hey guys does the standard hilbert space used to describe a spin of spin $L$ have a name? (ie $L=1/2$ you have that the space is $\mathbb C^2$ and the spin algebra is represented by the pauli matrices)
I am trying to formulate a lebniz integral rule extended to functions with finite number of discintinuities, I started with basically first principle for hte derivatives but I don't understand why my final limit is not converging
When I left university I worked as a colloid scientist in industry (for Unilever) but I eventually got bored by that and I left to run the IT department for a local company. Then I got bored again and left to work for what you'd call a start-up these days.
When I got to be 50 I decided I wanted time off so I retired, but the start-up still wanted me to maintain the systems I'd put in place. So now I spend the first couple of hours each day working for my old company then the rest of the day messing around on the Physics SE. That's what I mean by a part time computer nerd.
@s.harp "spin space" is a dangerous choice because people would tend to mistake it for spinor space. Just write $\mathbb{C}^{2L+1}$, not everything needs a fancy name ;)
@innisfree Your best bet among the regulars would probably be @DanielSank
@Secret The geometric connection between the Lagrangian and the Hamiltonian is that one lives on the tangent space of the configuration space and the other on the cotangent space. The Lagrangian has no meaning in phase space and the Hamiltonian has no meaning in the tangent space. You're not saying anything meaningful when you talk about trajectories allowed "by the Lagrangian" "in phase space".
the derivative of a force is the (minus) potential energy so the integral of a force is the potential energy but that is the work done by the force witch equals the change in kinetic energy. So the integral of a Force gives you both the kinetic energy as a function of time and the potential energy with a change in the sign .So when i integrate a Force what do i get??
@ManolisLyviakis No - for instance, the work might go into a different potential energy. You only have that the work is the change in kinetic energy if you are looking at the total force.
@ManolisLyviakis I don't know what you mean by "kinetic force"
Yes, there are forces which come from a potential, called "conservative forces", and those that do not, called "non-conservative forces". The latter are not really "associated with kinetic energy", although they tend to depend on velocity, like friction
@ACuriousMind when i translate it to greek i cannot derive to an exact meaning for example when tlaking about kinetic energy i can intuitively understand that it is about the kinesis-movement of an object ad its "energy"
@ManolisLyviakis It means the things whose derivatives are forces. If you're asking why they're called "potential energy", it's because things will always tend to move to regions of lower potential energy, so if something doesn't have minimal potential energy, it potentially can move (and do work in the process).
I.e. "potential" is meant to be the "potential to do work".
Need someone good at indices to help me, because in this question there's this strange term $\lambda \mathbf{x}\cdot \frac{d}{d\lambda}(\nabla{f(\mathbf{\lambda x})})$ that I don't know how to make it disappear. What conditions I am missing? (I have just done a back in the envelope calculation of the expression with indieces again, btu the term still persists)
So basically I read someone else's answer to a question regarding Euler Homogeneous function theorem
source:
http://quant.stackexchange.com/questions/8911/what-is-exactly-eulers-decomposition
Also here http://planetmath.org/sites/default/files/texpdf/40683.pdf
I tried to do the proof by hand ...
@Xasel Please learn all the standard concepts first before "resolving" to do something new. Also, note that using alternative definitions to prove things is in some cases just defining something such that the thing to prove becomes trivial, but this doesn' really answer a question about the original definition.
@0celo7 Depends on whether your definition is equivalent to the former one. If yes, then sure. If no, better not call it "curvature" else no one knows what you're actually talking about.
@DanielSank I still need help on that Entropy conceptual question. Since entropy is the number of accessible microstates (taking log), is it true that whenever it changes, at least one other variable that characterise the system must change with it?
@ACuriousMind :I appreciate your suggestion and I understand you concernt hat without learning about other ideas and building my fundamentals(built on should er of giant)..it may lead to crakcpot theory
@DanielSank is it true for the most general systems (e.g. up to QFT level and condensed matter quasiparticles) that whenever an entropy change took place, at least one other parameter that describes the system (e.g. volume, particle number, spin state, temeperature pressure etc.)must change with it, that is, you cannot just have S changed alone?
@DanielSank I found that puzzling because for energy, we can change it by simply having the system occupy different energy levels. Back in thermodynamics we have fundameental equations U(S,V,N) and S(U,V,N), thus I found kinda suprising that S cannot change alone while in some sense U can
@Secret Entropy is a state function, i.e. it can be computed from the parameters describing the equilibrium state without any additional knowledge, so you can't have entropy change without having a parameter change. Same goes for internal energy, I don't know why you think U "in some sense can" change.
