That's why I was so confused about what you were doing - the spectral theorem is the statement that every self-adjoint operator has an orthonormal eigenbasis with real eigenvalues
The "hard" part of the proof is that the eigenvalues and eigenvectors exist, i.e. know that a) the roots of the characteristic polynomial are the eigenvalues and b) every non-constant complex polynomial has a root
@0celo7 I prefer the proof through Liouville's theorem from complex analysis: If $f$ has no root, $1/f$ is a holomorphic function on the complex plane, and $f$ has a non-zero minimum. So $1/f$ is a bounded holomorphic function on the complex plane, and thus constant by Liouville's theorem.
The beauty of the Haar measure is that the method of constructing it doesn't matter, since it is unique up to scaling (which is a somewhat involved proof)
@ACuriousMind Trying to prove "a simply connected abelian Lie group is isomorphic to $\Bbb R^k$"
I'm thinking that $\exp:\mathfrak g\to G$ is surjective
oh...maybe it's a homomorphism, too
Now, we know that $\Bbb R^k$ is its own Lie algebra
And certainly $\Bbb R^k\cong \mathfrak g$ for some $k$.
so $G=\exp \mathfrak g=\exp \Bbb R^k=\Bbb R^k$
no
I never used simple connectedness
actually, maybe you get $\exp \Bbb R^k=\Bbb R^n\times T^l$
hmm
I know you can get a torus somehow
yes, apparently $\exp:\mathfrak g\to G$ is a surjective Lie group homomorphism
we also know that $\mathrm{dim}\,\mathfrak g=\mathrm{dim}\, G=:n$.
we can write $\mathfrak g=\mathfrak g_1\oplus\cdots\oplus \mathfrak g_n$ for any 1-dim subspaces.
So it's clear that we have $$\exp\mathfrak g=\exp\mathfrak g_1\times\cdots\times\exp\mathfrak g_n$$
but the only abelian Lie groups obtained in this manner are $\Bbb R$ and $S^1$
QED.
simply connected then eliminates the $S^1$ factors
ah
we invoke the classification of 1-manifolds to get the $\Bbb R$ and $S^1$ thing
@ACuriousMind That proof is probably wrong
Seems too easy.
I guess I should prove that $\exp \mathfrak g=G$.
Ok, $\mathfrak g$ is abelian iff $G$ is, so $\exp$ is a homomorphism by...BCH.
BCH is probably too strong there but whatever
surjective though...
clearly $\exp$ is a Lie group homomorphism
so we can use some theorems about that stuff
AH
Given connected Lie groups $G,H$ and a homomorphism $F:G\to H$, if the induced map $F_*:\mathfrak g\to\mathfrak h$ is an isomorphism, $F$ is surjective
Clearly $\mathrm{Lie}\,\mathfrak g=\mathfrak g$ and by definition $\mathrm{Lie}\,G=\mathfrak g$
and we know that $\exp_*=\mathrm{id}$
the identity is an isomorphism, so $\exp$ is surjective
I heated some copper to 1100C for lulz and it didn't melt :/
I'll do 1200 on Monday
user128101
@JohnRennie,Have you ever visited the defunct Darebury plant or Diamond House, at Oxford? The latter is open next October. If I asked a question, could you find some data from any of them? The usual formula relates power to acceleration, I'd need a curve matching light frequencies and velocities/energies of the electrons (and the angle of curvature of each deviation: 15° (2pi/24) or 7.5*?)
Here's yet another case study for why downvotes should carry more weight, at least on answers. If one looks at the reputation timeline, it is pretty clear that the majority of reputation of that user comes from a single other user upvoting each of their posts once (it doesn't really matter if it's more than one user, the important thing it that there is never more than one upvote). Yet no post has positive score.
One should not gain reputation by having no post with positive score.
@ACuriousMind May I ask why did you not vote this answer of mine down while you voted both question and the other answer down? Did my answer not deserve to vote down? Or you forgot to vote my answer down? I ask this, because I will feel honor if my answer isn't so bad in your opinion. ;-)
Lets say both engines operate between same temperature limits.
Carnot eff:
$$
e_\text{Carnot} = 1-\frac{T_L}{T_H}
$$
Otto eff:
$$
e_\text{Otto} =
1-\frac{T_4-T_1}{T_3-T_2}
$$
assuming that for Otto cycle points are located as below on a P-V diagram:
$$
\array{
3 & 4 \\
2 &...
@dmckee I stumbled across this question where especially the top answer is pretty non-sensical, shielding itself by saying "the following is simplified" from having to actually back up its claims.
oh god the second one talks about "generating virtual particles"
Really "science based" appears to mean "use your personal misunderstanding of pop science to generate technobabble instead of just making it up completely" especially combined with the "magic" tag...
@dmckee I stumbled across this question where especially the top answer is pretty non-sensical, shielding itself by saying "takes every possible path is a mangled statement" from actually answering the question.
oh god the second one talks about "paths which go 'back in time' and 'come forward' again".
