you can always add a photon of arbitrarily low energy to a system

so any eigenvalue of $P^2$ is continuously connected to any other

@BernardoMeurer what about the thing that says "pxe- e61: media test failure, check cable"

or maybe not, idk

@obe that's a virus

lmao

ofc it is, a virus on a brand new ssd.

@BernardoMeurer does that mean the ssd was incorrectly installed?

or will it go away once I put the windows 10 usb drive in the computer.

Ahhh, @AccidentalFourierTransform I realize now what irks you.

global warming

I wrote a long reply there how a non-Abelian GUT would solve the issue (since YM theories are believed to have mass gaps), then I realized I don't know how SSB actually works with that. How does a theory with mass gap get broken into one without?

dat shit cray

Hmm, the SSB is perturbative, though, while the YM properties are highly non-perturbative. I'm gonna go ahead and say the issues with the photon are artifacts of perturbation theory :P

this is what i love about physics: i come in thinking i know something and i come out realizing i don't know anything but i want to learn everything =)

I might ask a question about this but I'd need to do some research first and I should be reading about instanton corrections in M-theory instead of doing that...

anyway, I want to remind everybody that I have no formal training in QFT, so most of what I said can and probably is wrong in some way or another

so don't quote me on anything I said

@heather Good luck with that ;) You'll realize that at this point in time, "everything" is so much that you'd need a thousand lives to learn it all.

"Simplify::time: Time spent on a transformation exceeded 300.` seconds, and the transformation was aborted. Increasing the value of TimeConstraint option may improve the result of simplification."

C'mon Mathematica, you can do it!

@ACuriousMind yeah, sadly.

@AccidentalFourierTransform for no formal training you sure contribute to that tag a lot

@ACuriousMind well, I realised that by answering questions I get to learn a lot

so that's something I guess

@heather Oh, the converse would be even sadder! Imagine you got up one day and knew with certainty there's nothing more to learn out there. I love the feeling of finally grasping something I've struggled with too much to give it up.

@ACuriousMind what about a lifetime that perfectly fits the amount of things to learn?

@AccidentalFourierTransform Yes! Trying to communicate something to someone else is a wonderful test of both our understanding and our ability as writers.

now that would be nice.

@ACuriousMind it seems to me that that is the real reason most of us are here to begin with

@obe Ignore that

@obe There's no virus

well, except for the halp me plz people and the people like me who don't know much of anything to explain =) @AccidentalFourierTransform

@heather If it was up to me, "lifetime" would be infinite anyway ;) [That doesn't exclude never stop learning because all the other people alive would produce new stuff faster than you can take it in)

Bernardo, suppose I have a vector field $\vec{v}(x,y)$.

That doesn't mean anything for normal humans

@ACuriousMind true, that's a good point.

@DanielSank Okay

Now I give you a path in the plane defined by $x(t)$ and $y(t)$.

The integral of the vector along that path is

$$\int \left( \frac{dx}{dt}v_x(x(t), y(t)) + \frac{dy}{dt}v_y(x(t),y(t)) \right) dt$$

@heather you are very friendly and make us laugh all the time. You're one of the most important users here ;-)

@heather "You can listen to my lecture on Grassman algebras in fucked up fields", Ryan

@BernardoMeurer the thing in the integral is the dot product of $\vec{v}$ and the tangent to the path at each point.

@BernardoMeurer You're not a normal human, Mr Potato Overlord.

Stare at it.

"This user has been temporarily suspended by a moderator and cannot chat for 1 day 20 hours."

WE'RE ALMOST THERE

@DanielSank Staring

@BernardoMeurer grassman algebras? that sounds interesting. where can i listen to it?

@ACuriousMind I know about individual protocols for BIOS over the network boot, no I'm not normal :P

$dx/dt$ is the x component of the tangent.

etc.

@heather In two days when he's back

@AccidentalFourierTransform thank you =)

@BernardoMeurer oh, okay =D I await his arrival

@DanielSank I don't know line integrals yet, but Okay, I think I get it

@BernardoMeurer You're learning what it means right now.

I give you a path.

@AccidentalFourierTransform "I look forward to meeting you" - @0celo7

defined by $(x(t), y(t))$, i.e. for each $t$ you have a point in the plane.

@0celo7 we've met already

As you sweet the $t$ variable, you travel on the path.

