The h Bar

@bl00 you should compute that derivative

it's a good one

user218912

maybe later, need to hand in this pset tomorrow

user218912

@0celo7 if I want to show the vector field lagrangian is invariant under gauge transformations

3:19 AM

transform the gauge and see what happens, clearly

user218912

do I just write out the field strength tensor in expanded form and plug in the gauge transformation?

oh the field strength

well

user218912

so like

$A\to A+df$, right

user218912

yes

3:20 AM

and $F=dA$

so $F\to d(A+df)=dA+d^2f=dA$

user218912

wait what form is that?

what is there to show?

user218912

I have it with indices

$F$ is the pullback of the curvature to the base by a local section

$A$ is the pullback of the connection

(by a local section)

user218912

can we just use basic indices and no geometry.

3:22 AM

sigh

then $A_\mu\to A_\mu+\partial_\mu f$

user218912

yes

So what is $F_{\mu\nu}$

user218912

$\partial_\mu A_\nu - \partial_\nu A_\mu$

which transforms to?

user218912

$\partial_\mu (A_\nu + \partial_\nu f) - \partial_\nu (A_\mu + \partial_\mu f)$

3:24 AM

write that out

user218912

I have it on paper

so what's the issue?

user218912

wait I'm stupid

the parts with $f$ cancel by Clairaut

user218912

clairaut?

3:25 AM

partial derivatives commute on $C^2$ functions

user218912

oh

user218912

I was doing something completely wrong xD

user218912

I multiplied out the lagrangian

user218912

and got like 30 terms

what?

I see you didn't read Zee.

user218912

3:26 AM

it's in zee?

everything is in Zee

user218912

I thought everything was in shankar.

All the physics I know is from Zee or Shankar

@bl00 it's probably there too tbh

I know it's in Sakurai

user218912

sigh...

user218912

I'm so bad.

user218912

3:29 AM

well I guess after a billion cancellations and some factoring my method would have worked too.

user218912

but proving that $F^{\mu\nu}$ is invariant suffices.

if you knew some geometry this would be trivial...

user228700

@DanielSank Huh. It didn't make much sense to me. I mean, it sort of did but I kept wondering if this was legit...

user218912

it was trivial already basically.

user218912

just didn't see it how trivial it was

user218912

3:40 AM

@Obliv hey

@0celo7 wasting my time watching a playthrough of the last of us

what is up @bl00 @0celo7

user218912

working on stuff

user218912

just watched the new persona 3 movie

user218912

it was really cool and awesome.

@Obliv compute that derivative

3:41 AM

@Kaumudi When you learn general relativity, then statements like that will make sense.

Don't worry about it too much now... unless you want to :)

I will make one little comment...

@0celo7 would I have to differentiate both possibilities?

@Kaumudi consider yourself sitting in a car. If the car is not accelerating forward, you do not feel a push on your back. If the car accelerates forward, then you do feel a push on your back.

No acceleration->no push
Acceleration->push

Now suppose you are standing up on your two feet.

@Obliv try it

You feel a push up on your feet from the floor, even though you are not accelerating.

However, if you jump off of a house, while you are accelerating downward toward the ground, you do not feel a push on your feet.

user218912

@0celo7 that derivative is confusing

3:43 AM

No acceleration->push
Acceleration->no push

See, @Kaumudi, the situations are reversed!

How can this be? We know F=ma, so you should always feel a push when you are accelerating.

However, when you are falling off the house, it appears F=ma does not work!

Yet, you know that the trajectory of the falling person does follow F=ma where F is the force of gravity on the person, i.e. the person's weight.

@bl00 why?

@0celo7 never really computed a piece-wise derivative before.. how does this work if I'm not given the function definition in the form : $f(x,y) = ax^n + by^m + ... $?

user218912

@0celo7 idk how to do it...

user218912

@Obliv it is defined

:)

user218912

3:48 AM

wow sadist.

