The h Bar

Secret

15:00

@Danu
$$\sum_{k=1}^n k^{m+1}=\sum_{k=0}^n (k+1)^{m+1}-(n+1)^{m+1}$$
$$1^{m+1}+2^{m+1}+\cdots+n^{m+1}=1^{m+1}+2^{m+1}+\cdots+n^{m+1}+(n+1)^{m+1}-(n+1)^{m+1}$$

What's next?

ACuriousMind

...what the hell are you doing. Just apply the binomial formula on the $(k+1)^{m+1}$

Danu

@ACuriousMind ??

Secret

I have not thought about using the bionomial formula, the way he phrase it caused me to think he is just grouping terms in some way to achieve the required result

ACuriousMind

@Danu You wanted to get a formula for $\sum_k k^m$, right?

Danu

@ACuriousMind Ι know it's given by the binomial formula.

Did you think I didn't realize? :P

ACuriousMind

15:04

What does your "??" mean, then?

Danu

The message is like... half an hour late?

ACuriousMind

what

Secret just posted their attempt.

And I replied.

Danu

@ACuriousMind oh

Sorry.

I do not see certain messages ;)

2

ACuriousMind

lol

Danu

Welp, this is annoying :P

0celo7

15:05

#rekt

Danu

@0celo7 But who? :P

Secret

Pyring eyes So it seems random pictures are more annoying than JD huh? tsk tsk

Secret

Anyway, I can solve it now, thanks Acuriousmind

0celo7

@Danu Secret

Secret

I have already kinda guessed that. I have been monitorign his response for at least 3 months already

Danu

15:10

@Secret Random pictures?

Secret

Dec 12 '15 at 9:29, by Slereah

@Secret Please, stop

I do noticed one thing how everytime I post a pics of some maths stuff, the chat tend to go dead for a couple of hours

and how Slereah and Danu have dminished frequency of responses ever since Slereah's

This is one reason you see a lot less picture posts by me in the chat room this year, knowing it will put me into great trouble

Danu

@Secret I don't think you're getting into any "great trouble" as long as you do not violate any of the rules of the chat room. Don't worry about that.

If you're getting few responses to the pictures, it is probably because most of us have a very hard time understanding what to make of them. The way you think about mathematics is just different from the way I, for instance, do.

Secret

I am not talking about chat rules
I am talking about something .. more subtle...

Stealth banning

Stealth banning (also called Shadow banning and Hell banning) is a practice used by some online community managers to block content added by spammers and Internet trolls, as well as other individuals whose interests do not coincide with the managers'. The practice involves making a user's contributions invisible to all other users, but visible to themselves, making them less likely to create new accounts to add the same material. Often this blocks the problem user's contributions while making it look like they were "lost" due to a website error, thereby enforcing the community best practice of...

Danu

@Secret As far as I know, stealth bans are not possible here.

David Z

Yeah the SE team is opposed to them AFAIK

Secret

15:29

(Please do not respond to this message, the discussion of this topic will stop here) there's something similar to that, the "ignore user"

*Possibly opening pandora's box, but I think you guys need to know this piece of fact about me*

I am INCREDIBLY sensitive to all instance of perceived to be ignored (friends or strangers). Therefore I am always monitoring the flow of conversation that is currently taken place, as it will tell me whether I am being ignored by a user (possibly due to some of my stupid habits or other reasons)

Secret

Btw, DavidZ, if I recall correctly, you have worked in experiental particle physics, right?

Balarka Sen

@Secret What you have to understand is that most people do not understand what you're trying to communicate through the sketches you upload. Thus it is natural that some people would assume you're trolling. If you're not, and you want attention on the things you think about from the members of this and the math chat, why don't you write up what you want to communicate in an understandable language? That way people would understand what you're trying to convey more easily and would help.

This is just a suggestion.

Secret

I have tried to give some description (often follow the pics) in sentences and sometimes rewrite the thing in terms of maths symbols, but perhaps, it might be not clear enough

Danu

It's not even necessarily that people assume you're trolling---it may be as simple as them not knowing what to say about something they don't understand anyways, and thus remaining silent.

Balarka Sen

No, the pictures are absolutely very not clear. Why don't you follow the conventional way of communicating mathematics and write out the question or the idea you're trying to convey?

I'd at least be very happy to help - whenever I can - anyone who writes a clear, precise question/thought.

Notebook sketches are good for personal explorations, but not for communication of thoughts. That way you can't expect to face much more than general silence from your audiences.

Obliv

15:36

wow last night was weird. I dreamt that I was solving a physics question that I thought about literally right before I went to sleep. has this happened to anyone else before :O

is it even possible to solve questions in a dream

Balarka Sen

Yes.

Secret

@Obliv I have numerous examples of these that I can just dig up form my dream diary if you want to

Obliv

oh wow you made a dream diary? I always wanted to make one but Im always too lazy to start :p

Balarka Sen

Google Poincare's theory of subconscious.

