Speaking of analytic continuations, can they have physical meanings? Sorta related, it's always bugged me that people say we need to take the real part of a complex function to have a physical solution

I imagine that if there was some kind of averaging effect on a summed quantity (like damping), then the Cesaro summation would have physical meaning.

why does one like adjoint representations ?

not really a meaningful question, but just poking around

anyways let me try to do the su(2) question on paper

also another silly question

in what is this parameter counting thing?

say for O(N)

Welcome Agalloch ;)

@Alex hey dude :)

Hello @Cows, Hey @Avantgarde

The Mantle is good. Probably their best.

It's my favourite of their work, maybe of all work atm

Mgla's exercises in futility, and storm of the lights bane by dissection are classics though

They were kinda experimenting with The Mantle. The guitarist (Anderson) said so, and that he was surprised that people really loved The Mantle.

I've heard that one by Mgła. Not heard much Dissection.

I can feel the experiment - I learned 80% of the primary guitar in 'in the shadow of our pale companion' (i.e. switching between 6 guitars)

Nice that you play. I don't.

Probably my favourite song of the album

@Cows The adjoint representation reveals the representation theory of the dynkin type Lie algebras to us

I see

hmm

Are you doing Lie algebras or Lie groups atm?

I just decided to look at some aspects of SU(2) and SO(N) . Just wanted to check out some stuff that's been in my head for the past few minutes

how does one know that O(n) has dimension $\frac{n(n - 1 )}{2}$

I mean I can't count or something

lol

Well you can check the dimension of the corresponding coordinate ring

yeah, I don't quite understand that concept, nor the so called "naive counting" physicists throw out there

Can you help me understand how this is done?

hmm lol someone else was somewhat curious about similar things

Oh

One way to see this is to notice that $O(n)$ is defined as the set of matrices $A$ satisfying $A^T A = I$, $\det A = + 1$. this means that $A$ is a symmetric matrix with unit determinant. A symmetric matrix has $1 + 2+ \dots + n = \frac{n(n+1)}{2}$ free parameters, where we just added all the number of components in the top right-half of a given matrix.

However because $\det(A) = 1$ we can express one of these parameters in terms of all the others, so $\frac{n(n+1)}{2} - 1 = \frac{n^2+n-2n}{2} = \frac{n^2-n}{2} = \frac{n(n-1)}{2}$ is our answer.

The adjoint representation is just a way of turning the commutator $[X,Y]$ into a representation basically

You want to end up with $ad_X(Y) = [X,Y]$ as a linear representation, work backwards and figure it out

this is a very good explanation, I like it

finally!

Another way to see it is that $A^T A = I$ is a set of $n^2$ equations the components of $A$ must satisfy, however only $1 + 2+ \dots + n = \frac{n(n+1)}{2}$ of these equations actually matter, so you have $\frac{n(n+1)}{2}$ constraints on $n^2$ equations, i.e. $\frac{n(n+1)}{2}$ constraints on the $n^2$ components of $A$, leaving $n^2 - \frac{n(n+1)}{2} = \frac{2n^2-n^2-n}{2} = \frac{n^2-n}{2} = \frac{n(n-1)}{2}$ free parameters in $A$.

Another way is that $A = e^{iM}$ for $M$ anti-symmetric as the post mentions, and $M$ has $1 + 2+ \dots + n - 1= \frac{(n-1)n}{2}$ free components

nice!

ok let me take a small walk

I will look at SU(N) in a bit

(somewhat) trivial aspects of SU(N)

then should circle back to the SU(2) thing of importance

Are 20.000 volts dangerous when using around 0.5 mA?

Only one way to find out!

@Slereah You must like humans

Well if we take one MegaOhm for my body than we need 1.000.000*0.5 volt to enable current to flow.

I am so happy that I figured that out by myself right now

... if its even right

the behaviour of current through the human body is complicated

Because the resistance changes as current goes through it

Due to ion transport and whatnot

Yeah but the skin is the greatest resistance what I have heard and 20.000 volt aren't enough to let current flow through it...

am I right? I really don't know :D

It needs around 100mA to kill, though it can vary a great deal. A lower current can kill if you're unlucky.

