking)

Are you happy with this so far?

user228700

Sort of...

John Rennie

Oops, just spotted a sign error

user228700

Where?

John Rennie

@Kaumudi.H you need to be 100% on this, because it's our starting point

@Kaumudi.H I've just corrected it.

Looking at your book it gives the equation as: $$\frac{d^2x}{dt^2} + \frac{\gamma}{m}\frac{dx}{dt} + \frac{C}{m}x = 0 $$

Equation 1.2

user228700

@JohnRennie I'll tell you where this completely loses me, John, it's in the physical significance of all these friggin' constants. I remember what you said about the constants serving a purpose while solving it etc. but still.

John Rennie

06:25

Do you mean how we got that differential equation in the first place?

user228700

No, no, I understand that.

John Rennie

Do you mean how the $\gamma$ and $C$ are going to relate to things like the frequency?

user228700

YES, yeah.

John Rennie

That's what we'll discover when we feed our guessed solution back into the equation.

Shall I do that now?

user228700

OK, let's see, then!

user228700

06:27

I'll try on the side.

John Rennie

It's probably best if you do this on paper along with me.

user228700

34 secs ago, by Kaumudi. H

I'll try on the side.

user228700

:-)

John Rennie

We start with the guess $$x(t) = e^{at}$$

So: $$\frac{dx}{dt} = a e^{at}$$

user228700

Hang on, hang on.

John Rennie

06:29

OK?

user228700

OK, so I jumped a few steps ahead (:-P) and so, $e^{at}$ can't be zero?

user228700

No, no, it can't, what a stupid question. OK.

John Rennie

So where have you got to?

user228700

So the other expression is zero

user228700

$(a^2+\gamma a/m + c/m)=0$?

John Rennie

06:32

Hang on, let me write it out ...

Yes, I get the same as you. So we have a quadratic equation for $a$ and all we have to do is write down the solution.

user228700

I get some contorted expression for the two solutions for $a$

John Rennie

I get: $$ 2a = -\frac{\gamma}{m} \pm \sqrt{\frac{\gamma^2}{m^2} - \frac{4C}{m}} $$

Do we agree?

user228700

Nope. Hang on...

user228700

Yeah, yeah we agree. I'm dumb. Carry on!

John Rennie

OK, we'll do a quick sanity check. In the undamped limit we have $\gamma \to 0$. Yes?

user228700

06:38

Tends to?

user228700

Why?

John Rennie

If $\gamma = 0$ there's no damping ...

user228700

Right, and no damping = undamped, no? :-P

John Rennie

Yes

user228700

Right, so my question is why it only tends to zero, why it isn't zero.

John Rennie

06:39

You're over thinking my statement.

Anonymous

If it were 0 it wouldn't be DSHO :P

John Rennie

I just meant as we gradually reduce the damping to zero wer have $\gamma$ tending to zero.

user228700

@JohnRennie :-/ I'm sorry. OK.

user228700

Right.

Anonymous

If the $dx/dt$ term vanishes, then it would be your ordinary SHO

John Rennie

06:40

OK. So put $\gamma = 0$ into our solution and what do we get?

user228700

@Blue Yep, yeah, I get that.

Anonymous

You can think of it as the damping coefficient. More is the $\gamma$ more is the damping.

John Rennie

@Blue please don't distract K

user228700

@JohnRennie C is still significant?

user228700

@JohnRennie Lol.

John Rennie

06:43

Just take our solution: $$ 2a = -\frac{\gamma}{m} \pm \sqrt{\frac{\gamma^2}{m^2} - \frac{4C}{m}} $$ and put $\gamma = 0$

user228700

Yes, right, since there is a -4C/m term inside the square root if we do that, I was wondering if...ah, see, I forgot something important in the middle, never mind.

John Rennie

Focus young lady, this is just algebra

user228700

Yeah, yeah, sorry :-( So $k=0$ and $\omega t= \sqrt{C/m}$ ?

user228700

Oh, ah, that $\pm$ Didn't account for that.

John Rennie

There's a rogue $t$ in there, but yes I get $a = i\sqrt{C/m}$, and our solution is $e^{at}$ so the solution is $e^{i \sqrt{C/m} t}$.

user228700

06:48

Right, sorry about that $t$

user228700

Give me 2 mins?

John Rennie

But you've got the point I think. What we've shown is that $\sqrt{C/m}$ is the frequency of the undamped oscillator.

user228700

We have. WE HAVE!

user228700

Where did $k$ go though?

John Rennie

Your book uses $\gamma$ not $k$, so I did the same.

user228700

06:50

::Facepalm::

user228700

No, no.

user228700

::Head-desk::

John Rennie

:-)

This is the thing. It's actually really straightforward when someone explains it to you.

user228700

You, when you explain it. My teacher? Yeah, he also tried.

