The h Bar

0celo7

9:02 PM

@ACuriousMind What does this mean if the Hamiltonian is zero?

Kyle Kanos

@0celo7 That your cosmology has issues?

bolbteppa

It looks right to me

0celo7

@bolbteppa To me too. What does the book say are the consequences of this?

ACuriousMind

@0celo7 Well, it's an on-shell law, so it means that your theory is classically pretty boring

0celo7

@ACuriousMind Free particles are boring, I'll give you that.

ACuriousMind

9:04 PM

Yeah

Kinda the point here :D

I dunno what it means, you just have to keep it as a constraint just like any other e.o.m when doing quantization

bolbteppa

"we see tht $H_{can}$ vanishes identically. Thus, the constraint ($p^2+m^2=0$) governs the dynamics of the system. This is always true in theories with a (local) time reparametrization invariance." But they then derive a constrained Hamiltonian. Ahhh it's kind of making sense now...

ACuriousMind

Ah, I know what it means for the string, though

It gives you the mass-shell condition in terms of the modes.

But any other (reasonably solvable) constraint would do the same, so it doesn't do anything special, I guess.

0celo7

@ACuriousMind Oh, I could've told you that.

bolbteppa

Here's my guess - if you look in Lanczos' Variational Principles of Mechanics book he talks about how Hertz constructed a form of Lagrangian mechanics based exclusively on Kinetic Energy and it looks the exact same as the special relativity lagrangian, in classical mechanics it's equivalent to the normal formulation, but it eliminates potential energy, so I guess it's basically constrained mechanics, that's why they are pulling out constraints.

0celo7

p. 39 in BBS.

ACuriousMind

9:08 PM

@0celo7 Good, then you know as much as I about the significance of this :D

bolbteppa

Wow I think the Polyakov formulation of removing the square root is actually just a modern form of reversing Hertz' idea

0celo7

@ACuriousMind BBS is really confusing because they never tell you when stuff is on-shell and when it isn't. Like they say $T=0$ and then work with $T\ne 0$ for 20 pages.

ACuriousMind

That's usual for physicists

That's why Qmechanic always uses $\approx$ for on-shell and $=$ for off-shell identities

I wish more people would do that

0celo7

@ACuriousMind Speaking of shells, what does Qmechanic mean by "$\approx$ is modulo EOM"?

ACuriousMind

Heh

0celo7

9:10 PM

Wow

Why "modulo" though?

So $p^2\approx -m^2$?

And $G_{\mu\nu}\approx 8\pi T_{\mu\nu}$?

Stan Shunpike

Like, I feel like I understand pieces of my QFT book but not how they fit together. In particular, I roughly understand the QED Lagrangian I think, but can someone explain where propogators come into play? I thought a propagator was something that gave me the amplitude for a particle at one position and time to appear at another position and time.

Is that used in the QED Lagrangian?

ACuriousMind

@0celo7 Well, it's mathspeak. Do you know that $\mathbb{Z}/n\mathbb{Z}$ is also called $\mathbb{Z}$ modulo $n\mathbb{Z}$?

0celo7

@StanShunpike The propagator is what you get when you quantize the kinetic term, roughly.

ACuriousMind

@0celo7 Wait. Isn't that the equation of motion?

0celo7

@ACuriousMind The integers with every $n$-th one identified?

@ACuriousMind I'm confused by the notation.

ACuriousMind

9:13 PM

@0celo7 Right, so $4 = 12$ modulo $8$

0celo7

Yeah I get that.

Do we note use $\approx$ in the EOM?

Stan Shunpike

@0celo7 It isn't quantized already?

0celo7

That's an on-shell identity.

