// Function to calculate the propagator matrix for a magnet/rotator component. // The formula is exp(-iHt) = 1 - i*sin(theta)*H + (cos(theta)-1)*H^2, // and apparently it works as long as the eigenvalues of H are all 1, -1, or 0. // (I pulled this from my old Pascal code and I have no recollection of whether // I derived the formula at that time, or how I even learned of its existence.)
(it's not actually that hard to derive. since the eigenvalues are $\pm 1$, we have $H=P_+-P_-$ for projectors $P_+,P_0,P_-$ onto the $1,0,-1$ eigenspaces)
he has point charges that are alternately disconnected and connected to one another in various configurations and the question is what the ultimate distribution is.
lobic: it might help if you elaborated what you meant by 'squeezing.' it is conceivable that there is none, depending on what that means.
first A---B C, then C---A----B, then A---B C again, then A B C. i think C initially starts with a charge Q. we were focusing on step 2, which seems subtler than the others.
well I see it is from wiley that out instructor printed and gave us. My professor know nothing deep about knowledge in physics but he got professorship :(
lobic: the arc length integral can be thought of as a limit of riemann sums too. this might be clearest in the case where your curve is the graph of a function.
The arclength integral is rigorously proved with the mean value theorem, and so on an interval $[t_{i-1},t_i]$, if $c$ is in there, $f'(c)$ is between the minimum and the maximum of $f'$.
@leslie I was assuming he was doing the case of a graph. My bad.
typically when we do problems like "connect charges with wire" the point is that the two spheres have some radius, e.g., physicstasks.eu/300/electrically-connected-charged-balls, and thus the two charges won't be equal after reaching equilibrium
I've always felt that the notion of 'area under the curve' is more precise than 'length of a curve', because of the Riemann integral and the reasoning used there:
First, we assume the quantity we call area $A$ between the graph of a function and x-axis eixsts. Next, we show that it must necessar...
Consider the three conductor balls A, B. and C. As shown in the picture on the left, the initial load of the three conductor balls is shown on the left, and then the wires are connected and disconnected in order from left to right. In the last step, the ball What is the final charge of A?
What is the actual definition of arclength, @Lobic? The integral is the answer, not the actual definition. And I explained to you that there are upper and lower sums there. Did you read what I said?
if we assume the three balls all have the same radius, i can't see how the answer would be anything but Q/3. but evidently you have some kind of answer for this?
The integral is not the definition of arc length ;it can be proven under some condition that the ientegral gives the arc length ; the usual def is given Dave ullirchs answer and I’m asking why there’s no squeezing in that definition
In the auxiliary material of the physics textbook of Halliday, first chapter about electrostatics, there is an example that has the following statement and solution:
basically there are three identical conducting spheres. One has a charge $Q$, the others have no charge. The spheres are away fro...
@Lobic It is coming from the definition he gave. That is the length of the polygonal path using the subdivision. On each piece you use the mean value theorem to say the slope is $f'(c_i)$ for some $c_i$ in that piece. Write it all out.
in the background there's the geometric fact that a straight line path minimizes the distance between points. if you add a substep into a straight line path from A to B the length can only go up. this may or may not be related to what you are uncomfortable with. there's no "signed length" when doing these, the way there is "signed area" in computing random definite integrals.
as a test case: if you squeeze a circle horizontally by a factor of two, its area goes down by a factor of 2. but its circumference doesn't change by any easy formula
you have to go to elliptic integrals to get the circumference of an ellipse
But why is there no upper and lower bounds like in area I heard that when Archimedes proved that the ratio of circumference to diameter is constant he used both inscribed and exscribed polygons!
I have one small question: if $T\in L(V)$ then $\lambda $ is an eigenvalue of $T$ iff $\bar{\lambda }$ is an eigenvalue of $T*$, where $T*$ is adjoint of $T$. Is this statement true for infinite dimensional $V$ also?
I think I can prove it only for finite dimensional case. $(\implies)$ there exists a non zero $v$ such that $Tv=\lambda v$ so for any w in V, we have $\langle Tv, w \rangle=\langle v, Tw\rangle \implies \langle v, Tw-\bar{\lambda}w\rangle =0$
koro if you fix an orthonormal basis (e_n)_{n=1}^infty of an infinite dimensional hilbert space H, there is a bounded operator S on H satisfying S e_n = e_{n+1} for all n. it has no eigenvalues. its adjoint can be checked to satisfy S* e_n = e_{n-1} for n > 1 and S* e_1 = 0. it has the entire open unit disc for eigenvalues.
the spectrum of T* will be the complex conjugate of the spectrum of T. that's still true. but spectrum is not synonymous with set of eigenvalues (because an operator can fail to be invertible without failing to be one-to-one)
it seems like every day in november has had something touching on this issue. i wonder why. something in the water, i guess.
koro if you are interested in the infinite dimensional case you might peek at halmos' hilbert space problem book. it would not be useful for exam prep because many of the exercises are too hard to set on an exam. but it includes full solutions. lots of stuff about the operator S (up above) in there.
Been watching some algebraic topology videos and it reminded me of an old thought I had. It seems like the minimum number of charts needed to define an atlas for a manifold would be a reasonable invariant in addition to dimension. But I basically never heard about it in a differential topology course. Does this number have a name? I presume if it isn't used, its either trivial or too hard to figure out.
hm. for compact surfaces it seems like one oughta be able to figure it out in terms of the genus. maybe generally it connects somehow to homology groups.
the invariant doesn't have a name as far as I know, and you need to make a couple assumptions on which charts you wanna allow. in general, this seems close to impossible to grasp.
but, it does relate to some other invariants. e.g. if you only allow contractible charts, this yields an upper bound on the so-called Lusternik-Schnirelmann category of a manifold, which is an invariant that has been studied
if you allow charts to overlap in particularly nasty ways i dunno. if you want to overlap nicely maybe you can get something close to a minimal triangulation (for surfaces) or decomposition into simplices? which seems hard
and the Lusternik-Schnirelmann category is itself connected with other invariants like the cup length, which can be used to give lower bounds on the minimal number of charts with nullhomotopic inclusions needed to cover the manifold in terms of the cohomology ring
a positive result is that if you allow any open subset of R^n to be a chart is that n+1 charts suffice for an n-dimensional manifold
twice in two days we've had an issue where the cat sits on a napkin or paper towel that my daughter wants, and my daughter tries to get at the napkin/towel, resulting in unpleasantness. i don't know why they keep playing this out.
