The h Bar

12:35 AM

Oh my god

Apparently chapter 5 of that book is about the first order in perturbation theory in QED, which every QFT book proves is impossible, however, these maniacs point out it's non-zero if you allow for a third 'external' field... What the...

amazon.com/Quantum-Electrodynamics-Second-Theoretical-Physics/…

(Why wont a pic of the cover show up, sometimes it does?)

dmckee

12:52 AM

@ACuriousMind It might be worth pointing out in this context how much hidden structure there is even in Newtonian mechanics—it takes something approaching genius to get from Newtonian to Lagrangian and Hamiltonian mechanics without hints—but suddenly there are these completely new structures that expose hidden symmetries.

Rumplestillskin

1:26 AM

Hello one and all! Can anyone point me in the right direction to what is going on for Eq's (A7) - (A12) in the following manuscript? journals.aps.org/prd/abstract/10.1103/PhysRevD.79.084027

I am looking at the specific case where $t_{ij}$ has equal indices

I don't really understand how I can choose a suitable $t_{ij}$ in terms of $w$ and $w_j$

David Z

I can't access the paper

Do you have an arxiv ID for it?

Cows

4:54 AM

Ok so i had mexican food, thwn we watched jurasic world at the movies. Now at my date's place for some night cereal ;p . Btw the system analyst guy feom jurasic world was awesome. As one of the teeth gap minorities out there, I was excited to see that we were represented in the movie. There was also the part where he wore the head thingy people in clean rooms and labs wear. Pretty neat stuff hehe ;P . . . . At any rate going to be here for a while. :^P

Slereah

7:28 AM

morning

Mikhail

My seven layer integral isn't converging, how will I get a noble prize, now :-(

Slereah

My seven layer lasagna isn't converging

Mikhail

Oh no! You better go catch it.

Slereah

Mama mia!

Mikhail

I believe you'd say oye vey in Mexico

Novalium Company

7:46 AM

Guys, in the Joule-Lenz law, where $P = I^2R$, the power is basically equal to the amount of heat $Q$ given away by the wire for example, since the energy doesn't have anywhere to go? Is that right?

Slereah

Pretty much

I mean heat and light, really

Novalium Company

Yep, what about sound?

Why no sound is generated?

Slereah

why would it be

Novalium Company

xD You're right. There is nothing to vibrate and generate the waves.

Slereah

there's nothing that would compress the air around

Novalium Company

7:49 AM

I have another question. I'm reading a book on electronics and it says, more current will generate more heat, but what about the resistance, if you have high resistance, wouldn't that result in more heat as well, since the electroncs are bumping and giving out energy?

Mikhail

Yes, resistance goes up as temperature increases.

Novalium Company

So from the formula $P = I^2R$ we simply understand, that as current or resistance go up, the amount of heat given goes up as well. That's cool.

Mikhail

^ Lol, you can't do that. You need electron scattering.

When George Simon Ohm's theory of electricity was published in 1827 in his book "The Galvanic Chain, Mathematically Worked Out," it was called "a web of naked fancies."

Also why does my rep show up as 10k? is that some kind of bug?

Novalium Company

So overall guys, my assumptions on amount of heat given out from a wire (for example) and how it relates to the current and resistance are right? Well, basically the formula says it all so. @Mikhail it's not a bug, I guess.

Slereah

@Mikhail that's the same name as my website

Cows

8:43 AM

@Slereah hehe nice :D , that video is quite catchy. Hopefully I get to Fadeev-Popov ghosts soon :P

Slereah

8:59 AM

what video

Secret

Composition of step functions with step functions does not yield functions that are continuous in some nice way

And.. it gets worse

Let $f$ be a nonwhere continuous function as shown by the scatter plot above and $g$ be a step function. Then:

$g(f)$ is nowhere continuous except when it is zero

So whatever TheGreatDuck is trying already fails for step functions

Novalium Company

9:16 AM

Guys, I'm having a very simple problem but I seem to be stuck. In a circuit, if the resistance increases 2 times, then what happens to the voltage? How to write it mathematically with $V = IR$?

