« first day (2840 days earlier)      last day (2074 days later) » 

1:58 PM
$\langle \psi_{100} \vert x^2 \vert\psi_{100} \rangle = \dfrac{1}{\pi a^3} \int r^2 \cos^2 \theta \sin^2 \phi e^{-2r/a} r^2 \sin \theta dr d\theta d\phi = \dfrac{1}{\pi a^3}[\int_{0}^{\infty} r^4 e^{-2r/a} dr \int_{0}^{\pi} \cos^2 \theta \sin \theta d\theta \int_0^{2\pi} \sin^2 \phi d\phi] =\\ = \dfrac{1}{\pi a^3} \dfrac{4!}{(2/a)^5} \dfrac{2}{3} \pi=a^2/2$
It should be $a^2$, where did I screw up?
 
 
1 hour later…
Anonymous
3:11 PM
@JohnRennie By any chance do you know how to convert .ppm images to other formats like jpeg or eps or pdf?
 
Anonymous
I got a C program which produces a .ppm image
 
Anonymous
But I can't use that format in other places
 
Anonymous
(I mean that I could simply use "Snipping tool" or something like that, but I want to retain the image quality)
 
@Blue What format is ppm?
 
Anonymous
A Netpbm format is any graphics format used and defined by the Netpbm project. The portable pixmap format (PPM), the portable graymap format (PGM) and the portable bitmap format (PBM) are image file formats designed to be easily exchanged between platforms. They are also sometimes referred to collectively as the portable anymap format (PNM), not to be confused with the related portable arbitrary map format. == History == The PBM format was invented by Jef Poskanzer in the 1980s as a format that allowed monochrome bitmaps to be transmitted within an email message as plain ASCII text, allowing it...
 
3:13 PM
gimp? imagemagick?
 
@Blue Aha, Paint Shop Pro can open those and convert them.
 
Anonymous
I don't have prior experience with gimp :/
 
Anonymous
(And don't have Photoshop installed on my PC either)
 
@Blue didn't I give you a copy of Paint Shop Pro?
 
Anonymous
I can download gimp on linux it seems
 
Anonymous
3:15 PM
@JohnRennie Oh, I was reading that as "Photoshop" :P
 
Anonymous
oops
 
Anonymous
Yeah, lemme try
 
Anonymous
Switching to Windows... one moment
 
@Blue bloody users! :-)
2
 
I think it would just be open -> export. Don't really know about quality though. imagemagick (or however it's spelled) is a cli tool that a lot of people seem to use if you need to convert a bunch or something
Yeah users always go around breaking software
Have you guys that watch online lectures tried increasing playback speed? I thought I'd just miss a bunch of stuff, but I seem to have gotten used to it and can pause/slow down for interesting or difficult parts.
 
Anonymous
3:23 PM
@JohnRennie It seems you had given me a copy of MS Paint (since my PC only had Paint 3D) and not "Paint Shop Pro" :P
 
Do you want a copy of Paint Shop Pro? Or will you just use Gimp?
 
Anonymous
Lemme try to use gimp first, just downloaded it on Ubuntu
 
Anonymous
Will come back if I get stuck :P
 
Anonymous
 
Anonymous
Opened the image on gimp, where's the export option? @danielunderwood
 
3:31 PM
Ctrl+E?
 
It's playing hide and seek of course!
 
Or cmd+E
 
Ctrl+E should work as above, but it would be in the file menu above your image
 
Anonymous
Umm, so it doesn't have any .pdf or .eps export option...so I guess I should convert to jpeg first?
 
Anonymous
@Lozansky Thanks
 
3:34 PM
Try to give it a pdf extension. It lists pdf in the expandable box for me
It also lists eps if you prefer that
 
Anonymous
Damn, they wrote the full form of PDF instead
 
Anonymous
Lol
 
Anonymous
@danielunderwood Thanks a ton, it worked! :D
 
Wooo!
The real question is what the image is for. It looks like a QR code gone crazy!
 
