2:09 AM
13

I am a course instructor. In my courses, which are now online due to the pandemic, I have optional live sessions. In those live sessions, I give examples of how to solve problems (it's a math-based course), and answer student questions in a group setting. This design comes as close as possible to...

> I think that if I were to provide recordings of these live sessions, more students would not bother to attend. I strongly believe that attendance at these sessions is pivotal for understanding, and that watching posted videos of these sessions will not have the same effect...I specifically do not want to allow this, as I think it discourages proper time management and may lead to problems down the road.
Ugh, I hate this
why is it so hard to accept that attendance != understanding

3:05 AM

3:38 AM
I have not explored the chapter fully but i have one doubt
What is advanced version of ms∆t

4:13 AM
2

Both definitions are fine as long as you're careful with signs. Here's a derivation of your gravitational potential energy using the idea of the negative work done "by the field." (Note the initial negative sign.) $$U(r)=-W_\text{by field}=-\int\vec{F}_\text{field}\cdot d\vec{s}=-\int_{r=\inft... What does the initial negative sign stand for? Is it because the work done by gravity is positive and the change in potential energy is negative? And vice versa the external agent?  U(r) = - (+W_{by field}) ? and  U(r) = - (- W_{ext agent}) ? 7 hours later… 11:38 AM Sanity check: can I say that the renormalisation group equation (for the vacuum polarisation in QED) arises because the potential at some momentum should be independent of the arbitrary reference scale p_0 at which the renormalised electron charge is defined? p_0^2 \frac{d}{dp_0^2} \tilde{V}(p^2) = 0 is the RGE @NiharKarve yes @ACuriousMind ok, so why is there a factor of p_0^2 (or \mu in the more common RGEs) before the derivative? 11:54 AM @NiharKarve Convention so that it matches the form of a generic Callan-Symanzik equation (p_0^2 is the mass scale called M in that article's notation) Also I just realised that you can't just get rid of it because the derivative isn't always 0, whoops :-) Thank you, user6232128-san 12:16 PM \o @Slereah 1 hour later… 1:43 PM Hi all, I have a question from Griffith's Electrodynamics He starts off with J=sE , where J is current density, s is conductivity and E is the electric field for constant s, we get div(J)=s*div(E)=s*(p/e) (by differential form of Gauss Law) However, continuity equation states: div(J)=0 which implies p (i.e charge density)=0 Now if we apply this analysis to a current carrying wire...arent we running into a contradiction? "current carrying"---->electron flow, so how do we have p=0? 2:00 PM @satan29 J=sE is the formula for the current density produced by an external electric field applied to a material of conductivity s i.e. the source of the field is not inside the material, and hence \nabla \cdot E_\text{ext} \sim \rho = 0 but that \rho is not about all charge inside the region we're looking at, it's just about the charge responsible for this external field E_\text{ext} This is apparently the only photo I can find of Marie-Antoinette Tonnelat Lost in a Solvay conference photo what is this, a picture for ants? The full picture she is an elusive lady 2:32 PM That's someone else apparently yeah, sorry she certainly is elusive 3:03 PM I am trying to get familiar with the concept of corepresentations of magnetic groups. My question is what happens in the case where the magnetic group is unitary, hence, lacking time reversal symmetry. Do we still use the notion of corepresentations, is it the same as the irreps? I don't know how to construct them, or if they have any physical significance 3:14 PM I have never heard the term "corepresentation" in my life :P 4:06 PM @ACuriousMind It sounded like it has something to do with groups including antiunitary elements, such as time reversal @ACuriousMind whats up with the omniscient quest markers @RyanUnger a general plight of modern game design, yeah, I hate them too they seem particularly bad in this one yeah, they mostly don't even bother to give you anything resembling proper directions in dialogue so you have no choice but to follow the blinking light @ACuriousMind I have a rep theory question if i,j \ge 2, is$$\int Y_1^{m}Y^{m'}_i Y^{m''}_j =0
this seems like something a physicist would know
presumably you use the contraction rule
and then it's some mess

4:13 PM
@RyanUnger the $Y$s are the matrix elements in some representations?

