12:54 AM
@JohnRennie Hello, John, if you could spare a little time, could you take a brief look at this question? physics.stackexchange.com/questions/519489/…
Thank you very much!!

4 hours later…
4:35 AM

4:51 AM
-2

More often than not the correct answer here is "Have you considered learning philosophy or art instead?" Now this is kind of blunt but i stress that it correct answer most of the time.

2 hours later…
6:58 AM
@frt132 I didn't understand what you are asking in that question. I'll be in the chat room for a few hours this morning if you want to discuss it here.

1 hour later…
8:14 AM
I agree with the mission to teach kids science, but I don't think their motivation is the best.

8:32 AM
tsk tsk

8:57 AM
@JohnRennie as I recall, you're quite the GR fan -- chance you can help me understand some formality in that problem? I feel like it's very simple, but I'm fundamentally not understanding something. if I just follow definitions, it seems $\tilde{g}^{\mu\nu} \equiv \tilde{g}(e^\mu,e^\nu) \equiv g(e_\mu,e_\nu) = g_{\mu\nu}$. I don't know.

@dsm the problem is that I never formally learned differential geometry. I just picked up the bits and pieces I needed as I went along. So when people ask me questions about diff geo I generally have to admit I have no idea what the answer is. Sorry :-(

no worries, thanks for the response :) happy holidays

1 hour later…
10:11 AM
@dsm Things with different free index positions can never be equal! You should have $\tilde{g}(e^\mu, e^\nu) = g(g^{\mu\rho}e_\rho, g^{\nu\sigma}e_\sigma) = g^{\mu\rho}g^{\nu\sigma}g(e_\rho, e_\sigma)$

10:38 AM
@ACuriousMind Hi there, thanks for the input. Sorry, still confused. In his definition of the non-degen bilinear form (ndbf) $\tilde{g}$ on $V^*$, it's equal to the ndbf $g$ on $V$ but with the original dual arguments $f,g$ now their vectors $\tilde{f},\tilde{g}$. Aren't your arguments of $g$ in that second equality still the dual vectors $e^\mu$ and $e^\nu$?

@dsm No - the $g^{\mu\rho}$ are numbers and the $e_\rho$ are vectors, so $g^{\mu\rho}e_\rho$ is a linear combination of vectors (the image of $e^\mu$ under duality), not a dual vector.
If your basis is not orthonormal, it is not necessarily true that the dual of $e^\mu$ is $e_\mu$.

Ahhh, I think I see what you're saying. Can you clarify "the image of $e^\mu$ under duality"? The image of the dual under duality? Does that mean it's $e^\mu$ represented as a vector? e.g. for finite dimensional, transposing the row dual into a column, hence living in the vector space?
Nevermind, that makes perfect sense. Thank you greatly

11:03 AM
You're welcome :)

Also hello

@ACuriousMind Before I go, I'm also confused by one more thing on the same topic. I want to show that for a metric $g$ on $V$, the $(1,1)$ tensor associated to $g$ is the identity operator; i.e. $g_{i}$$^j$$ = \delta_i$$^j. Following in your footsteps... g_{i}$$^j = g(e_i,e^j) = g(g_{i\mu}e^\mu,g^{j\nu}e_\nu) = g_{i\mu}g^{j\nu}g^\mu$$_\nu But then I'm stuck. Is there an identity I'm not seeing? You have defined g^{\mu\nu} above as the inverse of g_{\mu\nu}. What does that mean for g_{\mu\nu}g^{\nu\rho}? that would be equal to \delta_\mu$$^\rho$, right?

Indeed

11:18 AM
does that help me with $g_{i\mu}g^{j\nu}g^\mu$$_\nu? the mixed indices are throwing me off or perhaps my approach to showing that about the (1,1) tensor of g is not the way to go @dsm Don't take the dual of both arguments of the mixed g Just take the dual of one of them. Ahh there it is, thanks! Slowly getting the hang of this -- should have done this self-study a long time ago. I'm off to dinner, cheers :) 12:02 PM Ryan hasn't been 'round in a while I think I need some alternate strategy to get that thesis I guess maybe I can try asking the librarians again Explaining the situation better 1 hour later… 1:12 PM @ACuriousMind I confused myself thinking about that second result during dinner. The metric g on V is the mapping g: V\times V \rightarrow \mathbb{R}. With that, why does it make sense to feed in a vector and a dual into that metric in order to get the (1,1) tensor g_i$$^j$? I'm confused why $g(e_i,e^j)$ is a valid expression, appearing to be mapping $g:V\times V^*\rightarrow \mathbb{R}$.
Fine with the first question you helped me out on, as we had properly defined that $\tilde{g}$ metric on $V^*$.