Anonymous
Heh, they're just some black and white clusters....like in $Z^2$ percolation models
 
Anonymous
3:41 PM
Not a QR code XD
 
Anonymous
 
Anonymous
I'm trying to make a it a bit more colorful like this ^
 
Anonymous
Or maybe like this:
 
Anonymous
 
Anonymous
(that might be a bit too much :P)
 
3:46 PM
Kind of looks like you generated random clusters lol
 
Anonymous
Indeed, random matrices/lattices
 
It might help if I knew what percolation models were. May give me something to google a bit later. This may be a silly question, but does it have anything to do with like coffee percolation? Like liquid going through granular materials?
 
Anonymous
@danielunderwood Sure, that's one application. Site percolations are prevalent in several areas of condensed matter physics (especially phase transitions) and computer science (network theory). Just last month we were discussing that lightning can be modeled as a type of site percolation...so...umm...tons of interesting uses
 
Anonymous
 
Anonymous
For a brief overview you could watch that ^
 
3:55 PM
Hmmm sounds neat. I suppose that could be used to model circuits as well?
 
Anonymous
@danielunderwood Circuits? How exactly? I haven't come across similar models of electrical circuits before
 
Anonymous
Circuits aren't really random graphs so as to speak :P
 
Ahh I didn't know it was meant for random graphs. I was thinking of modeling the flow from node to node in a network
 
Anonymous
Yeah, it works for random graphs only :)
 
Why is the convention to use eigenstates of $S_z$ as basis for the $1/2$-spin?
 
4:01 PM
Gotta pick something.
 
Why is the convention to use $E \mathbf{\hat{z}}$ for a constant electric field?
 
I suppose it's the same reason why one picks $z$ as the coordinate of the axis in cylindrical coordinates.
 
I guess it is the easiest to work with
 
@Blue any chance some of the graph traversal algorithms come out of percolation theory?
 
Anonymous
@danielunderwood I don't know what you mean by "come out of" but I definitely used graph traversal for cluster counting in a random matrix (using the quick union and path compression algorithms)
 
Anonymous
4:08 PM
I was studying the nature of clusters around the percolation threshold
 
Anonymous
(the behaviour is quite chaotic in that region)
 
welp, did some light debugging on the RL model...
I can sorta see what's wrong with it...but not apparent if it's because I implemented something wrong or if the behavior is to be expected with the current algorithm...=/
something does seem wrong in the updates tho
 
4:47 PM
@Blue that was a neat video. Looks like I may have a new channel to binge watch. Is this part of your research?
 
A lot of movement on the NLP font at my company it seems..heh
 
You guys have a font just for NLP? That seems like a different approach :D
 
hmmm typo'd
 
No no. You put it in a special font then do image processing!
 
lol
 
5:02 PM
I'll admit that one time I tried to do sentiment analysis on hashed documents. It did not work
I thought that hashing would be a good way to deal with differing lengths and the network would learn the hash...guess not
 
o.O
hmmm this guy is late for our meeting...
on a good note, microsoft is interested in interviewing me lol
The position seems pretty junior tho...kinda doubt it's better than my current position...
 
I mean ridiculous ideas work from time to time lol. I didn't know about RNNs at that point though
mmm microsoft
 
Anonymous
@danielunderwood Yes, indeed it is a part of my (first) undergrad research project (and yup that channel is good....network dynamics is an interesting and new field). I have been working on this thingy for around 6 months now. Sort of writing a paper on it. Will share once completed :)
 
@DavidZ Last time you said you don't remember seeing the definition of thermalization in a textbook, either. I just saw the definition of thermalization in the research introduction mathphy.ugent.be/wp/fields/research:
@DavidZ As a simple example of how all these concepts come together, let us consider the aftermath of a heavy ion collision. Lots of energy is pumped into a very small region. Chaos reigns but due to interactions, this very non-equilibrum situation evolves towards thermal equilibrum: this is called thermalisation of the quark gluon plasma.
Thermalisation typically goes from high energy scales which equilibrate first (because they carry more energy per degree of freedom) towards the low energy scales.
 
