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3:01 PM
$\partial_\gamma M^{\alpha \beta \gamma}=0$
@ACuriousMind Is this 6 equations (or 12)?
 
Depends how many dimensions?
 
greek letters= spacetime?
 
Spacetimes can have arbitrary dimensions
Also does $M$ have symmetries
Otherwise it's 64 equations
Well
16
 
yeah 4*4=16 for $\alpha$ and $\beta$
but this is supposed to be only 6 equations
 
What is $M$
 
3:04 PM
(conservation of rotation and boost angular momenta)
 
oh, so antisymmetrical
Antisymmetrical tensors have $(16 - 4) / 2$ independant elements
So 6 yeah
 
Thanks
and sorry for the *extremely* dumb question, without reading the basics first.
 
@Slereah Pf.?
 
Total number of components minus the diagonal divided by two since off diagonal is antisymmetrical
 
good
 
3:11 PM
0
Q: How to find speed of light in glass

J.SeLight travels slower in glass than in air, and if the index of air is $1$, then the index of glass should be higher. I got that the index of glass was $1.12$ with the formula $n= \frac{v}{c}$ which was given to us by our teacher. How do I calculate the speed of light? If the index of glass is eq...

is that a homework question?
no research effort type
 
Hello @YashasSamaga
 
hi, what's your question?
 
@YashasSamaga, Actually I have some exam questions.. Not the homework questions!
Q1. On what condition a body is weightless in earth's gravitational field?
 
1. freefall
2. revolving in an orbit
I shouldn't be giving answers this way :/
I am not going to give answers anymore. You should think. You are asking very simple questions. I don't know how you are going to pass your exams.
 
@YashasSamaga, how is free fall possible in earth's gravitational field?
 
3:16 PM
Do you know what Earth's gravitational field means?
 
@0celo7 of course I shower daily. But I do not wash my hair daily, that is bad for your hair! You should wash it once or twice a week, three tops
 
Yeah.. @YashasSamaga, the space around the earth upto which its gravity is felt.
 
@Ramanujan The gravitational field of the earth extends till infinity but it becomes negligible after sufficiently long distances as it falls over as $\frac{1}{r^2}$. Moreover, the effect of the sun's gravitational field begins to dominate over the earth's gravitational field at distances far away from the earth.
@Ramanujan In Newton's times, people hated the idea of 'action at distance' very much.
People couldn't believe that a force could be felt magically through air.
People were used to pushing and pull but were surprised that gravity can pull things downwards through air/vacuum.
Faraday introduced the concept of field for the first time.
The idea was to assume that the an object creates a field around itself.
Other objects interact with the field and feel a force.
Earth creates a field > object interacts with the field > object feels a force
 
@YashasSamaga, I just wanted to know can a body fall freely on the earth?
 
I am coming there.
 
3:21 PM
@AccidentalFourierTransform that's disgusting
 
Gravitational field is given by = $G\frac{M}{r^2}$
 
you have horrible hair
 
@Ramanujan mass * field_strength = force
 
you smell btw
 
@Ramanujan so the strength of the gravitational field = accleration due to gravity at that point
 
3:22 PM
just FYI
 
13
Q: How do higher-order optical chiralities look like?

Emilio PisantyThe optical chirality of the electromagnetic field is a conserved quantity, analogous to the energy density, linear momentum density, and angular momentum density, which describes how chiral the EM field is at each point, and it is given by $$ C=\frac{\varepsilon_0}{2}\mathbf E\cdot\left(\nabla\t...

Does this mean there are infinitely many conserved quantities (in classical EM)?
(The 2nd paragraph)
 
my hair smells great
 
@Ramanujan In the presence of a gravitational field, you feel a force.
@Ramanujan If you don't h ave an opposing force, the gravitational force will cause the object to acclerate.
 
user228700
@JohnR: One month until I am held responsible for all my actions in the eyes of the Law :-P
 
@Ramanujan The object is said to be in free fall in such a scenario.
 