1:27 PM
I want to justify it to myself by saying well, the second argument is rewritten as $e^j = g^{j\mu}e_\mu$, so all is well and we're feeding in vectors. But I guess that's the crux of my confusion. Writing that dual as the sum of vectors, it's still a dual, yes?
When I have some dual vector $f$ and I want to express the components of it in terms of its vector $\tilde{f}$, it's easy to show $f_\mu = g_{\mu\nu}\tilde{f}^\nu\equiv g_{\mu\nu}f^\nu$, but that's for the components of the dual. Am I making sense? New in this area of things.

@dsm If you wanted to be fully consistent, you would have to use a different symbol for the (1,1)-tensor, just like you did for $\tilde{g}$.

Ah, ok. For example, defining a metric $\bar{g}: V\times V^* \rightarrow \mathbb{R}$ to be $\bar{g}(\tilde{f},h) = g(\tilde{f},\tilde{h})$?

yep

Ahhh, I see. Thank you kind sir

2:08 PM
0

The title says it all. While surfing through the questions page in Physics SE, sometimes I stumble upon a few closed and duplicate questions which are begging me to get salvaged. Simply put, they have major issues which cannot be ignored. My first thought is to edit them right away. But then I t...

2:37 PM
What the fuck is non-degenerate bilinear form?

3:05 PM
@AbhasKumarSinha Which part of en.wikipedia.org/wiki/Degenerate_bilinear_form don't you get?

@PM2Ring dual Space...

@PM2Ring hello, sir. How are you?

shysics

@NovaliumCompany hello buddy?
studying shysics? XD

I just realized Marry Christmas is a wish to marry a person named Christmas
wrong letter nvm

3:14 PM
@NovaliumCompany marry? I think it's merry

@NovaliumCompany really? Where you live? Are not there any Christians there?

i'm christian.
I think
anyways
u still programming ai?

@NovaliumCompany ah okay...
@NovaliumCompany Yes bro, that's passion

coooool
i wanna see som ur stuff :(

3:15 PM
@NovaliumCompany coooooooooooooooooooooooool

but u never added me on that dicsord

@NovaliumCompany Secret ;)

...
i'm serious tho

@NovaliumCompany I dun use it, I'm making a Regression based Neural Net to generate atoms/materials of desired properties...
@NovaliumCompany ok bro, let's talk about serious philosophy and AI

i program stuff
i'm busy sori
i have to go in fact
(btw, if your making general AI, i want in)

3:18 PM
@NovaliumCompany I don't use discord, if you've any other IRC, then I'm in

make a chat room here?

@NovaliumCompany I've one, empty and new, if you like...

@NovaliumCompany Bit fancy but still okay... :P

### Artificial Intelligence

This is the year 2109 on Earth. The reign of Terminators and A...

@yuvrajsingh Hi. I'm not too bad today. Are you going ok?

3:25 PM
No.
I have pain in my ear, when I move my lower jaw.@PM2Ring

@PM2Ring what was the breakfast? :P

@PM2Ring Is that the same Indian food which tastes a bit bland, yellow color and stew like..?

@yuvrajsingh That's no good. I had a bad headache when I woke up, so I had some codeine after I ate breakfast. Now I'm feeling a lot better. ;)

@PM2Ring Are those foods in Aus too?

3:30 PM
@AbhasKumarSinha Yes. My kitchari isn't bland, it's rather spicy, with mustard seeds, chili, cumin, coriander, etc.

@PM2Ring You seem to talking like Indians...
:P

@AbhasKumarSinha Well, many Australians like Indian food. We have people from all over the world living here, and a lot of variety in the food ingredients we can buy. I've been vegetarian most of my life, and learned some Indian cooking when I was in my 20s.