@StanShunpike The Lagrangian is a classical object.

bolbteppa

"The fact that only 3 of the 4 components of the generalized 4-momentum are independent reflects the 3 degrees of freedom of a particle. The independence of the 4 functions does not yield 4 physical degrees of freedom, because of the freedom in choosing the parameter" - Gourgoulhon Special Relativity P. 372

ACuriousMind

@0celo7 The idea is now that you have an abstract "space of objects" of your theory, like $T$ and $0$, and they're not the same. But, if you use the EOM, then you can show that $T=0$, so, in the space of objects "modulo EOM" $T$, and $0$ are the same object, just like $4$ and $12$ are the same "modulo the EOM $8=0$".

Stan Shunpike

9:16 PM

@0celo7 Okay, didn't realize that. No wonder I'm confused. So are you saying that if we quantize the kinetic term using....field operators?....then we get a propagator?

ACuriousMind

Am I right or am I absolutely butchering this here, @Qmechanic?

0celo7

@StanShunpike You can quantize canonically or with path integrals. @ACuriousMind Is what I'm saying correct?

@ACuriousMind I see.

ACuriousMind

@0celo7 That's the first choice you have to make when quantizing, yes

bolbteppa

Okay that's correct, wow...

0celo7

Do I write the EOM with $=$ or $\approx$? The EOM are on-shell after all.

@ACuriousMind I knew that was right. I meant the stuff before that.

About the Lagrangian being classical and propagators coming from kinetic terms.

ACuriousMind

9:18 PM

@0celo7 Well, they define the shell, I wouldn't write them with the modulo sign, I think

Though, technically, it would be of course correct

0celo7

@StanShunpike Also the mass terms. The kinetic and mass terms make the propagator.

Qmechanic

@ACuriousMind : Nah, your explanations sound right to me.

0celo7

@Qmechanic Do you write $\approx$ for the EOM? Why or why not?

ACuriousMind

@0celo7 Uh, what of that? "The propagator is what you get when you quantize the kinetic term"?

0celo7

@ACuriousMind That is correct, right?

ACuriousMind

9:19 PM

I'm not sure what that is supposed to mean

I usually don't quantize terms, I quantize theories

0celo7

@ACuriousMind If you have a free field theory, and you quantize it, you get only propagators.

Stan Shunpike

By theory, do you mean the entire Lagrangian?

ACuriousMind

Ah, yes, that is correct - the thing we call the propagator comes from the free theory, usually

0celo7

In a sense by quantizing the free parts (kinetic and mass terms), you get the propagator.

bolbteppa

It is possible to write a lagrangian for special relativity exactly as you do in classical mechanics and when you take the Legendre transform you get the constrained Hamiltonian they use in string theory, weird...

Qmechanic

9:22 PM

@0celo7 The equivalence relation f \approx g <=> f-g is zero when using the EOM.

ACuriousMind

@StanShunpike If I'm given one, yes. But, e.g. quantization of a 2D CFT can be done without an action/Lagrangian.

0celo7

@Qmechanic I don't know exactly what you mean. How would you write the Einstein Field Equations: with $=$ or $\approx$?

Qmechanic

@0celo7 With $\approx$.

0celo7

@ACuriousMind All you need are the OPEs, right?

Stan Shunpike

@ACuriousMind What do you quantize then? Are there equations of motion?

ACuriousMind

9:24 PM

@0celo7 Yeah, that's one way to say it

@StanShunpike Well, you know the theory has an infinite-dimensional symmetry - the Virasoro algebra. You can get very far just by doing representation theory of it

And modelling a specific CFT instead of a generic is then just a question of selecting the representations that actually occur in your theory

Stan Shunpike

@ACuriousMind That's awesome.

bolbteppa

Nothing makes any sense anymore...

ACuriousMind

Where you still need to do a lot of computation, but essentially, the whole dynamics of the theory are fixed by the symmetry

Stan Shunpike

Question: I am reading about Wick's theorem on Wikipedia and they have this notation : :

What does that mean?

ACuriousMind

@StanShunpike Normal ordering

Stan Shunpike

9:26 PM

Ah, okay

thanks

tpg2114

Okay, this is awesome:

grindtv.com/lifestyle/in-style/post/…

3

I want to see slushy waves, that sounds crazy

ACuriousMind

@KyleKanos Well, that's a start, but not that helpful because that's about someone who had his own account hacked. I'm just some random guy who thinks another user's been hacked.