@ACuriousMind Can you please check my wall of text above :/
@ACuriousMind I can summarize.
@ACuriousMind Since $\exp_*=\mathrm{id}$, the map $\exp:\mathfrak g\to G$ induces an isomorphism on the Lie algebras. Furthermore, $\exp$ is a homomorphism between these groups (with $\mathfrak g$ viewed as a Lie group wrt. addition) due to BCH. Thus $\exp$ is surjective by a theorem in Lee.
Write $\mathfrak g=\mathfrak g_1\oplus\cdots\oplus\mathfrak g_n$, then since $\exp$ is a homomorphism we have $G=\exp\mathfrak g=\exp\mathfrak g_1\times\cdots\times\exp\mathfrak g_n=G_1\times\cdots\times G_n$.
Each $G_i$ is a Lie group in its own right, and is one-dimensional.
By the classification theorem of 1-manifolds, we have $G_i=\Bbb R$ or $S^1$.
Simple connectedness eliminates the $S^1$ factors, so we have $G\cong \Bbb R^{\mathrm{dim}\,G}$.
@JohnDuffield think youre sniffing on the right track there & have long mused the same. have you ever thought of blogging esp wrt physics/ EM/ QM theory etc?
@vzn : I did a bit of blogging a couple of years back. I was just one of several writers on something set up by a journalist called James Delingpole. But then things got screwed up when somebody changed the hosting, and I packed it in to have more time for other things. One thing I wasn't keen on was that it was like talking at people rather than talking to people.
@JohnDuffield huh no kidding. whats it on? blogging can help with that, eg posting work in progress/ TOCs/ outlines/ summaries etc, highly encourage you to restart it to some degree. not lying, its hard at times esp in age of facebook, the 800lb gorilla.
@DanielSank "multipings"? if you write stuff in chat without acknowledging earlier pings then think its a fair assumption you missed them, have no other way to tell/ know. am trying to limit pinging you to the key session admin stuff. you agreed in principle to slot but not on a date & we have to nail down the date, thx for that :) would still like to do some mgt re prior session documentation but will let it slide for now since you seem very busy, will try to match your pace
@vzn : yes. But it would be improper of me to talk about it. Besides, if it didn't work out I could end up looking like a chump. I'm not really mysterious. I'm an IT guy who got interested in where physics was going ten years ago when our teenage children gave up all their science subjects.
@JohnDuffield IT huh small world, what area? are you retired? hey lots of ppl have unpublished book manuscripts no shame in that. so your kids dislike science or did something different?
@NoahP : I don't know what to say Noah. I think about things like pair production and the wave nature of matter, I don't see a lot of difference between matter and radiation.
@0celo7 The operator $\mathrm{d}^\ast$ is constructed by mapping a preimage of $\omega$ somewhere. That's a choice, and you have to show that it doesn't matter which preimage of $\omega$ you choose, i.e. that the all give the same $\mathrm{d}^\ast[\omega]$.
And there's another choice when the preimage of $\mathrm{d}\xi$ is chosen
I have no idea what you mean by partition of unity
Yes, $\mathrm{d}\xi$ goes to zero so by exactness it comes from an element in $\Omega(M)$, but there might be more than one element that maps to it, so one has to check that the final cohomology class $\mathrm{d}^\ast[\omega]$ does not depend on which of those elements was chosen
The answer to the first question is no. The following example is standard in probability theory, see e.g. Billingsley "probability and measure", example 30.2.
$$f(x) = {1\over \sqrt{2\pi}\ x}\ e^{-{(\ln x)^2\over 2}}\ \sin(2\pi \ln x) \ {\bf 1}_{[0,\infty[}(x)$$
You can check that all the momen...
@Slereah On second thought, it does exclude terms like $\exp(\epsilon)$, since that has polynomial terms. The difference between the two things (let's call the integral $I(\epsilon)$) is that the first one says that $$\forall q\in\mathbb{N}:\lim_{\epsilon\to 0}\frac{1}{\epsilon^q} (I(\epsilon)-f(a)) < \infty$$, which does not imply the second.
What I wanted to say is that it doesn't have polynomial terms - it is an allowed contribution precisely because you don't see it when you go "up to order $\epsilon^q$", since it is zero upon expansion. It's a non-perturbative contribution that you cannot see order-by-order
Since the big condition is that the Fourier transform is $1$ around $0$, building a suitable mollifier should be as easy as defining some function that is $1$ around $0$, take the inverse FT, and then that should be a suitable mollifier
I'm guessing that the inverse FT of a Schwartz function is a Schwartz function, too
Lets say both engines operate between same temperature limits.
Carnot eff:
$$
e_\text{Carnot} = 1-\frac{T_L}{T_H}
$$
Otto eff:
$$
e_\text{Otto} =
1-\frac{T_4-T_1}{T_3-T_2}
$$
assuming that for Otto cycle points are located as below on a P-V diagram:
$$
\array{
3 & 4 \\
2 &...
