« first day (4857 days earlier)      last day (61 days later) » 

12:05 AM
I guess: how do boundary conditions transform upon transforming coordinates
for the case of a 1D string obeying boundary condition $\phi(x = 0) = \phi(x = L) = 0$.
at least visually, it seems like the above happens. where the lagrangian density which once was localized to $[0, L]$ upon translation to the right is now localized to $[a, L+a]$. And so it seems that the Lagrangian has indeed changed by more than just a total derivative. Namely, we have subtracted the old lagrangian and added the new lagrangian, to obtain the new lagrangian. And the new lagrangian is not just a total derivative term.
 
@SillyGoose What do you think the Lagrangian here is?
and where do you think this boundary condition enters?
 
something like $\mathcal{L} = \frac{1}{2}(-\omega^2\phi^2 + m^2\partial_t^2 \phi^2)$
 
@SillyGoose and where does this Lagrangian encode the condition that your string $\phi$ is confined to $[0,L]$?
 
Hm I guess maybe by specifying that $\mathcal{L}(0, L) = 0$ and is also $0$ beyond the boundaries. But I am not too sure
 
see, part of specifying a physical system is not just writing down the Lagrangian; if you remember my post about taking seriously the idea that you need to define the spaces these functions live on, you'd define the configuration space as $Q = [0,L]$, and the Lagrangian lives on $T Q = [0,L]\times \mathbb{R}$
you cannot even state what "translation symmetry" is supposed to be in this framework - there is no "translation operator" on $[0,L]$
 
12:19 AM
but if I have a nice string with fixed end points, and I face it so the string looks like how I drew, then shouldn't i be able to move to the left and right
and be able to mathematically model how the system transforms as I move to the left and the right
 
@SillyGoose what do you mean "move to the left and right"?
you can choose to shift your coordinates to the left and right, but that's not what symmetries are about
 
well I am not claiming that translation should be a dynamic symmetry here, just that I should be able to model what it means to translate my system
 
see this answer of mine for why symmetries have nothing to do with coordinate transformations
@SillyGoose but if you want to do that, you need to encode the restriction of the interval into the Lagrangian somehow, because obviously a string on $[0,L]$ with the string Lagrangian is not the same physical system as a string on $\mathbb{R}$
you're just getting confused because you're not treating the "boundary condition" carefully enough
 
1:10 AM
The 1D time independent Schrodinger equation with Yukawa potential is solved in Arfken-Weber-Harris assuming a power series solution $\psi(x)=a_0+a_1 x+a_2 x^2+...$ i.e. expanding around $x=0$. But isn't this wrong because $x=0$ is not a regular point due to the form of the potential? Am I missing something trivial?
 
@Sanjana The Coulomb potential is also singular at the origin but has a power series solution of the same form, isn't it?
 
1:31 AM
@naturallyInconsistent Yes...I didn't pay much attention apparently when I solved the equation with Coulomb potential. I was wondering why this method works...If $x=0$ is a singular point then we should use Frobenius method as clearly mentioned in every single textbook on math phys. and also in wiki introductory para
 
Hm so $\mathcal{L}: TQ \rightarrow \mathbb{R}$. Consider $TQ$ as $\mathbb{R} \times \mathbb{R}$. Then, is the correct imposition of the boundaries to do $\mathcal{L}(x, t) = 0$ when $x \leq 0$ or $x \geq L$?
 
@Sanjana Actually, technically speaking, the generalised Laguerre polynomials are somewhat already a solution that is in the form of Frobenius method. They start with $r^\ell$ rather than 1. Also, the differential equation that both Yukawa and Coulomb potentials impose on quantum theory are weaker than the singularity in Frobenius method, and so there is really no problem, I think.
 
1:50 AM
@naturallyInconsistent Hmm...I got a Q/A here... You are right I think, the asymptotics rules out the pathologies
Never wondered about these stuff while doing Hydrogen atom from Griffiths :p
 
@Sanjana His strategy worked!
@Sanjana Buzz said that one would have to figure out if $\sigma$ finds all possible solutions. On this point, it is actually provable that only one solution is admissible, if there are any. There is a standard trick that converts the radial ODE that the Schrödinger equation turns out to be, into a 1D Schrödinger equation. This allows us to use a theorem that applies to the 1D Schrödinger equation, whereby we study the form of the equation and find that normalisable solutions come alone.
The other solution is necessarily non-normalisable.
 