Secret

R'=2R

V'=IR'=2IR=2V

Jun 12 at 18:00, by Balarka Sen

vzn will paradigm shift you

Novalium Company

Increasing the resistance 2 times, increases the voltage 2 times... give this man a medal.

Increasing the resistance drops the voltage and viceversa.

@Secret Sorry but that must be wrong?

Secret

wait, which variable do you held constant?

Novalium Company

I don't know, my textbooks says that increasing the resistance, lowers the voltage and I was just wondering if I increase the resistance 2 times, what happens with the voltage and how to write it mathematically.

I suppose current is constant.

From there V/R = V'/R' where R' = 2R?

Ahh...I got confused.

Secret

You start with V=IR, and I is held constant. Thus doubling the resistance gives R'=2R, thus V'=IR'=I2R => V'=V/R*2R => V'=2V

Novalium Company

9:29 AM

Meh, ok then.

When current is constant I guess that's what happens.

ACuriousMind

@Mikhail Chat shows the total rep you have across all sites

@dmckee Indeed

Novalium Company

I proved it in another form (which I guess is the same, just another way written):
If $I$ is constant then $I = I'$. From there we have $\frac{V}{R} = \frac{V'}{R'}$. If R' = 2R then we have $\frac{V}{R} = \frac{V'}{2R}$ solving it equals $V' = 2V$. Is that a right way to do it?

ACuriousMind

@NovaliumCompany Sure :)

Worth noting that many real-life electricity sources are constant voltage, though

Novalium Company

Btw when we have AC, then that means the Voltage is alternating as well?

If the voltage is constant, then the resistance is alternating :D?

ACuriousMind

Ehhh, no, sorry

Maybe let me try to wake up first :D

Novalium Company

9:36 AM

Sorry :D

ACuriousMind

Okay, so: In AC, both current and voltage are alternating

Novalium Company

Ok, what about if the voltage is constant? Then maybe the resistance will have to alternate?

ACuriousMind

@NovaliumCompany Well, in a hypothetical situation where the current alternated but the voltage doesn't, yes, the resistance would be alternating

Novalium Company

@ACuriousMind Hm, ok. By the way, were you the one that advised me to read instead of watching videos?

ACuriousMind

That's not what happens in standard "AC" though

@NovaliumCompany I don't like videos and am all for reading but if it works for you, don't mind me.

Novalium Company

9:41 AM

No, indeed I listened to your advice and it's been working perfectly. I actually understand the stuff much better because they are going through my mind. The things Vitasium says in his video is actually true.

It may need more effort to read, but the results are better.

Soo yeah, thanks for the advice :D

ACuriousMind

Alright, happy to help :)

Novalium Company

Do you prefer e-readers or books?

ACuriousMind

I aesthetically like books more, but I usually find reading on my tablet more convenient because then I don't need to carry the physical books with me

Novalium Company

I bought a Kindle 5 and it's pretty awesome. It's like reading a real book, it's light and eyes don't hurt.

Ok I'll go for now. See you later :)

Slereah

10:03 AM

amazon.com/Geometrodynamics-J-Wheeler/dp/B0000CLJ20/…

Amazon don't got it

dang it

Secret

128

Q: Should I punish my teenage sister, whom I have full custody of, for lying to me in order to secretly see her boyfriend?

I'm 24 years old and have full custody of my 16-year-old sister. She has been living with me for the last 3 years. I found out today, from her, that she has been lying to me about spending time with a female friend in order to spend time with her new boyfriend. She confessed this to me because sh...

hmm...

We seriously need a culture that is more honest and open. That would have solved 70% of the world's problems

Secret

10:23 AM

The End of An Infinite Series

►

2 days left before the Turning Point is reached

Not surprising

On 27/6/2018, at least our state of Australia will be roasted by unions reaching boiling point

We don't know if there will be anyone alive afterwards

And if that day does not melt us,then November surely will. Trump is going to visit Australia on November, and the radicals in the country are ready to protest nation wide

It's still like 2 more years before 2020, and things are already spiraling out of control

ACuriousMind

@Secret That strikes me as an exceedingly grim thing to say. Can we lay off the doomsaying a little?