@Obliv btw here $df(x)$ is the gradient

@bl00 wait what are you saying $f(x,y) = (x,y) \text{if (x,y)}\in \mathbb{R}^2 - A$?

@Obliv You can't use the usual formulae, but the derivative is certainly still defined the same old way

user218912

@Obliv yes I think that's right.

@0celo7 isn't f(x) a vector function? A gradient is for scalar functions I thought

user218912

3:49 AM

no

user218912

it's for both.

@Obliv generalized gradient

@bl00 eh technically he's right

user218912

oh I thought he said nabla operator.

user218912

sorry I mixed that up with the specific case of gradient.

@Obliv well you know what the Jacobian matrix of a vector function is, right?

3:52 AM

$\frac{\partial f}{\partial x} = \left\{ \begin{array}{11} (1,0) & \quad \text{ if (x,y)}\in \mathbb{R}^2 - A \\ (1,2x) & \quad \text{if (x,y)}\in A \end{array}\right\}$ ?? @0celo7

something like this? then do the same thing for partial y?

hmm

maybe that works lol

@0celo7 matrix of 1st order partial derivatives right

well

that works on the surface

but it's wrong

problem is that $f$ is not continuous everywhere

so you can't write down the partials like that

@Obliv do you give up

also

$\Bbb R^2-A$ is closed

so you can't take partials on it!

so yeah, that's wrong

@0celo7 I don't follow. what does it being closed have to do with anything

you need a function defined on an open set to take the partial derivatives.

you can't just take partial derivatives of piecewise functions

you have to use the definition of the derivative

4:01 AM

@0celo7 you mean the limit definition?

yes

that looks too messy to write with latex. gonna give it a try

@0celo7 okay I think I'm doing this wrong.. Are you sure you can just take the derivative?

f(x,y) = (x,y) doesn't this imply the image being dependent on two variables? okay i give up give me the derivative lol

wait

did I not give the derivative

it says it right underneath, no?

just on the origin though

can't you take the derivative on the domain?

what?

4:09 AM

uh nevermind. So this function was only differentiable at the origin?

it's differentiable...definitely on the interior of $\Bbb R^2-A$

the origin

on $A$

nowhere else probably

do you wanna see how it works?

@0celo7 sure.

Since $f(0)=0$ and $\mathrm d f(0)[h]=h$ for each $h\in\Bbb R^2$, we must check the limit
\[\lim_{h\to 0}\frac{||f(0+h)-f(0)-\mathrm d f(0)[h]||}{||h||}=\lim_{h\to 0}\frac{||f(h)-h||}{||h||}=0.\]
We will compute this limit sequentially. Take a sequence $(h_n)\subset\Bbb R^2$ which has limit $0$ (but is never actually $0$). There are three cases: (i) a finite number of terms are in $A$, (ii) a finite number of terms are in $\Bbb R^2-A$, (iii) there are an infinite number of terms in both regions. We will treat case (iii), the other cases are similar. Let $(h_{n_j})$ be the subsequence contai…

what is $df(0)[h]$

the Jacobian at $0$ applied to the vector $h$

user218912

4:13 AM

@Obliv watch this from the persona 3 movie. spoilers btw.

Oh my. I will have to take a look at this again tomorrow. have a c++ exam tomorrow morning so I have to rest.

good night nerds @bl00 @0celo7

not a nerd

user218912

night.

@0celo7 Could you help with this question physics.stackexchange.com/questions/286870/…

user218912

oh I was reading a&m

4:25 AM

@Bosnia why would you ask me?

I'm a mather

Oh sorry then. I just wanted to share it here.

user218912

I dunno how the 5 appears either.

ah

you have to take the charge over a magnetic 5-brane

which will be dual to the 6-dim CY compactification

user218912

wow it's that obvious?

what is $c$?

