@DavidZ that's why you can't say $b$ has an inverse $g$ that maps $b$ to an identity element because the inverse has multiple elements that it can map $b$ to. they have to be unique yea.
@Obliv Yeah, but off the top of my head you could generalize this to multiple identity elements in a few ways, like given any $b$ there might be a $g_1$ that maps it to one identity element, $I_1$, and another $g_2$ that maps it to another identity element $I_2$. Or it could be that given any $b$, there is a $g$ that maps $b$ to one element of a set $\{I_1,\ldots,I_n\}$. Either of these is a potential generalization of the normal definition of an identity.
Presumably if you had that, you wouldn't have a group, though.
@ACuriousMind It seems if SUSY is kaput (LHC phase 1 didn't reveal anything) then so is String Theory. But are there a lot of people studying SUSY outside of string theorists?
I am guessing there might be axioms in the mathematical structure that might force them to be equal (if you want your structure to be associative), unless some nice axioms have to be knocked out
I cannot do the calculation on top of my head within 10mins to test my guess, though
@barrycarter You cannot have two different identities. If $l$ is the left-identity and $r$ is the right-identity, then $l = lr = r$ by their defining properties.
I'm mostly miffed that whoever wrote beamer (you know, the most prevalent, I mean the only, presentation class) couldn't be bothered to check that it compiled cleanly.
It's one thing for me as a user to allow warnings, but if the package comes mis-fitted into TeX, that doesn't say great things about the code environment.
In any case I can see the point: it stands to reason that beamer should specify the exact font sizes it wants, to the extent it's able to do so, and if those font sizes aren't currently available, the compiler can make appropriate substitutions. In the future, if the right font size does become available, then it's easy enough to just drop in and use.
@DavidZ Yeah, it's this whole thing where TeX has like three font sizes, period, and everybody just pretends that it's not a problem, and that if you want something else then really it's your fault and actually LaTeX knows better than you
@yuggib Yeah, I once was an inch from failing an exam (and a flat zero, too) because I insisted on adding áccénts to the printouts, so one of them got plonked into a &A%-like combination on the lecturer's machine, sending my carefully-trimmed line one character over the fortran line length limit
> Integrability in nonperturbative Matrix Models depends on understanding m-dimensional TQFTs deformed by four-quark operators, as realized in bulk currents.
@JohnRennie Hail! I'm not sure the invariance of the line interval was that helpful to me, sorry. I've instead taken a discrete approximation approach using relativistic addition and time dilation.
@0celo7 You keep asking me questions that are non-sensical because you don't give the right information. For instance, the map $F:\mathcal{X}(M)\to\mathbb{R}, X\mapsto \int_\gamma X$ for some curve $\gamma$ sends the zero vector field to 0, but certainly is not dependent on $X_p$ alone (assuming $X_p$ denotes the value of $X$ at a fixed point $p$).
Hey guys, sorry, I know I've been here for a couple days with this and I still can't get a correct answer out of it. Here is the problem, onedrive.live.com/… and here is my calculation onedrive.live.com/… .
@JoeStavitsky I have no idea what your problem is. I straightforwardly calculated the potential energy difference between the two points and converted that into the speed this has as kinetic energy and got one of the answers there.
@0celo7 Okay, then this is the claim that $\mathcal{X}(M)\to\mathbb{R}^n\times\mathbb{R}^{n\times n}, X\mapsto (X_p,(\nabla X)_p)$ is injective.
@JoeStavitsky Potential energy difference. Maybe you are calculating the electric potential difference?
@0celo7: and if that map is injective, then that means a point in the target space determines an element of $\mathcal{X}(M)$ fully (if it determines one)
@JoeStavitsky All you have to do is derive the acceleration that the electron feels and multiply that through a distance of 6 cm. The acceleration would be the force the electron feels divided by its mass.
@JoeStavitsky That gets you $\frac{v^2}{2}$ though
(Not agreeing with what you're saying overall, but I agree with that statement.)
@ACuriousMind Ok, there's obviously something conceptual I don't understand with tensors in general.
@ACuriousMind For instance, the tensor $\nabla_XT$ ($T$ some tensor field) depends, at a point $p$, on $X$ only through $X_p$.
Apparently this is equivalent to the fact that $\nabla_XT|_p=0$ if $X_p=0$.
I don't see why this is.
What I would do is expand $X$ in a chart, then it's clear $\nabla_XT|_p$ depends on the components of $X$ at $p$, not the derivatives or anything else.
(Use the $C^\infty$-linearity of the connection in $X$.)