Danu

@Obliv Sure

HDE 226868

15:38

@Obliv I seem to recall reading a semi-apocryphal anecdote about Srinivasa Ramanujan solving a problem in a dream.

Danu

^very famous

But I think historians agree that he just said that to please his religious momma :D

Balarka Sen

I don't think I personally ever got any serious math done while asleep. I have had exciting ideas right before I fell asleep, but not in a dream or anything.

Secret

@BalarkaSen In that case, let me try to phrase what I am up to in the conventional maths language and see if it helps on conveying the idea (that way, you guys can construct the picture using the descriptions), perhaps that might help (I'll do it later, as the current dream discussion I hav a lot to say about)

Danu

I have had ideas in a half-awake state: In fact I pointed out such a case only a few days ago.

Balarka Sen

@Secret Sure.

HDE 226868

15:41

Dream incubation

Dream incubation is a practiced technique of learning to "plant a seed" in the mind, in order for a specific dream topic to occur, either for recreation or to attempt to solve a problem. For example, a person might go to bed repeating to themselves that they will dream about a presentation they have coming up, or a vacation they recently took. While somewhat similar to lucid dreaming, dream incubation is simply focusing attention on a specific issue when going to sleep. == Description == In a study at Harvard Medical School, Dr. Deirdre Barrett had her students focus on a problem, such as...

Secret

@HDE226868 That does not work for me, as it turns out my dreams have a opposite mechanism. here more you desire something to appear, the more likely it will nto apepar

HDE 226868

@Secret See, I simply don't remember my dreams, so all those issues go away.

Secret

some people said people who can remember dreams has a poor quality of sleep

Obliv

Wow I had no idea this was common. The brain is very interesting indeed. I wish I could do physics in my sleep more often so I can get twice as much done :d

but like is 'focus' even a real thing in dreams? What are our iq's when we're sleeping?

Secret

@obliv I have dreams that are so vivid that I managed to replicate simialr to what Kekule did when he discover the bezene ring structure. For example:

yesterday, by Secret

Some of my dreams are like Kekule's (the guy who discovered benzene structure)

Here's an example of a dream I have last night one year ago that actually helps me learn something new

The Dark Side

15:45

So, just wondering if I'm right about this extrapolation of the original discussion. Not the curvature tensor, but if we have some general rank-4 tensor $T_{ijkl}$ in 4D spacetime which is anti-symmetric w.r.t. all four indices, there is only independent component e.g. $T_{1234}$, because e.g. $T_{2134}, T_{1243}$ etc. are all related to it.

So, no. of independent components = ${}^4 C_4$, which is equal to 1.

barrycarter

@JohnDuffield I summon thee (though this probably won't work)

The Dark Side

i.e. no. of independent components = no. of distinct combinations of 1, 2, 3 and 4, with no repetition allowed since that will kill it (i.e. make it 0).

Am I right? @GR guys

Danu

Weird quote of the day: "That school was extinguished as the losing party in a battle among physicists is always extinguished, like the losers in an Arab change of government."

Secret

@obliv From my 5 years of dream diary, I conclude there are 5 mechanisms in my dreams:
1. Inverse dream incubation: The more you desire something the less likely it will appear
2. Shock factor: Mechanism 1 will be overridden if the experience is smoehow shocking enough (scary, new, happy etc.)
3. Maths amplification: The lucidity and detail of the dream increases the more maths probelm (thsi include solving phsyics questions) I did during the waking period
4. Spacetime: Time tends to run normally and clocks tends to display correctly, but space of familar regions tend to 2x larger than thei…

HDE 226868

@Danu Context?

Danu

15:50

@HDE226868 The recent question on HSM about thermodynamics links to a book, where this quote can be found (page 40-41).

Obliv

@Secret Yeah the only dreams I can remember are the weird ones. I'll just keep those ones to myself..

HDE 226868

@Danu Oh, yeah, that one.

The Dark Side

e.g. $T_{1214}$ etc. are all zero, $T_{2134} = - T_{1234}$, $T_{3142} = - T_{1342} = + T_{1324} = - T_{1234}$, so again it is related to 1234 component. This appears to work!

Secret

I have dreams from routine to surreal, I have seen jesus, satan, anime characters, rea life friends, celebrities etc.

I have had many types of dreams, some as compelx as inception

The onyl type of dream I have not had so far is a sequal dream, i.e. a dream that continues the story left from the previous dream

barrycarter

Hi, @JohnDuffield

@JohnDuffield I assume you were joking and that you are ready to discuss your comments here?

The Dark Side

15:54

@Danu - Puhleez, does that logic make sense?

Secret

o and btw, I usually cannot wake up from nightmares with a sweat (like most people). I need to either solve the dream, or someone else wake me up , in order to wkae up from nightmares

John Duffield

Hi @barrycarter. No, I wasn't joking.

barrycarter

@JohnDuffield Who told you that you couldn't discuss physics here and by what authority?