4:30 AM

@0celo7 are you asking me?

both of you

$c$ is just the speed of light

then your answer and the book answer have different units

so something is already wrong there

Yeah. I'll see if someone answers.

user228700

4:57 AM

@DanielSank OK..? (Aw man, I'm falling off a house? :-P) This difference is...a consequence of GR?

5:08 AM

@Kaumudi Yes.

Gravity is not a force

For the purpose of simple mechanics problems etc., you can treat gravity as a force, but in the very end, when we really get into the nature of gravity with general relativity, gravity is not a force.

@privetDruzia Привет

[Division by zero] Two known ways to divide by zero:
1. Having the + isomorphic to the left or right null semigroup. The Zero power laws will then ensure a natural "Rock Paper Scissors" structure to emerge. This can be tied as a subalgebra with the real numbers, except by modifying the idempotent condition to one such that any element in the subalgebra adds to zero (thus the natural ordering of the zero power laws will mean that zero become the limiting superior element, i.e. the always win hand of a rock paper scissors game)

user228700

@DanielSank Hm, OK. I will pursue this line of thought later on. Thanks :-)

user228700

@JohnRennie: Morning! :-)

@Kaumudi Morning :-)

One thing that one must bear in mind that, due to the nature of the zero power laws and that one is the successor of zero, a division by zero algebra can NEVER have an additive identity. The best you can have is an element that is an additive identity for all except one element

To illustrate this, begin with the ground level of the zero power law, which is our usual relation in defining the additive identity:
$$0+1=1$$
Multiply both sides by 0
$$0^2+0=0$$
Therefore, the moment 0 is determined to be suceeded by 1, $0^2$ will be suceeded by 0 hence 0 cannot be the identity for at least one element in the algebraic system

All of the above findings will be updated into the blog shortly in a more coherent fashion for future reference

Q: How exactly do I distingish an interpretation from computation?

5:27 AM

it is getting annoying... it may not be a good question, but it get downvote without telling me why, while other questions obviously breaking the rules, no one downvotes them. the same happened last time I answer a question (where I only stated a fact).

-1

Feynman wrote this in his Quantum Mechanics and Path integrals To summarize: we compute the intensity ( ... ) of waves which would arrive in the apparatus at x and then interpret this intensity as the probability that a particle will arrive at x I found myself a hard time distinguishing...

soft-question

@Shing: I think you're mixing up two different uses of the word interpret

@JohnRennie I would love you to tell me how, this has been bothering me for quite a long time.

In everyday use the word interpret just means to assign a meaning to

So Feynmann is just saying that we take $|\psi|^2$ and assign it the meaning of probability density

However when we talk about interpretations of quantum mechanics this is a vastly bigger subject.

It basically means assigning a meaning to the whole theory of quantum mechanics, not just one little bit like the physical meaning of $|\psi|^2$

It wasn't my downvote, but due to this confusion of the two uses of interpret I don't think your question is very meaningful.

Okay, I see, I will try to edit the question.

thanks for the help

just looking at the other current questions.... I am the only one resulted in one downvote. (another one was given by me)

some of the other questions obviously breaking the rules...

it feels bad as if being treated unfair.

5:46 AM

I wouldn't attach too much significance to downvotes, especially when no comment is made. I often get downvotes on my answers with no comment even whenthat answer has lots of upvotes. Maybe the downvoter was just in a bad mood.

Generally speaking the high rep users will explain in a comment why they think a question needs to be improved, and it's the high rep users who are generally most reliable.

Thanks for telling me. I guess it takes some time to get used to.

user228700

6:44 AM

@JohnRennie: Are u familiar w/ multiplicity of bonds?

@Kaumudi as in single bond, double bond, triple bond?

user228700

Yeah, I think so...

user228700

I'm unable to figure wth my book is trying to say:

user228700

user228700

(ii) point

6:47 AM

That's just saying that all else being equal single bonds are longer than double bonds and double bonds are longer than triple bonds.

user228700

Ah, OK...

user228700

Thankoo! :-)

user228700

7:52 AM

Anybody familiar w/ the parametric equation of a straight line?

isn't that just y=mx+b, or are you talking about vectors?

