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7:01 PM
nice i liked that game
havent finished it tho
 
It's pretty good and I'll definitely play through it again some time when it's not so fresh in my mind
 
0
Q: Several questions about Newtonian World Index Algebra

ICCQBEWhich of the following quantities are meaningful and which are not? a) $A_{ij}=B_{ik}A_{kj}$ (Meaningful, because $k$ indexes cancel each other. How can I describe this rule, I don't know.) b) $A_{ij}=B_{ik}C_{kl}$ (Not meaningful, because free indexes on the LHS are $i$ and $j$ but on RHS $i$ ...

That's the question. I don't know the name of the subject is "Newtonian World Index Algebra" my teacher told that It's name.
@ACuriousMind
 
Sorry, that kind of questions that just asks to solve an exercise is off-topic at the main site.
 
It's fine I tried my chance :D
 
I don't think anyone except your teacher calls doing math with indices "Newtonian World Index Algebra". That notation (implying summation by repeating indices) is usually called Einstein notation
 
7:11 PM
Yeah, absoultely. My teached told about E.S.C and It's agreement text
Well, now I know what It's name.
Do you have any idea about my question?
 
Yes: there is not enough information given. The answers depend on whether you're trying to define the l.h.s. of these expressions by the r.h.s. or whether $A,B,C,D$ all exist a priori and these are just supposed to be relations between them
 
I don't really have no idea about it :/
Only thing I know is by calling them "meaningful" or "not meaningful" is just checking them if the expression is valid or not.
For example, $A_{i}=B_{ik}$ is not a valid (not meaningful) expression
Because it must have same free indexes on both left and right hand side.
On LHS there is only one free index, $i$.
On RHS, there are two free indexes $i, k$.
 
Yes, that's definitely always correct.
 
They are not same, so It's not a meaningful one.
But when trying to solve others, as I expressed what I can't decide, I can't solve.
 
ACM I got a close vote for my question can you see and tell me what’s the reason that user has chosen for voting to close my question?
 
7:26 PM
@ICCQBE As I said, for things like your c), it is not decidable without knowing whether it's supposed to be a definition of a new quantity or a relation between existing quantities
@Knight Can you not click on the close(1) and see for yourself?
If you can't, I won't tell you since other users who do not ask a mod about it also don't get to know that.
 
@ACuriousMind Oh that’s a great thing and I agree with that
@ACuriousMind I fear lest I vote to close my own question:)
 
@Knight you can retract your vote anytime!
 
@JohanLiebert I want to know what reason that user used for voting to close. Because I should defend my question as soon as possible
@ACuriousMind I’m unable to see it, clicking on it opens the reason tab and asks me “why you want to close this question”
 
@Knight sorry! I don't have the close vote privilege so can't tell you myself.
 
@Knight Yes, do you not see a blue '1' beside one of the reasons?
 
7:33 PM
@ACuriousMind Yeah I see, opinion based
Thank you
@JohanLiebert Are you wearing your knighty?
 
@Knight Do you wear knighty? As far as I know only women wear that. Though I might be wrong!
OK good knight!
 
@JohanLiebert Well you see you never told me that you’re a boy
:-)
I’m just joking
 
7:52 PM
@ACuriousMind do you know Levi-Civita tensor?
In 3-dimensions It refers to $\epsilon_{ijk}$
If $(ijk)$ double permutation, then $\epsilon_{ijk}$ equals to $+1$ and if It's single permutation, then $\epsilon_{ijk}$ equals to $-1$ and if it has got two same indexes then $\epsilon_{ijk}$ equas to $0$.
I have a question about defining Levi-Civita tensor on 2-dimensions
 
In two dimensions, you can just have an anti-symmetric quantity $\epsilon_{ij} = -\epsilon_{ji} = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$. Livi-Civita isn't a real tensor though
 
dsm
8:14 PM
@danielunderwood Sure it is, the $n$-dimensional version being $\epsilon\equiv e^1\wedge e^2\wedge\cdot\cdot\cdot\wedge e^n\in\Lambda^n\mathbb{R}^{n*}$; an antisymmetric rank-$n$ tensor with a one-dimensional basis.
 
@dsm What daniel presumably means is that it's not a "real tensor" in the physicist's (well, general relativist's) sense, since $\epsilon_{ijk}$ does not define a coordinate-independent tensor field on a manifold
When physicists say "tensor", they often really mean what a mathematician calls a "tensor field"
 
wikipedia quotes it as being a pseudo-tensor, on the grounds of it not changing sign under a reflection
 
I just learned about Generative Adversarial Networks (GANs). Damn that's cool
 
dsm
Ah, gotcha
 
🤣🤣🤣 gets me every time
 
8:29 PM
"proof" = "sketch of the proof" is so true it hurts :P
Also, how does one pronounce 'Hrvoje'?
 
I think I've seen a version of that with "Hilbert space (physics) = finite-dimensional Hilbert space (math)"
which, while not -quite- true (given how much we deal with the Schrodinger equation) is still a good description of our attitudes towards Hilbert space
 
@ACuriousMind I think it's the group theory bit that really gets me
 
Ah, so continues my cycle of "I know what a tensor is" -> "That's not what a tensor really is"
2
 
8:46 PM
@danielunderwood if it helps, that rabbit hole does mostly have a bottom
(unless you want to go and dive into category theory, of course, in which case no rabbit hole has a bottom at all)
 
There are no different rabbit holes, they are all isomorphic to the universal rabbit hole object
2
 
9:49 PM
hmm. trying to figure out how to tackle the following problem I just devised in thermodynamics
An application of equilibrium thermodynamics outside the context of gases is that of a rubber band, which (ideally) behaves like an entropic spring
The typical experiments for that are either with respect to the variables (T,f) (temperature and tension) or (T,L) (temperature and length). (compare with the gas cases of (T,P) or (T,V).)
For the first one, I hang a weight on a rubber band (constant temperature and tension). For the second, I place the weight on a scale and use the rubber band to reduce the scale weight (constant temperature and length).
What I'm thinking about now is: What happens if I start from the first case and then pull the weight down slightly, to see oscillations?
Clearly that's not constant length, but I'm not seeing it as constant tension either.
looking around online, rubber bands seem sufficiently non-Hookean that I probably should leave this well enough alonen
 
10:35 PM
I'm still so confused about the Frenet-Serret Frame
OK, so the above equation give sthe unit vector for the tangent
OK, so the acceleration vector is the sum of the acceleration tangent times the unit vector for tangent and normal
But what does the equation below tell me?
Q is the radius of the curvature which is the distance from the particle P to the center of the curvature of the path at that point
 
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