The h Bar

12:16 AM

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

user400188

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.

Cows

1:18 AM

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

Cows

1:40 AM

also another silly question

in what is this parameter counting thing?

say for O(N)

Avantgarde

Welcome Agalloch ;)

Cows

@Alex hey dude :)

Alex

Hello @Cows, Hey @Avantgarde

Avantgarde

The Mantle is good. Probably their best.

Alex

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

Avantgarde

1:44 AM

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.

Alex

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)

Avantgarde

Nice that you play. I don't.

Alex

Probably my favourite song of the album

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

Cows

I see

hmm

Alex

Are you doing Lie algebras or Lie groups atm?

Cows

1:50 AM

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

Alex

Well you can check the dimension of the corresponding coordinate ring

Cows

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

2

Q: Dimensions of various lie groups by counting parameters of matrix representations

Show that the groups $\text{GL}(N,R), \text{GL}(N.C), \text{SL}(N,R), \text{SL}(N,C), O(N), SO(N), U(N),SU(N)$ have dimension $N^2, 2N^2, N^2-1, 2(N^2-1), \frac{1}{2}N(N-1), \frac{1}{2}N(N-1), N^2$ and $N^2-1$ respectively. Attempt: Elements in $\text{GL}(N,R)$ are just $N \times N$ matrices w...

Oh

bolbteppa

2:15 AM

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

Cows

this is a very good explanation, I like it

finally!

bolbteppa

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

Cows

2:32 AM

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

Physics Meta

5:02 AM

0

Q: Restricting huge reputation gains from simplistic questions

This is kind of related to John Rennie's post 'The secret to getting a massive reputation is...', but I'm focusing on the responses to not-so-good questions, unlike that post. This is about how basic answers to bad questions get way too many upvotes and reputation. While I agree with the claims ...

discussion reputation up-votes

watchme

9:23 AM

Are 20.000 volts dangerous when using around 0.5 mA?

Slereah

Only one way to find out!

watchme

9:44 AM

@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

Slereah

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

watchme

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

John Rennie

10:00 AM

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

watchme

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

So "how much goes through"?

John Rennie

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.

watchme

Oh I didn't realize that higher voltage causes a higher amount of ampere....

I am so new

Novalium Company

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?

watchme

Dude you can buy the world

Novalium Company

10:06 AM

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...

John Rennie

@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.

watchme

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....

John Rennie

10:21 AM

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.

watchme

Ok yeah I thought so... thanks John!

I have a question to you if I may ask:

Do you consider yourself to know "a lot" in comparison to some others?
I know you might ask "who are 'some others'", but I think you know what I mean, so all those good physicists around the world which teach at universities or are "on the top".

Because arond 260k rep isn't little ^^

John Rennie

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.

Jim

11:59 AM

@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

John Rennie

@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.

Jim

@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

John Rennie

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.

Jim

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

John Rennie

12:15 PM

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

Jim

That makes a lot of sense to me

watchme

12:26 PM

@Jim so the resistance is just very huge

Jim

@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

watchme

Hmm... that might give me a different view on wall sockets.

In the US 120V (In my country its 230V) is the highest amount of volt a normal wall socket can give, right?
Or doesn't this amount matter as long as 1800W aren't exceeded?

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

Jim

12:42 PM

@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

watchme

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

@Jim ups

Jim

@watchme yeah, it's just a small one

watchme

And this wall socket is set to 250k voltage....right?
Because I might have gotten your sentence "These things are connected to the wall socket, which provides 120V at 15A maximum." wrong.

Because when a wall socket is always set to a given voltage (no more, no less), then 120V is to little for a tesla coil needing 250k, right?

Jim

@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

12:59 PM

Oh ok^^

That's interesting... So it increases the current to be nearly 15 Amps in order to lift the voltage up to 250k

I am really new to electricity so I get things a bit slow sorry :D
I didn't know that this is possible, so to have a specific amount of watt and just "play" (!?) around with it to adjust volt and amps.

rob

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

watchme

1:16 PM

@rob thats very interesting...

With losses you mean the efficiency right

rob

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

Semiclassical

1:43 PM

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)

enumaris

4:06 PM

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 -.-

dmckee

4:32 PM

@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.

Semiclassical

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

danielunderwood

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

Semiclassical

@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

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