The h Bar

@ACuriousMind Uh, interesting. A way to define connections on a bundle $E$ is as linear maps taking E-valued 0-forms (that is sections of the bundle) into E-valued 1-forms (with some specific properties I won't mention). Is the exterior covariant derivative you mention related to the generalization of the map above but from $k$-forms to $k+1$-forms?

I read about such structure on Frankel this morning

@Mr.Feynman yes

Perfect timing then, thanks :P

9:54 AM

What does the Ergodic theorem claims/say about the expectation value in classical mechanics>

@ACuriousMind to be precise, in this case you mean E valued tensors, right?

That is elements of $T^p_qM\otimes E$

what is $E$

The bundle space

of which bundle

The one on which we are defining the connection

9:56 AM

I didn't really talk about bundles because the notion of a gauge connection is a bit more general then a single bundle with connection

the gauge connection lives on a principal bundle and defines a connection on all associated bundles

And that I mentioned here

10 mins ago, by Mr. Feynman

@ACuriousMind Uh, interesting. A way to define connections on a bundle $E$ is as linear maps taking E-valued 0-forms (that is sections of the bundle) into E-valued 1-forms (with some specific properties I won't mention). Is the exterior covariant derivative you mention related to the generalization of the map above but from $k$-forms to $k+1$-forms?

@ACuriousMind oh...

if you want you can interpret every form as being valued in some associated bundle (the ones that do not transform under the gauge group being valued in the bundle associated with the trivial representation of the group)

I asked that because a connection lets us differentiate section of a bundle, and if that bundle is a tensor bundle we are differentiating tensors. On the other hand, when working with the notion of connection on a bundle $E$ (that apparently is less general than what you're talkimg about), we care about differential forms valued in that bundle $\in\Lambda^k(T^*M)\otimes E$ and not $E$-valued tensor fields, i.e. elements of $T^p_q M\otimes E$

yes, you can only differentiate general tensors when you have a connection soldered to the manifold - that's what the Levi-Civita connection does, it defines $\nabla_\mu$ as an operation

but a connection not tied to spacetime indices doesn't do that, it can only extend exterior derivatives

I see, that's why the induced connections on tensors bundles require linear connections (connections on $TM$)

The comparison between $\partial_\mu$ vs $d$ and $D_\mu$ vs $d_A$ helps a lot

10:30 AM

The following formula for entropy $S=K_bln\Omega$ is valid in which cases? For a thermodynamic system in equilibrium, for which the multiplicity of the macrostate of eq. is known?

@imbAF yes, this is an equilibrium formula, see also this answer of mine

@ACuriousMind thanks for the link

6:51 PM

I have a small question can I ask here ?

Yes

why do we not do vector addition for cases like T-mg where t is tension in string and mg and just do normal subtraction and netforce between on one charge by two charges which are opposite to the charge we are concerned with whereas both are vectors ?

@Razz Your question is not very clear... an expression like $T-mg$ actually usually represents a single component of a vector equation for forces. In this case, since the $mg$ term doesn't have any sines or cosines, it's probably referring to the $y$ (vertical) component of the vector equation (the equation btw should also have a RHS...)

like when we are to find the acceleration of a body in a pulley systems sometimes we do T-mg = some value , and we just subtract them rather than subtracting them vectorically why is that so is my question ?

we know values of T and mg and rhs probably has the "a" acceleration quantity so that we have to find a

7:09 PM

@Razz I think you're confused because you know that when forces are considered as vectors we have "the sum of forces is equal to the mass times acceleration" right? $$\vec{F} = m\vec{a}$$ but consider that in this case you have $F_{y(tension)}=T$ and $F_{y(gravity)}=-mg$ so you see you are still adding them, but one has a minus sign because it's pointing in the negative direction of your $y$ axis

@Razz you are subtracting them "vectorially", it's just that in the usual setup of a pulley where the rope is straight down the tension and the gravitation force point in the same direction

it's not necessary to consider them as "full vectors" when they're just parallel

ohhhh

so im still subtracting them vectorically

thank you for the help !

Peace

en.m.wikipedia.org/wiki/….

How do we know this theorem is true for all physical vectors

For example, it is not clear to me how this could apply to a vector like angular velocity or angular momentum

angular momentum is an axial vector :)

The proof in this article is definitely using a fancy method and there should be a simpler way to derive this (though maybe less general)

7:30 PM

A rudimentary method is present in intro to mechanics by kleppner and kolenkow, also in David morin, but I'm not able to digest this theorem applies to all "measurable vectors"

Why not, all interesting Euclidean vectors are just position and its derivatives aren't they? At most, multiplied by a scalar... regarding axial vectors, I'm not convinced this will hold

@nickbros123 what about "physical vectors" is different from the vectors this theorem is proven for?

Q: What actually is the vector of angular momentum?

7:48 PM

Some stuff about axial vectors, aka pseudovectors:

2

If an object spins around a central point, it gets angular momentum which is a vector with an orientation dependent on whether its clockwise rotation or anticlockwise, i get that. But what the vector part actually is, is confusing me. Does it mean that the object spinning gets a force going upwar...