I'd guess you're safe with 0.5 mA

But 20kV will easily kill you as the resistance of your skin is a ot less than 20KOhms

But has Volt really a big influence? isn't it more about Ampere?

So "how much goes through"?

Yes, it needs a current of around 100mA to kill. But the body is quite conducting so it doesn't need a huge voltage to generate that sort of current.

Wow, my school will give me 30$ a month next school year for having exellent grades (straight As). What should I buy :D? A quantum computer maybe?

Dude you can buy the world

I know right xD

Ok, off with the jokes, gonna watch a movie, later :)

Any suggestions for a good movie?

Maybe something fantasy or sci-fi?

But I think I've watched them all...

@watchme the current through the body is given by I = V/R, where R is the resistance of the body. So as you increase V you increase I.

The resistance of the body is rather variable because it depends on what bits of your body the electrodes are connected to. However it's low enough for 20kV to be fatal.

Do you know the video "Should a Person Touch 200,000 Volts? A Van de Graaff generator experiment!"?

@JohnRennie

here, a additional resistance shurely had to be added right, as the body resistance cannot be enough....

I haven't seen that video, though I have played with Van der Graaff generators. A VdG generator effectively has a very high internal resistance, and that limits the current that can flow through you.

I'm actually not a physicist. I trained as a physical chemist and worked in colloid science for 12 years.

My knowledge of physics is very broad but rather shallow i.e. I know a little about lots of subjects.

That 260k rep is largely from answering fairly simple questions.

@JohnRennie Because it's you, we can forgive being a chemist

@watchme We have a Tesla coil in our labs that gets up to about 250 000V. I haven't touched it myself but others have. It won't kill you; the current is too low. But "should" you touch it? No, it hurts quite a bit

@Jim the thing about being an industrial physical chemist is that it forces you to be good(ish) at lots of different areas of physics. So you end up with a really broad knowledge of physics. Far broader that research physicists who have specialised in a single area.

@JohnRennie You don't have to tell me, the faculty at my campus is primarily biophysicists. They know almost no actual physics

Half of them have never used tensors before

To be fair I had little understanding of tensors before I started learning GR as a hobby.

I'd come across them in continuum mechanics and with things like polarisibility, but nothing more than this.

In fact, under my authority as Supreme Jim of Physics.SE, for all his meritorious contributions to the Jimpire, I hereby bestow the title of Physicist on John Rennie with all of its associated rights and privileges. So let it be, henceforth

@JohnRennie They show up a lot in particle physics and advanced optics too

Neither of which are commonly used by physical chemists :-)

That makes a lot of sense to me

@Jim so the resistance is just very huge

@watchme absolutely. Both devices are intended just to supply voltage and, when you think about it, the current can't be very large. These things are connected to the wall socket, which provides 120V at 15A maximum. That's like 1800W max. At 200000V, you're looking at a max of 9mA. But I know through experience that putting 1800W through a tesla coil will blow some circuits, so drop the current a lot more

Not to mention, with at least a tesla coil, the arc goes through the air, which has a very high electrical resistance itself

I am relatively new to this stuff, I probably should read about that topic

@watchme The sockets have an upper bound on both voltage and current, which places a maximum on the power you can draw. The sockets are set at a given voltage. No more, no less. You can draw up to a set amount of current (there are high power sockets that allow more current) by decreasing load impedance. Beyond that, it's wired so that the fuse/breaker will trip and shut off the circuit

^ obviously, I mean AC voltage, which technically fluctuates

and your tesla coil is really connected to a wall socket?

@Jim ups

@watchme yeah, it's just a small one

@watchme No, the wall socket outputs 120V. I was not wrong about that. The internal circuitry of the coil ramps the voltage up to 250kV by drawing more current from the wall and transforming it into voltage

but the socket only provides 120V

@watchme This is basically what a transformer does: exchange voltage for current while keeping power constant (apart from losses).