I general $\langle v, a\rangle =\langle v, b\rangle $ does not imply that $a=b$. An example is $\langle 2i+2j,i\rangle =\langle 2i+2j,j\rangle$ but clearly $i\ne j$, where $i,j$ are unit vectors along X and Y axis respectively. But then I am confused as to why $T^\ast$ is linear transformation.
This would matter for what the number is, but whichever decision you make, the result is still an invariant right, since connectedness is preserved by homeomorphism
koro T* v satisfies (Tu, v) = (u, T* v) for all u. the 'for all' is important. not sure how you have defined T* but i agree that it may be hard to verify linearity before knowing that this recipe defines only one thing.
We have $\langle Tu, w\rangle =\langle u, T^\ast w\rangle$ we want to show that $T^\ast (w_1+w_2)=T^\ast w_1+T^\ast w_2$ So $\langle Tu, w_1+w_2\rangle =\langle u, T^\ast (w_1+w_2)\rangle\implies \langle Tu, w_1\rangle +\langle Tu, w_2\rangle$
it would be helpful to separate out what is true by definition and properties of the inner product from the unproven property of T*. one step at a time.
(Tu, w1 + w2) = (Tu, w1) + (Tu, w2). this is linearity of ( , ). then pop the T's over using the definition of T* w1 and T* w2. then use linearity of ( , ) again. then use the definition of T*.
This gives: $\langle u, T^\ast (w_1+w_2)\rangle=\langle Tu, w_1\rangle +\langle Tu, w_2\rangle=\langle u, T^\ast w_1+T^\ast w_2\rangle $. I don't understand how this implies the equality of the entries in second slot considering that such implication is in general not true.
now let's do unbounded operators, where T* may have a different domain than T.
just kidding. but if you do look at unbounded operators it's the same calculation, you just mentally pay attention to domains in the background.
so much of unbounded operator stuff is symbolically the same as bounded operator stuff and often with the same but just more carefully written proofs. this is why the physicists can get away with abusing it.
the thing is that even covers by connected charts seem completely incomprehensible, because we don't even reasonably understand open subsets of $\mathbb{R}^n$ (even for $n=2$, where some things can be said, this is not completely tractible)
so perhaps the better thing to do is add some homotopical condition that makes this susceptible to homotopical and homological analysis (e.g. having nullhomotopic inclusions, which gives the Lusternik-Schnirelmann category) or make a topological assumption like looking at covers by Euclidean balls to make it susceptible to geometric topological methods
in the latter case, you can at least argue some vaguely cool stuff like: if a compact, connected manifold is covered by two Euclidean balls, it's a sphere
Yeah, I admit that as stated I have absolutely no intuition about what the "problem" or "obstruction" is that prevents a manifold from being covered by fewer charts
Good evening everyone. I have read in a lesson on connectedness the following theorem :
An open set $G \subset \mathbb{C}$ is connected if and only if for any two points $a, b \in G$ there is a polygon from $a$ to $b$ lying entirely in $G$
That's sorta funny to me Thorgott because it makes me think of the opposite situation of "holes in a manifold". The idea of "holes" is obvious from a picture, but takes a while to formalize and prove that its invariant, but then turns out to be relatively computable. I have no idea what the minimum # of charts means in a picture, but its trivial to formalize and see that its invariant, and yet seems hard to compute.
A similar thing I was thinking of when ruminating on the Hopf fibration was the minimum number of local trivializations required to cover a fibre bundle as a means of categorizing how "far" the bundle is from a global product.
we generally don't just care for invariants, but only for good invariants, and this puts into perspective just how good homology is, even if it seems much more esoteric in construction at first (of course, there's underlying structural reasons for that, etc.) compared to something as straightforward as a minimal number of charts
as far as fiber or even just vector bundles are concerned, the minimum number of elements of a trivialization will be just as hard, if not even harder
they're good, because they're structurally relevant in encoding important information about spaces
meanwhile it's not clear at all what the least number of charts required to cover a manifold actually encodes (other than the very thing that defines it)
but I'm not a homotopy theorist, so I really shouldn't be defending homotopy groups here
I recently explained to a friend how homotopy groups are really just instantiations of the fact that spheres are naturally cogroup objects in the pointed homotopy category. I felt dirty afterwards
@KevinDriscoll Well I think it is a little hard to tell what is good universally and stuff. Like Thorgott implied that the minimal number of charts is a bad one, even though it's the basis for other ideas in homotopy theory.
I think that's justt because he's not a pro homotopy theorist, like he said
But a pro homotopy theorist might not be aware of other invariants from e.g. knot theory.
no, I'm saying LS is different than the minimal number of charts in a good cover because "cover where all intersections are contractible" is strictly weaker than "cover where all elements have nullhomotopic inclusions"
LS of $S^1$ is 2, but minimal number of charts in a good cover of $S^1$ is 3
In the auxiliary material of the physics textbook of Halliday, first chapter about electrostatics, there is an example that has the following statement and solution:
basically there are three identical conducting spheres. One has a charge $Q$, the others have no charge. The spheres are away fro...
