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1:26 AM
@ACuriousMind @DavidZ can you have a look physics.stackexchange.com/questions/567891/… at suspicious activity. How odd that 4 people should put up junk in so little time.
 
1:46 AM
@ZeroTheHero I'll take a look, sure, but it would definitely be best to cast a flag for mod attention if you didn't already.
 
300 point bounty for anyone that knows how to apply group theory to simplify quantum mechanical wavefunctions!!!!!
9
Q: How is group theory used to deduce that these integrals are equal to 0?

nougakoThe number of all two-electron integrals: $$ \tag{1} \langle \phi_1 \phi_2|\phi_3\phi_4 \rangle = \int d^3\mathbf r' \int d^3\mathbf r'' \, \phi_1(\mathbf r'') \, \phi_2(\mathbf r') \frac{1}{|\mathbf r' - \mathbf r''|} \, \phi_3(\mathbf r') \, \phi_4(\mathbf r''), $$ for $N$ number of basis funct...

 
2:08 AM
@DavidZ thanks mon.
 
:-)
 
@NikeDattani it cannot be done by group theory unless you give additional info on the functions.
you need better focus.
(for your question I mean).
 
@ZeroTheHero It's not my question (it was actually asked on Physics.SE but got no answers and no upvotes for 3 years!).
But it's not correct that more info is needed, the user is asking how the point-group symmetry can be used to eliminate integrals.
For example if the molecule is in D2h, then it has symmetry that can allow you to deduce that certain integrals are 0
It's similar to saying that the integral of an odd function is 0 (if integrated symmetrically around the origin)
Except a generalization of that concept: Water is in C2v because 360/2 = 180, and if you rotate H2O by 180 degrees around the vertical axis (Cnv, n=2 -> C2v), then you get the same molecule. The user is asking how this symmetry can be used to deduce which integrals are 0.
 
 
1 hour later…
3:46 AM
@NikeDattani If you do know the point group then the discussion is so general as to require a course on group theory.
 
@ZeroTheHero What do you mean?
 
I adapt better to reality than virtuality.
 
well this depends on the representation carried by the functions. If they can combine to the indentity representation then the result is non-zero but if not the result is 0. So unless you the group and the functions it’s a little moot.
 
@ZeroTheHero "So unless you the group and the functions it’s a little moot." ?
 
unless you know the group and the functions (so one can determine the representations they carry), it’s too broad. It’s like asking why can $\int f(x)g(x) dx=0$?
there are all manners of reasons why this can or cannot be 0. It is telling that nobody answered that question since it is not immediately answerable without additional information.
it is just too broad.
 
 
2 hours later…
5:30 AM
@ACuriousMind I was singing gucci gang, that's not non-sense.
 
 
3 hours later…
8:58 AM
Wouldn't it be awesome to name your business with an emoji? For example, my business is called ":D". So when someone wants to name the business, they just do the facial expression.
 
 
2 hours later…
10:38 AM
@Charlie congrats
 
10:57 AM
Can someone tell me what does this emoji mean?
???
 
@bolbteppa ty :D
@JohanLiebert I don't think it's an emoji, it's just one of those stickers you can send on facebook
 
I too think that, but while image searching this thing on google it's being called as an emoji. Though i didn't find the meaning of it.
@Charlie 👆
BTW, do you know its meaning?
 
I'm not sure what kind of meaning you're expecting
it's just a bear with wings and a wand
 
So these things are meaningless? Then why do people use them so often?
 
probably very context dependent
 
11:05 AM
Links not working :\
 
@Charlie Never mind. I was just curious if these things had any meaning at all. It seems like i would have to believe for now that the sender of this emoji was in fairy gown! Lol.
 
There's apparently a new more serious "An exceptionally simple theory of everything" (motls.blogspot.com/2007/11/exceptionally-simple-theory-of.html) style paper out: "A near-MSSM from the 10D superfield and nothing else", (motls.blogspot.com/2020/07/…)
 
11:17 AM
How many ladies in bikini does it have
 
11:29 AM
Does anyone know how I can use Mathematica to get a list of the solutions to tan(x) = x in some range of x?
 
Use either Solve or NSolve, depending
 
Bearing in mind I know little of Mathematica what would the incantation look like?
 