Secret

This overtime caused me to learn to have a great control of my dreams, ofte just to twist a bad ending into a good one

David Z

@barrycarter this sounds like the sort of thing you may want to handle in another chat room

@barrycarter the moderators

barrycarter

15:55

Wow

OK, how do I create a chat room or something

David Z

Go to chat.stackexchange.com and look at the bottom right

there's a button to create a new room

barrycarter

@JohnDuffield chat.stackexchange.com/rooms/37563/relativity

Secret

As for your question, this might be a good start https://www.quora.com/Is-it-true-that-the-higher-your-IQ-is-the-more-you-dream

To me and to most neuroscientist, they tend to understand dreams as the brain jumbling experience in the waking life and memory together, thus might not necessary have deep meaning

Danu

16:19

@TheDarkSide ?

Obliv

Can energy be expressed as a vector?

The Dark Side

@Danu That a rank 4 tensor $T_{ijkl}$, which is antisymmetric w.r.t. all four indices, and in 4D space, has only one independent component, as per the above logic.

Can't help it, the logic got buried in the parallel conversation.

36 mins ago, by The Dark Side

So, just wondering if I'm right about this extrapolation of the original discussion. Not the curvature tensor, but if we have some general rank-4 tensor $T_{ijkl}$ in 4D spacetime which is anti-symmetric w.r.t. all four indices, there is only independent component e.g. $T_{1234}$, because e.g. $T_{2134}, T_{1243}$ etc. are all related to it.

34 mins ago, by The Dark Side

i.e. no. of independent components = no. of distinct combinations of 1, 2, 3 and 4, with no repetition allowed since that will kill it (i.e. make it 0).

30 mins ago, by The Dark Side

e.g. $T_{1214}$ etc. are all zero, $T_{2134} = - T_{1234}$, $T_{3142} = - T_{1342} = + T_{1324} = - T_{1234}$, so again it is related to 1234 component. This appears to work!

Considering my earlier troubles with counting components, it made sense to ask for a small confirmation, if you have the time!

Obliv

16:37

nvm I know it can. It just isn't useful since we assign momentum to a mass as a vector which includes a velocity vector anyway.

ACuriousMind

@Obliv Energy is a scalar in non-relativistic physics. It becomes the zeroth component of the four-momentum in relativity.

Obliv

@ACuriousMind What energy is being assigned in four-momentum? Is it mass-energy + whatever other energies it has?

Danu

@TheDarkSide That's definitely true.

and the reasoning is correct

Obliv

@ACuriousMind is it $E = {p^0}{c}$?

ACuriousMind

@Obliv In the rest frame, the four-momentum of a single particle is $(mc^2,0,0,0)$, the form in all other frames follows. "Potential energy" does not really make sense in relativistic physics.

Obliv

16:51

@ACuriousMind yeah you can't really describe PE in a momentum vector that describes motion. Why doesn't it make sense? Can't it be described using 4-position? Does the math not conform or is the idea of PE abandoned?

Secret

@BalarkaSen Ok here goes:

Given a matrix $A \in M_n(\mathbb{R})$ in some basis $\mathcal{B}$. For example for $n=3$, A has the form

$$\begin{pmatrix}a & b & c \\ d & e & f \\ g & h & i\end{pmatrix}$$

The column vectors are labelled as $\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3 \}$ and row vectors $\{\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3\}$

We know that the usual inner product in $\mathbb{R}^n$ is defined as the dot product

$$\mathbf{x} \cdot \mathbf{y}=\sum_{i=1}^nx_iy_i$$

where $\mathbf{x},\mathbf{y}\in \mathbb{R}^n$

ACuriousMind

@Obliv You...just don't need the concept. Think about what "potential energy" is: It's just a name we give the scalar function that is the integral of a conservative force field in 3D. How would you generalize that?

Secret

@BalarkaSan i.sstatic.net/wUEEc.png

ACuriousMind

Of course, you have"potentials" for e.g. the electromagnetic field strength, but no one calls the four-potential "potential energy", and it doesn't make sense to say it is conserved because it is gauge-variant, hence not a physical object.

@Secret ...why do you want such a "pictorial representation"? What is it supposed to do? What about just writing down the matrix or drawing the parallelepiped is not enough?

In other words, why are you trying to jump through hoops to draw the $w$ when you can really straightforwardly just draw the $v$?

Obliv

@ACuriousMind You're right. It isn't needed to describe the motion of a mass. But, aren't force-fields still used in relativistic mechanics? In classical mechanics it's useful to know PE so that you can determine the velocity after the acceleration through the field. Couldn't you do the same with a force-field in relativistic mechanics to find the four-momentum after the acceleration?

having a value for the PE associated with the field would be useful, I imagine.

ACuriousMind

17:07

@Obliv The description of a "force field" is different. You can't just write $m\ddot{x} = F$ because what is the derivative supposed to be taken w.r.t.? you have to decide that it is taken w.r.t. the proper time along the world line of the particle. That is, the force on a particle can be different *even though it is at the same point in spacetime, i.e. all force fields now naturally depend on the four-velocity of the particle.