John Rennie

Once we got our equation for $a$ lots of other things are obvious. For example the sqrt is imaginary if $$ \frac{\gamma^2}{m^2} < \frac{4C}{m} $$

and it's real if $$ \frac{\gamma^2}{m^2} \ge \frac{4C}{m} $$

user228700

06:53

Yep, yeah.

John Rennie

So we immediately get the condition for critical damping

Which is just $$ \frac{\gamma^2}{m^2} = \frac{4C}{m} $$

user228700

Wait, OK, so this is the thing: none of my devices have any charge left, I'm panicking for other reasons and I'm also standing here charging my devices, panicking, which is why I was so distracted and dumb (and I'm sorry) before.

user228700

So now I have to calm down and this will take me a solid 10 minutes of work.

user228700

And then after that, I gotta write a letter or something, to go out tonight, to the event. To the warden, remember I told you about me having to write a letter to go out.

John Rennie

I'd take a few minute out to write the letter now. Then have a coffee, then come back to this.

It's a very common reaction to a page of maths that looks impenetrable. We all react that way.

user228700

06:57

Without the option of coffee. I don't have coffee. Nobody has coffee.

user228700

@JohnRennie No, no, it's not the Math :-P

user228700

Anyway, right, I will be back, all calm, in 10 minutes.

John Rennie

WHAT? Give me your address and I'll go onto Amazon and buy you some coffee :-)

No coffee? YOU COULD DIE!!!! :-)

user228700

:-)

user228700

NO COFFEE! OK, the agenda was the stop panicking :-P

user228700

06:59

10 minutes. 10.

user228700

07:12

@JohnR: I'm back :-)

John Rennie

:-)

user228700

@JohnRennie Yes.

John Rennie

So where did we get to?

user228700

I suspect that you were going to talk about the three cases that arise.

John Rennie

It's probably not worth going any further with this because I suspect it's obvious to you now ...

The point I wanted to make is that now we have all the info we need to deduce the various things about how the DSHO behaves.

user228700

07:15

Obvious would not have been my word of choice :-) Though you are correct that I will be able to move on now.

user228700

@JohnRennie Yes, indeed.

John Rennie

OK. Ping if you want to ask anything further. I'll go back to kicking my servers :-)

user228700

I have been saved from the hells of absolute incongruity and utter confusion! YES! Thank you :-)

Anonymous

@JohnRennie Don't break them :P

Anonymous

@Kaumudi.H farrago XD

John Rennie

07:18

To be fair, if you look back at your book you'll see that it does exactly the same as I've just done, only with less explanation.

user228700

@JohnRennie Yes, I also noticed that the key difference is that you didn't assume the value of $\omega$ in terms of $C$ before you started, while my textbook did.

John Rennie

@Blue I'm working remotely, otherwise I'd be strongly tempted to start throwing the servers out of a high window :-)

user228700

This was incredibly confusing to me at the time.

user228700

@Blue :-P

John Rennie

@Kaumudi.H yes, the book jumped ahead. If I was writing the book i wouldn't have done that! :-)

user228700

07:20

:-) You're only 56. Quit kicking servers and sit down to write a textbook for us dummies!

user228700

I'll be going now. Must have lunch, write that letter, figure out a way to reach the event, get dressed, and, well, leave!

user228700

(Lunch, if you're wondering, is the same as breakfast today: bread with nutella)

user228700

Thanks so much :-)

user228700

I'll see you later. ::Waves::

Wrichik Basu

08:16

What's the relaxation time?

@Blue ^^

Anonymous

@WrichikBasu Time between two successive collisions

Anonymous

Of electrons

Anonymous

Approx time

Wrichik Basu

@Blue so, the more the relaxation time, more is conduction, right?

1, 3 and 4 are correct.

If the mass is more, the collisions would tend to increase, resulting in less conduction. So, 2 is incorrect.

@Blue ^^

Prathyush Poduval

09:23

@Kaumudi.H If you're interested, David Morin's classical mechanics book has a very good chapter on oscillations. He does all those differential equations and goes through each case thoroughly

Wrichik Basu

09:41

@JohnRennie can you help me in the above sum ^^

Please see the last pic posted by me here.

John Rennie

I think it's just 3

The question asks what is proportional to the applied field.

The mobility is a fundamental property of the material so it wouldn't be proportional to the voltage.

Likewise the effective mass of the charge carrfier.

Wrichik Basu

@JohnRennie I see.

John Rennie

And the relaxation time is the time the electron takes to come to a halt when the potential is removed.

The relaxation time is certainly important in conduction, but I don't think it is proportional to the applied field.