3:23 PM
If you wash your hair once a week I guarantee it does not smell great
 
@Kaumudi.H 18th birthday next month?
 
@YashasSamaga, isn't there air resistance on earth
 
@Mostafa yes
 
hello @JohnRennie
 
user228700
@JohnRennie Yep! U forget a lot and don't give me "Irrelevant data", I remember even some of that!
 
3:24 PM
@Ramanujan If you consider air resistance, then no object will ever freefall.
 
@Kaumudi.H I'm terrible with birthdays. I even forget my own occasionally.
 
@Mostafa The answer to that question describes one such series
 
@0celo7 I wash it twice a week. And I use a conditioner that smells great. And I dont know why Im explaining this to you.
 
0
A: Quadrupole moment of point charges?

Prasad ManiGo here. What you are describing is called a linear electric quadrupole.

 
user228700
@JohnRennie ...and that I know. IIRC, it was I who reminded you of it a week before :-P
 
3:25 PM
 
are such answers allowed?
 
@EmilioPisanty Is this possible according to Noether's 1st theorem? I mean, Zilch is obtained using the duality symmetry of Maxwell's equations (From your other question). Can this (finite) set of symmetries give rise to infinitely many conserved quantities?
 
flaggable as low quality?
 
@Kaumudi.H anyway, happy "next months birthday" in case I forget :-)
 
AFT's hair
 
3:25 PM
@AccidentalFourierTransform because you're embarrassed by your terrible hygiene
 
user228700
@JohnRennie Lol, OK, thanks :-P
 
@AccidentalFourierTransform your hair looks like this I presume
 
@Kaumudi.H so what do you plan to do that you can't (legally) do now?
 
lol I wish
some day Ill get dreadlocks for sure
 
oh god
why?
what is wrong with people
 
3:27 PM
but I want to be original. Maybe one big, fat dreadlock
 
user228700
@JohnRennie This might sound boooring but I might attempt to procure my driver's license before heading off to college seeing as there won't be much to do in the summer.
 
@YashasSamaga, would not it be at the center of earth, where g=0?
 
@Ramanujan Yes
 
@AccidentalFourierTransform uhhhh...that won't look great. If you want to be original, better go for an unusual color
 
an unusual color is the opposite of original
 
3:28 PM
@AccidentalFourierTransform Only if it is an unusual color that is usual among people with unusual colors :P
 
@Kaumudi.H Not boring at all - go for it!! :-) I didn't get my car licence until I was about 24, because up till then I rode a motorcycle.
 
@YashasSamaga, why are wooden slippers used under railway track?
 
@Ramanujan no idea
 
@ACuriousMind sorry for pinging you for that question about indices...I thought you were here.
(I still feel guilty about it)
 
so what colour would be truly original? something around $200\ \mathrm{nm}$?
 
3:29 PM
@JohnRennie wtf
@AccidentalFourierTransform x rays
 
@Mostafa np, I saw it got resolved without me ;)
 
@Mostafa there's a corresponding hyerarchy of symmetries
 
@Ramanujan sleepers not slippers :-)
 
user228700
@JohnRennie Ah, OK :-) It might be some trouble getting both in the span of 60 days but hey, I literally came up with that just now and now I'm taking it seriously so thanks!! (:-P)
 
3:30 PM
maybe you'd die of head cancer
@JohnRennie I need chemistry help
 
@JohnRennie. Why are they used?
 
@Kaumudi.H Yeah! Learn to ride a motorbike - I'm sure your parents would approve :-)
 
@Mostafa Zilch does not correspond to the duality symmetry (that's the optical helicity if I remember correctly). Zilch corresponds to a more complicated symmetry transformation.
 
@ACuriousMind is this flaggable?
0
A: Quadrupole moment of point charges?

Prasad ManiGo here. What you are describing is called a linear electric quadrupole.

