@bolbteppa Example: Both pressure and tensile strength are scalars with units of Pa, but are completely different physical quantities. Therefore if something has units corresponding to some physical quantity M, it does not mean said units actually implies the physical quantity is M. An extra step is needed in checking how the physical quantities are related in an equation in order to show that it is M, and that M has a physically meaningful interpretation
The issue of that one, as the chat log between acuriousmind and bolbteppa above showed, is that torque is r x F. The cross product complicated the comparison as energy has no direction like properties yet torque has (it's coordinate vector is dependent on the basis used and has more than one component, unlike energy which has one component and is a scalar). This the the bottleneck that bolbteppa is arguing about with acuriousmind. My example above does not have this bottleneck thus served as an
Optimised counter example to bolbteppa's point
Another good but not optimised counterexample is in chemical engineering: Both mass flow and mass production has units of kg s-1, However because of the conceptual framework of general relativity, one can knock down this counterexample by treating mass production as in a way mass flow due to time evolution (think of this 4 dimensionally and consider how any matter has a worldline in relativity)
(I.e by combining mass production and mass flow into a 4 vector)
@Secret Looks like you're falling into the exact same trap, "Tensile strength is defined as a stress" en.wikipedia.org/wiki/Ultimate_tensile_strength there is a distinction between stress (a tensor) and pressure (a scalar) qr.ae/16LXh2 & en.wikipedia.org/wiki/Pressure#Scalar_nature taking the magnitude of a tensor and comparing the scalar quantity to definedly scalar quantity seems to be the source of the confusion again
@PhysicsGuy do you actually understand any of the modular elliptic curves stuff in an elementary enough way that could be scribbled in the margins of this chat?
I mean is $y = x^2$ modular? How about $x^2 + y^2 = 1$?
Is the Weierstrass p function modular? How do you explain that to a 5 year old?
@bolbteppa Specific heat capacity and specific entropy are both scalars, and both have units of JK-1kg-1 yet heat capacity and entropy are not the same thing
On a related note heat capacity and entropy has the same unit of JK-1 owing to the virtue of Clausius entropy before people realise later that entropy is something more general
We're going too hard about coming up with things with same units but not actually the same.
I mean, look at "momentum" and "product of the length of the Trans-Siberian railway and average velocity of HMS Beagle during it's voyage with Charles Darwin."
Acuriousmind is trying to convey the important message in this discussion that whether a unit has a meaningful physical interpretation depends on its roles in the model and cannot simply be obtained from dimensional analysis. But in some discussions, if one or more party hinge on a bottleneck, they usually will get stuck there until the bottleneck is resolved. This is why I am trying to address what the bottleneck have issues with, in hoping to unblock it so that the important message can
flow through. Even if the important message in question have no direct relationship that the bottleneck is questioning about, we usually need to resolve it first before we can proceed. Of course, this is a risky business because there are cases where there is actually no counter examples exists, which will make conveying the important information highly nontrivial
@0celo7 in addition to what I said, I am also more used to label the units of torque as Nm instead of its equivalent J. This is because in a physical perspective torque can interact each other in more complicated ways (e.g adding torque need to take account of direction and orientation but adding energy only need to take account of its magnitude and sign). They are simply physically so different to be considered as the same thing despite formally speaking they have the same units J
[Dealing with cases where unblocking bottlenecks are nontrivial] it is up to the party on whether to continue on the discussion or argument, which depends on whether everyone still think the discussion can lead to somewhere
(For me because I think like a computer, the answer is often obvious to me)
@BalarkaSen Your example illustrates nicely what Acuriousmind trying to convey. A bunch of units does not mean whatever corresponds to it is physically meaningful. The context on how the units came by matters
@JunaidAftab Google results so far send me to bogus downloading website, so likely none
@ForeverMozart Any example of a geometry such that given a point P that is within a ball of radius $\epsilon$ centred at another point A are actually far apart from each other (e.g. the metric when applied to the interval AP grew without bound?)
That is I am interested in an opposite example of degenerate points (e.g. how in the minkowski metric, two points on a light cone have a metric of zero)?
maybe I do not understand what you mean by interval
you know that the continuous image of a compact set is compact, which means closed and bounded if you are in $\mathbb R$... so the metric is bounded on compact sets.