what? spherical harmonics
I can reduce it to some crap with 3j symbols/CG coefficients
but I have no good reason to hope that this is actually true

ah, sorry, I don't know much about specific values of CGs/3jm

I think in general though $Y_i^{m'}Y_j^{m''}$ could have an $\ell=1$ mode
not that I have a good reason to believe so

sure, it's about whether the $i\otimes j$ representation has a $\ell = 1$ subrep

see, algebra
so is that true

4:18 PM
sometimes, I guess? Algebraically, you'd have to look at the Young diagrams, but the CG coefficients encode this same information
@RyanUnger Right, I remember now: A rep $\ell$ exists in $i\otimes j$ iff $\lvert i - j\rvert \leq \ell \leq i+j$
so that there is zero if your $i$ and $j$ differ by more than 1

yeah I know that
the thing I'm worried about is when $i=j$ of course
oh but then I think parity saves me or something

then it will not be identicallyzero, I'd say

I think $Y^{m'}_iY^{m''}_i$ is even
for any $i$
because they're both even or both odd
so the triple product is odd
hence the integral is zero
so the issue is when $j=i+1$
because then the product is odd and there might in fact be an $\ell=1$ mode

4:34 PM
You probably want to do this with 3j coefficients, since spherical harmonics form a basis the product of two of them is a linear combination of spherical harmonics which can be expressed in terms of 3j's, then integrating against the third leaves one remaining coefficient which involves one kind of 3j

I'm trying to find a table of 3j coefficients
I found an online calculator
which disproves the conjecture
and awful

5:02 PM
Can every timelike curve be sufficiently approximated by a (possibly infinite) sequence of lightlike curves?

Define "approximated"
It can converge to it but otoh the proper time of a sequence of null curves is always zero

Isn't the proper length of any sequence of lightlike curves always going to be zero?
Damn, beaten to it :-)

Hi...Yo

@Slereah does it make sense ever to perform that approximation?
Actually I think Caroll uses it somewhere

I think it's used in a proof for like
The continuity of the proper time function or something?
p. 15
"In fact, the timelike curve in (i) can be chosen arbitrarily close (in the $C^0$ topology) to $\gamma$."

5:12 PM
I just checked, Caroll uses it to confirm that timelike geodesics are maxima of the proper time
And not minima

Well locally, yes
Also yes, there's always a smaller (or larger, depending on your sign conventions) proper time between two points, since you can always connect two causal points via a sequence of null curves
So there's always a causal curve of length 0
But beware of saying that this approximate the curve, since they have different properties

Yeah, I wasn't really sure what to call it, I was just imagining jagged diagonals in a 1+1

Depends on the topology you define them on

Is the Poisson book good

I hear it's alright

5:21 PM
The name sounds cool but I think some people have deemed it the worst book they'd ever read

It's not the Eye of Argon
It's not even this : amazon.com/gp/product/B001KEFSE8

To be fair, it is tagged "high-school"

the worst book they've ever read
probably haven't read too many books

Wait, you're the famous 0celo?

5:38 PM
It seems the whole $C^0$ topology business is chapter 2.3 of BEE, if you want to check it out

What's BEE?

Global Lorentzian Geometry by John K. Beem, Paul E. Ehrlich, Kevin L. Easley
Also chapter 7 of penrose's differential topology

Thanks for the refs
I've been trying to up my GR game

@NiharKarve who?

Er, you?

5:45 PM
are we gonna do a whole comedy bit

Does anyone know some 'video programming' here

It may be more fitting

I want to see how the ball in this question would move
A small ball is suspended from a point O by a light thread of length l. Then the ball is drawn aside so that the thread deviates through an angle θ from the vertical and set in motion in a horizontal direction at right angles to the vertical plane in which the thread is located. The initial velocity that has to be imparted to the ball so that it could deviate through the maximum angle
2
π

in the process of motion is v
0

=
cosθ
xgl

. Find x.

### Game Development

Game development and other polite discussion. Game development...

5:56 PM
Thanks
But i don't think anyone would be interested
Since my message seems

why do we get it then

That i am bugging them

So this is pretty stupid and niche, and possibly not interesting to anyone here, but I'm pumped about it. I spent the last several months trying to change one of my guitar hero guitars into a left handed one by basically ripping up the front of it and swapping all the plastic around. I made some bad design calls which led to more soldering than needed, which led to me breaking more wires, which led to more soldering. I thought it was basically toast, but I actually got it working today.
So I'm pretty amped that a stupid project that snowballed out of control actually got finished.

6:14 PM
Perks of being an engineer, eh
Now you can grind Guitar Hero without reservations

6:40 PM
I get really into stupid projects like that too. The funny thing is, I was so used to the right handed version that I keep reaching to the wrong place for the whammy bar that I swapped to the correct place for lefties.

1 hour later…
7:55 PM
1

Symmetry First, the apple appears to break into 3 parts, not two. Second, the break is obviously not as "clean as a knife" if you slow the video down and watch it carefully. However, the break is much cleaner than if you tried to tear the apple into chunks with your bare hands (unless you happe...

Someone should introduce this guy to Bob Mortimer.
A.k.a. Edward Scissorhands, apparently.

2 hours later…
9:33 PM