NSolve is the same, but it does it numerically, if there are no exact solutions found
 
In[2]:= NSolve[tan (x) - x == 0, x]

Out[2]= {{x -> 0.}}
Thanks, Mathematica, I already knew about that solution.
 
Well they're not wrong
Remember that there's a domain argument
 
11:36 AM
In[3]:= NSolve[tan (x) - x == 0, x, Reals]

Out[3]= {{x -> 0}}
At least it's consistent :-)
 
3
Q: Solution of tanx = x?

Pushpak DagadeHow do I find the solutions of tanx = x upto any number of decimals? (Of course, there is the graphical method but it just helps in finding the approximate value...)

 
I was kind of hoping Mathematica would do it for me. Oh well. I'll fire up my copy of Excel ...
 
Mathematica is a tricky beast
It can probably be done but I don't really know its inner workings beyond the basics
 
How can you read papers on phone without eyestrain?
 
Increase your physicist power
 
11:47 AM
Zoom in, don't sit in a dark room reading a bright phone screen @CaptainBohemian
 
"The presence of an orbifold,... a typical stringy feature, is capable of circumventing the usual conclusion that you can't get a chiral spectrum with groups like $E_8$" if you read the older Lisi article above, you'll realize how stunning this sentence is, the pain this simple issue seems to have caused :p
 
I am surrounded by a couple of incandescent bulbs.
I feel it's like I get eyestrain so fast after reading papers on phone.
 
There's nothing more inherently straining about a phone screen than a book imo assuming you're in reasonable lighting and the text isn't too small
The alternative would be to use a laptop or some other device. I hate reading things on my phone personally
 
 
1 hour later…
1:26 PM
As I zoom in, the text boundary will exceed the boundary of the phone screen.
That will make reading hard.
 
Then I guess option 2 is your only choice lol
 
I think the main reason that reading on phone incurs eyestrain easily is the phone screen is too small.
 
do you have a laptop or pc?
 
1:45 PM
yes, but I feel laptop too heavy to carry around.
 
2:18 PM
I feel the words of papers on Research Gate are really small even after magnification.
 
Short of carrying around a magnifying glass I think the alternatives are limited.
 
too small to read on computer or phone screens comfortably but may not be if the papers are printed out.
but I don't have a printer.
 
2:43 PM
Hi guys can you help me in understanding how he did this?
I am messed with pressure works
Sorry if my question sound off
@JohnRennie hi sir
 
@Yuvraj hi :-)
 
Heading to lunch?
Or still time? @JohnRennie
 
@ZeroTheHero I'll report back to you when it gets answered, and then you can see whether or not what you're saying right now makes sense :)
 
@Yuvraj Lunch is finished :-)
I'm just watching the video you linked.
 
@Yuvraj Are you asking about the cryo-cooler thing that he's using?
 
2:51 PM
@Yuvraj the video looks straightforward. What did you want to ask about it.
 
@Yuvraj It seems like it's just using the ideal gas law. Compress the gas inside, and it heats up as volume decreases and pressure increases a bit. Then you let it cool down (the compression chamber seems to be designed to reject heat); then when you expand the gas back to it's original size, it cools down because of the ideal gas law, and since you rejected heat when it was compressed, it's cooler than it started. Repeat that process to maintain low temperatures.
 
Presumably they use helium due to its high thermal conductivity.
 
@JohnRennie I was thinking because you can get extremely low temperatures while staying a gas. There might even be more reasons.
 
True ...
 
Since obviously with a method like this, if you're gonna liquefy the air, you wouldn't want the working gas to also liquefy during the process (probably?), so I'm guessing it needs to have a lower condensation point.
Actually... maybe it wouldn't be too bad if the phase changed in the helium... I don't know anymore
 
3:27 PM
@ZeroTheHero "If they can combine to the indentity representation then the result is non-zero but if not the result is 0." I could be mistaken, but I think this is all the OP is looking for. Perhaps Nike wants something more, but the OP just seems interested in what they would do assuming they know the symmetry of the functions.
 