Emilio Pisanty

how on Earth is this a duplicate?

2

Q: Mirrors into Infinity

Could someone please name the phenomenon regarding the stretch of reflections into infinity between two opposing mirrors, and also explain why the reflections curve away instead of meeting at a perspective point within the reflections?

optics reflection

user228700

@Secret I actually dunno, that's the problem. (BTW, that ^ is the slope intercept form of a straight line)

user228700

The parametric form is often just given as:

user228700

$(x-x_1)/(cos\theta) = (y-y_1)/(sin\theta) = r$

user228700

8:03 AM

Where $(x,y)$ is any point on the straight line, $(x_1,y_1)$ is a known point, $r$ is the distance b/w the known point and $(x,y)$, and $\theta$ is the angle that the line in question makes with the positive direction of the $x-$ axis.

@Kaumudi what do you want to know about it?

user228700

The thing is, there are two such points! One above, and one below and this hasn't been accounted for in this equation...of course, when we're talking about just one line, either of the two would give the same line, but there was a question where I had to account for both, as if they were vectors.

user228700

So now I'm very confused.

user228700

This was that question:

user228700

8:13 AM

(C-5)

user228700

I solved it and everything, but I had to use this directions thing...

user228700

OK, hang on, I think I understand. When solving problems related to a given direction, I thinknI might have to account for both. On the other hand, if it's just the straight line equation they're asking for, I don't have to put a ± in front of $r$. Alright, I think this is correct but I dunno. Thoughts..?

Why would you use a parametric equation for the straight line? e.g. for the first step you know $\Delta x^2 + \Delta y^2 = (5\sqrt{2})^2$ and you know $Delta y/\Delta x = \tan(135) i.e. \tan(3\pi/2)$.

That looks like a typical orientation problem, you use the angles given to compute the trig functions, and then the distance of travel is basically the r in the parametric form of that equation. Sicne the angle go over all 360, $\pm$ is accounted for by the direction and hence r must be positive

I recall back then I just draw crosses for each orientation point, and then compute the angles when going from one point to another to resolve the x and y components of each travel displacement

user228700

@JohnRennie Yes, we could do that too. Using the parametric equation seemed like a shortcut tho.

8:18 AM

@Kaumudi seems like a long cut to me

user228700

@JohnRennie Not really. U know the angle and u know the point. Substitute and then BAM, u've got ur new point.

user228700

No need to solve anything B-)

yup, that's how I do it back then, but I never made use of the parametric form, jsut directly find triangles, notice which quadrant they are in, and then compute trig

Ok, now add in the time you spent scratching your head wondering what the parametric equation meant in this context ...

user228700

@JohnRennie But at least now I know! :-P BTW, tho, I spent like an hour on this problem, "scratching my head"; I discussed this at the MSE chat so ur method wins but still, now I know!

user228700

8:21 AM

@Secret Riight. Well, this question is given in the chap. for straight lines so I went ahead and did that.

user228700

OK, thanks, guys! :-)

Swapnil Das

What's the rigorous definition of deviation in ray optics?

@JohnRennie Could you help sir?

user228700

9:56 AM

Is anybody familiar w/ the properties associated w/ the centroid of a triangle? I wrote down/drew this but now I've got no idea what it means:

user228700

(G being the centroid. I've no clue as to what E and D are :|)

Q: Why would an Omicron ISE 10 argon ion sputter gun be limited to 100V?

10:16 AM

@SwapnilDas Ray optics isn't a subject I know a great deal about, but I don't think there's anything complicated about the concept of deviation. It just means the departure from a straight line. Typically the light ray will end up at an angle to the original ray so the deviation is an angle.

David Z

Thoughts as to whether this is on topic?

