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1:07 AM
I’d like to find a function which has the level curves attached in the image, i.e. centered in the first quadrant (between 0 and 1) and with an elongated peak. Maybe some kind of a Gaussian surface, i.e. $f(x,y)=Ce^{-(x+y)^2/D}$. I've tried Desmos, but to no avail.
 
 
1 hour later…
 
1 hour later…
3:23 AM
\o @RyanUnger
3
 
3:38 AM
@schn that certainly looks like a Gaussian - try fiddling around with $f(x,y) = A \exp\left(- \left(\frac{(x-x_o)^2}{2\sigma_X^2} + \frac{(y-y_o)^2}{2\sigma_Y^2} \right)\right)$
A is the amplitude, $(x_o, y_o)$ is the center, while $\sigma_X$ and $\sigma_Y$ control the spreads
 
4:15 AM
Do we implicitly assume homogeneity of space when we talk about translational symmetry of Newton's laws of motion?
(I think yes, but at the same time, I am at a loss as to how the laws would differ if the space wasn't homogeneous.)
 
123
Hi Guys... Yo
 
@schn actually, you'll need an $m(x - x_o)(y - y_o)$ cross term as well, since it's rotated
 
5:14 AM
@Jim I'm just here to confirm that I didn't give any TEDx Talk before.
 
5:45 AM
Happy Birthday Coronavirus. The first case was diagnosed in Nov 18.
 
 
2 hours later…
7:33 AM
Are questions ever randomly bumped to active?
 
Yes, the SE sometimes bumps old questions. Exactly how the algorithm for choosing what questions to bump works I don't know.
 
7:53 AM
How to go about showing that the total probability is conserved in Lindblad master equation?
I am trying to compute the derivative of the trace of the density matrix and show that it's zero but it's getting very messy.
 
123
Hi @JohnRennie did you checked series capacitor problem?
Also I am confused with force direction of electric field and magnetic field. Does In EM wave direction of Force of Electric field and Magnetic Field is same???
Electric field Force F_electric direction in diagram plot perpendicular to the direction of change motion. And Magnetic field B plot perpendicular to electric field E. It means force of magnetic field F_magnetic = q (v x B) it shows force of magnetic field is perpendicular to both charge motion and magnetic field direction. Which becomes the exact same direction F_electric (direction) = F_magnetic (direction). Not magnitude only direction
Pls help me.
 
@123 I thought we had answered that. What is there left to do?
 
123
8:22 AM
Yes @JohnRennie you cleared to me in series charge is conserved. But regular calculation does not support charge conservation.
Pls also help for understanding electric field force direction and magnetic field force direction.
 
I would have to see how that calculation had been done.
 
123
I can not upload pictures, I don't have upload button. It is any other way to share my calculation.
 
Upload the picture to imgur.com/upload and paste the link here
 
123
OK let me check imgur.com because I don't have account on it. But I used onedrive
 
You don't need an account on Imgur to upload pictures
 
123
8:31 AM
There is some error showing on imgur.com page don't open. I shared you one drive link. Pls Check @JohnRennie
 
OK, I've got those. I'll have a look now.
 
123
I have tried series combination with longer calculation. But found same result don't have this long calculation now.
Thanks @JohnRennie pls also help me to understand F_electric and F_magnetic directions
When you have time. It is confusing to me and how set specific polarization of EM wave to exact know the direction of both fields
 
I think the series calculation is wrong.
 
123
Pls share with correct calculation. I have tried with longer calculation step by step separately for each capacitor shows same result.
 
There is no calculation. The charges on the capacitors cannot change because there is nowhere for the charge to flow.
Note that connecting the two charged capacitors in series is not the same as connecting two uncharged capacitors then applying a voltage of 1600V.
In your calculation you are working out the charge if you apply 1600V to two uncharged capacitors in series.
 
123
8:42 AM
Yes Your explanation is seems right. But in book numerical I got the result what I have written. Also reffred two different solution manuals both have same results.
In my opinion series join two capacitors have potential differences which lead to flow of charge until charge become same on each capacitor. Then new potential is established at which charge is same on each capacitor.
 