Since the depend on the velocity, none of them will be conservation, hence none of them is the derivative of a scalar function

e.g. the Lorentz force is given by $m a_\nu = e u^\mu F_{\mu\nu}$, that thing on the r.h.s. is not the derivative of anything, although $F_{\mu\nu}$ itself is the derivative of the electromagnetic four-potential.

Secret

@Acuriousmind I am trying to encode the geometric meaning of the entire matrix onto just one object, so that just by looking at this object, both the row vector and column vector information is clear visible, thus illustrating how they are related when doing a coordinate transformation

If successful, this will give a picture of a matrix, similar to how vectors are first introduced is like an arrow, and thus it is then possible to show people who have little to no maths background (or too scared of maths) what a matrix can do before start showing them the deeper maths, which can scare them …

ACuriousMind

@Secret The "entire matrix" is already encoded in the parallelepiped of the column vectors.

Trying to draw the row vectors "on top" of that is redundant information.

Balarka Sen

@Secret I am a bit busy right now but I looked through what you wrote and I don't understand what you mean by 1-forms can be represented by contours. A 1-form on R^n is a linear map R^n --> R. Do you mean to say the collection of level sets f^-1(c), where c is a scalar? These are indeed hyperplanes, so that's the only way I can interpret what you are saying.

Secret

yes, a contour is just a level set for functions of two variables

Balarka Sen

ok. so what do you really mean when you say a pictorial depiction of $A$ which encodes the dual of it's row vectors? I mean all the row vectors automatically capture the dual space of the row space.

Secret

17:18

The reason I want to draw the row vectors is that it allows the meaning of why matrix multiplication is linear clear (because each vector only pierce one-forms a number of times that depends linearly on the corresponding vector components).

In addition, I am trying to visualise an adjoint, which only make sense if one can see what happens to both the row vectors and column vectors. Past experiment of mine on this seemed to suggest it is some kind of transformation that involves some kind of rotation and sheering, and that if a matrix is unitary, the 'rotation' happens to cancel each other …

@BalarkaSen I mean a pictorial description of $A$ (which is represented by the parallelpiped that is formed by the column vectors) which also encodes the information of the row vectors, not its dual

Balarka Sen

Oh, so you want to encode both the paralellpiped drawn using the row vectors, and the column vectors?

Secret

yes, one single parallepiped (or something geomtric) that one can directly see the information of btoh the row and column vectors

as well its determinant, trace , invertibility etc.

and the ability to use ti in actual computation

For the transpose (basically adjoint in $\mathbb{R}$ for 2 x2 matrix, I found that geometrically the transpose is reflecting the off diagonal components of the vectors along the line y=x and then redraw the picture that results

Balarka Sen

But row space of a matrix is just column space of it's transpose. I am sure you can translate that into a meaningful picture.

(by the way, if the question was simply this, I am confused why you talked about covectors above)

Secret

I need the covectors in order to work out what happens in the transpose (and ultimately, adjoint) . Just looking at the column vectors they seemed to be doing something random when comparing the two pictures of those matrices (except for the 2x2 case)

Prelimary attempt at that back in the old version seemed to hint for the 3x3 case, it has something to do with rotating the triad formed by the column vector in a certain way

Later on, it seems to involving 'rotating' some but not all of its components around

Obliv

@ACuriousMind Alright I'll just have to study relativistic mechanics from the ground up to comprehend what you're saying. For there to be a different force on an object in the same location in space-time simply because it has a different four-velocity makes no sense to me. I thought force-fields were position dependent,usually, and not velocity-dependent. That's wild.

ACuriousMind

17:32

@Obliv You know a very common force that's velocity-dependent: Friction.

So it isn't wild at all.

Secret

@BalarkaSen actually, this whole conversation here is more like a sharing of my idea that I have been workign on more than a question. But hoepfully, now that it is presented completely in traditional maths language, I have clarified what I am doing with those pictures

ACuriousMind

@Secret I still have no idea what you're doing in those pictures except "trying to draw complicated pictures that carry no useful information".

Secret

I am trying to give a notion of "vector as an arrow" to general matrices so that matrices can be more accessible to the general public (before showing the more interested ones the maths behind)

On a more selfish side, it helps me to made some educated guess when computing matrix equations because a geometric computation method is then avaiable to be used when doign roughwork

Obliv

@ACuriousMind Yeah I wasn't thinking about forces due to physical interaction like that. Isn't gravity,E&M force,weak&strong position dependent? (I actually don't know much about the latter two very well)

ACuriousMind

@Obliv In classical Newtonian physics, gravity and electromagnetism are indeed conservative forces with a proper potential energy. The weak and strong forces do not exist on classical scales, and to speak of "forces" in the relativistic quantum theory where they appear is rather ill-defined.

Well

electromagnetism is not conservative, electrostatics is

The classical magnetic force is already velocity dependent: $v\times B$!

Obliv

17:39

What happens to the definitions of gravity & electrostatics in relativistic mechanics?

