12:47 AM
Why is the gradient of the scalar potential $U$ choosen with a negative sign?

1:08 AM
Why does it follow from $\nabla \cdot \mathbf A =0$ that the polarization is transverse to the propagation?

2:03 AM
One way to see the minus sign is to consider the electrostatic case where the minus sign naturally arises in looking at Coulomb's law, with $F = qq'/r^2$ (Gaussian units), the work done by this force from $r_1$ to $r_2$ is $W = \int_{r_1}^{r_2} qq'\frac{dr}{r^2} = qq'(\frac{1}{r_1} - \frac{1}{r_2})$ so if you start at $\infty$ then $W = - qq'/r_2$.
The negative of this, $U(r_2) = - W = + qq'/r_2$, represents the potential work that will be done from a starting point taken conventionally as $\infty$, so we call $U(r_2)$ the potential at $r_2$.

2:14 AM
For the second one, note in the Fourier expansion of $\mathbf{A}(t,\mathbf{r}) = \sum_{\mathbf{k}} \mathbf{A}_{\mathbf{k}} e^{i (\mathbf{k} \cdot \mathbf{r} - \omega t)}$ that $\mathbf{k}$ determines the direction the wave moves in, it acts like a velocity in the direction of motion as you can see from the units alone or an argument like this.
Thus since $\nabla \cdot \mathbf{A} = 0$ implies $\mathbf{k} \cdot \mathbf{A}_{\mathbf{k}} = 0$ we see each $\mathbf{A}_{\mathbf{k}}$ is spanned by two vectors orthogonal to a basis vector pointing in the direction of $\mathbf{k}$ which is the axis of propagation of this mode

2 hours later…
4:24 AM
Hi All....

4:45 AM
hello
#thursdaymorning #ThursdayMotivation

Motivation for #Engineering .

Best Engineers who changed the world!
"Science can amuse and fascinate us all, but it is engineering that changes the world" - Issac Asimov

5:15 AM
didn't you just tweet that

yep
just tweeted
@NiharKarve yu watchin' my tweets?
I'm still worthy.JPG

7 hours later…
12:37 PM
Do you think this question could be tweaked to be Physics appropriate?
1

So this started out a question related to a Dungeons and Dragons session I had once, where I was playing a druid and used the spell Bones of the Earth and the fate of a Purple Worm, now known affectionately as Chad. Our DM of the session has a penchant for the Rule of Cool, and tends to ignore ce...

One of our members has already started some crude calculations:
Off topic, but.... For a rock with a density of 3000 kg/cubic meter, the mass of each column is about 2×10^6 kg. If the rocks are subject to constant acceleration before magically stopping, and the whole column travels 30 feet (about 10 meters) in less than a single round (6 seconds), then d = 0.5*a*t^2 + v0*t, or 10 = 18*a, and a is approximately 0.5 meters per second squared. Then the force on each column is about 10^6 newtons. It exerts this force on the purple work for 30 feet, until it magically comes to a stop, imparting 10^7 joules of energy per column. — Adamant yesterday
so I'm guessing it's possible, but not sure if it would be welcome?

@AncientSwordRage It's too vague currently to be about physics, but anything more precise will turn it into something just asking for a calculation like that comment does, so it would be likely off-topic under our homework-and-exercises policy

@ACuriousMind hmmm, there's no middle ground?
Because it's possible that the OP doesn't know what equation to use

@AncientSwordRage You're asking for the force, in newtons, produced by a spell?
2

@AncientSwordRage Yes, but "What equation do I have to use to compute the result?" isn't sufficient to make it on-topic

12:48 PM
We're a bit stricter than many other sites - not every question that is arguably about physics is on-topic here
I don't really see a way to turn this into a question that would be well-received here

but, we do argue :-)

I agree with ACM. First off, if you take away the physics, it's just asking for a calculation (OT via homework).
But besides that, there just isn't enough information there to produce a physically plausible question
> the spell moves six cylinders of rock 5ft wide from ground level to 30 ft tall instantaneously and how much force that would produce (imagine a piston moving that fast)
regardless of the "spell" context, that's already not physics

well, the answer clearly is that it produces infinite force and anything hit by the cylinders should be atomized, not launched into space ;)

you can't judge force without knowing the time it takes for the change in velocity (starting and stopping)

1:05 PM

0

I get that this is the physical limit - an infinite mass of Uranium would produce a runaway reaction. But an infinite mass is impossible.. Really, what is the minimum % of enrichment of Uranium, that when stacked into a mass, runs away before becoming a black hole?

oooh, boy
shy→why?
but then
> runs away before becoming a black hole?
any thoughts about how to handle this?