I probably should stop worrying about this, but I like(d) NikolajK.

Stan Shunpike

9:42 PM

@tpg2114 That's one of the coolest things I've ever seen.

It reminds me of the slow-motion volcanoes on Io

tpg2114

@ACuriousMind The profile thing looks suspicious, but I'm not sure I'd call the question he asked a homework question (although it is not a question that I would expect from that user)

I actually spent a few hours digging through books and trying to work out the math on that question and the things that you could do if it were homework don't hold up in that configuration

@StanShunpike It's really awesome looking, but I also wouldn't want to brave the ridiculously cold weather needed to see it in person

I'll stay in the south where it's only almost ridiculously cold.

ACuriousMind

@tpg2114 Well, he also removed my homework-tagging without any comment, and has not replied. I'd expect a user that's been here that long to actually state why this is not a trivial/HW-like problem in the question.

So, if it's a hacker, it's at least a hacker with an interesting question, eh?

Kyle Kanos

@ACuriousMind I know the circumstances are different.

But maybe linking that Meta page to one of Nikolaj's posts might be useful in case it was hacked?

tpg2114

@ACuriousMind Possibly, but I don't really make expectations about what people know/don't know anymore. All I can say is the only compelling evidence is the profile. The rest of it, who knows?

I didn't think it was at all homework when I first saw it. I thought "finally, an experimental set up question"

I still don't know how it got 3 close votes as HW like. I can't figure out why it would be, but until it closes for that reason there's not much motivation to question it

ACuriousMind

Well, it throws a bunch of numbers at me, and it asks for the calculation of values that also typically occur in HW problems

David Z

9:54 PM

@StanShunpike yes, it was popular enough on social media sites that I expected a question would pop up here. For a few hours yesterday, something like half of all submissions to reddit's /r/askscience were about that picture.

tpg2114

And experimental science

ACuriousMind

@tpg2114 And I would expect an experimental scientist to demonstrate that this is not as easy as a HW problem, just as I expect the theoretical questions that are just about calculating something to demonstrate this is actually tricky and interesting

I also close theoretical questions that just throw formulae at me and ask how to get a certain quantity out of them

tpg2114

I can see that, but as an "expert in the field," I looked at it immediately and knew it was interesting and challenging so it didn't need any disclaimers to me

Which is why I don't even look at most of the questions on the site, I can't possibly evaluate if they are good/bad because I have little to no exposure to that field

ACuriousMind

@tpg2114 So, the answer to your question is that three (uh, two) people with my mindset looked at it that were not as well-versed in the field as you.

tpg2114

That conclusion seems to offend you but that's not what I'm saying. What I am saying though is that I don't think every question needs a disclaimer explaining why it is interesting/challenging/complicated/etc

ACuriousMind

9:59 PM

Well, surely not, but if they haven't one, they need to be prepared to go through the closing/reopening cycle.

Because after closure, surely someone would have said precisely what you just said to me, and voted to reopen

tpg2114

Sure, which is what I would do if it ends up closed. But with only 3 votes, it would be premature I think

ACuriousMind

And I'm not offended, I am actually quite clueless about thermodynamics/fluid dynamics/whatever this is

I retracted my vote, btw.

tpg2114

But all of this is tangential to the original point I was making -- that question isn't inherently a sign that the account was compromised :) The profile, however, is questionable

user54412

@tpg2114 Which just makes this all stranger

Kyle Kanos

Could also be he left it open and some friend messed around with him

tpg2114

10:02 PM

That question won't really be able to be answered here though. It's too complex of a setup for an analytical expression so little short of experimental measurements/simulation in that or a similar configuration will address it

wtf

Bad Copy-Pasta?