ACuriousMind

@Obliv Well, including gravity means you have to do general relativity, not special relativity.

And "electrostatics" doesn't exist relativistically. What is a static electric field in one frame is a mixture of electric and magnetic fields in another.

Balarka Sen

@Secret Well, I don't completely understand what you're trying to do, but note that people are actually interacting with you this time.

Secret

well I have tried, it is kinda hard to explain it not in face to face scenarios (due to lack of gestures that help me out)

Balarka Sen

I understand you're trying to work out properties of matrices just using the parallelpiped picture, but I am not convinced this is actually going to be too useful. Good for motivation maybe, but I find the linear maps picture much more motivating, if you are talking about pedagogy.

Obliv

Would one need to learn classical mech before learning hamiltonian/lagrangian mech?

ACuriousMind

17:49

@Obliv Hamiltonian and Lagrangian mechanics are classical mechanics, are you asking whether one needs to know the Newtonian approach?

Obliv

yeah

it's just a reformulation of newtonian mechanics right? Seems like it's just more refined

Secret

I have the linear map picture encoded in that as well, because I have used that to compute matrix multiplication before (I wll show this shortly after illsutrating the computation of transpose (which is one of the disadvantages of this picture compared to the algebraic approach)

ACuriousMind

@Obliv They are the more theoretical and formal approaches to classical mechanics, yes. Whether you need to know Newtonian mechanics depends solely on whether or not the thing you learn it from expects you to know it, I would say.

The Dark Side

@Danu Thanks :)

Secret

18:13

@BalarkaSen
(Geometric computation of a transpose)

Update:
OMG, I stumble upon a new result
Now I can determine whether a matrix is symmetric VERY QUICKLY

sum of row vectors= sum of columns vectors => symmetric matrix

That will surely help on increasing the algorithm that determines symmetric matrices, because we don't need to transpose it anymore

increasing the speed of *

ACuriousMind

No, it won't.

Determining whether a matrix is symmetric or not just requires you to compare all upper off-diagonal with exactly one lower off-diagonal entry, these are $\frac{n(n-1)}{2}$ comparisons, and nothing else. Your "very quick" way is summing $n$ $n$-vectors twice, i.e. $2n(n-1)$ addition operations, and $n$ comparisons.

Secret

ok, makes sense

ACuriousMind

Also, it isn't even true.

$\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix}$ has the same sum for rows and column vectors, but is not symmetric.

Secret

O great, forgot those permutation matrices ( I often need to check for counterexamples when doing these kinds of things, but I have nto done the usual routine today for sme reason)

Mike Miller

The space of matrices such that the sum of the columns is the sum of the rows is codimension 1, since it's determined by one equation. The space of symmetric matrices is way smaller, codimension n(n-1)/2.

You're just lucky for n=2.

Secret

18:32

alternately, it can be simplified to just laid a graph paper onto a vector, but remembering that each unit length on the paper is a vector

This give a notion of a "vector valued" length, which might be useful in introducing vector valued one forms when they go more advanced

ACuriousMind

Sorry, but you've gone into "not making sense" mode again.

kevin Tah N.

hey friends!

Obliv

hey @kevinTahN.

kevin Tah N.

nice day for some hbar chat :D

Secret

then I am not really sure what is not making sense here (except for my comment about vector valued one forms, which I do admit I have not study it in detail yet)

@Acuriousmind What sounds illogical or nonsense in the above slide?

kevin Tah N.

18:41

Does anyone know what Paul Davies is up to these days as far as physics is concerned. He seems like one of those old guys who has done it all and is now engaged in some cool stuff in secret lol

Secret

@Acuriousmind Perhaps this will help clarify what I am trying to say?

(If not I am stopping here)

ACuriousMind

@Secret What do you mean by "vector valued length"? What do the pictures there mean? You say to count the times the vector "crosses the origin of each matrix picture", but how do you know how to do that "tiling" there? Why do you count "crossing the origin"? What if it's so short it never crosses the origin? Doyou implicitly assume only integer entries here?

@Secret Well, that does make sense. After all, the $i$-th column of the matrix is what the $i$-th basis vector gets mapped into, but as you clearly see in that picture, you don't have to count some "crossing the origin", you just have to count how "many" of your current basis vectors are in your vector.

kevin Tah N.

ok, I could not find anything interesting in the arxiv on Paul Davies, officially moving on with my life now

ACuriousMind

this does not explain at all why you drew those strange overlapping parallelograms in the first picture.

Obliv

a picture is worth a thousand words, @ACuriousMind surely you can decipher the meaning? :D

Secret

18:52

@Acuriousmind The centroid of each of the parallelograms marks the vertices of the graph paper (I was too obssessed to using just one picture to derive everything)

(In the old notation, the matrix looks just like that box with the two black vectors shwod up there (plus their negatives fillign up the remainign to sides), thus the tiling will then become obvious)

I agree it is kinda messy, thus this graph paper slide shown here will be a clearer illustration

Therefore, as shown in the graph paper, counting how many lines the graph paper the vector crosses is equivalent to counting the numb…

Ali Caglayan

Do ionic substances behave similar to a plasma in liquid phase?