1:28 PM
@EmilioPisanty the answer is obviously (100+infinitessimal)% enrichment
meaning it's impossible. I'd just ask for a clarification. If they feel there is a minimum, then there must be an amount of enrichment that they know will runaway to a black hole

1:46 PM
@EmilioPisanty If you think that's strange, wait until you have to discuss with a bunch of physicists how invisibility spells work :P For instance, if light just passes through the invisible object, can you sabotage radiation shielding by making it invisible? (That actually became very relevant in the context!)

2:07 PM
@EmilioPisanty The spell causes an effect, that effect would be associated with some physics. I mean if you phrased it without mentioning the spell, it would be the same physics question.
@ACuriousMind gotcha

2:22 PM
@AncientSwordRage It's interesting that you're trying to make the case based on the off-the-cuff comedic reaction instead of the detailed physics case of how this is a flawed question.
There's a clear trend of people asking questions about fictional physics where they fall into this rabbit hole (and attempt to pull you in with them) of coming up with one band-aid fix after another, without addressing the core problems. It makes us extremely wary of engaging with any of it. Please don't do it.
There's plenty of relevant threads on meta if you want to dig into it
physics.meta.stackexchange.com/search?q=fictional results, and links within those results
18

No, questions involving magical situations are not generally on topic here. Our help center lists "questions about fictional physics" as one of the off-topic categories. The reason is basically what Kyle Kanos said in a comment, ...Pretend the laws of physics are broken, what does physics say...

as an example

@EmilioPisanty Comedic reaction?
@EmilioPisanty Thats a really relevant and interesting meta question, really appreciate you singling that out
Ok I re-read the sci-fi question, and I see where the comedic response comment comes from
I guess the question is close to "if a cylinder of rock 5ft wide from ground level to 30 ft tall instantaneously moved 30ft how much force that would produce?"
So then I suppose you'd have to say, things don't move instantaneously, so I'd tweak the question to say it takes 6 seconds.
None of that is unphysical
AFAIAA

3:13 PM
If we have an object that "transforms in two different representations" and so has two sets of indices. For example a quark has spinor indices and also those indices for the fundamental rep of $SU(3)$, when we act with say the spinor representation does this affect the $SU(3)$ components of the quark too? Or do they act entirely independently?

@Charlie There's a big QFT theorem stating that, for a reasonable theory, any symmetry group you have to deal with will be $P \times G$

ah that's coleman mandula right

With $P$ the Poincaré group and $G$ the "vertical" symmetry groups

I haven't actually got very far into SUSY yet, but the example of quarks came up and I just wondered

So you don't have to worry about SU(3) rotations acting on the spacetime indices
it's just a trivial product

3:16 PM
@Charlie if they didn't act independently you wouldn't write two different sets of indices for them!

Yeah I suppose not

Well you know, it could happen
There could be a symmetry that acts on both
But it's not the case in the theories we study

But I thought an important point of SUSY and CFT was that they are non-trivial extensions of the Poincaré algebra, so they aren't just $P\times G$
idk how unique the extension is, but it was my understanding that they are essentially exceptions to C-M theorem

Sure, but the super extension of that theorem is the same, basically
Instead of the Poincaré group, it's the super Poincaré group

oh

3:19 PM
SUSY escapes Coleman-Mandula due to the "super-" part, the equivalent there is Haag-Lopiszanski-Sohnius

you don't have to worry about symmetries like $\phi(x) = \alpha(x) \phi(f(x))$
All the gauge symmetries are "vertical", as we say
ie they only affect fiber per fiber