 
@Ramanujan Have you tried to google it instead of asking random people in here?
3
 
user228700
3:31 PM
@JohnRennie :-P On the contrary, actually. My father has asked if I want to learn to ride his bike.
 
@YashasSamaga why do you want to flag stuff so hard?
 
@YashasSamaga When I first saw that I had a moment of hope that someone was asking about the experimental methods of doing reasonable precise bench-top measurements.
 
they do not give you points for valid flags :-P
 
The naive duality symmetry is $A^\mu \mapsto A^\mu +\theta C^\mu$
 
@JohnRennie :(
 
3:32 PM
Zilch corresponds to $A_\mu\mapsto A_\mu-\xi^{\alpha\beta}\partial_\alpha G_{\beta\mu} ,\quad C_\mu\mapsto C_\mu+\xi^{\alpha\beta}\partial_\alpha F_{\beta\mu}$
 
@0celo7 Chemistry? I can try to answer ...
 
@AccidentalFourierTransform Has been 2 years and I haven't figured out how to flag questions and answers.
 
@YashasSamaga You know that there is a "link only" deletion reason in the review queue, right? So that's certainly worth a VLQ flag.
 
There are some very nice ways of using interferometers to get three to four digits of precision in a very simple measurement. (You are limited by your thickness measurement, because the instrument has dynamic range to spare.)
 
3:33 PM
@Ramanujan You have to fix the rails to something to stop them moving. Since the rails have to pass over all sorts of different types of ground the simple solution is to put a base under the rails. That's what the sleepers are.
 
@JohnRennie what the hell does the 3s, 3p, etc. notation mean
 
But, alas, it was someone's homework. And misunderstood at that.
 
Higher-order members of the hierarchy would come from higher-order derivatives of the fields.
@Mostafa
 
@0celo7 spherical harmonics?
 
I think the number means the energy level, and the letter denotes some $\ell$ value
but why the hell couldn't they make things simple
 
3:33 PM
you know the Legendre thingy $P^n_\ell(x)$
 
@AccidentalFourierTransform Well, for the s, p, d. The numbers are the radial part.
 
@0celo7 yes
@0celo7 because history
 
@Ramanujan As @JohnRennie says, they balance the train's weight to the ballast below, keep the rails spaced evenly (because of expansion, etc.), and other stuff.
 
@EmilioPisanty Thanks a lot! Do higher order conserved quantities of the zilch tensor have physical significance? (like chirality)
 
@EmilioPisanty I have selection rules $\Delta n\ne 0$ and $\Delta\ell=\pm 1$
 
3:34 PM
@dmckee ah right
 
what does that mean in that strange chemistry notation?
 
@0celo7 It's a historical artefact. I think they refer to the appearance of the related lines in spectra.
 
@0celo7 it means you can only jump one step in the s,p,d,f,g,h,... series
you can move right or left by one slot
@Mostafa what do you mean by "physical significance"?
 
Physical interpretation
(Like chirality, for example)
 
@dmckee particle scattering question for you ...
This question:
1
Q: Random directions of the trajectories in a cloud chamber

S. KohnIt is easy to construct a home-made cloud chamber. In observation, one finds that the directions of the trajectories are quite random. Does this mean that all the particles detected are secondary cosmic rays?

 
3:37 PM
@Mostafa they can be expressed as functions of $\mathbf E$ and $\mathbf B$
so in one sense you're sort of done
 
points out that in a cloud chamber we see only the secondary electrons not the cosmic rays themselves. Why don't the cosmic rays (high energy protons) leave a trail in the cloud chamber?
 
@Mostafa what are you understanding "the physical interpretation of optical chirality" to be?
 
Peresumably it's because they don't scatter strongly off the air atoms at very high energies, but if so why not?
 
@JohnRennie We see mostly muons I would think, so yeah. They're secondary.
 
@EmilioPisanty How does one TeX 3s?
 