@PM2Ring Thanks, someone should add this to the long list of reasons in favor of teaching physics students differential geometry from day #1 :) lol not that I'm dogmatic about it, it's just how I would have liked to learn

@ACuriousMind my issue is particular for angular momentum and angular velocity

@Amit Maybe. ;) But at least they should say something like "If you think this cross-product thing is a bit suspicious, you're right! Unfortunately, you need a fair bit more maths to understand what's really going on here. We'll get there, eventually..." ;)

@nickbros123 Actually I think this formula may work trivially for both angular velocity and angular momentum, at least if we take the particular example of transforming to a frame that rotates at an angular velocity s.t. the axial vector points in the same direction as $\omega$, you'll have $\omega\times$ something in the same direction which of course vanishes...

8:03 PM

@Amit This $\omega$ ur talking about is the angular velocity of the frame?

@PM2Ring Yeah, but OTOH if anyone said that I would actually be more frustrated... my reaction would have been I wanna know NOW :)

@nickbros123 Yeah, I'm quoting from the theorem

Ah sorry the theorem had the capital version, $\Omega$ :)

@PM2Ring To put it slightly more coherently, I felt for example that the cross product was complicated enough back then anyway, as far as memorizing all this gibberish with using the determinant to calculate and so on... so it could be nice if a bit more background was given so that I could understand why I was doing what I'm doing lol

@Amit what if this doesn't turn out to be the case though

Instead they just throw a determinant with partial derivative operators at you and let you crank. Well I don't know if that's how they do it today... lol

@nickbros123 Idk :) First one needs to really understand the derivation, then consider whether it works with pseudovectors.

@ACuriousMind Hi, earlier today when I asked about the Boltzmann entropy, you provided the following link : physics.stackexchange.com/questions/141321/…

I have only 2 questions

@Amit I agree, it is a bit mysterious and frustrating. It's not good to do "magical" symbol manipulation without knowing what's going on. However, the teacher can explain that rotation is a process that happens in a plane (mapping points in the plane to other points in the plane), and a normal vector to that plane is a convenient way to keep track of the orientation of that plane.

8:12 PM

Ok, i think i understand this, so if these angular velocity / angular momentum quantities are coordinate independent and do add like vectors, could one argue that these "psueo vectors" also fall into the general category of physical vectors and conclude the theorem works? I don't know what is going on on the derivation given on the Wikipedia page

@PM2Ring Yes. I do remember one Professor who at least showed in a very simple way why it's not a vector, by showing that it doesn't change sign under a parity transformation, I remember thinking that's quite cool

1. Are you making (for simplicity) the assumption that the macro state of eq. has multiplicity of $N$?
2. You make the claim that for a thermodynamic system in eq, all the microstates belonging to the macrostate of eq. have the same probability. Does the type of system plays a role? Meaning, whether it's isolated, closed or open for your assumption to hold true? I know for a fact that that's the case for an isolated system. @ACuriousMind

@Amit this course would ask for a pre req of Linear algebra, single and multivariate calc wouldn't it? I think for a general intro to mechanics, not the Goldstein or Landau types, but the morin type, one can just accept things for the time being

did your question about probability of getting a certain measurement value ever get answered? @imbAF

A fun angular momentum conservation toy:

Tippe top

A tippe top is a kind of top that when spun, will spontaneously invert itself to spin on its narrow stem. It was invented by a German nurse, Helene Sperl in 1898. == Description == A tippe top usually has a body shaped like a truncated sphere, with a short narrow stem attached perpendicular to the center of the flat circular surface of truncation. The stem may be used as a handle to pick up the top, and is also used to spin the top into motion. When a tippe top is spun at a high angular velocity, its stem slowly tilts downwards more and more until it suddenly lifts the body of the spinning top...

Q: Probabilities of measuring an eigenvalue and the different cases

8:18 PM

@SillyGoose I was able to deduce the correct expression for all 8 cases myself. I even did a thread, so others could give their insight on the matter, but no one did

2

This thread has the following 2 goals: Knowing how to properly write the below considered cases. To help individuals who might want to know the correct expression and calculation for a specific case. It will be a lengthy one, but I believe it benefits, me personally, but also other members of t...

quantum-mechanics probability

@imbAF 1. Not sure what you mean by "for simplicity", but yes, $N$ is the multiplicity of the equilibrium macrostate. 2. It's called the ergodic hypothesis and generally assumed to hold true for the systems we consider in equilibrium statistical mechanics

I was wondering why it need be split up into 8 cases. Was your question not about the procedure in general, but about the actual explicit forms for each of the 8 cases?

@nickbros123 I've found a book now that says "Statics, kinematics, and dynamics are based on 13 types of physical vectors, namely:" ... and continues to list Angular velocity and angular momentum as two of them

For example, I feel like there wouldn't be a need to distinguish between pure and mixed cases because one could represent the state (whether pure or mixed) as a density operator and go from there. then, the pure case would pop out of the more general case by setting $p_1 = 1$ or what not

@PM2Ring Witchcraft, I don't think mechanics is ready for this

8:21 PM

@ACuriousMind the multiplicity of the macrostate can be anything other than N, it can be a nr. way smaller than N etc. The ergodic hypothesis is different than the ergodic theorem right?