@watchme Right. Real transformers absorb some fraction of the power they convert.

doing some googling, the main source of loss for a real transformer is due to joule heating of the copper wires (hence the phrase 'copper losses')

and now I find myself curious about the state-of-the-art as far as superconducting transformers go...

i imagine that a major limitation is the usual issue of 'high-temperature' superconductors still needing to be really cold (liquid nitrogen not liquid helium, but still)

wow lol, I got so much rep from that one question...

Such a simple answer too, didn't involve much effort on my part...

ah crap...deepspeech has no windows support -.-

@watchme In general, no. Skin resistance depends on the size of the contact patch, how sweaty you are, where on the body you measure and so on. The number suggested for safety calculation in the college physics text I was teaching from this spring was only $300\,\mathrm{\Omega}$.

And as John said, $100\,\mathrm{mA}$ is potentially deadly.

Lower for AC in the wrong frequencies.

I've seen guidelines that suggest treating $60\,\mathrm{V}$ DC as dangerous when working with chemical cell batteries.

oof, i knew it was going to be humid today but daang

or maybe it's just the heat in general. No clouds, 90 degrees outside, 60% humidity

Hey that's my weather all summer...except on the occasional days it decides to be 100 degrees instead

@ACuriousMind In retrospect, there's an even simpler version of the whole A and B thing I was on about. If A and B are observables, then A+B is as well; but that doesn't mean that you can deduce A+B from measurements of A,B by themselves.

@danielunderwood yuck

@Semiclassical True enough

The case of spin measurements is a bit special, though, since e.g. $S_x+S_y$ is the (proportional to) the spin along the 1,1,0 direction

@Semiclassical Sure - measuring the spin along an axis given by a vector $v$ is measuring the (normalization of the) operator $v_i S^i$

So you can measure $S_x+S_y$ directly in that case by changing your measurement axis. But for one thing that's special, and for another thing one has $(S_x+S_y)^2=2I$ as well

which recovers $S_x S_y+S_y S_x = \frac12[(S_x+S_y)^2-S_x^2-S_y^2] =\frac12[2I-I-I]=0$

and if the directions weren't orthogonal, you'd still end up with the symmetric product being a multiple of I anyways

so the symmetric product of spin components can't help but be pretty boring.

(you can say that <A+B>=<A>+<B> by linearity but that's a one-off)

"You're only infinitesimally correct."

Of course it is because you can divide it by another number and get $\frac{dx}{dy}$!

anyone here familiar with how to schedule a python script to run at fixed intervals on a windows machine?

Could you just set up a scheduled task? Or are those even a part of windows still?

I can

but it seems like I need to make the script into a windows executable?

right now I'm running the script from command line by directly calling python

it's not like a .exe file or something that would just run if I double click on it...

I don't think the scheduler is smart enough to let me go "start anaconda and then execute this command line:" is it?

Ahh I've never really messed with scheduled tasks...maybe a batch/ps script can be run as one?

I dunno what that is

I tried turning my code into an executable but it failed

seems like both modules to do that don't work with python3.6

It looks like you can add args to the task, so just use the python.exe as the program and your script name as argument

hmmm

I created an executable using CX_Freeze...but it appears to do nothing when I execute it

it just opens and closes a command prompt window

I would also need to navigate to the appropriate folder tho...

I don't think the script can be run that fast...

yeah it's not doing anything -.-

it doesn't even show up in task manager

annoyingzzzz

Usually opening and closing a command prompt is because of an error. If you open a command prompt and do .\program.exe, it'll show you the error. I haven't had a great experience with the freezing/packaging into exe though. You can create a powershell script to cd into whatever directory and run the script, use powershell as the program, and your script as the arg

Or you could ditch windows and use cron lol

can't ditch windows

powershell script...hmmmm

sounds like that could work

hmmmmmmm

I think powershell would probably be easiest. You may also be able to use wsl if you're on win10 and your company would allow that. I have no idea if cron works within wsl though

what's wsl?