ACuriousMind

@Secret Then it has nothing to do with the paralellograms at all!

You're just decomposing the vector in the basis given by the graph paper.

And for some inscrutable reason you have decided to draw those parallelograms over it, but they have nothing to do with it.

Secret

yes and yes

for the latter yes, I will keep a note to avoid irrelevant stuffs in the future

however the main point is clear. If I show a high schoolar the above graph paper then they will know easily what a matrix is doing, and thus the linearity of matrices (an important property of them being linear maps) can be illlustrated nicely

Mike Miller

19:13

I certainly wouldn't know what a matrix is doing from that, but I'm not a high schooler.

Obliv

for the given equation $Mdv = -dM{v_{rel}}$ multiplying by $\frac{1}{dt}$ on both sides yields a formula for the force that is proportional to the relative velocity * change in mass w.r.t . You could divide by $dx$ to get KE = change in mass with respect to position too right?

wait nvm that's not true. $\frac{dv}{dx} \ne \frac{v^2}{2}$ right?

0celo7

@MikeMiller Where (in America) are matrices taught in high school

@ACuriousMind survived the hike, only almost died 10 times

Secret

@ 0celo7 Americans probably don't, but they do in singapore, china and hong kong

I first learnt about matrices in my grade 8

0celo7

Germans don't do it in 8th grade either

Secret

you guys are definitely less pressured than we east asians do back in high school

Mike Miller

19:27

Probably at Harvard Westlake.

Secret

Btw there is nothing else to share about that matrix visualisation project of mine since perhaps one of the significant results, is buried in this picture of MINE drawn 2 years ago and right now it does not seemed to make much sense to me other than back then I claimed I found a way to compute the kernal of a matrix geoemtrically

Mike Miller

I suspect you will find your current images similarly incomprehensible in two years.

Secret

I really need to put some effort on making my pics less cluttered and put more descriptions and translate it back to maths language completely

I am not that worried about the new ones, since I now have written a maths translation of it (as part of the sharing with Baleraksan) and I have now made my graphics not via doodling but softwares to make them cleaner

0celo7

@MikeMiller I'm a Freshman and I have NO clue what that pic means ;)

Secret

@0celo7 You are not alone for This pics
https://i.sstatic.net/ATi0t.png

right now it looks weird (but perhaps I can make sense of what my 2 years ago self is thinkign some time later)

But for this pics, after some explanation, Acuriousmind undeestood it
https://i.sstatic.net/e7Xm9.png

DanielSank

21:10

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ No doctor. The only thing they could do is needlessly prescribe antibiotics.

@Danu It was amazing.

I used to hold weekly dinner at my place for friends when I was an undergraduate.

One weeks, an Italian friend made risotto. I woke up the next morning writhing in pain on my bed.

I felt like there were sharp objects in my gut. The pain was so intense that I was actually writhing, like something you see in a horror movie. After a few minutes of that, I suddenly had the urge to run to the bathroom and evacuate whatever was in my stomach. Then I went back to bed. I woke up the next day at around 17:00, went and ate some rice, went back to sleep, and then woke up the next morning feeling more or less ok.

For years after that I could not approach risotto.

user54412

21:56

@HDE226868 I've done that, as have a number of mathematicians I know. Nothing profound that I wouldn't have figured out the next day, mind you. But I've definitely woken up and had my first thought be "oh that's how it works."

Bass

Hi guys, in the amplitude of the one-loop photon self-energy, Maggiore writes a $\mathrm{Tr}$ symbol, most likely a trace of gamma matrix products. However, I have never seen any trace operators in the Feynman rules of QED. Did I miss something?

@ACuriousMind any idea where that might come from?

user54412

@AliCaglayan That's a pretty broad question, mostly because "plasma" describes a large parameter space. The plasmas I work with, for instance, are more like gases with some extra electromagnetic features. Other people work with plasmas where particles scatter off electromagnetic fields millions of times better than they collide with one another because of how diffuse they are. That certainly doesn't sound like a liquid.

HDE 226868

@ChrisWhite I've had many epiphanies like that, but often when I'm about to go to sleep, not when I wake up.

Ali Caglayan

@ChrisWhite I'm probably thinking denser plasmas like you said. Perhaps the plasmas found in stars?

ACuriousMind

@Bass The one-loop correction comes from a fermion loop, and you have to sum over the possible spins of that fermion/anti-fermion pair. Those sums can often be written as such traces of gamma matrices

user54412

22:09

@AliCaglayan I suppose there's probably some parts of stars (certainly not the outer layers we see) that sort of work.

Bass

@ACuriousMind Makes sense intuitively that this sum is needed. But where exactly in the perturbative expansion does it come from? The n-point function can be expressed as $\langle 0\rvert T\{\phi_I(x_1)\dots\phi_I(x_n)\mathrm{exp}(-i\int d^4\mathcal H_I)\}\lvert 0\rangle / \mathrm{exp}(\dots)$. Then Wick's theorem etc.

