I don't think we've covered the dot product in physics...am I allowed to use it on the homework?
Damn, I have to talk to the TA...
@ACuriousMind I just sent the TAs an email if I can use the dot product for this assignment. My reason: I don't like coordinates. They probably think I'm insane :/
@0celo7 Ah. Well, take any geodesic $\gamma$. $i^*$ tells you what the tangent vector at $p$ maps to at $i(p)$, so you know which geodesic through $i(p)$ it maps to (tangent vectors determine geodesics uniquely). Call that geodesic $\delta$. Now, $i(\gamma(\tau)) = \delta(\tau)$.
@0celo7 It's a curve $[0,1] \to M$. You may partition $[0,1]$ into finitely many closed intervals $I_i = [b_i,e_i]$ (with $b_0 = 0$, $e_N = 1$) such that $I_i \to M$ is a geodesic.
> I strongly encourage you to break it into components, as it's very helpful in understanding the forces and what exactly they're doing to the beam they're attached to. Breaking forces into components will be a very handy skill later this semester so it's best to practice it now. It's also easier for me to grade and give you partial credit if you break it into components.
@ACuriousMind
"Practice," seriously?
I just completed this course in high school and got the highest score on the standardized test for it...
@DanielSank I'm curious for your opinion on something. Regarding fluctuation-dissipation, it's always been presented to me as some weird and wonderful result. But isn't it just a natural and immediate consequence of microscopic reversibility?
user54412
Or put another way, isn't it only weird because we often neglect one or the other facet of a problem? When we pretend systems are purely dissipative, we're ignoring the back reaction of the thermal motion in the heat sink on the upstream parts of a system.
user54412
But there is no pure upstream, since again everything is reversible. It's no wonder one can design thermodynamics-violating machines if one assumes a black box that can affect its environment without in turn being affected. Such a device is nothing less than a passive Maxwell Demon.
I'm not 100% sure reversibility is enough to get fluctuation-dissipation. I have to think about that.
@ChrisWhite I think it's only "mysterious" because, as you say, we often forget one side of the coin.
@NeuroFuzzy Asking if @0celo7 wants a cookie?
@ChrisWhite I think there is a bit more going on.
Reversibility itself doesn't seem like enough to get the full relation of the FD theorem.
user54412
"forgetting one side of the coin" is right up there with "interchanging two limits" in the list of things physicists commonly do that sometimes bite them real hard
@ACuriousMind I see your point, it is true "rest" and "zero acceleration" make a difference. I am quite interested in that topic honestly, but it sounds strange to me. To my knowledge, Newton's I law is to define* and induce* inertial frame. and if one can define inertial frame in different ways, and different ways do give you different results or predictions, then it doesn't make sense. Physics or nature should be independent of how you define things.
@Shing Well, even with "rest", there is not a single moment in the moving solution for Norton's dome that violates the law "Things upon which no force act, are at rests/uniform motion" since in the interval $[T,-\infty]$, no force acts on the particle, and it is not accelerating, and in the interval $[\infty,T)$, it is accelerating, and the force on it is non-zero because it is no longer at the top of the dome.
The problem here is not in the laws, it is in the shape of the dome - if you can construct a situation such that the conditions for the uniqueness theorems for the solutions of differential equations are not satisfied (in this case Lipschitz continuity), then you get "classical non-determinism".
@Slereah and @JohnDuffield "If you ain't got nothing nice to say, then don't say nothing" You are both getting a short timeout from me, and I suggest you simply ignore each other in the future (use the "ignore" function of chat, perhaps?).
Rate the following sentence for its accuracy on a scale from 1 to 10: "In conventional quantum field theory the elementary particles are mathematical points."
@ACuriousMind Not all embedded submanifolds are closed.
In my notes, we first show that Lie group homom's are constant rank (thus the kernel is an emb. submnf.), and then there's a remark that it also follows from the kernel being closed (and then applying the closed Lie subgroups theorem thingy)
@0celo7 The manifold has to be connected, but then it follows from "Every cohomology class contains one harmonic form" and "The zeroth cohomology vanishes".
It's about counting topologically different "holes" and "surfaces" that fit into the space, heuristically
(Co)homology is a topological invariant, but tells you surprisingly much - cohomology on smooth manifolds tells you how many forms there are that are closed but not exact, and, more generally, how many different bundle structures you may construct