It's kind of like linux within windows. My understanding is that they implemented some of the core unix functions to tie into windows, but I'm not entirely sure how it's done

docs.microsoft.com/en-us/windows/wsl/install-win10

I've have to talk with IT and the security team about that

I'd*

Actually that may not work with cuda either. Probably more trouble than it's worth if you can get a powershell script working

also powershell doesn't seem to have access to python or something cus typing python in the powershell gets me a "no command" error

I need to maybe write a script that would work for anaconda prompt

Just stick the commands of what you want on lines and hope it works lol

That means your python isn't in path, you could add it to path or just do the whole path like C:\python\bin\python.exe or whatever

I see...

I don't think I can edit my path variables

this is such a pain in the @$$

oh there's a much easier way

just run python.exe as the executable and then the argument is my script and the start_in is the path to the script...

Ahh I didn't know you could choose the start path

found a stackoverflow answer that was very helpful

heh

What should I know about a battery if I want to know how much time it will last if the circuit it powers draws 2A per second for example?

@NovaliumCompany How much energy/usable charge is stored in it. For convenience, manufacturers usually already give that to you in the form of specifying it in units of "Ampere-hours".

@NovaliumCompany You should know that “2 ampere per second” doesn’t make sense here.

Ah, right. Also that :P

I should consider it in joules?

@NovaliumCompany No, Loong's point is that "Ampere per second" is not the unit of any meaningful physical quantity in this scenario

Think about it: What is the definition of current/Ampere?

Actually, forget it, I don't know what exactly I am asking and I'm from another computer.

I know about Ah, Wh... the problem is that I don't know what exactly I want to know xD

@NovaliumCompany Don't be discouraged - a large part of understanding is figuring out what exactly it is that you don't know

I'm not, I'm sorry for wasting your time with stuff that I guess I am familiar with.

I just don't have a specific problem xD

I just saw a battery next to me that says 9.6V and 1.2Ah and my mind started wondering...

(stuff like "2 amps per second" does show up when it comes to inductors, since there the voltage is proportional to the rate at which the current changes. but that's not this sooooo yeah)

I know, I just didn't think about it :D

So the Ah represents the charge that's stored in the battery?

And knowing the voltage we can find Wh, which represents the energy in joules that are stored and ready to be used by a circuit?

@NovaliumCompany Yes. Note that it is current times time and so has dimensions of charge.

I just view it like Ah = A * h = C/s * 3600s, we cancel out and we have C * 3600?

(Also note that, for real-world batteries, you may get more or less charge out of it depending on the exact circumstances of operation - it gets a little more complicated when you consider that real-world batteries are not idealized storages of charge)

1 A-h = 3600 C yeah

In a circuit, P = IV, so we can know the joules per second that the circuit consumes. And we can calculate for how much the battery will last if we know the Wh of the battery?

well, it'll depend on what the battery is connected to. If you connect it to a large resistance, then the current coming off the battery will be very small.

I think Wh may be less useful since the voltage of the battery will generally drop over time, though I'm not entirely sure what's best in practice

Well Wh is how much energy is stored in a battery, isn't it?

yes. but to go from "total energy" to "time until battery runs out" is a bit tricky since only an ideal battery has a constant voltage

Well, why it isn't?

Oh yes.

in reality, the voltage will go down over time and therefore so will the current drawn from it

Yep, so as V goes down, P in P = IV will go down as well.

hence the power of a battery in a circuit would be expected to decrease over time if you left it plugged in

So real batteries are a bit annoying :P

xDD

on the second page of the datasheet for this particular battery, they give voltage vs. time curves for some typical applications: data.energizer.com/pdfs/1209.pdf

so that'll take a while :)

(Looking at datasheets always makes me feel smart) :D

ikr

well, right up until you realize you don't actually understand what it's saying

xDD

or until you spend an entire day going through datasheets and your brain just goes numb lol

Oh nvm

So as a circuit draws energy from the battery, the provided voltage from the battery drops with time?

Yes. I think the reason is that the cell becomes less efficient. If you have some interest in electronics, you may want to check out this book. It had a ton more stuff than I learned in lab courses/work smile.amazon.com/Practical-Electronics-Inventors-Fourth-Scherz/…

@NovaliumCompany It depends on the exact type of "battery" you're using.

Is there a type of battery that doesn't drop over time?

@ACuriousMind Of course, we can see how it drops from the datasheet of the battery?