Hah

He just made a comment "I have a bad case of not dreaming" to me

And then immediately deleted it

This is not really doing anything to make this feel less creepy

That's why the copy-pasta was bad, the comment link was dead and went to the question itself

tpg2114

My advice? Tap out of it. You reported it to the mods, they escalated it to the team, there's not much you can do anyway

If it is compromised or trolling, engaging that account will just be messy, and if it's hacked could lead to us having this same conversation about you :)

ACuriousMind

10:07 PM

Yeah, it will be better for my sanity if I ignore it, probably

Oh, well, as long as I don't have any real problems in my life :)

user54412

@ACuriousMind That could be said of much of life, you know.

tpg2114

@ChrisWhite And almost all of the internet

ACuriousMind

Wait, there's a difference? :P

0celo7

10:20 PM

@ACuriousMind Meh. A bit formal. physics.stackexchange.com/questions/167466/…

Stan Shunpike

10:46 PM

@ACuriousMind going back to the quantizing and Lagrangian stuff, for gravity then, is the classical action the Einstein-Hilbert action and one then tries to quantize through either canonical methods or path integral?

ACuriousMind

@StanShunpike Yeah, I think so. I've never really cared for the details because even if you manage to quantize consistently, that stuff is non-renormalizable, anyway, and hence not even a good fundamental theory if you don't run into problems along the way.

0celo7

@ACuriousMind Thoughts on that post?

ACuriousMind

...reading...

@0celo7 It's a good answer to the question, I think.

0celo7

@ACuriousMind That piece with the characteristic functions makes sense? (Like I said, a little formal.)

ACuriousMind

@0celo7 Oh, to me it does. You might want to link Wiki to it or something, perhaps

I always try to link explanations of the technical terms I use if the question is not obviously at a level where you can presuppose them

Stan Shunpike

10:55 PM

Is there any compelling evidence for supersymmetry?

0celo7

@StanShunpike Nope.

Stan Shunpike

Not even indirect?

0celo7

@ACuriousMind Should I link "support" or just make a footnote?

user54412

@StanShunpike This was the dream in the early 60s, with the ADM splitting of spacetime. They figured out what the canonical momentum conjugate to the metric was. Everyone was just thinking "all we do is promote these quantities to operators, and we have quantum gravity."

0celo7

Made a footnote. Easy enough to explain.

user54412

10:58 PM

The formulation was much more successful at allowing numerical relativity (evolving spacetime dynamically), but even that took 40 years to do after the equations were written down.

tpg2114

@ACuriousMind On your edit to that : question, you forgot the original had a ^2 on it :) I edited it back in

infinitesimal

@DavidZ could you please permanently pin this already starred comment as the most innocent comment I've ever read :) chat.stackexchange.com/transcript/message/20287615#20287615

ACuriousMind

@0celo7 Well, if you're having fun explaining things, footnotes are better, of course

@tpg2114 Oh, damn. Good catch

0celo7

Yay footnotes!

tpg2114

@0celo7 I'm glad you're learning to change your posts before I recommend it :)

@ACuriousMind So is there a general symbol for tensor products? Like we wanted to do a triple or quadruple inner product on third/fourth rank tensors... There's no symbol on my keyboard with 3 or 4 vertical dots

0celo7

11:04 PM

@tpg2114 I have voiced my opinion, no need for it to stay here.

@tpg2114 $C$ for contraction.

tpg2114

@0celo7 As in $C(\mathbf{A},\mathbf{B})$ like a function? Or an operation like $\mathbf{A}C\mathbf{B}$?

0celo7

@tpg2114 The former.

DanielSank

Guys, what's the charged particle momentum operator in nonrelativistic quantum in SI dimensions?

tpg2114

Interesting. I'm glad I hardly ever need to work with higher than rank 2

0celo7

@DanielSank Newtons x seconds

DanielSank

11:06 PM

@0celo7: Uh, thanks.

:|

0celo7

@tpg2114 Do engineers not use index notation?