Danu

@DanielSank "amazing" :P

Bass

Where is the sum over the spins?

Danu

@Bass "Just do the functional derivatives"

(not serious)

I mean to say that it's not so obvious to me that there should be any more "real explanation" than (1) the intuition from the diagrams (2) the fact that this is what you get if you grind through the calculation

In fact, if you'd like I can give you the complete derivation of the trace. I did it once, explicitly, in TeX.

Bass

@Danu Well, cool, if you have it, why not :)

Danu

22:17

Actually, it's really not that bad.

It's about 5 lines of ugly shit.

Bass

@Danu The thing I don't understand is not how does the sum over spin states produce a trace?, but where in the perturbative expansion is the sum over spin states?

Danu

@Bass In the calculation the trace comes out completely separately from any interpretation like "sum over spins"

So idk if ACM is referring to anything more than diagrammatics

ACuriousMind

@Danu Uh, that's a completely different calculation from the one Bass asked about

He's not asking about a four-point function, he's asking about the perturbative one-loop correction to the photon propagator in QED

Danu

Wait lol jk

That's what I get for barging in on conversations

I see fermion trace, I think I know it :P

Bass

@Danu I'm already lost at the notation. No idea what $\delta$, $\eta^a$, $S^{xy}_{xy}$ is in that context :D

Danu

22:22

@Bass And yes I also realized I should give you the whole document

@ACuriousMind Hmm... So what diagram is the 4-point function?

Bass

@Danu The combination of all diagrams with four external fermions.

Danu

It's a fermion loop, no?

I want to say that this trace is somehow the same thing.

my final result is a trace over $G_F(x,y)G_F(y,x)$

which looks like a loop to me

@Bass That's not what I meant, but yes I know that much ;)

ACuriousMind

@Bass Well, what is your Feynman rule for internal Fermion lines? I.e. what does the gamma matrix from the propagator associated to it act on?

@Danu I'm not sure I understand the question?

Danu

@ACuriousMind So this thing I get from this calculation looks like a loop to me, and it features a trace---I want to say that this is related to the fermion-loop-trace

In fact yes, I think I am calculating just that

Bass

@ACuriousMind Just the fermion propagator. I.e., for any internal fermion line with momentum $p$, add $\tilde{S}(p)=i / (\gamma^\mu p_\mu-m)$.

Danu

22:27

Note that the 4-pt function I was calculating had an electron-positron pair created at one position and then annihilated at the other.

I think my calculation is the right one after all, no? @ACuriousMind

Or at least what I was doing is one instance of this general truth, that a fermion loop induces a trace

Perhaps the simplest case

Bass

@ACuriousMind So the sum over spin states is contained in the fermion propagator? Would make sense to me.

ACuriousMind

@Danu Looking closer at it, yes, it is, but that's not a four-point function :P

Danu

@ACuriousMind Well yeah it's a 4-fields at 2 points-function

@Bass my calculation is the relevant one after all

ACuriousMind

But you indeed computed a fermion loop there

Danu

I'll give you the whole document

@ACuriousMind And by hand! :D

@Bass what's your email?

(can also create separate room if you don't want to say it in public)

Bass

22:30

@Danu [email protected]

Danu

lol

ACuriousMind

@Bass Yes, when you multiply the fermion propagator to the gamma matrices of the vertices and do some tedious algebra, you get this trace, which you could transform into a sum over spin states by some further tedious algebra. I think. I've not done that here, but you get the same "kind" of trace when you sum over all the polarizations of external fermion legs for other diagrams

Bass

We're talking about that one, right?

ACuriousMind

That's the one you should be computing, yes.

Danu

Yes, so what I did was calculate only the loop

No external photon legs

ACuriousMind

22:33

Computing one- and two-loop corrections in QED was what made me really realize I don't like Feynman diagrams and perturbation theory :P

Danu

I've never calculated a loop correction.

Cry everytiem.

I feel so illegitimate, never having actually done any renormalization :P

ACuriousMind

It's the regularization that's so bad, not the renormalization ;P

"Now we compute this diagram in $4-\epsilon$ dimensions"

Danu

Yeah yeah

ACuriousMind

"Oh, for this diagram, we are in 4D again, but we will give the photon a mass"

Bass

@Danu Thanks for the PDF. Am I right that the notation is more from path integral quantization than from the canonical one?

DanielSank

22:36

@Danu Food poisoning is an amazing thing.

ACuriousMind

Really, all those regularization prescriptions feel so random when they're just handed to you.

DanielSank

@ACuriousMind we have a nice post about this. Finding...

12

Q: Complex integration by shifting the contour

In section 12.11 of Jackson's Classical Electrodynamics, he evaluates an integral involved in the Green function solution to the 4-potential wave equation. Here it is: $$\int_{-\infty}^\infty dk_0 \frac{e^{-ik_0z_0}}{k_0^2-\kappa^2}$$ where $k$ and $z_0$ are real constants. Jackson considers t...