@danielunderwood I can't think of any that doesn't drop, but how they drop varies considerably (compare the exponential decay of a capacitor against the rather long approximately constant phase of a standard commercial battery)

well, they don't provide enough data to do so completely

See e.g. electronics.stackexchange.com/a/128248/152799 for some curves

looks like I don't have sufficient privileges on this computer to run a script while logged off......

but for instance you can estimate the time for the voltage to drop from 6.0 V to 5.2 V to 4.4 V to 3.0 V from the diagrams given

Is this why, for example, flashlights start to get dimmer and dimmer with time ?

Like the voltage the battery provides as the load consumes energy?

I'll also note that an alkaline battery is as much chemistry as it is physics

@NovaliumCompany Yes

Ok cool. Thanks everybody. I'll go now so see you some other time :)

Btw, instead of thinking of a battery as a voltage source whose voltage drops, you can also think of it as an ideal voltage source with a growing internal resistance.

Never thought of that. Is that useful for certain calculations or something?

Not sure, ciruitry is not really my strong suit :P

Seems nice for intuition if nothing else. The battery is in series with the load, so if the internal resistance goes up then the total resistance of the circuit is also increasing

Though I'm not sure said internal resistance counts for instance as a source of joule heating.

@Semiclassical Joule heating is a feature of real-world resistors, not of ideal circuit elements.

well, some googling suggests that the answer is yes, but that in modern batteries the internal resistance is small enough that the heat generated this way is negligible

@ACuriousMind Hm, if you want to use the Nernst equation, voltage drop might be more suitable.

@Semiclassical Link? I find that claim dubious since there is no actual voltage source with the voltage of the "full" battery in the battery, so there's no actual voltage drop across any actual resistor

But we're really getting into territories I don't know anything about :)

this is the one I was looking at, though on closer inspection I think it's a bit out of context: books.google.com/…

see also the first few sentences of the abstract of this paper on heat generation in lithium-ion batteries: ieeexplore.ieee.org/stamp/stamp.jsp?arnumber=6563807

in both cases it certainly seems as if they regard the internal resistance as an actual resistance, and therefore subject to joule heating

But this is also assuming I'm correctly interpreting what amounts to chemical engineering papers

and uh

not sure I buy that

Heh

this paper looks nice, as a more realistic account of how batteries discharge: pdfs.semanticscholar.org/64bc/…

it doesn't seem to entirely match with the curves in the Electronics SE link, which is a bit disappointing

crappppp

accidentally overwrote an important file -.-

that's not good

@enumaris Good old rm -rf

what's that?

not familiar with the -rf tag

@enumaris Equivalent to -r -f - removes everything in a directory, in particular the content of all other directories in it

hmmm

That is, execute that in the wrong place and you wipe your entire disk :P

how permanent is that wipe though

It's very permanent :P

does it just remove all the pointers to the file

it actually overwrites the data on the harddisk?

I think it drops references and it's up to the file system to decide?

Well, it deletes the files. What "deletion" means is an implementation detail :P

generally when you "delete" a file, it just removes the pointers to that file doesn't it...the data is still there on the hard drive. Unless you defrag or do something else that moves the data around, it should be recoverable...

That command is basically the archetypal instance of your "oops, I deleted the wrong file".

in my case, I overwrote the file, so I think it's actually gone...

possibly not all gone, possibly just corrupted

@enumaris Sure. But recovering that data is not trivial and once it's deleted, it's up for grabs by any other write action.

yeah

If the important file you deleted happened to be open or recently open in an ide, you may be able to restore it. I managed to delete a whole directory with an rm -rf once and that saved me

that's when you quickly unplug your hard drive

and take it to someone to restore as much as possible

Now I just live with automated backups every hour

I'll just reproduce the file

Easier: Back up your important files!

it just requires me running another 18 hour long script

I'll make a backup of that file after I run it this time lol

If it happened to be a trained model file or something similar, I always stick a timestamp on it to make sure not to overwrite new ones

It's a dictionary of something

but this issue does remind me of another issue

"A dictionary of something" - clearly an extraordinarily important file! :P

well it's the dictionary linking named entities together for all 3 million comments that I processed the other day lol

in order to reproduce that file I need to reprocess all the comments

I guess deleting it was a dict move.

lol

doing this has revealed another issue...