@DanielSank Holy crap. I read that backwards.

tpg2114

Levi-Civita symbol is usually the most complicated I get. Although I do have to deal with that rank 4 stiffness tensor now that I'm doing structures

@0celo7 I prefer it greatly, I don't like working in vector notation at all

But in fluids and structures, it's a mixed bag what you get. The turbulence community usually goes with index notation

DanielSank

@0celo7 What does it look like backwards?

0celo7

@DanielSank I thought you were asking for the SI units of momentum.

DanielSank

Ah.

Actually, your answer is helpful because it reminded me that I can figure it out myself.

Sorry for being lazy and stupid.

By dimensional analysis it's $q \vec{A}$.

0celo7

11:09 PM

@DanielSank What are the units of $A$ again?

I don't know any of this stuff with constants in it.

@tpg2114 If you want new notation, simply define $(A,B)$ as the contraction of $A$ and $B$ over all indices.

user54412

@tpg2114 As always, you might want to take his opinions with a grain of salt, but I somewhat understand Ron's argument for indices here:

user54412

23

A: Mathematically-oriented Treatment of General Relativity

The Physiccs work in this field is rigorous enough. Hawking and Ellis is a standard reference, and it is perfectly fine in terms of rigor. Digression on notation If you have a tensor contraction of some sort of moderate complexity, for example: $$ K_{rq} = F_{ij}^{kj} G_{prs}^i H^{sp}_{kq}$$ ...

DanielSank

@0celo7: The way I figure it is that $B = \nabla A$, so $A$ has units of magnetic field times length.

tpg2114

@0celo7 Yeah, I knew one could just say "And the inner product (a,b) is..." but I didn't know if there was a standard shorthand like \cdot or :

user54412

Trying to extend things like . and : will lead to notational clumsiness down the road

0celo7

11:12 PM

@DanielSank I was thinking that, but then I realized I had no clue if that equation really has a constant in it that we set equal to 1 :/

DanielSank

So then if I guess that $p = qA$ I can check that as [charge][length][magnetic field].

I type 1 coulomb * 1 meter * 1 Tesla into Google and Bob's your Uncle.

tpg2114

@ChrisWhite I could see that. Again, super happy I only deal with rank 2 tensors! And I'm also happy my tensors only have indexes in one position and not more cause I don't have any idea what that means

DanielSank

@0celo7 Yeah, this is why I absolutely hate it when people say that they are e.g. "setting hbar to 1".

You are never setting $\hbar$ to 1. you are redefining all your variables to have different dimensions.

If you "set $\hbar$ to 1" you should really think of e.g. Schrodinger's equation like this:

$i (d/dt)|\Psi\rangle = (H/\hbar)|\Psi\rangle$.

In other words, your Hamiltonian is now a different thing (which I like to denote as $\Omega$) which has dimensions of frequency.

/rant

user54412

@tpg2114 It's essentially an artifact of the metric not being the identity matrix. In SR there's a possible sign change, between different orthogonal coordinate systems there are rescalings, and in GR everything changes between upper and lower indices.

0celo7

@tpg2114 You know that a vector transforms with the Jacobian under a coordinate transformation. A covector transforms with the inverse Jacobian and carries a lower index.

tpg2114

11:18 PM

I can't think of a situation where I would ever need that in my work... Even when we do our mappings from physical to computational spaces

0celo7

@tpg2114 en.wikipedia.org/wiki/Geometrothermodynamics

I have no idea what use that has, but engineers use thermo, right?

user54412

As long as your coordinates are orthonormal there won't ever be a difference.

tpg2114

@0celo7 We do, but we are generally just operating in relatively simple coordinate systems and rarely changing between them

user54412

my adviser takes spherical coordinates to be orthonormal, whereas I don't, so my indices matter, but I don't have any weird $r^2 \sin\theta$ factors in my equations

0celo7

@tpg2114 You can use metrics as a powerful tool to change coordinates in your derivative operators.

tpg2114

11:22 PM

@0celo7 We do that all the time computationally. We map our physical grid to a uniform computational one

0celo7

It's probably better than plugging and chugging with the transformation laws.

tpg2114

And we end up with contravariant velocity vectors and gradients

0celo7

@ChrisWhite He's just using a vielbein approach, right?

tpg2114

@ChrisWhite Can't your make any set of orthogonal vectors orthonormal?