2:24 AM
anyone know what this is?
listed on an equation sheet for a future thermo exam
im a nub, does exp(a) mean $e^a$
or maybe $\times 10^{a}$
 
@Obliv yes, this, and always this. We are not savages using 10^x
 
It's this
 
2:44 AM
@Sanjana where in AWH?
 
@SillyGoose wait why isn't it $k\log\Omega$ where $\Omega$ is the multiplicity
are they equivalent definitions
man I'm so bored of being dumb I just wanna download all of physics into my brain. i'll commission someone to develop such a program
 
3:00 AM
@SirCrackpot LOVE !!!!!
 
@Obliv $\ln\Omega$ is what Boltzmann used to bootstrap and start the discussion. It is very woeful because it can only deal with one single energy macrostate, and it clearly was not meant to be believed. Instead, the point is that Boltzmann managed to use it to derive the Boltzmann distribution $e^{-E/k_BT}$ behaviour that is so much more important. After derivation of the canonical ensemble and grand canonical ensemble, we are supposed to note that they are more physically plausible and so
 
@Obliv no, the definition i posted reduces to the equation you wrote when $p_i$ are all the same, i.e. for an isolated (microcanonical) system in equilibrium
 
assume them instead. Similarly, we can upgrade to other forms of entropy. $-p_i\ln p_i$ has the exact same behaviour if you simply pick $p_i=\frac1\Omega$ and thus it is a generalisation. In thermodynamics it is called Gibbs entropy. Shannon arrived at the exact same expression from computer science and von Neumann suggested for him to also use the same name for it since the expression is exactly the same.
 
when you move into non-isolated systems, e.g. canonical ensemble, grand canonical ensemble, etc., then try to use your equation for entropy there; it won't really make sense
 
@naturallyInconsistent u mean $p_i = \frac{1}{\Omega_i}$?
 
3:04 AM
@Obliv no. I meant $p_i$ is constant and independent of $i$
@SillyGoose I know what you mean, and I agree with you, but technically, canonical ensemble was derived from Boltzmann entropy i.e. from microcanonical ensemble. It is extremely annoying to have to write technically correct statements in stat therm
 
im just gonna review some stuff lol
I always giggle when I scroll past this section
proof that physics is just magic with extra steps
so number of microstates for a given macrostate i.e. the multiplicity of a given macrostate / by the total number of microstates in the system = probability for that specific macrostate
$$\frac{\Omega_i}{\Omega_{total}}=p_i$$ where $\Omega_{total}$ is the sum of all of the microstates per macrostate. $\Omega_i$ is the multiplicity of just one macrostate $i$
that's what I thought $p_i$ was @NaturallyInconsistent
for a continuous distribution of macrostates we have $\sum_i p_i \to \int_i p_i dp$ I guess
@JohnRennie I haven't seen this before, my gosh that was hilarious.
not even in a judgemental way, like just actually funny
 
Little pictures like that are simply brilliant
 
You've got support from your fans too @JohnRennie I'm going to put this quote on my profile
 
3:50 AM
@Obliv Sigh, the Boltzmann entropy is strictly only defined for the microcanonical ensemble. You don't have any other macrostate.
 
4:11 AM
I did not understand that
we haven't delved into ensemble's or boltzmann/gibbs
I guess I'll wait till we do, things should clear up by then
 
It is getting tedious to complain about that, so maybe you should generate the complaint yourself so that I can go back to meowing.
 
4:33 AM
Ok "@Obliv This is what you get for reading a middling text like Schroeder. Perhaps if you spent your time learning from a good book instead of fumbling around with that garbage you wouldn't have these headaches. I like to sneepu"
Yes I know cat. Away with you
 
Very good. Meowth
 
5:21 AM
20 hours ago, by Jason Harwell
Hi John, thanks for your clarification earlier; if convenient would you be able to explain to me if my terms are correct and if there is considered to be any difference between Zero-Point fluctuations, Vacuum "Fluctuations," and Virtual Particles? My current best understanding summarized is:

"Vacuum Fluctuations: the temporary changes in the energy content of empty space, or the vacuum, as described by quantum field theory. Even in a vacuum devoid of particles, there are still fluctuations in the fields that permeate space. These fluctuations can lead to transient changes in the energy den
@JasonHarwell Hi Jason. Ping me when you're around and we can discuss it.
 