Secret

Well, maybe doomsday sounds like an exaggeration, but a law that basically ban protests and unionisation was passed in NSW just 3 days ago, thus many are preparing for Right to Strike protests, and also a recent survey showed a rise in people engaging in socialist and activist activities

ACuriousMind

Protests don't mean everyone is going to die.

Secret

10:40 AM

Well, that might be true. Despite the radicals keep going on the TV and news recently due to some education and anti-ban protest campaigns, the state as a whole seemed pretty ordinary and unaffected

so yeah, sorry for the unintentional hyperbole...

Slereah

79

Q: Is 💩 (Unicode 'pile of poo') considered NSFW?

Can I use the 💩 character as a legitimate status indication on, say, a web application, or in a desktop application? Or will it offend/embarrass people? NSFW = "Not safe/suitable for work", although it's also become generally used regardless of the environment (for example: public places), and ...

indicator special-characters

Secret

http://www.nswccl.org.au/council_for_civil_liberties_condemns_regulations_allowing_for_bans_on_public_gatherings
There are no reports of this in any main media however, thus apologies I cannot find a more objective source

💩 is orders of magnitude away from NSFW

Physics Meta

1:46 PM

-1

Q: Light passing through Colour filters

White light is first passed through a green filter. Then it is passed through a red filter. It then finally falls on a white paper. Could you please say what the colour of the paper will be?

discussion homework-and-exercises

vzn

2:39 PM

lol Sh*t Is Hitting the Fan Over the New Poop Emoji nymag.com/selectall/2017/11/…

bolbteppa

4:58 PM

Nuts: Maxwell's equations are simply conditions that some wave function satisfying some Schrodinger equation has to satisfy :o Turns out the Schrodinger equation in momentum space is $$i \frac{\partial }{\partial t} \hat{\mathbf{f}}(t,\mathbf{k})_a = k(\delta_{bc} - \frac{k_b k_c}{k^2}) \hat{\mathbf{f}}_{\mathbf{k}}(t,\mathbf{k})_{c}$$

for $\hat{\mathbf{f}}$ the coefficients in $\mathbf{E}_{\mathbf{k}} = N(k)[\hat{\mathbf{f}}_{\mathbf{k}} + \hat{f}_{-\mathbf{k}}^*]$ and $\mathbf{E} = \int \mathbf{E}_{\mathbf{k}} e^{i\mathbf{k} \cdot \mathbf{r}}d\mathbf{k}$

(That $a$ should be a $b$)

Coulomb propagator sitting happily in there as part of the Schrodinger Hamiltonian

Turns out the normalizaton $N(k)$ has to have a factor $\sqrt{k}$ if you want that Hamiltonian to actually be the energy of the EM field, and so if you try to Fourier invert the $\hat{\mathbf{f}}_{\mathbf{k}}(t)$ to find the coordinate space wave function, that $\sqrt{k}$ becomes the square root of a differential operator, which is an integral operator over a region, i.e. non-local,

so you simply can't find a coordinate space wave function to locate the photon exactly, only within a region, destroying normal QM thinking of wave functions already, however the momentum space interpretation still makes sense

Apparently can use Heisenberg uncertainty relations to argue all this would happen too

Bob van de Voort

5:51 PM

Is there anyone that could maybe have a look at this question?
https://physics.stackexchange.com/q/410790/160526

0

Q: The Purcell effect, it's influence on the lifetime and quantum yield (of fluorophores)

So I've been looking into the Purcell effect and how it interacts with fluorophores (fluorescent molecules). The Purcell arises when you have a dipole in a cavity or even just near a dielectric or metal interface. It can then enhance the emission of the dipole in comparison to free space. Now fo...

optics resource-recommendations quantum-electrodynamics quantum-optics

Mithrandir24601

6:22 PM

@BobvandeVoort I don't know the answer to this of the top of my head but for the sentence "saturation in the sense that they can not absorb a new photon if they're still in the exited state" - this sounds like it could be modelled by the Jaynes-Cummings Hamiltonian

Which should simplify things considerably

It's really annoying because I'm sure I've came across something exactly like what you've mentioned but I can't seem to find it again :/