Anyone here?
After 4 months of working on my product, I've come to a place where I'm almost done. Unfortunately, the way that I started the whole thing was incredibly wrong. I did no validation of the idea, no feedback loops... I just went straight to building something that I now realize people probably won't need (although I haven't validated that assumption). I've come so far so I'll at least release a promotional video with a website to sign up for the product launch.
I'm secretly hoping there will be no demand for the product because I have no idea how I'll be able to manufacture it in time. It's a really complicated electronics device that requires a lot of 3D printing, buying modules and components, soldering and so on.
So the best way I think I can get out of this is to either try to sell the idea (with the CAD files, schematics...) or leave it be and use it as a portfolio (CV?) for jobs and stuff.
 
if you don't know how to manufacture it, what exactly are you launching and how is it not fraudulent? :P
 
3:42 PM
@ACuriousMind That doesn't seem to stop people on kickstarter lol
 
@ACuriousMind The promotional video/images will be faked. I know how to manufacture it but I'm stuck at an annoying bug in the RF ring controller. I have the software, hardware, everything, but the controller that's used to control the device through radio frequency doesn't work properly (since it involves a custom PCB I designed).
 
I see. I don't think I'm interested in helping you defraud people, then.
@JMac The good kickstarters usually are upfront about risks such as "we have to figure out how to make this first"
 
It's not defrauding, I'm not taking their money. I'm just showing them a concept video of a product. The webpage doesn't even have a "Order now" button.
 
@JingleBells Then I don't understand what you're "launching" if people can't buy it and you don't seem to plan to actually bring this into production
 
Dropbox was started by creating a fake video of unexisting software just so they can confirm that people wanted such thing.
@ACuriousMind Me neither, that's why I'm here :P give me some wisdom
 
3:46 PM
@ACuriousMind The thing is, "we have to figure out how to make this first" is basically saying "We don't even know if this is feasible, but fund us so we can figure that out for you!". It's pretty crazy how often that works for things that don't even make sense.
Like I have no problem with the idea of kickstarter, it's just that some people get money for some really dumb things on there.
 
It's a small-scale demonstration of how well the invisible hand of the market works :P
 
I'm actually going to put it into production, only and only if there's high demand. Otherwise, the suffering is not worth it.
or I can sell the idea (with the CAD files, schematics, etc)
 
the problem with selling an idea is finding a buyer :P
 
true :P
 
Why are the products of inertia, i.e. $I_{xy}$ for a rectangle 0?
 
3:53 PM
So, I'm going to release it, that's said. From there, I have three outcomes, the idea gets sold, people demand a lot and I start production, or (the most likely scenario) no one needs it and I leave it for job portfolio
 
@StanShunpike what do you mean? A rectangle is a 2d object, how is it supposed to have a moment of inertia?
 
If demand is 50% or higher then I start production. Is that a good metric to represent "high demand" or it's impossibly high?
 
@StanShunpike do you mean the off-diagonal terms in the inertia tensor?
 
@JohnRennie I was asking about the cryo cooler
@JMac
@JohnRennie how we make this?
 
A cryocooler is just a specialised refrigerator. All fridges work by compressing a gas to heat it, letting the hot compressed gas cool then expanding it to cool it.
 
4:07 PM
Ok is there any other kind of specification we should take in account while making this?
 
Realistically you can't make a cryocooler. It's specialist engineering that requires helium gas and a very powerful compressor.
 
I mean for them"
 
Isn't this a moment of inertia for rectangles?
 
@Yuvraj What do you mean by specification? I assume it depends on the specifics of what they are trying to accomplish. The things they have to take into account would likely depend on their goals, like any size or power requirements; and generic things like user safety.
 
@StanShunpike That's not what a physicist thinks of when they hear "moment of inertia", it's one of these engineering definitions where the engineers use the word differently from physicists
 
4:13 PM
oh damn
what do physicists call it?
 
@StanShunpike Strictly speaking the moment of inertia is a tensor because it relates a vector to another vector.
 
I don't know, not many physicists have to think about cross section's resistance to bending :P
 
It's a rank 2 tensor so it's normally written as a matrix. A 2x2 matrix for 2D and a 3x3 matrix for 3D.
 
