Why do I see a lot of books titled like "Vectorial mechanics for engineers". Why would a book on maths be 'for engineers' I feel like this is akin of "Quantum Physics for babies"
@0celo7 If I understood correctly from MSE, a broken geodesic is like a broken line graph version of a geodesic on some manifold, do the cusp joining any two segments solve the geodesic equations?
Let $\gamma$ be an infinite broken geodesic in the hyperbolic plane, that is a curve formed by consecutive geodesic segments. Assume also that each of these segments is longer than a certain positive constant $C$. For any two consecutive geodesic segments $[p,q]$ and $[q,r]$, we can consider the ...
Hmm. I don't remember the proof of that. I think you use a variational technique to show that a critical point of the energy functional is a geodesic, and there's a way of getting smoothness too.
Oh, no
A geodesic solves an ODE so smoothness follows from ODE theory.
Levitating a superconductor on a Mobius strip
In the video the superconductor levitates upside down, how is it possible when the gravity and acting downward and by definition superconductor repels the magnetic field.
We live in an expanding universe - so I'm told. But how can that be possible? Everything imaginable is attracted by a bigger thing. So, why can't gravitation stop the expansion of the universe? I know the "Big Bang" theory, but is it possible that the expansion of the universe is caused by the at...
Einstein wasn't wrong. And the moot point is that science is not a democracy. I see correct answers from people whom I know to be experts, and incorrect "popscience" answers from people who aren't. The latter gets more upvotes, and the former votes with his feet. The expert poster becomes the ex poster.
Because the 18-year-old who thinks Einstein was wrong downvotes every time.
@BernardMeurer : no. I just get more determined. Here's why:
"I say that because there’s a standing joke in our house, that I’m the only one who can change a light bulb. But somehow it isn’t funny. If you selected a hundred people at random and tested their technical and scientific knowledge, I think the average score would be lower than that of a comparable group from fifty years ago. Yes, we’re more specialist these days, and some things are more difficult to understand.
But it seems there’s more people around who just don’t understand the basics, who have only the vaguest concept of how things work. They wouldn’t know where to start if their car broke down. It’s like there’s a low-rise, low-brow tide that doesn’t feel healthy, that slowly, insidiously, is getting worse."
@BernardMeurer thought if you were into hackerspaces you might have heard of em. liked the network chess kit & the jet turbine off of ebay etc sparkfun.com/news/2155sparkfun.com/news/1890 ... same guy does battlebots =D
@0celo7 Okay: Fix an isomorphism to $\mathbb{R}^n$. Then for any $x = (x_1,\dots,x_n)$, we can take the maximum $M$ of $||x||$ on the unit ball w.r.t. the maximum norm, and this from this we get $||x|| < (M+1)||x||_\text{max}$. We can also take the minimum $m$, and get $m/2 ||x||_\text{max} < ||x||$.
If you don't like my "cube is compact because it's $[-1,1]^n$", then use the Euclidean norm and use the standard ball, which you should know to be compact.
Oh, I'm not doing this silly routine. If you willfully don't want to use any other theorems, then of course you can't show the statement any other way than by brute force.
@0celo7 It's silly because you asked me "is there any other way to prove this", and when I give an answer, you add arbitrary restrictions to what is allowed in the proof
@0celo7 I didn't think about how you know this. Euclidean Heine-Borel, finite Tychonoff, it doesn't matter. But you know this. If you want to pretend you don't for some strange reason, then fine. But that's not my problem.
@ACuriousMind Let $S=\{v\mid ||v||=1\}$. Let $v_n\to v$ be a sequence in $S$. Then $|||v_n||-||v|||\le ||v_n-v||\to 0$, so $\lim||v_n||=||v||$. But $||v_n||=1$ for all $n$. Done.
I can offer up something similar to this, which is an isomorphism between something called the Tsirelson bound and the spacetime metric. This is not exactly the emergence of spacetime from quantum mechanics, but it does illustrate how spacetime could be seen as quantum mechanics in diguise.
Supp...
^this answer, and probably some other answer by the same user as well, could use some closer scrutiny
There is a consistent pattern of not actually answering the question posed, and using a lot of formulae and terminology while actually misusing the terminology. The answers are well-written and sound good, but if you look at them in detail at least some of them just fall apart
I've problem choosing between two classes. One is about Banach algebra and spectral theory; the other is PDE class. I'm a theoretical physicist. Which one should I choose.
Thanks; any help will be appreciated.
The only argument I can extract from it is that the Tsirelson bound has the same sign pattern like the Minkowski metric, hence the two must be "deeply related"
I've problem choosing between two classes. One is about Banach algebra and spectral theory; the other is PDE class. I'm a theoretical physicist. Which one should I choose.
Thanks; any help will be appreciated.