123
8:59 AM
I think different potential is allowed in series join. V = V1 + V2, but different charge on each capacitor is not allowed. Due different charges on each capacitor charge flow from one greater charge capacitor to lower one. If wire has no resistance flow of charge is conserved Otherwise not (but calculation not showing conserved). In book we are taking an ideal conditions. I don't know the right explanation. Aaah... ;O
 
 
2 hours later…
10:56 AM
need a hint for (i)
 
 
2 hours later…
12:47 PM
Do we actually even know how mirrors really work?
 
1:01 PM
Do they not just... reflect light?
at what level are you talking about?
 
quantum mechanical level
like what exactly happen in the moment a photon 'hit' the surface of a mirror
 
I believe that field is called quantum optics
it uses quantum field theory afaik
 
hm thank you
do you think someone investigated this already?
 
@ACuriousMind thank you for the link
one answer stated 'each atom in the mirror will absorb and re-emit photons'. How can this be proven? Mathematically or by observation?
 
1:20 PM
I just came to knew why Computer Science & Engineering is called Computer Science Engineering and not Computer Science & Engineering because & is not a valid character while naming files.
 
@undefined we don't really "prove" things in physics :P "Observing" what happens to an individual particle when a beam reflects off a mirror isn't really feasible
 
yeah better not use 'special characters' in file names, RewCie :)
 
this is just the explanation that our otherwise well-tested theories offer, I don't think you can really "observe" this in detail
 
yep
 
@ACuriousMind I'm trying to wrap my head around this for many months now. But sadly I still can not fully understand it. And I feel like every explanation I find is not detailed enough (or I simply don't understand them). But I can't get rid of the feeling that there is something missing
 
1:24 PM
@undefined The "detailed" explanation would be that you have to compute the quantum mechanical interactions of the photons with the mirror surface and then find in some classical limit that it reproduces the simple classical reflection behaviour we observe for the beam
every physical explanation you get that somewhat comprehensible for a layperson is usually simplified to some extent, this is not unusual
 
the question which led me to the 'how does mirrors even work' is the question 'How does a reflected photon keep its entanglement?' (I know there are some questions with answers about this) but I feel like they are not satisfying and/or that something is missing.
 
why would it "lose" any entanglement?
 
It does not, and I know it doesn't
but when I try to combine pieces from various answers from questions about this topic, I always feel like I dont understand it at all and/or that something is missing
like the sentence from Emilio 'each atom in the mirror will absorb and re-emit photons'. If the atoms absorb and re-emit the photon, why/how does it not lose an entanglement? How does this absorb and 'processing' in the atom really works?
for me it feels like I have a bunch of puzzle pieces in front of me but I can't put them together to reveal the beautiful whole picture
 
What is the difference between Information Retrieval System and Real Time system? Why do we study them?
 
Hey everybody
 
1:36 PM
Hey
 
What is the equivalent resistance of the shown part of a circuit?
I kinda confused by the middle branch :/
 
You need to use Kercheroff's law
 
Let me try
Would the current through the 9 Ohm resistor would split?
 
VSLI System is at 7th sem.... I wanted it to be the first one.
@M.ÇağlarTUFAN No idea, I've forgotten them. I studied them in high school.
 
@ACuriousMind have you met Professor Weigand?
 
1:40 PM
@RewCie Hmm okay.
 
@NiharKarve I attended his QFT and string theory lectures. He's a great speaker if you can keep up with his fast-talking.
 
See Cyber Security syllabus. It's sooooo coooooooool!
 
@ACuriousMind awesome!
 
1:54 PM
What Firewall Design is also a course
!!?
 
Jim
@RewCie obligatory xkcd:
user image
3
 
@Jim LMAO! This will be totally fun!
Thanks for leaking Exam Papers we now know how to prepare :P
in Computer Science noob tries to get better (ft. RewCie), 14 hours ago, by Feeds
posted on November 17, 2020 by Andy Greenberg

The blockbuster game of deception has security holes that let cheaters run wild.

 
2:16 PM
I don't know what to do, I'm excessively excited as well as excessively nervous....
 