@JohnRennie that's what I thought. and it's super annoying because all the articles (i guess written by engineers) only show a couple of parts of the tensor and ignore the rest
@ACuriousMind so i must be getting them mixed up. no wonder im confused lol
 
But the physics moment of inertia is about the spatial mass distribution, i.e. it has the mass density inside of these integral and it really only makes sense in the top dimension (i.e. for 3d objects in 3d space, for 2d objects in 2d space, etc), which is why I was confused about the idea of a rectangle having a m.o.i.
 
4:15 PM
@JohnRennie I am trying to create a simulation in python of a 4 bar linkage system
where the linkages have rectangles
that's it
i'm just trying to figure out what math/physics i need to do that
 
@StanShunpike we normally choose the principle axes for our coordinate system, and in those coordinates only the diagonal terms are non-zero.
 
i have an article that does it with rods
 
@JMac like like there are different gases in the atmosphere.
 
so i was thinking if i figure out how a rectangle rotates
then i could just switch the formulas out and rework the rest of the math
 
So if one want to make a gas x cryo cooler should have some specific modification to liquify the gas
 
4:19 PM
@Yuvraj no, a cryocooler just cools whatever gas is around it. That's why in the video they ended up with a mixture of liquid nitrogen and liquid oxygen.
All the component gases in the atmosphere will liquify and mix together.
 
@ACuriousMind Even as an engineering term, usually people specify the difference between mass moment of inertia, and area moment of inertia, because you often use both.
 
4:31 PM
I've just learned that the inch is defined as precisely 25.4 mm. So in fact imperial units are really just SI.
 
Did you expect it to be $1/\pi\ \mathrm{m}$ or something? :P
 
the width of the King's thumb
 
I suppose I assumed there would be a standard inch kept somewhere. Henry VIII's thumb or something, preserved in the Tower of London.
 
His obesity could have been a problem for a standard thumb.
 
5:04 PM
I've read through this section on indistinguishable particles several times now, I think I can reduce my confusion to one point: Why, when we measure an observable $X$ on the two-particle space $\mathcal H_1\otimes \mathcal H_2$ do we use $X_1\otimes \Bbb 1_2+ \Bbb 1_1\otimes X_2$ and not $X_1\otimes X_2$. This is the first time I've met measurements on combined systems, do we generally use $X_1\otimes \Bbb 1_2+ \Bbb 1_1\otimes X_2$ for measurements or is this specific to identical particles?
 
@Charlie The operator $X_1 \otimes X_2$ would correspond to the product of the two observables (If $\lambda^{(1)}_i, \lambda^{(2)}_i$ are the eigenvalues of $X_1,X_2$, then the eigenvalues of this operator are $\lambda^{(1)}_i\lambda^{(2)}_j$). If that's what you want, you can measure that, too, but usually you're interested in the sum since most observables behave additively - the momentum of a system is the sum of momenta of its parts, etc.
 
ohhh
ty, that makes sense now
 
5:34 PM
Thanks guys for the help. I have made my plan. ACM thanks
 
user434058
5:55 PM
@JingleBells The English language doesn't have enough expletives to express how much I hate Gucci Gang and his BS music :(
 
7:27 PM
@FakeMod I don't like Gucci Gang either, nor do I listen to his music. I said it because it reminds me of a meme where a kid sings the song to his grandmas.
 
8:05 PM
This is a bit of a stab in the dark, if we take a pair of identical fermions and a pair of identical bosons, the entire system here is (I think) within $\mathcal H_S\otimes \mathcal H_A$ where $\mathcal H_{A/S}$ are the two particle symmetric and antisymmetry states. Is the ability to decompose $\mathcal H_S\otimes \mathcal H_A$ into $\mathcal H_S$ and $\mathcal H_A$ analogous to being able to take a tensor space $T=V\otimes V$ and separate out its elements into "symmetric" and "antisymmetric"
tensors. I.e. $\{t_{\mu\nu}\in T|t_{\mu\nu}=t_{\nu\mu}\}$ and $\{t_{\mu\nu}\in T|t_{\mu\nu}=-t_{\nu\mu}\}$.
and all the tensors that are neither, but they aren't important here
 
8:18 PM
@Charlie No. $V\otimes V$ decomposes into (anti-)symmetric parts as $V_S \oplus V_A$, not $V_S \otimes V_A$.
 
8:47 PM
hmm ok
will need to think about this for a bit
 

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