CFT escapes Coleman-Mandula because it is not an internal symmetry - Coleman-Mandula does not forbid that the spacetime symmetry group is larger than Poincaré
it just forbids that the spacetime symmetry and the internal ("gauge") symmetries combine non-trivially

ok, I still haven't really touched actual gauge theory yet. When we say an internal symmetry we mean the action of a group on the infinite-dimensional representation of quantised fields right

additionally, CFTs do not have massive states, so the prequisites of C-M are not fulfilled

An internal symmetry is one that affects the fields themselves and not the coordinates

3:22 PM
ah I did see something written about CFT not having well behaved massive states too, ok

the simplest case is the electric potential

@Charlie No, we just mean an action on some representation that's not the representation of the spacetime symmetry group

$\phi \to \phi + c$ is an internal symmetry

in your quark example, the $\mathrm{SU}(3)$ symmetry is internal

ah
but in the case of say a quantum scalar field, an internal symmetry would act on the infinite dimensional operators $\hat\phi$ that act on the Fock state space, that's just a special case though

3:24 PM
The "vertical" part comes from the idea that the change doesn't change the projection

I'm not sure what "projection" means here

ie if you have a section $\phi$ and tranform it to $\phi'$, both $\phi$ and $\phi'$ are on the same point

the small amount of differential geometry associated with gauge symmetry I learned is something rapidly fading from my memory lol

$\pi(\phi) = \pi(\phi')$

uhm, and this is a section of a principal fibre bundle right

3:26 PM
Well, this is a general argument
Also gauge fields aren't sections of the principal bundle
That's the gauge itself

they're the connection 1-forms right

Yes
They're part of the tangent space of all that

ok yeah

The connection form is in $T_vP$
The vertical space of the principal bundle

wait it's in $T_vP$?

3:29 PM
$TP$ is the tangent space
Since $P$ the principal bundle is itself a manifold, you can construct a tangent bundle over it, same as any manifold
the $v$ indicates the vertical subspace of that tangent bundle

yeah, I sort of remember this

You only consider the subspace of $TP$ that's tangent to the fibers of $P$ itself

$T_v P\cong \mathfrak{g}$ canonically, so it's just Lie-algebra valued from another point of view

the vertical subspace is the kernal of the pushforward along the projection map, then you make a complementary choice of horizontal subspace and this gives you a way to connect fibres over the total space
oh yeah that sounds familiar too

ie if you have a group $G$ as a fiber of $P$, then their own tangent space will be isomorphic to their Lie algebra, yes
So the vertical bundle will be the set of $\mathfrak{g}$ at all the points

3:32 PM
just as a separate thing, what makes quarks special (in that they also have the $SU(3)$ indices) compared to electrons that don't?

that's why the connections are just algebra-valued

@Slereah ok this I remember too

So all in all the actual $TP$ will look locally like $\mathbb{R}^n \times \mathfrak{g}$
@Charlie You could put $SU(3)$ indices on electrons if you want to, but that gauge symmetry just acts trivially on them

how come?
maybe I'm incorrectly thinking of electrons and quarks as basically being the same thing with different couplings to different fields

Electrons also technically have symmetry indexes, technically, I suppose
but since they're one-dimensional (for EM at least), people don't bother writing them
Also if you really want to, you can just write every particle as part of the same field
A giant mess that's invariant under $SU(3) \times SU(2) \times U(1)$
times whatever
Then specific subspaces of that "field" will correspond to the usual fields

3:51 PM
oh yeah ok

It's a bit blurry because in some sense, you can convert quarks to leptons and vice versa
but if you select a proper basis, that's how you find your "hard" boundaries

1 hour later…
5:16 PM
hi all was doing a lot of research on minev et al experiment, quantum trajectory theory, etc & turning up a lot of cool material. then started looking into objective collapse theories. is anyone around here conversant in these? maybe am coming up with question for site, but also think it may be kind of esoteric for this crowd :| :o
Objective-collapse theories, also known as models of spontaneous wave function collapse or dynamical reduction models, were formulated as a response to the measurement problem in quantum mechanics, to explain why and how quantum measurements always give definite outcomes, not a superposition of them as predicted by the Schrödinger equation, and more generally how the classical world emerges from quantum theory. The fundamental idea is that the unitary evolution of the wave function describing the state of a quantum system is approximate. It works well for microscopic systems, but progressively...