6:01 PM
mmmm
 
6:58 PM
oof. I think I finally appreciate the nonlocality of Bohmian mechanics at the level of a toy example
and it's kinda ... yuck
(I have in mind page 11 of arxiv.org/pdf/1305.1280.pdf as I say this)
Basically, if you entangle two electrons in a singlet state and send them into stern-gerlach detectors, then the allowed trajectories depend not just on the geometry of the individual detectors but also depend sharply on which detection occurs first
which...ughhh
 
7:45 PM
Is a Wick rotation the same as choosing between $-,+,+,+$ and $+i,+,+,+$?
The difference between having $\sqrt{1 - \beta}$ and $\sinh{\beta}$ as Lorentz transforms?
Actually $\sinh \beta$ is wrong, but it involves the hyperbolic trig functinos
 
John Wick rotations?
4
 
That's when physics seeks its revenge
 
Is $l$ limited to (non-negative) integers or half-integers?
 
Because it's too angry you wrote down something dimensionally inconsistent like $\sinh \beta$
Wow I just magically found my quantum professor's 200 pages of notes in a random google drive folder. Neat
 
'John Wick is a gun fu action thriller' hmm
That's probably the best movie category name ever and I never heard about it before
 
8:02 PM
You never seen Equilibrium?
the original gun fu
 
vzn
@Semiclassical there is another approach... contrarian yet probably viable...
 
Looks cool
@Semiclassical the paper says Bohm treated spin in 1955, probably done better than the way that paper does it
 
@enumaris gun kata. Get it right :P
(I haven’t actually seen Equilibrium, I just know random trivia)
@bolbteppa eh, at the technical level I doubt introduction of spin is different: replace scalar wavefunctions with spinor wavefunctions
the Norsen paper isn’t innovative, it’s just s good review
I should be more careful, though: the tack Norsen is taking is the same as what Bell advocated, ie that spin in the dBB framework is to be understood not as a property of the particle but rather of its guiding wave
There are people who have advocated for viewing spin as a particle property in the dBB framework , but that’s not the typical choice
 
Given a spin state measured/prepared by a SG, is it always possible to determine the orientation of the magnetic field?
Guess I can afford to be more precise
I have a beam of $s=1/2$ particles in the state $\chi = \dfrac{1}{\sqrt{5}}\chi_+^x+\dfrac{2}{\sqrt{5}}\chi_{-}^x$ and a SG apparatus oriented in a direction in the $zx$-plane and I want to determine the direction $n$.
 
8:20 PM
Is chi the state of an electron that will be going into the SG apparatus, or the state of such an electron after having gone through the SG?
 
After
 
@Semiclassical I think Gun Kata is a subset of Gun Fu...kinda like a square and a rectangle
 
@enumaris makes sense
 
:D
 
@Lozansky maybe consider the problem first in a simple case, eg $\chi=\chi^x_+$
 
8:26 PM
@Semiclassical So in matrix notation, the spinor is $\chi = \dfrac{1}{\sqrt{10}}\begin{bmatrix} 3 \\-1 \end{bmatrix}$ and if we by $\theta$ denote the angle the direction $\mathbf{n}$ measured from the $z$-axis, we should have $S_n = S_z \cos \theta + S_x \sin \theta = \begin{bmatrix} \cos \theta && \sin \theta \\ \sin \theta && -\cos \theta \end{bmatrix}$
 
If you’re saying that you send in a beam of electrons and all of them come out in the state $\chi_+^x$, then that tells you not only the detector but the state of the electrons going in
Which seems stronger than you perhaps intend
 
@Semiclassical Then I guess they are parallell with $x$
 
What is parallel with x?
 