@ForeverMozart I am thinking of something like say my space is that of $\mathbb{R}^2$, but has a extended metric e.g. $g(x,y)=\frac{1}{x^2+y^2}$, so that any point within a neighbourhood of another will be "infinitely far apart". Is there a name for all kinds of topology similar to this (where points in a neighbourhood have an unbounded distance) and are they useful in proving theorems or similar things?
@bolbteppa if you are talking about Shannon entropy then yes it is dimensionless. However if you are talking about Boltzmann entropy (the type commonly used in statistical mechanics), then it has the units JK-1
Here we have an example of the converse, different units can mean the same physical quantity
Hence another possible source of ambiguity that dimensional analysis alone cannot restrict
@ForeverMozart Sorry typo the extended metric should be $g(x,y)=\frac{1}{(x-y)^2}$
@Secret Landau Statistical Physics Page 25 explains why entropy is dimensionless and why if you interpret it classically invokes dimensions in the 'definition' & why you have to define this fundamental concept of nature to within additive constants which immediately should indicate that the concept does not make sense classically and why all of classical statistical mechanics is logically flawed
I recommend you read real books ;)
Any book that uses that definition is introducing you to statistical mechanics classically in the beginning, but this illustrates the breakdown, worth thinking about, very cool how dimensional analysis illustrates such an important concept
Roughly from memory, equilibrium quantum statistical mechanics is just multi-particle time-independent quantum mechanics in stationary states in the energy representation
Doing the combinatorics of particles fitting into those stationary states at energy levels is where you end up with entropy arising
Those 25 pages give a good intro if you have some lagrangian/hamiltonian mechanics & QM down
But the wordy sentences might take 2 years to decode :p
I have a fundamental disagreement with the explanation of a real world phenomenon. That fundamental disagreement renders the current explanations of that phenomenon completely wrong.
The simplest way I could think of to bring everyone's attention to it was by asking a question on PSE (Physics St...
@DanielSank I've flagged it. It's always disappointing when site members fail to observe the principles of common courtesy. At some point the mods will have to start handing out sanctions for this.
@ACuriousMind I see, so instead of having the condition "there exists a cyclic vector" one could also say $X,P$ should be an irrep of the standard representation? Which representation of which group is that?
@Bass Not exactly. To every state of the Weyl C*-algebra of Canonical Commutation Relations (the group you want to consider, that is essentially the Heisenberg group) there is an associated representation with a cyclic vector (the GNS representation). Another problem is that of irreps, where every non-zero vector is cyclic. The finite-dimensional CCR have the property that every irrepresentation is *-isomorphic to the standard (Schrödinger) representation.
@Bass Position and momentum form the Lie algebra of the Heisenberg group. What cyclicity means is that the representation is (topologically) irreducible as the representation of the universal enveloping algebra, not of the Lie algebra consisting of $1,X,P$ itself. Among other things, I believe that the existence of a cyclic vector ensures that the representation is unitarizable as a representation of the Heisenberg group.
@ACuriousMind The unitarity is an automatic feature of the Weyl C*-algebra (it follows from its definition), essentially a priori from the choice of representation.
I've mostly studied representation theory of groups and algebras where no C* structure exists on the abstract algebras, so I'm not sure what exactly demanding the representation to be a *-representation does
Yes, that's usually the case - since the algebra forgets the topological structure of the Lie group, one needs to find out "by hand" which representations of the algebra actually correspond to unitary representations of its group
And - in the case of sl(2,R)/sl(2,C) - I've seen that the cyclicity is essential in showing that the algebra representation induces a unitary representation of the corresponding group.
@Bass well, they are linear operators on that vector space; but there is not a notion of scalar product on polynomials (as far as I know). So that is not so useful for quantum mechanics
@yuggib Ah of course. You need a basis of $L^2(\mathbb R)$ like Hermite polynomials.
Next question: I'm currently confused about the standard SSB "Mexican hat" example $V(\phi)=-10\phi^2+\phi^4$. [Wikipedia](https://en.m.wikipedia.org/wiki/Spontaneous_symmetry_breaking#A_pedagogical_example:_the_Mexican_hat_potential) says *This potential has an infinite number of possible minima (vacuum states)*. What confuses me is that the "minimal fields" (the $\phi$'s with a minimal Lagrangian) are called "vacuum states". If the fields are operators, not states, then how can a *field* be a vacuum *state*?