@imbAF I don't know what you mean "can be anything other than N"

I just use $N$ to denote the multiplicity

Ok

But on point 2

"Your question seems to be not clear. Try to improve your question" lol at this comment on thee vector of angular momentum post

@nickbros123 It later proves the kinematic transport theorem for "physical vectors"... so it is very specific that it does apply to these two

I am familiar with the ergodic theorem, which makes a claim about the expectation value of a physical quantity, but the ergodic hypothesis, what does it claim, so that your assumption of the microstates being equally probably to be true?

Regardless of the system in consideration

8:24 PM

What is unclear about the Wikipedia article I linked?

@SillyGoose I was trying to find the expression for all 8 cases. That's it, I wanted some sort of order in the expressions

It literally says in the introduction that the ergodic hypothesis is the assumption that all microstates are equiprobable

It does say that

It doesn't however say anything about the type of system in consideration or whether that's important at all

Which is what I asked

I don't know what a "type of system" is

What is a good way to see that real systems occupy mixed states frequently? Or is this not true

8:27 PM

well an isolated system, a closed system and an open system, can we make the claim that in all 3 cases, once the system is in eq. the microstates that correspond to the macrostate of eq. are equally probable

no, it's a hypothesis precisely because this doesn't hold for all systems

Ok so than, the gibbs entropy is limited to some types of systems

@nickbros123 Yeah but I can extend my whining beyond the introductory courses lol. Anyway, I don't wanna belabor this point, we can also get into the subject of how various "Math for physicists" courses are done...

if you have an ergodic theorem then the ergodic hypothesis also holds (since the ergodic theorems about time and space averages imply that a distribution that is equiprobably at one point in time will remain so at all times

Basically those who are considered as hamiltonian systems

8:30 PM

@SillyGoose whenever you have two systems that are entangled and you measure one of them, the state you should assign to the other after measurement is a mixed state

"a distribution that is equiprobably at one point in time" how to understand this? In mathematical terms or in terms of the phase space??

so since "real" systems interact with their enviroment all the time and hence become entangled with it, you should probably believe that most of these systems are in some mixed states

@imbAF I'm not sure what's not mathematical about the phase space

cuz the concept of distribution comes up

?????

a distribution that is equiprobably= phase space probability density constant in some region of phase space? Is this a correct interpretation ?

8:33 PM

@SillyGoose There was also a Q with a funny title now: "What is the real part of $Re(...)$" :P

@imbAF I really don't make any sort of distinction between "probability distribution" and "probability density" in this context - I use the two terms interchangably

aha

but yes, an equiprobable distribution/density is of course one that is constant where it is non-zero

Well I don't xD and that threw me off

what do you think the difference is?

8:36 PM

I will give you the difference but for a discrete random variable. The same may be applied in the continuous case. In the discrete case, you have the pmf which assigns a probability to some value that the random variable can take. Now the totality of all the probability values for all the values that the random variable can take, is the probability distribution, how the probability is being distributed, since it has to ultimately amass to 1.

So, in a plot, one point is the PMF, all the points represent the probability distribution

This is how I see the difference between The probability distribution and the probability distribution function

I don't really understand how that's supposed to work

The function you're plotting is the probability distribution function!

And that function is not a single point on that plot, it's the entire plot!

Unless you have a reference for this, I don't believe anyone else shares this strange distinction and you'd probably do well if you just stopped drawing it since it will just confuse other people you're trying to talk to about this

Because the probability is distributed in the plot

But perhaps I shouldn't think of it in this way

9:02 PM

@ACuriousMind One thing that confuses me about the ergodic hypothesis is the following: "the time spent by a system in some region of the phase space of microstates with the same energy is proportional to the volume of this region". But for a sistem in eq, the system spends time in that region of phase space where the pdf is non zero, and it spends infinite

amount of time, for as long as it's in eq. That would imply infinite volume of that region, but it's not the case, since we said that outside of this region the pdf is zero

9:28 PM

Could you also elaborate how : "since the ergodic theorems about time and space averages imply that a distribution that is equiprobably at one point in time will remain so at all times". I don't see the link between the two

@nickbros123 One way to convince yourself this theorem makes sense is to choose a vector such that what they call $\left( \frac{d}{dt} \right)_r$ vanishes, that is, a vector that is constant with respect to the rotating frame. Then it's easy to see the contribution to the derivative in the non-rotating frame is only from the angular velocity $\Omega$...

@nickbros123 There's also a much simpler derivation of it here: en.wikipedia.org/wiki/…

antimony

9:50 PM

hahah @Slereah i'm all for deluxe energy. actually was thinking, i wonder how many crazies see the term "Gibbs Free Energy" hahah

"if this gibbs guy is getting free energy, how come we're still paying for our power bills"