The thingy
The apparatus
:P
 
8:31 PM
I have omitted $\hbar/2$ from the matrices btw, hope that's ok
 
But it would also have to be the case that the electrons are already in the $\chi_+^x$ state upon entering the SG-x apparatus
 
yo folks
does anybody know what "(IP)" means in the author list here? arxiv.org/abs/1808.03427
 
In general, an SG-x apparatus will provide both $\chi_+^x$ and $\chi_-^x$ as output states
 
Well, I guess we don't really care what states they were in prior to entering
Hmm
 
8:33 PM
Oh
Maybe we are preparing and measuring at the same time? Idk
 
My point is that, unless your input beam has already been suitably prepared , you really will have two output beams
And since any spin operator is an observable and therefore hermitian, the second state should be orthogonal to the first
 
@Semiclassical OK let's say we want to prepare a beam with that state
Via an SG apparatus
 
Then you want a spin operator such that $\chi = \dfrac{1}{\sqrt{5}}\chi_+^x+\dfrac{2}{\sqrt{5}}\chi_{-}^x$ is an eigenstate
 
Ah
So $S_n \chi = \lambda \chi$?
Wait
 
Well, that’s one choice (and a satisfactory one)
 
8:37 PM
Uhh
$\lambda = \hbar/2$?
 
that’s one, yes
What are the allowed values for a component of a spin-1/2 particle?
 
$\pm \hbar/2$?
 
Right. So $S_n\chi =(-\hbar/2)\chi$ will also suffice
In that case, of course, the output beam of interest will be the one deflected down rather than up
 
@EmilioPisanty Institut Pascal?
 
What does that suggest about the two SG apparatuses which would create these outcomes?
 
8:41 PM
I don't know why they'd list that if they're all the same though
 
@danielunderwood that's not even the start of their affiliation
 
@Semiclassical Two different directions of the magnetic field?
 
Yes, but more specifically ?
(Or, more simply: what’s the simplest SG device that will to take a beam that deflects up in another SG and instead deflect down?)
 
How is the definite integral for a finite change in volume changed to a differential form with T1 and P1 given T and P respectively?
 
@Semiclassical It's turned upside down :P
 
8:51 PM
Pls help me as well
 
Yep :)
 
@gateprep eww thermo
 
$V_2 - V_2 = \int_{V_1}^{V_2} dV = \int_{P_1}^{P_2} \frac{\partial V}{\partial P}|_T d P + \dots $
 
oO
@Semiclassical Can you explain time-dependent perturbation theory? :P
 
@danielunderwood probably, yeah.
they probably put in an explicit (IP) in the Author field? and I imagine the find-authors-and-link-them algorithm excluded anything in brackets
 
8:54 PM
So there will be two SG devices: one with orientation $n$ where $\chi$ is the state of the upward-deflected beam, and another with orientation $-n$ where it’s the state of the downward-deflected beam
 
@bolbteppa this I cant understand justify pls
 
@Lozansky not in any sufficient amount of detail ;p
 
@bolbteppa cant get the last method
 
I don’t remember time-dependent perturbation very well if I’m honest
 
@gateprep this is multi-variable calculus, better to learn that before reading a thermo book
 
8:56 PM
I remember there is such a thing as Fermi’s golden rule, but I haven’t had to make a calculation with it in years
 
Time-independent is not too bad
Yeah
But once you throw in time, stuff gets messy
 
@bolbteppa name this property
which property to google search@bolbteppa
 
Even time-independent perturbation theory gets nasty when you allow degeneracy or go to higher orders
 
multi-variate chain rule
 
8:58 PM
@gateprep Equation (11)
 
But at least in those cases the nastiness is of finite extent, so to speak.
 