Wrichik Basu

@JohnRennie Oh, then that won't affect.

John Rennie

The point is that the question isn't asking which of the four affect conduction. It's asking which of the four are proportional to the applied field.

Wrichik Basu

09:45

@JohnRennie yes.

I think only drift speed will be the answer. As you've said.

What about this one? ^^

I think 1 and 2 are correct. But the temp is high. That makes me think a bit.

3 may also be correct, as the energy gap is proportional to $kT$ where $k$ is Boltzmann's constant.

@JohnRennie ^^

John Rennie

I have no idea what the state of silicon is at 5000K. What is the boiling point of silicon?

Wrichik Basu

3265°C

Less than 5000K.

John Rennie

Ok so the silicon atoms are probably present as single atoms so they won't be $sp^3$ hybridised so (2) is wrong.

And since the silicon is a gas not a liquid or solid there are no energy bands. So (3) is wrong.

How does kT at 5000K compare to the ionisation energy of silicon?

Wrichik Basu

@JohnRennie I don't know. I didn't even know that Si would be liquid at the Sun.

John Rennie

Well according to Wikipedia the first ionisation energy is 786.5 kJ/mol. Since that's per mole not per atom we need to comapre it to RT, which is 8.31*5000 = 41.55kJ.

So the temperature isn't high enough to ionise the silicon.

Wrichik Basu

09:57

@JohnRennie ok.

John Rennie

So we have a gas of neutral silicon atoms, so (4) is wrong.

That just leaves (1)

Wrichik Basu

@JohnRennie understood.

In a semi conductor, does effective mass of majority career affect conduction?

Sanya

shouldn't the energy difference between 3p energy levels always be rather small :o

John Rennie

@WrichikBasu Yes. The force on a charge carrier is $F = qE$ where $q$ is the charge (normally $e$) and $E$ is the electric field. So the acceleration of the charge carrier is $a = F/m = qE/m$, where $m$ is the effective mass.

Wrichik Basu

@JohnRennie ok.

@JohnRennie what does the last one mean? ^^

What are scattering centres?

John Rennie

10:10

Scattering centres in a semiconductor are places in the lattice where electrons or holes collide with the scattering centre and lose energy. So scattering centres increase the resistance/decrease the conductivity.

Wrichik Basu

@JohnRennie I see. So, correct options are 1, 2 and 3.

John Rennie

Can you explain your reasoning for case (2)?

Wrichik Basu

@JohnRennie very sorry, I misread it as less. 2 is incorrect. More effective mass means more collisions and less conductivity.

John Rennie

More effective mass means the charge carriers accelerate more slowly in response to the electric field.

And option (3)?

Wrichik Basu

@JohnRennie if electric field is more, then conductivity increases as charge carriers are acted upon by a greater force as per $F=qE $.

John Rennie

10:18

@WrichikBasu The current certainly increases with increasing electric field ...

Wrichik Basu

@JohnRennie yes, yes, that's what I meant.

John Rennie

So does the conductivity increase with increasing electric field?

Wrichik Basu

@JohnRennie just a second. I think conductivity is unaffected by electric field. The resistance of a wire remains constant irrespective of current. So, conductivity should also be constant for a material.

@JohnRennie if I'm correct in my above reasoning, then one last thing: is coloumb potential a central potential?

John Rennie

Correct, the conductivity is unaffected by the field. So only (1) is correct.

@WrichikBasu it depends what you mean by the Coulomb potential. The potential due to a single charge is a central potential.

Wrichik Basu

10:36

@JohnRennie ok.

In this, I know that the last two are correct. I'm having problems with the first two.

John Rennie

That's a surprisingly complicated question because at elementary level we use an approximation called the mean field in which the potential is central. But this is just an approximation and the potential is actually not central.

I don't know what answer your question sheet is expecting.

Wrichik Basu

@JohnRennie I see. I'll mark the first as correct and submit. Once I get the answer, I'll tell you want they are expecting.

Anyways, thank you a lot due sparing out time for these questions. :-)

I'll always be grateful to you Sir! @JohnRennie

John Rennie

11:07

@WrichikBasu only the first answer is correct. The potential is not central because it includes the repulsion between electrons and this force isn't directed towards the centre. Angular momentum is only conserved if the potential is spherically symmetric, and since angular momentum isn't conserved there are no $n$ and $l$ quantum numbers.

ACuriousMind

12:32

@0ßelö7 It certainly defines a physical theory (for integer $p$), but I don't know any actual system that's described by that.

0ßelö7

@ACuriousMind I'm interested in noninteger p

ACuriousMind

Then that's not physical at all, I think

0ßelö7

:/

0ßelö7

12:56

@ACuriousMind what makes integer $p$ a physical theory?

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