@PM2Ring Fact, there are 60% of Indians vegans,....
@PM2Ring What are some Good Australian foods, I can try to learn...?

@AbhasKumarSinha Like I said, Australia now has people from all over the world, so our food culture is very cosmopolitan. A century ago, the (white) Australian cuisine was heavily influenced by the English & Irish, but after World War 2 we got lots of immigrants from Europe, and in more recent years from Asia and Africa.

@PM2Ring So,it doesn't has any food of it's own? Mostly those which are carried by people?

3:50 PM
@AbhasKumarSinha The Australian Aboriginal people were mostly hunters & gatherers, not farmers. There is some interest in "bush foods", like flour made from wattle seeds, and various kinds of fruits & berries, like lilly pilly, but they aren't widely available.

@PM2Ring oh okay...

Australian soils in most regions aren't very fertile, so they need a lot of fertilizer if you want to grow crops. Also, the Aborigines didn't have any animals that could pull a plough, which makes farming pretty difficult. ;) So the kind of farming that developed in Asia & Europe didn't happen here until Europeans arrived with their cattle & horses.
Bush tucker, also called bushfood, is any food native to Australia and used as sustenance by the original inhabitants, the Aboriginal Australians, but it can also describe any native fauna or flora used for culinary or medicinal purposes, regardless of the continent or culture. Examples of Australian native animal foods include kangaroo, emu and crocodile. In particular, kangaroo is quite common and can be found in Australian supermarkets, often cheaper than beef. Other animals, for example goanna and witchetty grubs, were eaten by Aboriginal Australians. Fish and shellfish are culinary features...

@PM2Ring Australia seems to have a very unique variety of flora and fauna

4:07 PM
@AbhasKumarSinha Very! One interesting food is the bunya nut. The cones of the bunya tree are large & heavy. It isn't wise to park a car under those trees when the nuts are ripe: a falling cone can pierce the roof of a car. :)

@PM2Ring Whaaaaaaaaaaat!? Bunya Nut, oh okay... I didn't knew, that I must expect this too :O

@SufyanNaeem If you don't understand anything, then ping me after a while, I'll present you a simplified version of QM... — Abhas Kumar Sinha 4 mins ago

@PM2Ring Is that a smelly thing which can't be eaten...?

Hmm...

@AaronStevens :)
@AaronStevens hmmmmmmmmmmmmmmm.....

4:11 PM
@AbhasKumarSinha That cone contains the bunya nuts, which are quite edible, as described in the article I linked.

@PM2Ring oh okay.... Have you tried picking it yourself? :P

@PM2Ring oh okay... You've tasted them?

@PM2Ring hello :-) How are you?

@AbhasKumarSinha No, I've never had the opportunity. A good friend told me about them. He also saw a car that had a bunya cone crash through its roof.

4:20 PM
@PM2Ring whaaaaaaaaaaaat! That's dangerous.... :|

@AbhasKumarSinha Yes. There were signs warning people not to park under the bunya trees, but some people aren't good at reading signs. ;)
One Australian nut that I like is the macadamia. My high school had over a dozen macadamia trees. We weren't supposed to pick the nuts, but during the macadamia season the school playgrounds would be littered with macadamia shells. :)

@PM2Ring loool, I must remember that, before I come to Australia....
@PM2Ring same happens to us during summers, but with mangoes...

@AbhasKumarSinha It's ok, they don't grow everywhere.

@PM2Ring oh okay....
@PM2Ring It must be summer in the Australia? You got mangoes? :P

@AbhasKumarSinha I wonder if they will take you up on that

4:30 PM
We had some mango trees at my last house. But the possums would steal the mangoes. You can grow mango trees in Sydney, but the climate here is a bit too cool for them, so they don't produce a lot of fruit.

@AaronStevens didn't get it....

@AbhasKumarSinha We do grow lots of mangoes in Australia, in the warmer parts.

@PM2Ring oh okay.......
@PM2Ring northern probably....