Thanku
@bolbteppa
 
Time-dependent PT feels far more unbounded in its nastiness :/
 
Yeah
We've only done 2nd order corrections to the energy and 1st order to the state
 
That’s pretty typical
fun fact: one of the problems that was assigned to us as HW was to do perturbation theory for the hamiltonian $H=p^2+\alpha \cos\theta$
with $H_0=p^2$ being the intention
 
9:02 PM
Can use time-dependent perturbation theory to show the coordinate of a wave function is meaningless in relativistic quantum mechanics which is shocking and nearly destroyed QFT
 
@Semiclassical Fine-structure?
 
nah. more like a particle on a ring in a uniform horizontal electric field
 
Oh
Makes me think Stark effect
 
he amended that, however, when it was noted that resolving the double degeneracy of the level $E_0=n^2$ would require $n$th order degenerate perturbation theory :>
 
Was this intro level QM? oO
 
9:05 PM
grad level, but he hadn't realized the problem was as difficult as it actually was
 
Oh, yikes
I think the problems in Griffiths can be more than difficult enough :P
 
you can resolve the degeneracy of the states $\psi_{\pm 1}^0 =e^{\pm i x}$ just fine
first order degenerate perturbation theory will suffice for that
if you actually want to see it, look at this DLMF page on the Mathieu equation (which is what this problem basically is): dlmf.nist.gov/28.6
the eigenvalues, as functions of $q=\alpha/2$, are the functions $a_n,b_n$
and the point is that, for instance, if you look at $a_5,b_5$, then they agree up until the $q^5$ term
so if you wanted to resolve the degeneracy, you'd need to do fifth order degenerate perturbation theory...
which, uh, no
gtfo
 
Maybe he was trying to make it a weed-out class :P
 
Weeding out all the non-Nobel prize laureates
 
9:11 PM
funny thing was that I ended up having to deal with that same problem again in my research
though in that case i only needed to know the splitting numerically
and for that you can take advantage of the periodicity to turn it into the eigenvalue problem of a big matrix
(technically it's an infinite matrix, but you truncate it if you want useful results)
So I more-or-less was able to side-step the problem once more :P
 
@bolbteppa Eveh then eqn 1.3 and 1.4
seem different
dont u think so
One has T1 and the other has P2 as the limit
 
@Semiclassical You can't outrun your problems forever :P
 
depends. you can outrun some of your problems if you keep doing it. problem is, that tends to tire you out
and then new problems show up instead =)
(just because you outrun one problem doesn't mean a new one won't pick up the scent)
 
@Lozansky Can u come to my rescure?
rescue*
 
@gateprep $V_2 - V_2 = \int_{V_1}^{V_2} dV = \int_{P_1}^{P_2} \frac{\partial V}{\partial P}|_T d P + \int_{T_1}^{T_2} \frac{\partial V}{\partial T}|_P d T = \int (\frac{\partial V}{\partial P} d P + \frac{\partial V}{\partial T} d T) $ so $dV = (\frac{\partial V}{\partial P} d P + \frac{\partial V}{\partial T} d T)$
 
9:20 PM
@Semiclassical And unresolved problems have a tendency to haunt you indefinitely :>
 
There's a lot going on here tbh, "partial derivatives", "line integrals", "differentials", "chain rule", all concepts from multivariable calculus
 
i think part of the trouble is that, when you're doing multivariable calculus in the context of electrodynamics, there's a strong 3D geometry context to it
in thermodynamics, you don't have that geometric context. as such, you need to be able to confident with multivariable calculus in a a more abstract way
 
Speaking of those thermal quantities. Are $dV$ and the like technically differential forms?
 
you don't have to deal with stuff like div grad curl in thermodynamics, but you have to deal with the multivariable chain rule a ton
@danielunderwood I believe you can view them as such, yeah
the main reason to hesitate on it is that, in thermo, you only really run into differential 1-forms
you don't deal with generic n-forms
As such, viewing it in terms of forms can be useful, but it's not essential
 
But you can't really divide 1-forms can you? Like $\frac{dV}{dP}$ wouldn't really make sense. Though I suppose you can't do that with differentials all willy-nilly anyway
 
9:29 PM
yeah
i mean, differential forms as such are objects which act on vector fields
and it doesn't make sense to divide by them, any more than it would make sense to go from $f(x)$ to $(1/f)(x)$
(you can divide by $f(x)$, but that's the output of $f$ not $f$ itself)
i have heard that there is some virtue to the forms description
but for most purposes in thermo it's overkill
 