Slereah

7:16 PM

"Just to set the record straight, in this article we will be using the convention that the Lorentz scalar product that defines a Minkowski space of dimension n has one plus sign and n-1 minus sign"

Reee

bolbteppa

Is there an intrinsic reason why you even need to set up second quantization for the free electromagnetic field at all

Slereah

7:55 PM

Well I mean you don't have to

but using the particle basis for the hilbert space makes things much easier

You could just use the coherent state basis instead

Mithrandir24601

@bolbteppa From a practical point of view, single photon detectors can only 'detect' within a certain frequency range, so this kinda makes sense practically. Sounds like you're talking about something much more theoretical though :P

bolbteppa

Seems like you can't avoid the fact that the free Dirac equation is multi-particle because of anti-particles, with second quantization just a basis change, yet you can avoid all this for the free EM field and actually treat it like baby QM, though only in momentum space (already radically messing up normal QM), but if you want to do interacting qed you must put the EM field in a second quantization basis

Slereah

Well you don't have to, unless you're doing perturbation QFT

(ie most of QFT)

Since that's the whole perturbation QFT idea

bolbteppa

qed alone is nuts

Slereah

Since we don't know what's the actual Hilbert space of QED (if one exists) can't really use much better

bolbteppa

8:04 PM

I guess you could formulate it in terms of multi-particle wave functions, but you have to go multi for interacting qed because the Dirac wave functions in the current are multi-particle

Slereah

Well you don't have to use the notion of particles at all

It's just practical

You could just use the Hilbert space $L^2(D(M), \mu)$ instead

It has no obvious particle interpretation

(the link being the Fourier transform)

Slereah

8:39 PM

You know I wonder if there's some non-linear stuff that has a Backlund transform as a Hilbert space addition

worldscientific.com/worldscibooks/10.1142/2345

maybe related

bolbteppa

9:22 PM

Getting more nuts

Lets see if this makes sense:

Ahh

Slereah

So do the nuts make sense now

bolbteppa

9:50 PM

Seems like Bjorken and Drell in volume 1 set up the Dirac equation as the Schrodinger equation, where we just want to solve Schrodinger equations because we're doing QM, and they note anti-particles force a multi-particle perspective. Then when moving onto interactions with the EM field they still just treat it as a Schrodinger equation problem, for potential $e \gamma^{\mu} A_{\mu}$,

but now the Maxwell equations function as a condition this potential has to satisfy, so you get two coupled equations, Dirac and Maxwell. Solving Maxwell by a Green function and then plugging this into the Dirac equation you get back a single Schrodinger equation problem, and you know nothing more really, only kind of know this will be a multi-particle problem.

From this you 'solve' the Dirac equation with a Green function, and then expand the solution with free particle Dirac solutions, and free photon wave functions, and it gets huge, but it's all just a Schrodinger equation really

Since it's all just a big Dirac equation, you can alternatively transition to the Heisenberg picture, and then change basis to get second quantization since you know it's multi-particle and not live in fear of this, and then go to the interaction picture, but you get results like Compton and Mott pretty quick by using the first method, a lot less confusing!

Looking at the transition from Schrodinger to second quantization, seems like the first step is basically actually going to the Heisenberg picture if you view it a certain way, such a mess making sense of this

Bob van de Voort

10:22 PM

@Mithrandir24601 Thank you so much, I'm going to look into that, thanks for the idea

@Mithrandir24601 Ps if you do remember, please let me know ;)

Mithrandir24601

@BobvandeVoort Yeah, I might stumble upon it again in the future if we're lucky! No problem! :)

Jack Clerk

11:41 PM

Excuse me fellows, but here is my question: physics.stackexchange.com/questions/413525/…

ACuriousMind

May 19 at 13:51, by ACuriousMind

@user187604 For the record: Please don't advertise newly posted questions in chat unless you have a specific reason to believe it is of interest to someone here. The volume of questions our site gets is so high that, if everyone did that, the actual conversations in here would be drowned out by such advertisements, so we prefer that users don't do that at all.

Jack Clerk

ok

pardon

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