@Bass Strictly speaking, a field configuration is indeed not a vacuum. But each solution $\phi_v(x)$ to the classical potential corresponds (to first order) to a vacuum such that $\langle 0 \vert \phi(x) \vert 0 \rangle = \phi_v(x)$. It is reasonably clear that the same $\lvert 0\rangle$ cannot yield two different VEVs for the same field, so each classical solution induces a different vacuum
On a physicist's level of rigor, one can show that the two vacua belonging to different solutions have "zero overlap", i.e. define completely different Hilbert space.
I'm not sure if there's a rigorous discussion of the correspondence of different classical minima and vacuum states somewhere, but I suspect not, given that we don't really know how to associate a Hilbert space to interacting theories
On another level, you might just observe that the LSZ formalism needs zero VEVs of all quantum fields to work, so one must expand $\phi$ around a classical solution, and the deviation from the classical solution becomes the quantum field. Each choice of solution defines a different quantum field then.
"An object is paced far away from all the objects that can exert force on it.A frame of reference is constructed by taking the origin and axes fixed in this object.Will the frame of reference be inertial?"
would some one like to discuss this question with me?
@ACuriousMind Isn't $\langle a|\phi|a\rangle=0$ for all fields $\phi$ and all states $\lvert a\rangle$ because the fields are composed of CA operators, which means that $\phi\lvert a\rangle$ is orthogonal to $\lvert a\rangle$?
@ACuriousMind All quantum fields I know of are like $\phi(x)=\int dk(a(k)e^{-ixk}+a^\dagger(k)e^{ikx})$, i.e. composed of creation and annihilation operators. Right so far? If not, which fields are not?
see first of all if a frame is inertial then it should be that if a force on any object with respect to the inertial frame is zero then the acceleration on it must be zero or the other way around if the acceleration on any object with respect to the inertial frame is zero then the force on it must be zero.My book says that earth is not an inertial frame and it proves it by fact that lets say a book is placed on the table then acceleration of book is zero w.r.t. earth but if we accurately
measure the sum of normal and gravitational force by earth then it doesn't comes out to be zero.Now coming to my question,lets say hypothetically we brought that object to a play ground and placed it over there and lets say it is not experienceing any force there.then if now we see a book placed on that table again the same normal and gravitational force will be acting on book and still their sum will come non-zero but book is still at rest.So,acc. to me it is not true that if we take an object
to a place where no force acts on it frame will be inertial
Wait, I'm not sure I even understand your problem.
Or, rather, I don't understand your book's argument why the Earth's frame is not inertial. Earth's frame is not inertial because there are forces like the Coriolis force that aren't exerted by anything but are results of the frame being non-inertial.
Stonoes theorem only say there's a one to one correspondance between strongly continuous one parameter groups and self adjoint operators in a hilbert space. How does it rule out the possibility of having correspondance to groups with more than one parameter?
@KartikWatwani Yeah, but that example doesn't make any sense to me. If the table is flat, then the normal force is exactly the opposite of the gravitational force pretty much by definition of the normal force.
@ACuriousMind I think that the author wants to say that if we find the normal force by some very very accurate way to very high accuracy it might be very very close to gravitational force,close enough to say that they are equal for now lets not go into why the author said that the net force is not zero.what is your take on the question?
@ACuriousMind "Goldstone theory": the toy theory to demonstrate the emergence of the Goldstone boson: $V(\phi)=-10\phi^2+\phi^4$. The minimal field: one of the $\phi$ that minimizes $V(\phi)$.
@KartikWatwani By Newton's laws, there exists an inertial frame. In that inertial frame, the object far away from anything moves in uniform linear motion, since no force acts on it. Then defining a new frame by affixing the coordinate system to that object just defines another frame that moves at constant velocity with respect to the original frame and is otherwise the same. So this new frame is also inertial.
@ACuriousMind In the wikipedia page, stone theorem for one parameter group is said Let $(U_t)_{t\in \mathbb{R}}$ be a strongly continuous one parameter unitary group. Then there is a unique self adjoint operator $A$ such that $\forall t \in \mathbb{R}$ $U_t = e^{itA}$
How does Stone's Theorem showed that there exists no self adjoint operator $B$ which corresponds to some multiparamter groups $(U_{t,s})_{t,s \in \mathbb{R}}$ such that you have something like $U_{t,s}=e^{i\mathbf{w}\cdot B}$ where $w=(t\hspace{1mm} s)^T$?
@Secret Who said Stone's theorem shows such a thing?! What does $\exp(\mathrm{i}(t\ s)^T B)$ even mean, I don't know what $v^T B$ is supposed to mean for a 2D vector $v$ and an operator $B$.