5:47 PM
Good morning/afternoon/night to everyone!
I hope you all are safe and healthy.

So, I would like to ask: where I can find updated values for Density Parameter $\Omega$ of cosmology?

You can try the particle data group

I will take a look then. Other places where I can find?

The cosmos, presumably

@M.N.Raia the Planck data has the latest value I know of.

Thanks @Slereah and @JohnRennie =)

6:14 PM
a parallelizable manifold doesn't necessarily have a global holonomic frame for its tangent bundle, right?

6:42 PM
Not that I'm aware
It has a global frame field

7:02 PM
@EmilioPisanty made it all the way to the top! (almost)
@ACuriousMind looks like there's plenty of people out there ready to support this point 🤷🏻‍♂️

@bolbteppa Thanks for the replies! Why does $\nabla \cdot \mathbf{A} = 0$ imply $\mathbf{k} \cdot \mathbf{A}_{\mathbf{k}} = 0$?

I've seen acm asked this question on the site a few years ago but there was never a definite answer given; does the finite dim. rep. of the Lorentz group uniquely determine the infinite dim. unitary rep. on the Hilbert space? @ACuriousMind did you ever find an answer to this?
in the context of qft ofc

7:20 PM
@Charlie the answer is in chapter 5 of Weinberg, I forgot to write it up
at least, the answer for the non-tachyonic case

ah ok then

I'm not sure the question is really interesting for the tachyonic case since the QFTs shouldn't have tachyons anyway
maybe I'll write it down this weekend, might be fun

I know precious little about what a tachyon even is so that's fine :p

ah, a nice wordy answer followed by a predictably enormous and comprehensive answer by qmechanic lol

7:27 PM
My question is simple: I need to look at the metric tensor matrix and calculate a tetrad basis.
But how to do this?
How is the modus operandi?
For diagonal metric it's direct
If there aren't a modus operandi which assumptions I need to work with to construct a tetrad? If you have a non-diagonal metric tensor how do you start to think on the building a nice tetrad basis?
for example in this image I simply do not know how the authors builded their tetrad basis?
for example in this image I simply do not know how the authors builded their tetrad basis*
And there's more: In the paper "Catalogue of spacetimes" Müller created two types of tetrads: comoving and static....

I'm looking for the following formula on the web. It is supposed to be the Hamiltonian of an electron in an electromagnetic field.
But all I can find is the formula without the last term, as given here:

The thing is: if you ask me to create a tetrad basis from a diagonal metric I simply divide the basis vectors by the squared metric tensor components. But I simply do not know how to start to think for non-diagonal metrics.

@schn The last term is the interaction of the electron spin with the magnetic field
you're probably just finding stuff that ignores the spin, i.e. it just looks at "charged particle in EM field", not "charged particle with spin in EM field"

@ACuriousMind I see, then why the minus sign?

uh, because it works? :P

7:41 PM
:)

@schn but more seriously - the Hamiltonian is often the energy, and the energy of a magnetic dipole moment in a magnetic field also has that sign classically

Right, so it probably cancels out the minus sign and so each term in the Hamiltonian is actually positive...maybe!

I'm not sure what you mean by that
since there's a dot product in there that term can have any sign you want depending on how $S$ and $B$ are oriented w.r.t. each oither
i.e. if it's positive, flip the magnetic field and it becomes negative

Well, I was thinking if the last term represents $U=-\mathbf m \cdot \mathbf B$ as linked, then the minus sign in the formula would cancel it.
Actually
never mind what I said.
Just (kind of) put $-\mathbf m \cdot \mathbf B$ in the Hamiltonian and there is the minus sign. Thanks!