3:39 PM
And the azimuthal distribution is quite wide.
 
3$s$?
 
@0celo7 ah, I'm never sure
 
@EmilioPisanty A sort of handedness. Like polarization, or vortex of an optical field
 
I usually $s$ the $s$s
 
@JohnRennie, what happens if we use oil instead of water in hot water bags?
 
3:40 PM
@Mostafa that's pretty washy, though
you could say the same about the optical helicity
 
Isn't it the meaning of chirality?
 
does cyanide smell like marzipan?
 
hmm no
why would it?
 
there is a strong marzipan smell in my lab
 
oh, because the almond
 
3:41 PM
@AccidentalFourierTransform people say it smells like almonds
 
yeah, I forgot about that
so, you got cyanide in your lab?
 
maybe
if you don't hear from me I either got banned or killed
 
@Mostafa there's other quantities that share the same meaning
so, for instance, you could take the optical helicity, $\mathrm{Hel} = \frac12 \left( \mathbf A^\perp \cdot \mathbf B - \mathbf C^\perp \cdot \mathbf E \right)$, and it's also "a sort of handedness".
So there's this big hierarchy of related quantities
and they're similar but different
 
@EmilioPisanty how many $4p$ states are there? I'm still confused
 
3:45 PM
and not particularly easy to understand
 
@dmckee as I recall the droplets in a cloud chamber form around ions created by the charged particle. So that must mean the (low energy?) secondary muons are better at ionising atoms than the high energy protons in the cosmic rays. Why would that be?
 
at least in physical terms
that's what those questions were getting at
and I think Danu did an amazing job at explaining what the deal is
 
assuming $p$ corresponds to $\ell=1$ states, then there should be three of them
 
@0celo7 three linearly independent ones
 
times 2 for spin
 
3:46 PM
^ plus that, yes
 
oh jesus
 
@JohnRennie The protons (and heavier nuclei) basically don't make to the surface level. There are 10 tons per square meter of stuff in their way.
 
@EmilioPisanty So, intuitively, they all correspond to the chiral properties of the field, right?
I'm asking because of the claim made here by D.J. Candlin: (p. 1394)
 
In NuSea (Fermilab E866) we put about 3 tons per square meter of shielding in the cavity of the main magnet to remove most of the non-muon junk and it worked just find even though the mean energy of the detritus was higer than a mere GeV like cosmic rays.
 
@dmckee if you watch a home built cloud chamber you don't see lines spanning the chamber. You see tracks starting at random points in the chamber then travelling a few cm before fading out.
I assume those are electrons created when a muon scatters off an air molecule.
 
3:49 PM
no, those are probably neutrinos
 
The question is why the muons themselves don't leave tracks.
 
or axions, or dark matter
 
@Mostafa Эта статья у меня есть!
got it
 
@AccidentalFourierTransform don't be silly this is a serious conversation.
 
lemme have a look
 
3:49 PM
@JohnRennie I would guess so. Muons have pretty low dE/dx, and a only slightly supersaturated detector might not notice them. But delta rays (scattered electrons) have much higher dE/dx.
 
If I run out of votes to close, I can't flag it? -_-
 
I'm used to working with detectors that have plenty of dynamic range and pick up muons with ease.
 
@YashasSamaga nope, 24 votes/day, that's it.
 
@dmckee is there an intuitive explanation of that? Assuming the scattering is mainly EM the only difference is the momentum of the particle. Why are high momentum muons bad at ionising atoms in the air while low momentum electrons are good at it?
 
@Mostafa ok, what claim?
 
3:52 PM
@JohnRennie Uhhhh. I figured that out once, but I'll have to get back to you on it.
 
The Bethe formula describes the mean energy loss per distance travelled of swift charged particles (protons, alpha particles, atomic ions) traversing matter (or alternatively the stopping power of the material). For electrons the energy loss is slightly different due to their small mass (requiring relativistic corrections) and their indistinguishability, and since they suffer much larger losses by Bremsstrahlung, terms must be added to account for this. Fast charged particles moving through matter interact with the electrons of atoms in the material. The interaction excites or ionizes the atoms...
 