Yes. We used to grow lots of bananas up north too, but it's hard to compete with the cheap bananas from China. So the area in New South Wales that used to be famous for bananas has now mostly switched to growing blueberries.

@PM2Ring do you have dragonfruit there?

4:46 PM
@AaronStevens I agree with the premise that teaching kids Newtonian physics first and then later telling them that Newton got it wrong isn't a great strategy. If you tell people that relativity & QM stuff is counter-intuitive, they'll believe you. ;)
@AbhasKumarSinha I've seen them in the supermarket, but I don't think I've ever tasted one.

@PM2Ring I find it strange to teach students "Newton got it wrong" at all. I mean yes, Newtonian mechanics is wrong; but from a practical perspective it is still very useful and essentially correct in many contexts. It's not like it's the only field where you first learn a simplified model, and then eventually learn a more complicated model that addresses where the simplified model stops modelling reality.

Huh
That was that, then, I guess

5:09 PM
@PM2Ring Yes Newton was technically wrong, but I think it's still useful to teach about Newtonian physics/gravity. You can then move onto showing where it fails, how the new theory fixes things, and how the old theory can still be useful.

@AaronStevens what in the world is that clown hat

Newton was not unconditionally correct, but Newton's mechanics are perfectly suitable within their domain

@JMac Fair point. And of course we still need Newtonian mechanics. But I think it's important to explain to kids (sooner rather than later) that these things are models, and they have limitations. And that introducing relativity & QM as weird & counter-intuitive makes it hard for people to develop an intuitive understanding of them.
@NovaliumCompany See meta.stackexchange.com/q/339891/334566 for all the hats, and how to earn them.

@PM2Ring Personally, the way I was taught physics I was totally fine when I was told newtonian mechanics wasn't perfect. We started with very basic assumptions anyways, like that objects were all point masses; then added details from there. Adding conditions where newtonian mechanics didn't always apply just seemed like the normal progression of how physics was built up, to me. Maybe I just got lucky and had a good program in my school.

There's stuff in Newtonian mechanics that's initially counter-intuitive, in particular, that a body will travel forever in a straight line with constant speed if no external force is acting on it. Fortunately, we can (sort of) demonstrate that pretty easily, with low friction horizontal motion. It's a little harder to demonstrate relativistic or quantum stuff in that hands-on fashion.

5:23 PM
The hat thing is very smart and interesting
I watched Jumanji 2 today btw, not as exciting as the first one :P (in my opinion)

About 10 years ago, Greg Egan wrote a novel called Incandescence, about an alien race living in a rock orbiting a collapsed star (it's not clear whether it's a neutron star or a black hole). These people never developed Newtonian mechanics, they went straight to GR. Some of the exposition is a bit clunky, but IMHO Egan mostly does a good job of making the scenario seem plausible.
2

@PM2Ring Sounds interesting. I've definitely heard of Egan before. It might be a bit out of my wheelhouse; but definitely a cool concept.

6:12 PM
I request for someone to summarize a question. . .
.... As it is rather difficult to find something to construct upon ...

6:24 PM
@ShaVuklia Are you referring to the last two equations?

2 hours later…
8:46 PM
@PM2Ring Do you mean Newtonian gravity, not Newtonian mechanics?
0

This is a variation on the stacking problem. A block is a 1D object of length L and uniformly distributed mass. (with some negligible thickness). A stack of size n is a series of n blocks placed flat one over the other(i.e. their lengths are parallel). A stable stack is one that doesn't topple un...

Is this on topic?

1 hour later…
10:14 PM
@ArtFowler I figured out my mistake, so I deleted it~
I have a conceptual question
parahelium has a symmetric spatial configuration, and hence a higher energy
as opposed to orthohelium
however, having higher energy also means that it's easier to ionise the atom
but intuitively, I would think that the antisymmetric configuration would be easier to ionise, as the electrons are further apart
this is not the case, and that confuses me
could someone help me out?
ah
"If the electrons are on the average further apart, then there will be less shielding of the nucleus by the ground state electron, and the excited state electron will therefore be more exposed to the nucleus. This implies that it will be more tightly bound and of lower energy."
I thought that being closer to the nucleus equated being more tightly bound, but I guess that's wrong then