0celo7

@tpg2114 Yes, by scaling.

tpg2114

11:24 PM

Right -- so if you can, why not just scale everything and then not worry about the whole upper/lower indices?

user54412

@0celo7 I suppose that's one way of putting it.

0celo7

@tpg2114 In $E^n$, yes, you can.

tpg2114

I guess my question is -- why choose to work in non-normalized coordinate systems?

user54412

@tpg2114 Everyone but the purest mathematicians probably would go that way, except us GR folk have non-diagonal metrics

user54412

We can't globally diagonalize them in general

0celo7

11:25 PM

We can locally, this is called Sylvester's theorem. (Or perhaps a corrolary thereof.)

user54412

Sylvester..? Never heard of him. Always called it a version of the spectral theorem myself.

user54412

Or better yet, the equivalence principle

Stan Shunpike

@ChrisWhite So supersymmetry has no evidence but because so many theories have been created based on it, people think it has merit?

0celo7

@tpg2114 There is an approach to GR called the tetrad formalism which does this.

@ChrisWhite en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia

I misrecalled. This theorem just says that the signature of the metric does not change.

user54412

@tpg2114 In the end, I could always normalize my vectors, but often I can't even orthogonalize them (without making all my other equations suck). Alas.

0celo7

11:29 PM

@ChrisWhite It is always possible to diagonalize, is it not?

user54412

@StanShunpike Did you mean to address ACuriousMind?

tpg2114

@0celo7 Maybe in theory, but in actual practice no

0celo7

@StanShunpike Pretty much.

tpg2114

If you have 5 or more eigenvalues, their exact values are impossible to determine

0celo7

@tpg2114 GR is in 4 dimensions :)

user54412

11:30 PM

@0celo7 Sure, at a point. But doing GR at a point isn't very interesting. Ultimately you need some way to connect tangent spaces at different points, whether that's through some weirdly-changing set of bases or through a non-diagonal metric is your choice.

Stan Shunpike

@ChrisWhite no I read something you posted half an hour ago

0celo7

@ChrisWhite I was talking about changing bases. You can diagonalize with a changing basis.

Stan Shunpike

@0celo7 okay thanks for clarifying lol

user54412

I know next to nothing about supersymmetry.

0celo7

@tpg2114 What we do is write $$g=\eta_{\mu\nu}\theta^\mu\otimes \theta^\nu$$

The basis $\{\theta^\mu\}$ is now variable.

user54412

11:32 PM

I did take a particle physics course where the prof swore that by the end we too would be SUSY believers. It didn't quite happen.

0celo7

This has some merits.

tpg2114

We definitely took a "Only learn enough about tensors that you need to make it through the fluid dynamics class" approach in the way we were taught this stuff

0celo7

@tpg2114 You've been cheated.

tpg2114

Or just very narrowly focused...

user54412

People keep talking about the merits of tetrads and vielbeins, but I've never seen them actually used.

0celo7

11:35 PM

@ChrisWhite Many people swear by the Cartan method of calculating curvature.

user54412

In some sense I'm actually going into such bases in the code I'm writing, but I just think about it as temporarily using a weird basis.

user54412

@0celo7 I think the difference is between textbook problems (where there's always more than enough symmetry) and numerical problems like I work on.

user54412

In code, simpler formalisms can be better than simpler arithmetic

0celo7

@ChrisWhite The difference is you have Mathematica (or equivalent) and textbooks don't.

Zee mentions in his vielbein chapter that this is pretty much useless now that we have CASs.

user54412

lol -- my suspicions confirmed

0celo7

11:39 PM

@ChrisWhite If of course still has merit when trying to solve a novel problem analytically.

user54412

Ah, analytic solutions! I believed in those once, in my youth.