> it seems unlikely that any special significance or application can be found for zilch.
(P. 1394; first paragraph of the page)
 
^^ is the place to starts.
 
@dmckee OK, at least I don't feel guilty for not know the answer to a simple question :-)
 
But it's complicated because it depends on the particle mass in an involved way.
 
3:53 PM
@AccidentalFourierTransform aha, thanks, I'll have a read through that.
 
physics.stackexchange.com/questions/315371/fluid-mechanics @YashasSamaga should this answered looks like he has explained his work out in the image.
 
I think I need a selection rule $\Delta j_z=0$ too
so that $\Delta m_s=0$
argh I hate physics
2
 
@0celo7 Burn the witch!
 
@Mostafa meh
 
I dont think that there is an intuitive/physically obvious reason. Its just that when you plug in actual numbers, muons have low dE/dx while electrons have high dE/dx.
 
3:56 PM
I mean, it's useful enough, more useful than the higher terms in the hierarchy, not as useful as the stress-energy tensor.
it depends what you mean by "special"
 
Jan 8 at 14:20, by AccidentalFourierTransform
man I hate physics
 
@ATHARVA It is unreadable and it is still a HW question.
 
but I find e.g. Robert Cameron's papers on the subject to be a pretty strong refutation of that claim
 
@YashasSamaga Ok.
 
4:08 PM
@EmilioPisanty Ok, thanks a lot. Your questions on this were (actually are) extremely helpful for me.
 
@Mostafa I can only claim pretty limited credit for those
Danu did most of the work
but yeah, read those questions, plus the chat they link to
and anything by Robert Cameron referenced there
 
4:26 PM
Every time I try to upload an image to imgur (while adding an image to a Q or an A), I get failed to upload image.
I have to try around 10 times before it succeeds.
 
What format? Imgur doesn't seem to like PNGs.
 
It never used to happen before.
It's PNG :D
but it used to work without fail before
on first try
My answers always have images
 
Try a GIF instead.
 
@ACuriousMind Why does $\mathbf J$ commute with $\mathbf S\cdot\mathbf r$?
 
It uploaded as PNG but I had to try around 5-6 times.
 
4:29 PM
@ACuriousMind If you say "rotational invariance" I will get upset
 
calculate the commutator and check that it, indeed, vanishes
if you want to know the reason, it is because $\boldsymbol S\cdot\boldsymbol r$ is a scalar operator
and so it has to commute with $\boldsymbol J$
 
I just said don't say that
 
rotational invariance.
 
Are you trolling me?
 
your question doesnt make much sense though. They commute because when you calculate the commutator, it is zero
thats why
if you want to know the reason for why we expect it to commute to begin with, or why we know that it has to commute, then it is because rotational invariance
 
4:33 PM
I'm trying to wrap my head around the difference between the $B$ and the $H$ fields, and reading up is just getting me more confused.
 
I dont believe in magnetic fields
I mean, I know they exist. I just dont believe they do
 
I don't believe path integrals exist
 
interesting question!
 
i.e. why does $ϴ$ ($ϴ$) look so horrible and so different to $\theta$?
 
4:37 PM
it's a fat theta
 
ah, yes, it's marked as a capital theta
U+03F4 : GREEK CAPITAL THETA SYMBOL
doesn't look very capitalized, though
ϴ
 
no unicode pls
 
⸘‽
 
@EmilioPisanty It's capital in that the outer thingy is round, not ellipsoid
 
$\Theta$
$\huge \Theta$
 
4:40 PM
...or at least, not as ellipsoid as the non-capital one :P
 
@Slereah yeah, that is how capital theta's meant to look like
 
$$\Huge{\Theta \omega \Theta}$$
I'm a goat
 
including a distinctly non-connected middle bar
 
what is that?
 