@0celo7 Well, suppose there were two fields $X,X'$ with $X_p = X'_p$ and $\nabla_{X'}T\neq \nabla_X T$ at $p$. This means $\nabla_{X'-X} T \neq 0$ at $p$, and $X'-X$ is a field with $(X'-X)_p = 0$. Contradiction.
@JoeStavitsky What's "interesting" about that, it's the correct unit?
@0celo7 Hmmm? I just showed that assuming two vector fields with the same $X_p$ give different $\nabla_X T$ leads to a contradiction, hence having the same $X_p$ is sufficient for having the same $\nabla_X T$.
@JoeStavitsky Are you sure it's not Coulomb^2/(Nm^2)?
Because Farad are Coulomb^2/(Nm), and then the units match.
@JoeStavitsky the dimensions of \epsilon_0 are Q^2T^2M^{-1}L^{-3}, so any combination of units with these dimensions can be used. Farads per metre is the most common, but you could equivalently use C^2⋅N^{-1}m^{-2} or CV^{-1}m^{-1}
@0celo7 The proof there shows that, given $X_p = X'_p$, $\nabla_X T \neq \nabla_{X'} T $ leads to a contradiction. Hence, given $X_p=X'_p$, we must have $\nabla_X T = \nabla_{X'} T$. Can you make more explicit what you don't understand?
Usually, functions map the entire domain, unless you are in some area where it is so common to encounter partial functions that it is never made explicit.
okay for a surjective function $f(x) = x^2$ that maps $f: A \to B$ in the domain $A \subset \mathbb{Z}$ and the codomain $B \subset \mathbb{Z^{+}}$, how can you say a left inverse $g$ doesn't exist such that $g: B \to A$ and a composite function $g \circ f$ is the identity on $A$? Like given $A = \{-1,-2,0,1,2\}$ and $B = \{0,1,4\}$, shouldn't the image of $A$ under $f$ give $B$ and mapping that back to $A$ would work? $\sqrt{B}$ would give back a subset of $A$.
oh wait the subset of $A$ is not the identity of $A$.
I see now.
@ACuriousMind If you could summarize the study of physics to one word what would it be?
why wouldn't you glance at the finish line before you start running @Slereah ? what if you're running towards a cliff without realizing. Or there were shortcuts along the way
have I got a mi-understanding here at all? The question states:
I then see that on the first one that the strangeness isn't conserved, on the right, the strangeness is $+1$ and on the left the strangeness is $0$, then on the second one the lepton number on the left is $0$ then on the right it ...
I apologize if this is a naive question, but I never really learned about this. I'm curious as to what happens to sound waves after they are "used"? For example, if I say something to you verbally, then a sound wave is transmitted and picked up by your ears, but what happens to the wave after tha...
@JohnRennie Well...it's not really clear what the question is, I'd say. You seem to have hit the point that confused the asker, but from what's written in the question, it's not clear how they concluded that the lepton number is 0 on the r.h.s., and hence not clear what an answer has to address
@ACuriousMind I thought it was on the homework side, which is why I commented instead of answering. I would vote to close except that, yes, I'm all out of close votes (again). But now I'm wondering if I should convert my comments to an answer.
If I release a piece of metal and it flies towards a magnet, the magnet has exerted a force over a distance, and thus done work. Does this mean the magnet gets weaker?
@JohnRennie If it just asked: "What is the lepton number of this process?", then it would be homework, I'd say. If it asked "Why is the lepton number of the $\mu^+$ -1?", it would be an okay-ish question. As it stands, it's just unclear to me.
@Obliv that's a quirk of the way the human eye works. You aren't actually making light of different colours you are just fooling the eye into thinking there is light of a particular colour present.
When a permanent magnet attracts some object, lets say a steel ball, energy is converted into for instance kinetic energy and heat when attraction happens, and they eventually collide. Does this imply that energy is drawn from the magnetic field and the magnet is depleted, making it weaker and we...
@barrycarter the centre of mass of the system is conserved. Assuming the magnet is bolted to the Earth then yes it will move but not by very much (the Earth being quite heavy).
Incidentally the drinking session last night went on longer than expected, which is why it took me until mid afternoon to make an appearance hereabouts :-)
@JohnRennie It's a textbook exercise type of problem. So definitely homework under the old definition and I would want to be off-topic not withstanding that it is exactly the kind of question that students get an "Ah! Ha!" moment from when they are studying the conservation properties of the fundamental interactions.
have I got a mi-understanding here at all? The question states:
I then see that on the first one that the strangeness isn't conserved, on the right, the strangeness is $+1$ and on the left the strangeness is $0$, then on the second one the lepton number on the left is $0$ then on the right it ...
