The h Bar

12:19 AM

Anyone here know anything about the Heisenberg model?

8

Q: Is Heisenberg model or in its simplier form Ising model a good approximation to study magnetic systems?

Heisenberg model $$\hat{H}=-\sum_{\langle i j\rangle}J\hat{S}_i\hat{S}_j$$ And in its simplified version, the Ising model $$\hat{H}=-\sum_{\langle ij\rangle}J\hat{S}_i^z\hat{S}_j^z$$ are widely applied in the field of condensed matter to understand magnetic systems. For example, there is cla...

Bryan Krause

2:05 AM

Do y'all want this one here?: biology.stackexchange.com/questions/94532/… (or alternatively, have a dupe to suggest I link to if we close it? I wasn't able to find a dupe quickly but also didn't search very hard)

Captain Bohemian

4:11 AM

what does weak thermal contact in thermodynamics or statistical mechanics mean?

Captain Bohemian

5:31 AM

According to Wikipedia - stochastic process stochastic suggests conjecture while random suggests chance or luck and the first written appearance of the term random process pre-dates stochastic process, which the Oxford English Dictionary also gives as a synonym. So conjecture implies chance and luck.

does complete set of commuting observables mean the set of commuting observables which don't include any degenerate eigenstate?

Philip

5:51 AM

@CaptainBohemian I don't think it's got too much to do with degeneracy. A CSCO is the total number of physical observables that are required to completely specify a quantum state.

Since they all commute with each other, you can have a simultaneous eigenstate of all these observables.

Captain Bohemian

@Philip is completely specifying a quantum state just to distinguish the degenerate states?

Philip

For example, in the hydrogen atom, the CSCO is $\{H, L^2, L_z, S \}$. All these observables commute, a state of the hydrogen atom can be completely specified by these 4 numbers. $|n,l,m,s\rangle$. However, all these states are still degenerate! :)

@CaptainBohemian As you see above, degeneracy just means that the states have the same energy eigenvalue, while being distinct. A CSCO is necessary to define what it means to be distinct, I think.

user434058

What!?!?

Philip

@FakeMod Is that wrong? :(

user434058

@Philip I don't have a wife yet, so I dunno ;-)

JingleBells

6:04 AM

g00d m0rn!ng

user434058

Mornin'

Physics Meta

9:07 AM

0

Q: Why don't we improve captcha?

I just wrote a self-answered question on the main site. The answer is really long, and frankly speaking, it takes some time getting all the diagrams right, researching, making sure the explanations are lucid etc. So, I started yesterday evening (IST) and then I stopped for the night and continued...

bug captcha

Sha Vuklia

11:34 AM

guys, if you are given how long a particle collider works, and you are also given the cross section

is it then possible to determine the amount of scattering processes that occur?

Nzall

1:36 PM

I asked superuser.com/questions/1566469/… on SuperUser, and they redirected me here for part of the question (and Biology for the rest). I'm not sure what aspects of this question I can ask on here though

ACuriousMind

3:03 PM

@Nzall You could certainly ask about what decibels are and how they're measured here, but we don't really do things like "how loud can my fan be?" here, this would be considered off-topic as "engineering"

Nzall

@ACuriousMind honestly, it would surprise me if there wasn't already such a basic "what are decibels and how are they measured" question

ACuriousMind

@ShaVuklia You also need specifics about the beams, so that you know how many particles are incident on a unit surface per unit time to get that from the cross section, since the cross section is basically the probability of the scattering process divided by the number of particles per unit surface

JMac

@Nzall Might be worth it to try asking in like a computer hardware forum or something. It seems a bit subjective for a SE question. Now I'm not really active on SuperUser, but I would think if you trimmed most of the question, and basically just asked "How is the sound level of laptop fans measured?", I would think that's on topic as hardware question, especially if you find those measurements in computer specs.

Nzall

@JMac one of the mods already explicitly said that's off-topic

Because it's not related to laptops or computing hardware in general

Also because those measurements aren't generally part of computer specs, since they depend on what settings you're running the computer on and how much effort you're hoisting onto the device

Generally speaking, what happens is a review channel like Jarrod's Tech or Dave Lee does their own testing using combinations of fan profiles, OC settings and loads

JMac

3:18 PM

@Nzall I think part of it is how subjectively you're asking it. If you had fan specifications that listed decibels, it would seem fair to me to ask how they measure that. If you're just getting values from random sources though, and not giving them, there's not much SE can tell you about how they measure those values I guess.

@Nzall Do any of those channels have their own forums/discords/subreddits? Bringing up the discussion there could be fruitful, especially if the creators manage to chime in on how they took the measurements, or if any of the followers know.

Nzall

@JMac that might be a good idea

Charlie

4:51 PM

If I could ask a naive question that's been bugging me, a four-vector is a tensor, and so transforms as one under arbitrary coordinate transformations, yet they are usually highlighted as transforming a particular way under the lorentz transformations, I don't see why this distinction is necessary, since all tensors should transform in the same way under the lorentz transformations, no?

JingleBells

5:19 PM

Is it okay to leave a website with for example www.example.com/test.php? Instead of www.example.com/test where there's a folder test with an index.php there? Which is better for search engine optimization (SEO)?