5:45 AM
Broadly, vacuum energy is the zero point energy of a quantum field so in this sense they are the same. Note however that zero point energy is a general property of all bound states so it is far more general than vacuum energy.
But vacuum fluctuations are not real. See this answer for more details.
 
@JohnRennie can you take a look at this question? I know it's been asked a million times but I can't help but feel there's a misunderstanding
in this answer along with this one, I don't see $1/r$ dependence in $\mathbb{R}$
actually idk what Qmech's answer even means
but I was thinking in the same vein as Sean except idk why he uses $\sigma$ for 1D and $\lambda$ for 2d
 
@Obliv I think you're mixing up a 1D object in a 3D space with fields in 1D and 2D spaces.
The first question is asking about the field from a line charge in a 3D space.
 
OH jeez
I was thinking of point charges idk why
welp, off to bed I guess. I wrote out a whole answer describing how to get the Gaussian Surface for point charges in various spaces.
 
Oh well ... :-)
 
I'd have to modify it to be cylinders >:(
 
5:59 AM
@JasonHarwell Virtual particles and not real in the sense that the particles we observe in colliders are real. For more on this see:
11
Q: Are virtual particles only a fictive tool in equations?

SofiaThere is no "action at a distance" in nature. Attraction of a piece of iron by a magnet, attraction between distant electric charges of opposite sign, have to be mediated by something. The virtual particles are proposed as an explanation. If they have an observable effect, it seems like they mus...

 
6:38 AM
what is the most wonderful idea you guys have learned through your education/life
 
@ACuriousMind Is the Cartesianness of the tensor product for Poisson manifolds due to the commutativity of the algebra
 
Too many wonderful ideas to pinpoint any single one
 
I was trying to solve this question and came across this idea that I can take any point as the axis for writing the Torque= d(L)/dt as long as the total force on the system is zero.
This is how I understood it - only the COM frame is inertial but I can take other points, like the centre of the rod in this case, as axis because it would give the same answer.
The proof of this is -
 
 
6:47 AM
@naturallyInconsistent what about an example
separately, do i just need to add these extra $\delta^{00}$ purely to get the indicies "right"?
 
As sigma Fi=0. The answer would be same for both. Am I getting this right?
 
these are some components of the stress energy tensor for a scalar field with the covariant index raised
 
Also, are you all on laptops. The chat seems to lack a lot of features on mobile browser. I couldn't find any good apps for it either.
*on laptops?
 
@SillyGoose when contravariance and covariance in standard GR tensor index notation finally clicked, the world became beautiful for months on end.
 
@Swan On a PC not a laptop, but yes I don't like the mobile chat.
 
6:56 AM
@Swan you are somewhat being excessively restrictive but at least you are not wrong. Note that this is in the instantaneous collision limit, so forces really are not your friends. Things will be easier if you just work directly with impulses and so forth.
 
@naturallyInconsistent hehe I feel like this semester taking this qft course has at least finally made me able to do index computations. before i was wildly confused. i hope to one day achieve such enlightenment :P
 
@SillyGoose When I finally kinda caught what it was that QFT calculations were about, it was somewhat disappointing, but at least the dayum door was finally opened. One cannot but feel depressed when one throws > 5 years at a topic and it just didnt work out. I do, however, feel that if I had access to Duncan early on, it would have worked by self-study.
 
i think if i recall when i took al ook at duncan it sounded like weinberg in disguise
hehe
so it sounds like a nice book
 
Thanks @JohnRennie and @naturallyInconsistent :)
 
@SillyGoose The opposite is true. Weinberg knows all the correct stuff; he discovered or invented most of the correct ways to approach QFT. However, he will not include the reasoning inside the text, and you have to deduce from his order of treatment why he chose to do things his way. Whereas Duncan is not at all interested in the trees, and instead wanted the student to see why the forest is constructed the way it is.
 
7:13 AM
"These are simply monoidal categories whose unit object is terminal. But the word “semicartesian” is repellently technical, and you’d be forgiven for believing that any mathematics using “semicartesian” anythings is bound to be going about things the wrong way."
 