@ACuriousMind OK, getting closer. All free fields have 0 VEVs. Does that go in the other direction too? Is every field with an exact 0 VEV a free field?
@ACuriousMind Ok sorry, not stone's theorem is saying that
I think I might be thinking about something along the lines of multiple parameter exponential family http://statweb.stanford.edu/~jtaylo/courses/stats306b/restricted/notebooks/multiparameter_partI.pdf
Because I noted how in quantum mechanics any self adjoint operator $A$ has some kind of infitesimal operator of the form $e^{itA}$ where t is the parameter. But why it only stops at one parameter, what prevent the existence of operators that give some kind of infitesimal operator that requires more than one parameter?
You're not making any sense to me. Of course you could define "multi-parameter groups" by e.g. $\exp(sA + tA^2)$, but...what would be the point of that?
@yuggib I see. They are so rare that I don't aware of their existence For the unitary evolution operator of a time dependent hamitonian, I can see one of the parameter is t in the hamiltonian expression. But for the other parameter that is found by $e^{i\tau H}$ the expression, is it also time, and if it is is it independent of t?
the evolution is a so-called two-parameter group $(U(t,s))_{(t,s)\in\mathbb{R}^2}$ that satisfies $U(t,s)U(s,r)=U(t,r)$. The two parameters stand for the initial and final time in the evolution (since the generator is time-dependent, it is not true in general that $U(t,s)=U(t-s)$)
ok I am a bit confused here. So we have a two parameter group $U(t,s)$ and t,s,r corresponds to 3 different values for the parameter t,s? (are they initial, intermediate and final time)?
I see. I think I saw something very similar looking to that in Feymann lecture http://www.feynmanlectures.caltech.edu/III_08.html (8–4How states change with time).
$U(t_1,t_2)$ so this unitary operator, which Feymann used to derive the hamitonian, is also a two paramter group (because of having initial and final times)?
Also for the expression $U(t,s)=U(t-s)$ is my intuition on explaining why it is so for time independent generators a non misleading way to think about it. i.e. analogous to path independence in vector field in integrals, the value of the generator only depends on the difference of the parameters, not how it get from some initial value to final value?
What is up with all the questions about tides and gravity as of late? I'm beginning to think that two of our low reputation members are in fact the same person. They both are active at the same time of day, they both appear to espouse some not quite standard (aka crackpot) notions, and they both appear to be immune to logic.
I'm sorry if I'm asking something that doesn't belong here, but:
Why are questions that are closed, still able to be up/down voted? This is quite disruptive to the quality of the scoring. As far as I've seen inappropriate questions are closed in a pretty fair way, yet some of them keep being vot...
@Danu Don't know. First thing is getting some better data - I think the effect is much bigger here than in other sites on the network, but we need to show it. I'd like to port that query into a cross-site version but cross-site querying on SEDE is... tricky.
The current situation
Let me start of with two a priori observations:
Due to the way StackExchange is designed the easier a question is, the more reputation one will earn. This simply a consequence of the fact that more people will visit a simple question.
Simple questions have often been aske...
@ACuriousMind thanks - then I'd guess we would just need to delete questions more consequently, because if I look at my flagging history, I have the feeling that there are a lot of those questions which still gave people credit for answering them and which stay - thus effectively giving people a reason to answer the most stupid questions in hope of a bit of +rep
@Danu In any case, heavier downvotes would definitely affect the site in other ways that we do need to contemplate. If, say, it was -3 or -4 rep perdownvote, what happens to a well-intentioned rookie who gets a lot more rep erosion from some beginner mistakes?
@EmilioPisanty that choice is theirs - I just don't see the problem with telling them "your answer does not meet the quality standard, thus it's best to remove it and be more careful next time"
@Sanya Well, we can agree to disagree. Getting the hang of the site is hard for a lot of people, who then go on to become very valued community members. Making it harder means making it more likely that they'll just give up before it clicks.
@EmilioPisanty Yeah well, I wasn't trying to annoy you or sth., you certainly have more experience here than I do - it just doesn't go into my head, but if reality is like this, that's okay
@Sanya No, it's a subjective issue and open to opinions. For me it's easy to forget that I had a hard time getting the hang of the site, and it's always good to try to think through the experience of everyone that will be affected by any proposed change. You seem to have had a relatively easy time (going by your rep and join date) but other people take a bit longer - just saying, keep an eye out for those differences. Peace out =).