Okay

you answered one of my first questions

At each point, the tangent to the path is $(dx/dt, dy/dt)$, right?

when I was a newbie in this site

Yes

Like, at each point there is a tangent vector $\vec{\text{tangent}}=\frac{dx}{dt}\hat{x} + \frac{dy}{dt}\hat{y}$.

Yep

Why do x and y get a hat?

The winter bash is over Daniel

Right, so the line integral is $$\int \vec{\text{tangent}}(t) \cdot \vec{v}(x(t), y(t)) \, dt$$

@BernardoMeurer Hat means "unit vector".

Alright

@BernardoMeurer lol

@DanielSank Okay, got it

Ok, well that's a line integral.

The curl of a vector field is where at each point, you draw a tiny circle and compute that integral, and divide by the perimeter of the circle.

The curl is the limit as the circle's radius goes to zero.

or, you know, $\partial_y-\partial_z$, etc.

like the rest of us mortals

@AccidentalFourierTransform Yes but to figure that out, you have to start with a reasonable definition.

And we're trying to get to Stoke's theorem.

my definitions are always reasonable

@DanielSank Cool, okay

@BernardoMeurer Ok so now suppose I tell you to sum the curl at each point within some boundary.

Eh

I tell you to compute the curl at each point inside that boundary, and sum them up.

Do a double integral over that line integral?

That integral gives you the curl at each point. Now sum that up as you sweep over x and y.

See the picture.

Okay

Ok notice this: at each boarder between two little chunks, you're going to integrate along that boarder twice, and each time going in a different direction.

The chunk on the left integrates on the boarder one way, but the chunk on the left integrates the other way.

see it?

Yes!

So, what's left in the end is just the integral along the outer boarder where there's no cancellation.

Therefore:

Woah

The surface integral of the curl of a vector field is equal to the line integral of that vector field around the outer boarder.

i.imgur.com/1BtRj6X.gifv

That's the baby version of Stokes's theorem.

me right now

So cool

Yeah, there are similar theorems, called "Gauss's theorem" or "the divergence theorem" and "Green's theorem" that you'll learn next semester.

They're very important.

This seems nice

It turns out that they're all actually special cases of the one samurai theorem: $$\int_{\partial S} \phi = \int_S d\phi$$ where $\phi$ is a differential form and $d$ is the exterior derivative, but I'm not going to even try to explain what that means yet ;-P

When you learn that theorem your mind will collapse into a black hole and become one with the universe.

really cool theorem you got there

You will never have to eat again; your body will be harmonious with Nature and the second law of thermodynamics will not affect you.

yes, I can confirm that

Lol

That's how you're skinny!

@DanielSank What's that funky squiggle next to the "s" in the bottom bound of the first integral

@BernardoMeurer $\partial S$ means "boundary of $S$".

@BernardoMeurer its a poorly drawn penis

@DanielSank Boundary in what sense?

@BernardoMeurer In the intuitive sense.

$S$ can be any dimensional object as long as it has a boundary.

What's an object here?

A space in $\mathbb{R}^n$.

@ACuriousMind u around?

i was thinking about the normal ordering thingy

@AccidentalFourierTransform I've answered it just now

yeah i know

the thing is

OP's normal ordering is the CFT ordering, not the QFT one, right?

I mean, its not the "a to the left" operation

right?

or is it?

it is

i dont know shit about CFT

expand the field into its modes which have the CCR of creation/annihilation operators, then the CFT normal ordering corresponds to ordering the modes

but I thought that normal ordering there was something different

There's some subtlety but the idea is the same

something about radial ordering or something like that

or that is the $T$ ordering?

@DanielSank no reply re last question on bloch spheres. can you/ someone answer a question? isnt (roughly) measuring a point on a bloch sphere (in an experiment) forbidden by standard QM?

@AccidentalFourierTransform That's the time ordering, time=radius

right

k thanx

It's only confusing until you realize that the "punctured plane" is just a cylinder :P

Never understood the radial thing before I saw that

ooh, i think i may have just earned my second gold badge =)

@DanielSank looked this up last 2 nites, am very chagrined that seem to have lost one of my fave refs, am gonna try to re-find it but think that will be very difficult. anyway one that comes near is "the meaning of quantum theory" by baggott, in the section "the postulates of QM" & there are other similar refs... again am not saying there is anything unusual there wrt other refs, just like how its formulated... global.oup.com/academic/product/…

kbye

@BernardoMeurer this is taking longer than expected

Greetings, hbar