@SirCrackpot : I also got it from Goldstein.
 
7:42 AM
suppose i have a complex scalar field $\phi(x) = Ae^{-ik_\mu x^\mu}$ where $k_\mu = (k_0, k_1, 0, 0)$ where $k_0$ and $k_1$ are real and positive.
the x-momentum density for this particular lagrangian is given by $T^{10} = \frac{1}{2}(\partial^1 \phi^* \partial^0 \phi + \partial^1 \phi \partial^0 \phi^*)$
I expect the momentum density (and total momentum) to be positive, as $\phi(x)$ is sort of like a wave moving in the positive $x$ direction, so whatever we call momentum should probably agree with that.
but when I actually do this computation, I get $T^{10} = \frac{1}{2}AA^*(-2k^1k^0)$, which is certainly negative
i am wondering if i am missing something conceptually. i checked that my expression for $T^{10}$ is correct
 
8:07 AM
hmm. Maybe the error is that I should be using $T^{01} = P^1$?
nvm abt that
 
@SillyGoose If $(k_0,k_1)\in(\mathbb R^+)^2$ then either your energy or your momentum is negative, depending upon your choice of metric signature.
 
oh really
i have (+, -, -, -) as my metric signature
what is the a way to think about making sense of the momentum being negative?
 
That part is most understood. If momentum is negative then it is positive in the opposite direction.
 
8:24 AM
@naturallyInconsistent u r confusing the energy with the sign of the frequency. this is not correct. the energy of this field is always positive
 
8:40 AM
That is technically correct. I was using the energy as colloquially the same as the frequency, as is commonly done.
 
sometimes ive seen energy be written in $\text{cm}^{-1}$
 
@naturallyInconsistent yes but in SillyGoose's question, the relevant notions of energy and momentum are the field energy and momentum
 
It does not matter. The fact of the matter is that the wacky fowl was using $(k_0,k_1)$ for parametrising the system and later using $(k^0,k^1)$, and this annoying difference will run into that issue. Not to mention that the wacky fowl has already said that the metric signature in question is mostly minus, thus the issue does not happen on the energy side and thus your objection is not relevant.
It is just a bare fact that I had enough experience being burnt by this problem as to warn the wacky fowl on the topic. Your reminder that I was making a tiny technical mistake is not very helpful to meow, but might be helpful for the wacky fowl and other readers, so I am happy about that. However, it is not something to then go on about field energy and momentum. It will achieve nothing good to talk about that.
 
 
1 hour later…
10:09 AM
@Qmechanic let's just say I benefited more from your answers :P
 
10:20 AM
this UVir is just the universal enveloping algebra of Vir no?
 
 
1 hour later…
11:31 AM
I thought that the Newtonian limit was $$ds^2=-(1+2\Phi)dt^2+\delta_{ij}dx^idx^j$$
Now I see this weak field limit given by $$ds^2=-(1+2\Phi)dt^2+\delta_{ij}(1-2\Phi)dx^idx^j$$
Clearly it's not up to expansions or so, as the coefficients are of the same order
What's the different assumption, then?
 
@SirCrackpot whatcha tryna do?
 
@naturallyInconsistent Not much, just ended up finding this alternate form and I'm trying to see why it's different
If we assume the same things only one of them can be correct
 
Will there even be any gravitational effects if only the time component has a spatial dependence?
At least the weak field limit looks like a consistent 1st order approximation
 
11:47 AM
Apparently this is also the reason why Einstein got a factor 2 off initially. I mean, you can clearly see that the second approximation is indeed the non-rel limit of Schwarzschild
Then I need to understand why some call Newtonian limit the former and why others the latter
I mean, I agree that the second is more physically reasonable
@naturallyInconsistent by the way yes
 
oh meowth.
 
Think of it in terms of extremizing proper time: if time flows differently in different points of space, you'll move to extremize it
I said that very handwavily
@Slereah do know why many sources call Newtonian limit the one where only $g_{00}$ is non-Minkowski?
 
I think it makes sense. In the Newtonian limit observers will see that all the length measurements agree and are not affected by gravitational curvature.
 