You mean you don't want to have to take a differential geometry class before thermo? :D
 
hah
i'll pass
 
I do have a book that's titled something like "Relativity, Cosmology, and Thermodynamics" but I kind haven't really looked at it yet. I have my hopes that it'll be like that
 
We had thermo in the very first semester
 
that'd be a good bet
 
9:35 PM
I think I took thermo in my third semester and I think I took it too early since I don't seem to remember much of anything
Other than "take normal physics and count things"
 
the forms mindset is more useful in electrodynamics, because for instance you can interpret $\rho \,dV$ as a 3-form
and because you can start doing hodge star business
 
I haven't even gotten to using the Hodge start yet. I do remember $\star dF = 0$ or something like that from MTW
 
ya
i mean, it's basically the relationship between $\vec{E}\cdot d\vec{A}$ and $\vec{E}\cdot d\vec{l}$
first is a two-form, second is a one-form
and $\star$ maps between them
 
@bolbteppa What about T1 and P2 in the suffix instead of T and P
As per the text that is place I am stuck up with
 
Wait are you talking about the whole dotted quantity or just the element? I was thinking that $dA$ was a 2-form and $dl$ a 1-form, but what you're saying sounds like the whole thing
 
9:42 PM
well, $\vec{E}\cdot d\vec{A} = E_x \,dy\wedge dz+E_y \, dz\wedge dx+E_z \, dx\wedge dy$
so it's a sum of two-forms
often, though, you're dealing with the area form for some surface e.g. the xy plane
in which case you might as well say $dA=dx\wedge dy$
 
Ahh I was thinking the dot would change the object
 
well, i'd say it does: $d\vec{A}$ is formally a 'vector' of one-forms
but that's more notational than anything. it's $\vec{E}\cdot d\vec{A}$ that is meaningful in actual computation
 
@gateprep I think it means $\frac{\partial V}{\partial P}|_{T_1} = \frac{\partial}{\partial P}V(P,T=T_1)$ where $V = V(T,P)$, and $\frac{\partial V}{\partial T}|_{P_2} = \frac{\partial}{\partial T}V(P=P_2,T)$
 
@Semiclassical As in an element of $T(T^*M)$ or something? The way I think of it, a "vector of one-forms" would be a 0-form/scalar. Unless I don't actually know what a 0-form is
 
nah
as I say, it's just notation
 
9:57 PM
Ugh differential geometry hurts
 
so that $\vec{E}=(E_x,E_y,E_z)\implies \vec{E}\cdot d\vec{A}=E_x dy\wedge dz+E_y dz\wedge dx+E_z dz\wedge dx$
 
uh...
 
So you're saying $\vec{E}$ is defined in terms of $\vec{E} \cdot \vec{dA}$?
Not that $E \cdot dA$ is just an operation on $E$ wrt a surface?
 
no, I'm saying what $\vec{E}\cdot d\vec{A}$ represents in terms of $\vec{E}$
 
Is there a symbol for cyclic permutation?
 
9:59 PM
$d\vec{A}=\hat{x}\,dy\wedge dz+\hat{y}\,dz\wedge dx+\hat{z}\,dx\wedge dy$ is just notation
@Lozansky beyond just (123...n)?
 
@Lozansky $\epsilon$?
 
Well, $\epsilon_{ijk}$ I guess
 
can someone clear up the doubt from physics.stackexchange.com/questions/5888/… ? Basically, would a black hole sink or float over water, assuming it has a density lower than water?
 
@Semiclassical ahhh I see what you're saying now
 
I have seen $\displaystyle \sum_{\substack{i,j,k\\cyc}} f_{ijk}$
 
10:01 PM
Ron Maimon's answer claims that no, and he defends his position against the accepted answer... but that's so over my head that I have no idea who is right. can someone shed light?
 