ACuriousMind

@Charlie the distinction is not necessary

Charlie

Someone in another SE chat would almost certainly be able to answer that @JingleBells

ACuriousMind

but the context in which "four-vectors" are usually introduced, i.e. basic special relativity, is usually one where the student may not yet be familiar with the notion of general coordinate transformations

Charlie

@ACuriousMind Ok glad I'm not going crazy

ahh I see

bolbteppa

5:36 PM

This is a bit confusing, you can't do a GR transformation in special relativity Minkowski space right

Four-vector

In special relativity, a four-vector (also known as a 4-vector) is an object with four components, which transform in a specific way under Lorentz transformation. Specifically, a four-vector is an element of a four-dimensional vector space considered as a representation space of the standard representation of the Lorentz group, the (½,½) representation. It differs from a Euclidean vector in how its magnitude is determined. The transformations that preserve this magnitude are the Lorentz transformations, which include spatial rotations and boosts (a change by a constant velocity to another inertial...

The wiki talks about SR vs GR four-vectors a bit

ACuriousMind

@bolbteppa What do you mean? I can certainly treat Minkowski space just as the spacetime $(\mathbb{R}^4,\eta)$ and do everything on it I can do in general GR

bolbteppa

The article talks about different bases, $A = (A_t,A_x,A_y,A_z) = A_t \mathbf{E}_t + A_t \mathbf{E}_x + ...$ where $\mathbf{E}_t = (1,0,0,0)^T$ etc... but also talks about e.g. a spherical basis $A = A_t \mathbf{E}_t +A_r \mathbf{E}_r + ..$ but if you take a derivative of the vector $A$ with respect to $r$ in this basis you're going to have to differentiate the basis vectors too and so you immediately get covariant derivatives... in special relativity...

ACuriousMind

You'll have Christoffels appearing in some places etc., but the four-vectors are still four-vectors

But you're right, I guess my "the distinction is not necessary" was too strong, @Charlie

There are quantities which are "four-vectors" under Lorentz transformations but not under general coordinate transformations, namely those involving derivatives

Charlie

but they're only "not four vectors under general transformations" because we're using a non-tensorial derivative right?

bolbteppa

I think it's that you restrict the group of transformations to be just the Lorentz group in SR but in GR it's a more general diffeomorphism group that the vectors transform under i.e. live in a representation of

Charlie

5:44 PM

when you say something is "not a four-vector", this by extension means it's not a tensor right?

ACuriousMind

@Charlie Yes, if you replace every derivative by the covariant one then it works out again, and since the covariant derivative and the usual derivative agree on flat space for the frames Lorentz transformations transform this won't go wrong, but it's still good to keep in mind that you have to do that

bolbteppa

The electromagnetic field tensor is not a four-vector, e.g. higher rank tensors

ACuriousMind

@Charlie Yes

Charlie

but it is a "four-tensor" right?

bolbteppa

Yeah

Charlie

5:45 PM

@ACuriousMind Ok I get that

does the regular $\partial_\mu$ derivative generally only work on spacetimes with vanishing curvature tensor then?

bolbteppa

2

Q: Coordinate transformation in the context of special relativity

I was wondering how can one change from Cartesian coordinate system to some other like polar coordinates or spherical coordinates, in the context of special relativity. For example, with the four-velocity, $$V^{\mu}=\frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau},\tag{1}$$ where $\mu=0,1,2,3$, $x^0=ct...

special-relativity lagrangian-formalism spacetime metric-tensor coordinate-systems

Basically if you even go to spherical coordinates it already looks like general relativity, the only thing is that the Ricci scalar is zero or whatever, but it's basically already GR and I definitely don't fully appreciate the subtlety between them

Charlie

I thought that spherical coordinates needed the covariant derivative because the basis vectors depend on radial distance

even if you're using them in a space with no cuvature

I am but a humble noob though

ACuriousMind

The thing is that the Christoffels aren't really about curvature, they're about how the metric changes with respect to the basis you've chosen. You can get non-zero Christoffels by choosing a basis that varies (the spherical basis vectors "change direction and length" between different points), or by having the metric vary (i.e. non-flat spacetime)

Charlie

ahh

bolbteppa

Ah I remember, going to e.g. spherical coordinates is just going from an inertial frame in SR to a non-inertial frame hence it's no wonder covariant derivatives can appear, same thing even happens in Newtonian mechanics, but this transformation can always be reversed, and the reference frame can move to infinity in principle, while for a GR field you can't always (globally) diagonalise the metric and the GR field has to go to zero at infinity

bolbteppa

6:31 PM

The fact that there's no difference between SR in a non-inertial frame and GR is one way the principle of equivalence is sometimes stated, so going to spherical coordinates in SR is basically GR in that sense but because you can simply invert the spherical coordinate transformation to get back to Minkowski space it's just non-inertial SR. In GR you can't diagonalise a metric 'globally', even though you can do it at each individual point, have to say I don't fully appreciate this yet

Philip

Has anyone here made an N body simulation on Python with large N (say around 100 or so?)

Physics Meta

7:15 PM

0

Q: Should 'newbie' questions be allowed to be deleted by the contributor without oversight?

'Vets' will probably know where this is leading to as most of us have experienced it at some point (it's not my first time!) Yesterday a new contributor posed a well-formed question (about a steady-state heat conduction problem). Some members and myself made some decent comments. We got no reply....

feature-request

Charlie

well I just paid the fine for treating tensors as matrices

that's 2 hours I'll never get back

Physics Meta

7:39 PM

0

Q: Care over closing "duplicate" questions

On a number of occasions, I have seen questions closed as duplicates although the linked question either did not have answers, or had only poor answers, and in some cases it had already been closed (often for quite different reasons). In such cases, I think that to close a question on the grounds...

discussion close-reasons vote-to-close

ACuriousMind

9:35 PM

@Charlie that's getting off cheap, buster!

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