I have one idea
The other possibility is that it's just wrong
It's possible that it's related to low speed limit as well as weak field
 
12:04 PM
What do you mean by non Minkowski
 
Look at my two eqs above
35 mins ago, by Sir Crackpot
I thought that the Newtonian limit was $$ds^2=-(1+2\Phi)dt^2+\delta_{ij}dx^idx^j$$
34 mins ago, by Sir Crackpot
Now I see this weak field limit given by $$ds^2=-(1+2\Phi)dt^2+\delta_{ij}(1-2\Phi)dx^idx^j$$
Oh now I have a more reasonable explanation maybe. The former is geometrized Newtonian gravity, while the latter is the non relativistic limit of GR
Which means that the Newtonian limit of GR is adding additional effects to the geoemtrized Newtonian gravity
 
 
2 hours later…
1:54 PM
@SirCrackpot the newtonian limit is weak field and low speed limit of GR and the metric has the form $$g = -(1+2\Phi) dt \otimes dt + \delta_{ij}dx^i \otimes dx^j$$ in that limit
for sure one book where you can read this is Weinberg
its possible that you are just missing the low speed part in the other limit as you were saying
if you just do weak field, say on minkowski background, but not low speed, you should just get the linearized einstein equations, which are still special relativistic equations
 
@ekardnam_ that and also some authors not distinguishing Newtonian and WF
Thank you for confirming, though
I had eventually accepted my hunch because I was lazy to do the math again
 
there are many approximations of GR unfortunately
And they tend to be poorly differentiated in most books
 
@SirCrackpot i remember quite clearly that to get the expression of geometrized Newtonian gravity low speed is a quite fundamental assumption at some point
i dont remember the steps or anything like that though, just that you need the low speed assumption at some point
 
2:15 PM
@ekardnam_ well, that rejects three components of the geodesic eqn
IIRC
 
@SirCrackpot yeah it should be something along that line of thought
but i would say its about the time component of the geodesic equation "su due piedi" (idk the english for that)
 
2:57 PM
do u think aristotle had a net positive contribution to physics by showing "what not to do"?
i doubt this view because aristotle's physics lasted way too long to be positive
but empiricism became big after it finished
 
fqq
3:18 PM
Aristotle had a net positive contribution to physics in general.
 
 
2 hours later…
5:01 PM
Aristotle would fail a Newtonian mechanics class
 
I mean so would we
Newton's formalism was pretty antiquated
and I would probably fail any ancient greek class
I don't even know the Iliad at all
 
@Slereah is that a proof that humans are points on a manifold and coordinates are only local
And most importantly, did Aristotle know the Iliad?
@Slereah Also, would Aristotle agree? I guess he would say "futuristic", not "antiquated" :P
 
5:18 PM
Aristotle had pretty different ideas on how the universe worked
He'd probably accuse Newton of being some vile pythagorean or something
Aristotle does discuss heliocentrism and other ideas like the Earth rotating
He strongly disagrees
Whether he'd be amenable to be shown wrong, who knows
 
the representations of the quantum algebra $U_q(\mathfrak{sl}(2, \mathbb{C}))$ are still indexed by the spin like for regular $\mathfrak{sl}(2, \mathbb{C})$?
 
6:04 PM
@Slereah We could use Beta Character AI to create a conversation between them
Do you know that site?
 
6:46 PM
@user85795 ?
 
7:40 PM
@ACuriousMind What's going on with tensor products in the category of Poisson manifolds, anyway
I am reading that it 1) doesn't have a projection operator at all [within the category of Poisson manifolds] and 2) that it does but this projection operator doesn't make it a full Cartesian product
closest I can find to an explanation is this line :
"It is easy to be impressed by the fact that while the Cartesian product is a categorical product in the category of smooth manifolds (thus the possibility of forming M1 × M2), there is no such product in the category of symplectic manifolds. This is due in the first instance to the fact that one cannot define a symplectomorphism between manifolds of different dimension, and so obviously one cannot define anything like a projection here."
Is it different for the category of symplectic manifolds and Poisson manifolds???
apparently even defining a category of symplectic manifolds is a harsh problem
 
7:59 PM
i didn't realize that minkowski is german
i have always prnounced it minkowski but now i know it is more like minkovsky
@naturallyInconsistent i see what you're saying now with the switching of upper and lower indices
 
@SillyGoose it's Polish
 
oh no
wait minkowski the person is german though right
but you are saying the name is polish?
 
