@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.
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.
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.
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.
@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.
How 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...)
"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
@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.
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
@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.
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.
@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).
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.
@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.
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
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.
@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
@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.
@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.
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.
@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.
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
How 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...)