Be careful. Fixing issues is known to lead to other issues lol

@danielunderwood That's why you have automated tests.

Unless those are the source of the issue! I actually at one point had some tests bypassed because the code seemed fine, but not the tests

Turned out I was comparing sets in the tests and hadn't implemented a hash function for the classes

In which case I really shouldn't have been using sets in the tests, but lesson learned I guess

@ACuriousMind better than a dictionary of somethings

I have 2 other issues to deal with now :D

the issues are reproducing

"99 bugs , 99 bugs in the code. Take one down, compile it again, 127 bugs in the code..."

well it's not bugs at this point

it's known features of the code that I have to change if I want it to behave and work well in a new environment

and then also administrator privileges on my machine...

I don't think IT will give me admin rights

-.-

Tell them special computer should mean special rights haha

I think they are worried I'll get "hacked" and then our whole data security is at risk

@enumaris Corporate policies are usually set by people without actual technical expertise; it is likely that IT is not worried at all, but they are instructed not to do it.

well we do have an infosec team

I think they are pretty worried

those guys are pretty uptight too :P

took me a month to get Python approved

and they only approve python 3.6.5 and no other version

I'll not go into any details, but our corporate lawyers recently circulated a policy draft that would basically have made standard development processes impossible and would have declared the release of our software product a security breach

heh

Never realized that I may have it pretty nice working at a small company

I also will need a virtual linux machine if I am to run one of the software I want to run...

but that seems unlikely as well...

Although mine did see ssh as a security issue

I think I can forget about ssh lol

A vm may be tough if you're using cuda unless there have been improved ways to do it. I think docker has some cuda support and there's a docker for windows, but it sounds like they wouldn't go for that either

@enumaris Heh. I could get a virtual machine set up with basically any OS needed. I just don't have any use for that because code in ABAP :P

@ACuriousMind do you still get to do any physics in your spare time or anything?

Assuming your job is just software and not physics-related

@danielunderwood The only physics I do these days is when I answer a question here

"get to do" implies I'm somehow deprived of it, though - I'm not, and quite happy with the way things are

Ahh I went from undergrad into software and a bit of electronics and miss the physics

Maybe I'll miss it one day, but not at the moment

To circle back briefly

But, well, I've only been at this for about half a year, stuff is still somewhat new and exciting

Alex mentioned yesterday that we could get dim of SO(N) by looking at the dim of corresponding coordinate ring. Does this capture the reduction due to symmetry?

just something I was wondering

not sure if I am making sense

I'm afraid I don't understand the question

Reduction of what, due to what symmetry?

let me try to get the response alex gave me

so I asked yesterday how one gets $\frac{n(n-1)}{2}$

I understood the other two responses, but alex mentioned

"Well you can check the dimension of the corresponding coordinate ring"

A coordinate ring is a concept from algebraic geometry and strikes me as a rather unnatural way to look at a Lie group.

yeah, I thought so too

after I looked at it some more I was wondering if maybe one over-counts if attempting to do it this way

It is generally true that the algebraic dimension of a variety that is also a smooth manifold is equal to its ordinary dimension as a manifold. Since Lie groups are smooth manifolds, you are not overcounting. But again, looking at the coordinate ring is probably really not the most useful way to compute the dimension here.

hmm yes

ok yes it works

i tried it

hehe

@danielunderwood oh dear, I had forgotten about that issue...if I boot to Linux I have to deal with the CUDA issue all over again -.-

I think it's probably not as bad to handle on linux...unless you're using linux in a vm

In that case, good luck

what if I dual boot to linux

linux! hehe

I am actually on my linux mint laptop right now :P

Dual booting to linux would probably be my preference if it's an option

I would say to be careful about the distro/version if you get a choice. The very first time I tried to use tensorflow was a pain due to my distro. I think that may not be as much of an issue now though since last time all I needed was a pip install tensorflow-gpu or similar

right...