8:51 PM
Claim: Take the definition of reality for granted. That is, reality is what you as a person observe in your day to day life, ignoring any and all philosophical issues or differences. Define any single or collection of things you observe as an object. Then, an object is (conjectured) to be defined by some collection of quantities invariant under a certain group of transformations.
Would you guys say this is sort of an accurate depiction of how people think of classifications in physics?
So then classifying objects is really just constructing invariant theories for each sort of mathematical object an object can be represented by?
For example, an electric field is described by a vector field, so we would construct a theory of vector field invariants
 
That is the general idea of how observables work yeab
Although things are complicated with covariance of course
 
what is a or multiple alternatives to this line of thinking?
(or perhaps there is an argument that it is unavoidable)
on the mathematical side of things, i mean do we even know if (non-trivial) invariants of a mathematical object always exist?
 
How are symmetries generally defined? If I have an operator $\mathcal{O}$, and I make an active transformation with the states transforming, won't I get $\langle \psi' | \mathcal{O} | \psi' \rangle$ which is in general different to $\langle \psi | \mathcal{O} | \psi \rangle$ where $\psi'$ refers to the transformed state. So the idea of symmetries being those that keep expectation values invariant seems incorrect
In this sense, is a symmetry transformation one that keeps transition amplitudes invariant i.e. $\langle \psi | \phi \rangle = \langle \psi' | \phi' \rangle$?
 
@DIRAC1930 this is how weinberg in vol. 1 defines a (Wigner) symmetry at least
Wigner symmetry; in contrast to a dynamic symmetry, which leaves the action invariant
 
9:07 PM
Aren't Lorentz transformations both Wigner symmetries and those that keep the action invariant though?
 
I think the conceptual idea is: If I rotate everything, physical results should be the same.
So you rotate i) all states and ii) all operators
for instance if i have a stern gerlach experiment, if i rotate just the state this is equivalent to rotating just the stern gerlach device. but i want to consider the case in which i rotate both the state and the device
 
But then won't I get a trivial result for everything since $U U^\dagger \mathcal{O} U U^\dagger = \mathcal{O}$?
 
that is the statement that physical results are the same which is what we want i believe
so $\langle \psi' \lvert O' \lvert \psi' \rangle = \langle \psi \lvert O \lvert \psi \rangle$
and $\langle \psi' \lvert \psi' \rangle = \langle \psi \lvert \psi \rangle$
if a transformation is to be a Wigner symmetry
 
So in the above, I need both?
 
here is Moretti's answer from which the terminology I use comes from: physics.stackexchange.com/questions/512442/…
@DIRAC1930 Depends on the action I suppose. If you have a situation in which total energy or total momentum is not conserved, then the corresponding translations (of the poincare group) would not be a dynamic symmetry for instance. But we still expect certain physical observations to be frame independent (by postulate).
I am not too far into QFT so I am not sure what class of actions one considers when doing textbook QFT
 
9:18 PM
I think this transition probability being invariant makes the most sense as a way to define a symmetry
 
I think the transition probability case encompasses the expectation value case
because $\langle \psi \lvert O\lvert \varphi \rangle = \langle \psi \lvert \phi \rangle$ where $O\lvert \varphi \rangle \equiv \lvert \phi \rangle$
since an operator sends a state to another state
so the expectation value of $O$ is just a transition probability upon redefinition
 
Yes but then expectation values aren't invariant. You have to also transform the operators corresponding to the 'measurement device' (like you stated previously)
But then this is how you would do anything
i.e. if I translated the system, to get the same measurement, I would also have to translate the measurement device
 
oh i see what you are saying hm
 
9:38 PM
@DIRAC1930 hm what do you mean by this is how you would do anything
@DIRAC1930 hm but $\langle \psi' \lvert \phi' \rangle = \langle \psi U^\dagger U O \lvert \varphi \rangle = \langle \psi \lvert O \lvert \varphi \rangle$ isn't it
@DIRAC1930 This is assuming that translations are Wigner symmetries. But a transformation on (projective) Hilbert space is not a Wigner symmetry from the outset. From the postulate that Poincaré transformations are Wigner symmetries and by Wigner's theorem, we get that we can represent these translations as unitaries in the first place. Once we represent such transformations are unitaries then it seems trivial that oh this will be invariant and so on.
i might be misunderstanding what you have said above though
 