4

tpg2114

Right up there with unicorns and pots of gold at the end of rainbows

user54412

Not to mention smart compilers and erasable whiteboards

0celo7

:(

tpg2114

Well, on the plus side, if you can find analytical solutions to problems that are actually of real interest to somebody

user54412

11:45 PM

Perhaps we shouldn't depress the young ones so much. They too deserve to go through the better part of college in bliss.

tpg2114

You will become very rich and famous

So it's a noble pursuit. But it's also one that may lead to an entire life spent in vain

0celo7

1

Q: Are empty black holes possible

When just considering GR without evaporation nor QM, is an empty (containing no matter or anything) black hole possible ? Let's say that there is only GR and nothing else (no matter or boson fields), and that at time t in some coordinate the metric is a black hole, how will it "evolve" ?

What?

user54412

This on the other hand is rather interesting:

user54412

2

Q: Why do we define the magnetic field around a conducting rod as concentric instead of radial?

Why is it that we define the magnetic field around a conducting rod as concentric instead of being radially outwards? Ok, we know from the Lorentz force law, that the magnetic force is the cross product of velocity and magnetic field, but we might as well have defined magnetic force to be the do...

electromagnetism magnetic-fields

user54412

I feel it could be answered a number of ways, like the Helmholtz theorem or something about differential forms

user54412

11:49 PM

and on that note, I'm currently trying to figure out the best way to plot magnetic fields from my simulations

0celo7

Is he asking about field lines?

tpg2114

@ChrisWhite I like plotting things... what exactly do you need to plot?

user54412

field lines? vectors? pseudocolor?

tpg2114

2D? 3D?

user54412

@tpg2114 So, I have MHD on a 2D plane: density, pressure, 3 velocity components, 3 magnetic field components

tpg2114

11:51 PM

Any other variables that you need to show at the same time?

user54412

but let's say I'm just interested in density, Bx, and By

user54412

It would be nice to do a density plot of density, and have magnetic field overlayed

tpg2114

Yeah, that's what I would recommend. A filled contour plot of density with the field lines

As contour lines

Or maybe as vector arrows along the field lines. You could color the vectors by something else if you needed to

Like pressure

Or whatever might actually matter along the field lines

user54412

what about magnetic field strength? should I try to visualize that?

user54412

if plotting field lines I could try for the density of the lines (but that seems difficult to coax plotting packages to do)

tpg2114

11:54 PM

Doesn't the spacing between the lines tell you the strength?

Or does it tell you something like magnetic flux

user54412

In textbooks, but I feel like most software will try to go for roughly uniform spacing

user54412

maybe not though

tpg2114

No, the values are what matters

If you plot the contours with equal value spacing, they will end up not equal in space

If they plot with equal spacing in space, you will get non-equal values for the contours

But I know for things like Paraview (and even matplotlib) you can specify the contour values to plot

So you could do arbitrary spacing, or linear/log spacing between two bounds, or whatever else you want

user54412

hmm, I think we're thinking of different things

tpg2114

Possibly...

I'm thinking of something like:

user54412

11:57 PM

I was thinking of field lines like this

user54412

tpg2114

(Side note -- we really need SVG support on stack exchange)

So streamlines

user54412

Contours would be perpendicular to this, right?

ACuriousMind

@ChrisWhite I think it depends very much on what you want to do. If you need to calculate something, then indices are very much the way to go. I find the index notation absolutely horrible to see what is actually going on in the equation, though, and definitions like "a tensor is a thing with components that transform like this" absolutely unenlightening.

user54412

put another way, if my field were Bx=1, By=0 everywhere, what would a contour plot of the field look like in your mind?

NeuroFuzzy

11:59 PM

@ChrisWhite if you could do contours that didn't intersect wouldn't mean the B field is a gradient?

tpg2114

It wouldn't exist because there's no gradients

I was thinking of something like:

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