0

We have an Omicron ISE 10 argon ion sputter gun, and we could only put about 100V on the anode before one of the controller buttons started to flash which apparently means the current is too high. No bursts in pressure were seen, so it is assumed that there is no breakdown occurring, but there is...

experimental-physics vacuum experimental-technology

It seems like it could be a reasonable experimental question, but I'm not sure.

William Budd

As a non-physicist, physics never fails to confuse me. This time it's Maxwell's Demon. Yes, I'm stooping to shameless self-promotion of my question here: physics.stackexchange.com/questions/286915/…

D. Ennis

11:05 AM

Can I just start a new topic here, or do I need to "get a room?"

heather

11:43 AM

@D.Ennis, just start a new topic.

12:24 PM

Vacua Morghulis

2

I know Ben, he's a funny guy :D

12:35 PM

@Danu lol

I don't get it

What does Morghulis mean and why is it a good name for a paper?

It's a pop culture reference, so no surprise there :P

...why the hell would you make your tile a pop culture reference.

You think Witten watches Game of Thrones?

Our work on division by zero algebra. It give us some idea how to think about possible genersliations of rock paper scissors commutative magmas. I am not sure if I have use the maths term correctly, through

user228700

@heather: Hi :-) Are u familiar w/ that property of triangle? I remember it being taught in middle school...

12:40 PM

@0celo7 I dare not speculate on that

user228700

Please. 🙏 Is anybody familiar with that property?

What property

user228700

3 hours ago, by Kaumudi

user228700

I have no idea what those two points $E$ and $D$ are supposed to be. $G$ is the centroid.

@Kaumudi D and E are just points on any line through G, i.e. the intersections of a line through G with the side

The property you wrote down says that no matter which line you choose, that sum of ratios there is always 1

12:44 PM

@ACuriousMind wtf one day you don't know any geometry and then you're basically Yau

user228700

@ACuriousMind THANKS! 🙏

@0celo7 To reassure you, I don't actually remember how to prove that statement. Haven't done stuff with triangles outside of Feynman diagrams for ages

@ACuriousMind I've never even heard of that statement.

I never learned Euclidean geometry

What even is a centroid?

centroid is where all moments of area of the shape balances

What the hell is that

12:54 PM

for an object with uniform mass density, it coincides with the center of mass

@0celo7 The center of mass of the triangle (if it had uniform mass density)

@ACuriousMind ...I don't remember what that is

@ACuriousMind do you have a good book for Euclidean geometry?

Uh

I should maybe learn it

I learnt this stuff from German school textbooks way back

12:59 PM

Not all of us were so lucky to grow up in Germany

1:22 PM

Hi

hey

I asked a question in morning and now that I have an answer I am having trouble interpretting it physics.stackexchange.com/questions/286909/…

I just forgot about the time zones. I asked it before 7 hours

Can someone help?

user228700

@0celo7: To put it simply, the centroid of a triangle is the point of intersection of the three medians from the vertices of a triangle. A median is the line segment connecting one of the vertices to the opposite side, bisecting that side.

How did this get four stars

23 hours ago, by DHMO

$\Huge\color{purple}{\LaTeX}$

@ParthMaske You are claiming that somehow the convergence of the series implies a discreteness of space/physical quantities. The answer is saying that that doesn't make sense - if the values physical quantities could take were discrete, no series of physical quantities could ever converge

@DHMO Because people had fun with \renewcommand and made it say all sorts of different things by redefining \LaTeX

1:28 PM

@ACuriousMind What did you do?

I didn't do anything!

@ACuriousMind You didn't have fun with \renewcommand?

Nope

alright

I just stood by and watched

1:29 PM

I see

Heh, this paper I'm reading uses "The theory is Higgsed" to say "the gauge theory is spontaneously broken"

@ACuriousMind Do you also have the feeling that chat has recently been "taken over" by a lot of people who have never been here before?