9:59 PM
@SillyGoose well, "German" is a hard notion to pin down as you go into history - a single entity that called itself "German" without further qualifiers only existed from 1871 onwards (founding of the German empire ("Reich")), and Minkowski was born seven years earlier than that in what is today Lithuania, what was then part of the kingdom Poland, which in turn was part of the Russian empire, and then Minkowski spent most of his life in Prussia (which turned into part of the German empire).
Now does that make him "Russian", "Lithuanian", "Polish" or "German"? :P
@Slereah yes, symplectic manifolds are "too rigid" - there's no good notion of morphism that allows "enough morphisms" while still respecting the symplectic structures in a meaningful way
@Slereah I think the relevant foundational question is: What even is our definition of state? Why are we looking at a tensor product?
in quantum theory, I can explain to you why the tensor product arises - essentially from the principle of superposition
but in classical theory, I don't really know how to argue for any notion of combining states that isn't just the Cartesian product of the state spaces
what do I care about "Poisson structures" etc. - classically a system always has a single state as a point in some state space $Q$ (regardless of whether it's Hamiltonian phase space or some other), and if you have a second system with states in $Q'$, the combined system is just listing the two states one after the other, $Q\times Q'$
there's no principle of superposition (equivalently no requirement for observables and states to be linear on each other in specific ways) in classical mech
More mathematically, you might talk like this: There is a "configuration space" $M$ (the space(time) manifold) and classical Hamiltonian mech and quantum mech are both, among other things, functors that assign state spaces to this. Our notion of combining systems operates a priori on the configuration spaces:
$M,M'$ combine into $M\times M'$. The state space functor of classical H. m. is $T^\ast$, and so $T^\ast(M\times M') = T^\ast M \times T^\ast M'$. The state space functor of QM is $L^2$, and so $L^2(M\times M') = L^2(M) \otimes L^2(M')$
 
@JohnRennie Thanks v. much John! Pardon it takes me a long time to reply as I'm pretty swamped with work. - I do understand that virtual particles are not real in the sense of being around for long enough before annihilating to act on anything. Thanks for confirming their difference from zero-point/vacuum energy. - Re: zero point vs. vacuum en.; ahhh makes sense! Perfect. Very clear picture thanks for clarifying. Much appreciate your time. :)
@JohnRennie for background I'm a science & data visualization designer & producer, aware that a lot of popular media on the subjects can give false impressions so trying my best to minimize that in my work.
 
10:53 PM
Is negative temperature ever meaningful? I thought it was impossible to get to abs. zero so why do we even allow it to be negative when talking about $\frac{1}{T} = \left(\frac{\partial S}{\partial U}\right)$
 
@ACuriousMind how is that not forbidden but reaching $0$ is? in math we often consider convergent series to be equal to whatever they converge to. i.e. $(9/10)+(9/100)+... = 0.999...=1$ so if anything approaching $0$ from the right makes more logical sense than wrapping around infinity to -$\infty$ which I have no idea what that means
 
read the part again where I state that it's really $\beta = T^{-1}$ you should be thinking about
 
11:10 PM
I don't understand how we can flip temperature to begin with? The only time I've ever seen numbers that increase become negative is in computer programs where there are overflow/wrapping
so do we just add $\infty$ to our temp to make it negative
 
what are you talking about? If you look at any number $\beta$ going from slightly above 0 to slightly below 0, its inverse $T=\beta^{-1}$ becomes ever larger, becomes undefined at $\beta = 0$, then starts going up from large negative numbers
that's just how math and division work :P
 
I just mean in practice, does negative temperature ever have a relevant purpose?
Like maybe in some advanced stuff?
 
that's a completely different question!
and you can answer it by reading Wikipedia
 
Oh, I see. I should have stated that instead of if it had meaning
@ACuriousMind D'oh, I forgot the two state paramagnet is also a system with negative temperature since the highest state of entropy is the ground state of energy.
it can also only occur for systems with fixed total energy
 

« first day (4857 days earlier)      last day (61 days later) »