It's a goat
 
4:41 PM
I...don't get it
 
Goats have freaky eyes
 
@Slereah oh, indeed
 
learn something new every day
 
Are you guys really talking about physics ?
 
@Cham That only happens in here by mistake.
 
4:48 PM
@Cham trust me, compared to some of the stuff that gets discussed here goats eyes are nothing! :-)
 
no, if you want to talk about physics youll have to goat-to another chat.
 
Ba dum tss
 
Take my wife ...
 
ok sorry for that
 
@JohnRennie Goat =/= sheep
 
4:49 PM
How very biblical of you :-)
 
You're not a...well. Better not continue that line of thought.
 
Oh you mean the take my wife comment? That's an old music hall joke from the Victorian era.
 
He prolly think it's a club light, But really it's the red dot. - Quavo
 
I just received an SSD, and by god, it is tiny.
 
Take my wife, please!
Thank you, I'll be here all night
 
4:51 PM
Shut up, Rodney
 
4
Q: Is the phrase "Take my wife – please!" a paraprosdokian?

iddoberI was reading the wiki page about paraprosdokians and I don't understand why the phrase: Take my wife – please! is classified as one.

 
I'm a physics teacher in my classroom right now, watching my students doing their astrophysics exam. And watching you guys doing physics. LOL !
 
Hey! I'm trying to calculate whether we could measure gravitational time dilation caused by lab scale masses (the answer appears to be no).
 
@JohnRennie You should tell me if you don't want to learn any topology
 
@0celo7 No respect at all!
 
4:52 PM
I'm tired of asking
 
@Cham Oh well, I guess we just diversify topics.
 
What was the original topic here anyway ?
 
@0celo7 In principle I would like to understand how to define a manifold, but in practice I lack enthuisiasm for the amount of work required. I think you should assume I'm a lost cause unless I suddenly discover I have a lot more free time.
 
Manifolds aren't terribly complicated to construct
 
@Cham Unicode. Long story.
 
4:54 PM
You just stitch together pieces of Euclidian space
 
But thanks for trying. Genuinely thanks that is - I mean no irony.
 
Manifolds are dull to define precisely in physics.
 
@Cham What does that even mean?
 
dense ensemble of "points" supporting a notion of continuity...
 
A manifold is a mathematical object.
That's not precise at all
 
4:56 PM
manifolds are stuff which look like R^n locally
 
yep.
 
@BalarkaSen and hausdorff and paracompact
 
my hand s a manifold
 
Well no
 
@0celo7 nah
 
4:56 PM
You can have a non-paracompact manifold
 
those are just nice assumptions
 
It's just gonna be fairly long
 
@Slereah Forget about partitions of unity
 
that's why I said "dull to define" : Hausdorff, etc...
 
Who cares about partition of unity
 
4:57 PM
@BalarkaSen Dude. Riemannian metrics imply both paracompactness and Hausdorff.
 
Partitions of unity are awesome!
 
Not all manifolds have a metric tho
 
All useful ones do.
 
@0celo7 I am well, well aware of the drawbacks of not having that assumption. That doesn't mean I have to include them in the definition of manifold.
 
Lie groups could define a manifold too.
 
4:58 PM
@JohnRennie it's any shape that locally looks like $\mathbb{R}^n$ at each point.
 
would you say the leaf space of plane foliation isn't useful
 
@DanielSank Locally looks like?
 
@0celo7 yep
 
I think @JohnRennie understands that much, he means talking about homeomorphisms, smooth structures, etc.
@DanielSank I'm saying that's not precise at all.
 
Meh.
@0celo7 so?
 
4:59 PM
can you rephrase this sentence in terms of toothpaste
 
I always found manifold definition to be extremely boring, in physics...
 
@Slereah Well, not per se. Leaf space of the Reeb foliation isn't as useful as the Reeb foliation.
 

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