ACuriousMind

This is not about the $\mathrm{i}\epsilon$ in the Feynman propagator, if you mean that one

Danu

@Bass Yup. I think canonical quantization is just blurgh :)

ACuriousMind

Ah, yes, you meant that one.

Danu

22:39

Even more tedious than the functional derivatives

DanielSank

@ACuriousMind It would be nice for someone to write an answer like mine, but which explains the rotation method.

i.e. "Feynman's prescription"

Danu

But that's not the renormalization/regularization we're talking about

ACuriousMind

^

DanielSank

@Danu oh, ok I shut up

Danu

Those are the easy ones that nobody even bothers to complain about

Sadly :(

There are much worse ones

Ultraviolet divergences, and they're a real pain.

DanielSank

22:40

@Danu Without even knowing what you're talking about, I bet you $1 they're all the same thing.

Danu

@DanielSank No.

DanielSank

@Danu oh that's different...

Danu

Hah

That's a funny juxtaposition

DanielSank

@Danu Too bad, you should have taken the bet

0celo7

@ChrisWhite Me too.

Trivial problems, mind you.

Danu

22:41

@DanielSank The "No" was with regards to them being the same thing ;)

Anyways, yeah

DanielSank

@Danu Nevertheless...

0celo7

I spread out my analysis homework over two days, I've come up with solutions in my sleep before.

Danu

@DanielSank The ultraviolet divergences require much more creative and/or scary methods...

...like doing your integrals in $4-\epsilon$ dimensions and taking the limit $\epsilon\to 0$...

Bass

@Danu My understanding of path integrals is at the level of Zee-just-integrate-over-all-possible-paths, so I first have to learn all that notation.

DanielSank

@Danu Actually now that I think about it more... I can probably bs an argument that they're related...

Danu

22:42

@Bass Oh, okay.

DanielSank

@Danu I always liked that one...

Bass

@Danu But thanks anyway, gonna study that some time later.

Danu

The worst thing though, that I've seen so far, is doing "integrals" where you don't have parameter independence anymore.

DanielSank

@Danu Basic idea

?

Danu

For things-that-look-like-integrals-but-are-actually-undefined you can't do shifts in the integration variable any more and shit

0celo7

22:43

@Danu You know, now that I know you have certain people blocked, I'm amazed you don't have me blocked.

ACuriousMind

@Danu I still somehow prefer just introducing a hard cutoff and just doing everything with that. You're perturbative anyway, what's the point in integrating up to arbitrarily high momenta :P

0celo7

Even @DanielSank has me blocked!

Danu

@DanielSank Basic idea: Try to preserve integral notation, do total crazyness under da hood

DanielSank

@Danu I really hate it when people do that.

Danu

@ACuriousMind But did you explicitly check Lorentz invariance? ;D

@DanielSank Yup.

DanielSank

22:44

For the same reason I can't stand "hbar = 1" nonsense.

Danu

Hahahaha

DanielSank

<3

Danu

But that's so tame & under control

0celo7

@DanielSank Sachs and Wu has $c=8\pi G=1$.

Danu

there is nothing that can really go wrong when dealing with serious people

ACuriousMind

22:44

@Danu I do everything in the Euclidean theory and then analytically continue to get the Minkowski result, naturally

Danu

Renormalization, however...

DanielSank

@Danu It adds difficulty to our lives for the same reason. It's not explicit about what's going on.

Danu

@DanielSank But the dimensional analysis thing is like... A factor literally 100 billion less problematic. The other one took 2-3 Nobel prizes to get straight :P

(and even now, of course, it's not rigorously justified)

It made people like Dirac completely lose faith in field theory!

Bass

@Danu The other one being the thing where bare mass gets replaced by experimental data?

Danu

But yeah, if you wanna see something disgusting, just look at physicists evaluating the "triangle diagram"

@Bass "the other one"? Sorry, I don't understand what exactly you're referencing :P

Mike Miller

22:51

@ACuriousMind what's the deal with large N duality

0celo7

@Danu If he's reading Zee, he'll see it.

Bass

@Danu The other one that took 2-3 Nobel prizes to get straight..

Danu

@Bass Regularization/renormalization is what I meant

ACuriousMind

@MikeMiller No idea what that is, sorry

Bass

@Danu Oh ok. (I thought you meant one specific kind of reg./renorm.) nvm

Mike Miller

22:53

darn

Bass

Bye guys. Thanks a lot.

Mike Miller

@Danu what's the deal with large N duality

Danu

@MikeMiller Large $SU(N)$?

that $N$?

Mike Miller

sounds right

Danu

Are you asking me why physicists care about it?

Or are you just sighing in exasperation? ;)

What I would like to know is why mathematicians care about it

Does anyone remember what the original reference is for this nonsense treatment of linearly divergent integrals that ruins reparametrization invariance?

Slereah

22:59

@Danu My favorite trick

ACuriousMind

@Danu I have no idea what you're talking about

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