Hmm, I've seen that too
 
If we wanted to do it without that notation, we'd need something like $E^i dx^j \wedge dx^k \epsilon_{ijk}$? I feel like some of the indices aren't where they should be
 
Don't fancy it :P
 
well, the hodge star helps: $E^i \star dx_i$
 
maybe @knzhou has an idea... ?
or @ACuriousMind
 
10:03 PM
Oh it looks like one of the definitions of the Hodge star is something like that indeed
 
since $\star dx_i = \frac 12\epsilon_{ijk}dx_j \wedge dx_k$. (you need the 1/2 because $dx\wedge dy=dy\wedge dx$ etc)
by contrast, the one-form is just $\vec{E}\cdot d\vec{l}=E^i dx_i$
though I think you can just write $\star(E^i dx_i)$
hence why I said it's basically just the relation between $\vec{E}\cdot \vec{dl}$ and $\vec{E}\cdot \vec{dA}$
 
Isn't $dx \wedge dy = - dy \wedge dx$ but the 1/2 comes in from antisymmetry of $\epsilon$?
 
Yeah, basically area is an anti-symmetric tensor, $dx^j \wedge dx^k = F^{jk}$ and any anti-symmetric tensor in 3-space is dual to a vector via the Levi-Civita tensor $F_i = \varepsilon_{ijk} F^{jk}$
 
yeah, I missed the minus
 
Same idea with cross products
 
10:10 PM
yeah
 
Cross product depends on two vectors and is anti-symmetric, i.e. is an anti-symmetric tensor, and so can be represented via a vector also in this special case
 
fun beans
 
that's a big difference between thermo and electrodynamics as far as multivariable calc goes: you never give a damn about cross products in thermo
 
Sounds like it's time to study more diff geom
@bolbteppa Is that related to the black magic where they get rid of metrics in favor of Levi-Civita?
 
(Though, you can view the occasional minus signs in the thermodynamic Maxwell relations as arising from the antisymmetry of 2-forms. So it does show up, but only in a restricted way)
 
10:17 PM
can't get rid of a metric in favor of levi-civita bruh
 
Basically things like $\delta_{ij}, \eta_{\mu \nu}, \varepsilon_{\mu \nu \rho \sigma}$ have the same form in all coordinate systems, so you would expect them to act as 'metrics' in certain cases, e.g. $\varepsilon$ is the 'metric tensor' for spinors but not normally
 
eww, no don't do that ._.
 
Yeah it's not pretty
 
vzn
@user54826 seems like a bizarre thought experiment. afaik there is a sense in which black holes are the "most dense/ large" "objects" in the universe. also they dont really interact with anything else in the sense of typical objects (eg "floating"), they only suck up all nearby mass.
 
Crap I actually don't know where I picked that up from now. Maybe it was from Zee's introduction to twistors? I didn't really understand that though
Yeah I guess that section was indeed talking about spinors
 
10:33 PM
uh huh...
 
10:57 PM
Is twistor theory something normal GR people use or is it only used in edge cases?
 
It's definitely not normal
Was probably the spinor section
 
11:10 PM
-1
Q: What is the point to ask a question that everyone knows the answers?

Sylwester LI've asked the question twice, How central forces work? I gives examples of bodies that have determined position and a defined angular velocity. But these are questions from forbidden knowledge, no clear, excommunicated to which no one knows the answer. Someone wrote that, I would edit the questi...

 
whomp whomp
 
@user54826 Ron's answer seems totally fine to me!
Often people are hesitant to answer any physical questions about black holes. They're like, well, this is the domain of GR, so the only "allowed" objects of discussion are black holes, point-like planets, and ideal point-like spaceships.
But if you were to actually mix other physics in, which you can, I think what Ron is saying is totally reasonable.
Even if Ron's answer is not ideal, it seems to me to be much better than the existing accepted answer, which basically just says "I refuse to answer because black holes are too weird to think about".
 
11:28 PM
My own attitude is more that I can’t conceive of a way to test it, so—absent any other motivation—I don’t find the question very stimulating
 
11:54 PM
One step further into quantum thermodynamics with the energy temperature uncertainty relation
 

« first day (2840 days earlier)      last day (2074 days later) »