@ACuriousMind That's fairly standard terminology

Also Higgs branch

@Danu I do note that the group of regulars seems to be shifting

@ACuriousMind It seems like a very acute change to me: It started after the elections ended.

@Danu I have heard it, but I hadn't seen someone consistently write it in a paper before

1:33 PM

And it's the same people in the math and physics chat rooms.

Not contributing anything essential in either cases IMO but that's just me being cynical :P

@Danu and we meet again, in a different room

o/

@Danu You need to accept that the times are a-changing ;P

@ACuriousMind I've accepted it :)

What kind of stuff are you trying to learn about nowadays?

@Danu Generating chiral matter and non-Abelian gauge theories in 4D from compactification of M-theory (well, 11D SUGRA, really) on $G_2$-manifolds

It's the 11D analogon of the CY compactifications of the 10D string theories

But a lot less well understood because we can't use the complex algebraic geometry we use to control the CY-folds.

Unrelated to this, I'm also taking a course on D modules and D sheaves

Which was sold to me as "differential equations for algebraists" ;)

1:38 PM

What are "D" modules?

The D is some ring of differential operators

In the simplest case, it's all operators of the form $\sum_i p_i(x) \partial_x^i$ for polynomials $p_i$

@ACuriousMind Okay.

So why do physicists care about $G_2$?

Also do you have a nice description of $G_2$? The one I know is very inelegant: The stabilizer of some sorta-random 3-form on a seven-dimensional vector space.

@Danu Because it's the possible special holonomy in 7D, and just like with the SU(6) holonomy of the 6D CYs, such special holonomies are necessary for the existence of covariantly constant spinors, which in turn are necessary for the compactified theory to retain some supersymmetry

@ACuriousMind Ah, okay. Spin structures somethingsomething

Aren't people looking into non-SUSY compactifications by now? Or is that just so wild that there's no point?

@Danu That sorta random three-form is actually encoding the multiplicative structure of the octonions. The slickest definition of $G_2$ is simply as the isometry group of the octonions as a normed $\mathbb{R}$-algebra.

1:44 PM

@ACuriousMind I see.

Was pretty shitty to work with, haha.

physics.stackexchange.com/questions/286970/… For starters, light signals will deterioate, thus pretty much useless to make a time capsule that way

Q: Relativity: 2 particles moving at different speeds and time taken to close the gap between them

0

Particle A is moving at speed 0.6c and particle B is moving at speed 0.2c in the same direction. at t = 0 the distance between them is 1 light hour. How long does it take for particle A to catchup with particle B in: Particle A's reference frame Particle B's reference frame My thought is tha...

special-relativity relativity time-dilation

People who don't know velocity addition formula...

0.6c minus 0.2c is not 0.4c !!!

@Danu It is generally, for reasons I have not yet been able to extract, favoured that the SUSY breaking scale should lie below the string compactification scale. But yes, it's also a problem that we don't seem to have much structural insights into non-SUSy compactifications.

Okay.

Apparently Donaldson's group is working on G2 nowadays.

@Danu Yes, I'm not looking forward to calculate stuff with that 3-form. Maybe I won't have to, though

1:47 PM

So should be some good stuff coming up ;)

lol, hopefully I don't get scooped by Donaldson :P

I doubt he's doing M theory compactifications :)

You never know with these crafty mathematicians ;)

So are you doing pheno stuff, like actually writing out Lagrangians?

@Danu I'd rather not do pheno (and neither would my advisor), but I've just started reading into this stuff, it's not clear yet what I will actually be supposed to do

1:52 PM

Alright. Who are you working under (if you don't mind telling me!)?

@Danu J. Walcher

we haha

I think that's supposed to be "We, the research group", not "we, your majesty" :D

@ACuriousMind I am a little bit confused. Do you think that the answer completely answers the question

@ParthMaske That I can't tell because I don't actually understand what your question is aside from you being uncomfortable with the modern resolution of Zeno's paradoxa.