@ACuriousMind Probably because somehow I mixed it up with energy levels, since you can change that by moving particles to occupy other energy levels. I am not sure what I overlooked here
currently my ideas are in rough stage(scrabbing in my notbook) I wil sooon compile them share with guys here...I simply start from basic set theory:I define infromations is a set definging a relation betweent wo sets
@ACuriousMind Let me see, if I move e.g. 3 particles up and down energy levels (let's suppose we have 3 energy levels each with degeneracy 3), then I will change the energy contributed by them to the system, if I say, place all 3 of them at $E_3$ while initially they are at $E_1$, thus changing U?
If a sphere were to expand outward faster than the speed of light, will the sphere be considered infinity big?
Kind of pertains to the universe, but its mostly hypocritical.
I'm not following you. Placing all of those at a higher energy level is a realization of a different macrostate (the overall energy changed). You seem to be conflating a classical thermodynamic system that's fully described by volume, pressure, particle number as thermodynamic state variables with a quantum system where that's just not true.
So yes, in a quantum system you can change certain state functions without changing the classical state variables, but that's just because the classical state variables aren't your full set of state variables anymore.
So are the new variables in the quanutm system the labels for the enegry levels themselves, is that what I am changing when I change U by sending the particles from $E_1$ to $E_3$?
@Secret The "state variable" for a quantum statistical system is the density matrix. You don't get explicit state variables like in classical equilibrium thermodynamics, although for equilibrium knowing the temperature and Hamiltonian is of course enough to know the density matrix.
The desired equation is not true in general for homogeneous functions. As a counter-example, take $f(\mathbf{x})=|\mathbf{x}|^2$. Then
$$\lambda \mathbf{x}\cdot\frac{d}{d\lambda}\nabla f(\lambda \mathbf{x})=\lambda \mathbf{x}\cdot\nabla |\mathbf{x}|^2 \frac{d(\lambda^2)}{d\lambda} =\lambda \math...
And the mistake is I differentiate the WRONG equation correctly
Looks like in the future when I hunt for careless mistakes, I need to check whether there are discontinuities when I go from step i t step i+1, not just the algebra
@0celo7 a lot of the occurrences of eu in English come from ancient Greek and tend to be pronounced you. For example eugenics and euphemism. I would have guessed that Keurig was a Germanic name, but it's no great surprise to hear it pronounced like the Greek.
Presumably before migrating to the US Herr Keurig would have pronounced the eu as in Euler.
@JohnRennie Both the orignal Greek and the current German pronounciation of Greek -eu- is like the standard Germanic pronounciation. So you you English insist on pronouncing Greek words like "you", you should also pronoune German names like that :P
@ACuriousMind Well we were Germanic originally. Well, a certain percentage of our DNA is Germanic depending on where we live and how horny the invading Anglo-Saxons were :-)
Speaking as someone who is one quarter Welsh and one quarter Scottish 50% of me kicked the Anglo Saxons' butts :-)
@ManolisLyviakis My father is Scots-Irish and my mother is half Welsh and half English. Probably best to just describe me as British and not go too deeply into the details.
Britain has been invaded lots of times. Defining the British race is largely a matter of working out which racial group invaded your particular area of Britain most recently.
If you go back 10,000 years we were originally Basques. That is it's believed the first people to occupy Britain after the ice age are the same racial group as the Basques.
Then you got the Picts, though no-one sems to know much about them.
Then the Celts, then the Romans, then the Anglo-Saxons, then the Vikings, then the French (Normans, who were really more Vikings).
Yes. It's believed that as the ice retreated at the end of the last ice age the very first colonists worked their way up the Spanish and French coasts to the UK. They have since disappeared everywhere except in the Basque regions of Spain.
My niece (age 14) was explaining to me the difference between a nerd and a geek but sadly I was only pretending to listen so I still don't know the difference. We nerds are like that.
@yuggib I cannot seemed to derive this result for distributions except when the distribution is a dirac delta
Consider some general tempered distributions $T(y)$, rapidly decaying test functions $\phi(x,y)$ and some smooth function $f(x)$. Then $\langle f*T,\phi\rangle=\langle T,\tilde{f}*\phi\rangle\neq\langle fT,\phi\rangle=\langle T,f\phi\rangle$?
No, I am just trying to wrap up one of the question I discussed with yuggib that was way before I have started read susskind. My main focus is still going to be EL and classical mech
Btw to get back to your question:
@0celo7 Line 19, why when the partial derivatives of H bounded will result in the integral of them to vanish?
I am not sure if it is proved (and I am not sure about the name of that theorem other than it is some bounded thing), but in your EL notes screenshot it is used from Eqt 18 to 19 without proof
So basically I read someone else's answer to a question regarding Euler Homogeneous function theorem
source:
http://quant.stackexchange.com/questions/8911/what-is-exactly-eulers-decomposition
Also here http://planetmath.org/sites/default/files/texpdf/40683.pdf
I tried to do the proof by hand ...
If a sphere were to expand outward faster than the speed of light, will the sphere be considered infinity big?
Kind of pertains to the universe, but its mostly hypocritical.