Do you know how to approach the problem of finding the work required to form a shell of radius $R$ and total charge $-q$ spread uniformly around a point charge $q$ at the origin?
@JohnDoe Are you seriously comparing the moderation of physics.SE, which is indeed more heavy-handed than that of math.SE, to a Nazi regime? Putting that in scare quotes doesn't actually make that less offensive, it just shows you're fully aware of how that sounds if taken at face value. The Nazi crimes are no laughing matter, and using them in comparisons so freely is both offensive to those that are being compared to Nazis and disrespectful of the victims.
The annoying thing about the move to trash is that it leaves the post visible (in the Trash room) and if creates a "moved to trash" message to advertise the fact. I would rather the offensive post just disappeared silently.
@acuriousmind What I mean is, we may have a specific definition of a particle by their mass/charge/spin; but could we change these properties thus changing the particle to a different one?
@Obliv That doesn't make much sense to me. What, exactly, do you mean by "changing"?
There certainly are interactions where an electron interacts with something and out comes a neutrino
But to say we "changed" the electron into a neutrino isn't really the right way to think about it. There is no continuity of identity - indeed, nothing to distingushing one electron from an other electron at all - that would support to say that any specific electron "turned" into a neutrino.
@acuriousmind perhaps there is no continuity of identity. But, then wouldn't that mean the electron just disappears and something else appears in its place?
@Obliv Because particles already appear and disappear around you all the time - every time light hits something, for instance: The photons disappear during absorbtion by atoms, and appear when they are emitted.
But there is nothing you could attach to a fundamental particle to give it "identity" - all fundamental particles are exactly the same, e.g. one electron emitted by a cathode ray behaves like every other electron emitted by it.
@JohnRennie Hi mister Rennie. Please don’t reply this message. I wouldn’t like to discuss with you (and anyone else) because I don’t have necessary instrument for discussion here (i.e. English language).
(Continue) I am going to leave the site for a long time. I am not posting this message for saying goodbye. I don’t say goodbye to anyone because I didn’t want to go. You dislodged me. Because you judged about me immediately and destroyed my validity. But I am not uncomfortable too much. Maybe it is better for all that I leave here. Because no one likes me.
(Continue) I am posting this message just because I want to say that you are mistaken about me. I confess that I made a mistake that day but not because of that that sentence was wrong or I am not right to have such impression. I made a mistake because I was not familiar with impression of people of a country about a person. Regardless of that mistake, I see no other mistake that I want to feel faulty.
(Continue) If I respect someone, this doesn’t mean that I confirm (or support or defense) all of his/her actions. I respect you. But I have problem with some of your actions. If I was you, I have left easy questions for poor users (like lucas) to answer. But you sometimes answers easy questions. In addition, I hate beard, but you have an awesome beard in your picture.
(Continue) I respect Gauss so too much more than you. But I always blame him because of his treatment with Abel. I greatly respect Evariste Galois much more than Descartes, Fermat, Pascal, Lagrange, Laplace, etc. or even more than Lazare Carnot. I love him. But if I was him, maybe I have not done what he did and finished his life.
(Continue) Long story short, I am too far from what you are thinking about me. I am a perfect anti-war. However I respect him (I don’t say his name because I think you get uncomfortable) so much and I believe that he was (and is and will be) a great man. (The end)
People always make grand announcements about wanting to leave the site but sometimes they just want attention, or something. I dunno. I'd just carry on with the prior discussion.
@Obliv They have, in some sense. Still, you couldn't tell if I "interchanged" the electron over there with the electron over here. You can't throw paint on an electron to track it in a mess of other electrons like you can with birds, is what I#m trying to say, and it is such tracking that would be needed so that I'd allow the notion of "changing the properties" to be meaningful.
I have a theory on the 5D and imaginary time, namely that when you are dreaming you enter the fifth dimension.
As portrayed in the sci-fi movie "Interstellar", the 5th dimension is shown as a world which you float in and you have no sense of time or physical space. When you are dreaming you don'...
I am a member of a Facebook group called, "the fifth dimension in practice" (it's in Dutch, sadly).
It's really amazing to see what kind of things a lot of people believe. Also interesting is to note how much they need each other's confirmation that whatever they are saying is actually correct.
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The current situation
Let me start of with two a priori observations:
Due to the way StackExchange is designed the easier a question is, the more reputation one will earn. This simply a consequence of the fact that more people will visit a simple question.
Simple questions have often been aske...