I was surprised to read this, anyone know exactly what it means? I thought the whole problem of qft was that 1. it can't be done fully non-perturbatively and 2. there are basically theorems that say perturbative qft doesn't even exist
 
@Charlie QFT is not defined by perturbative methods - you can write down non-perturbative definitions of e.g. what a scattering amplitude or the expectation value of some observable is. You can't really compute that, but you can write it down. Then you can do other stuff - often perturbation theory or lattice computations (which are not perturbative, juts approximative in a different way!) - to actually compute these things
Concretely, the asymptotic series over Feynman diagrams is a way to compute amplitudes, but it is not its definition
 
oh I see
 
in string theory, a close analog to the series over Feynman diagrams, a sum over worldsheets, is taken as the definition of stringy ampltudes
i.e. it's not that we have some non-perturbative idea that we approximate via a perturbative series, string theory just takes this perturbation series as the amplitude, there's no approximation there
 
oh ok that's interesting
 
2:26 PM
Many believe there "should" be some underlying non-perturbative theory (cf. "string field theory", "M-theory",...), but so far they've got nothing conclusive :P
 
:o
 
 
2 hours later…
4:42 PM
A good amount of users here with nothing going on... hey everyone haha
 
Hi BioPhysicist, how're the calcium signals?
More seriously, how were you able to do any experimental stuff during the pandemic?
 
123
Yo...
Pls explain : Direction $F_{Electric} = $ Direction $F_{Magnetic}$
 
@NiharKarve Most of my work involves mathematical modeling, so the only hit to my work has been the lack of productivity from staying home :)
 
123
Does both forces has same direction.
 
I think you will need to give more details
 
123
4:55 PM
$F_{Magnetic} = q ({v} \times \vec{B})$ it shows force of magnetic field is perpendicular to direction of both $\vec{v}$ & $\vec{B}$ due to cross product. Which is the direction of electric force.
 
I am still not following. Why does the magnetic force have to be in the same direction as the electric force?
 
@123 1. If you're talking about an electromagnetic wave again where electric and magnetic field are perpendicular, you should say so. How is BioPhysicist supposed to know that? 2. There is more than one perpendicular direction to a given vector unless you're in two dimensions.
(i.e. just because the magnetic force and the electric field are both perpendicular to the magnetic field, that doesn't mean they're parallel)
 
Can you tell just by looking at a metric that the manifold is flat? For example, this metric $$ ds^2=\frac{dxdy}{xy} $$
 
123
Hi @ACuriousMind ... magnetic force is not perpendicular to electric field.
magnetic field and electric field are perpendicular . I think this is what you want to say. or anything else.
 
It might be, it might not be - depends on the velocity of the charge its acting on - but I didn't say anything about that, so what's your point?
 
123
 
@geocalc33 you can tell by just looking if you can compute the Riemann tensor in your head, otherwise you can't :P
 
that's the Minkowski metric isnt it
 
@BioPhysicist that's pretty cool - I suppose the general public doesn't get to hear as much of the mathematical modelling aspects of biology. Are there many BioPhysicists like yourself? (I don't know many people who enjoy both fields :D)
 
@NiharKarve Where? There are many people who study biophysics, physics of living systems, etc.
 
@NiharKarve I almost have what I want, however, it's not elongated but very circular. Check this out: wolframalpha.com/input/…
 
123
5:05 PM
look at this link. $F_{Magnetic} = q (\vec{v} \times \vec{B})$ direction of propagation of EM wave is the direction of $\vec{v}$ and direction of $\vec{B}$ magnetic field is perpendicular to electric field. But $F_{Magnetic}$ is perpendicular to both $\vec{v}$ & $\vec{B}$ due to cross product.
 
@BioPhysicist oh, I just meant the relative size of your department
 
123
So direction of magnetic force & direction of electric force become same. because direction of electric field and direction of electric force is parallel.
 
@NiharKarve I'd like the Gaussian to be more egg shaped and symmetric around $y=x$.
 
@123 The $v$ in the expression for the magnetic force is the velocity of the charged particle the force acts on, not of the wave
 
oh ok. Well I am in the physics department at my school, but there are 4 biophysics labs within that. Then there are some biophysics people over in the biology department too. I am unsure about numbers though
 
5:08 PM
@ACuriousMind I wonder what the velocity of the wave is...
 
@RyanUnger depends on the ether :P
 
the speed of light is variable
 
123
@ACuriousMind Ooookay. It means $F_{Magnetic} = q (\vec{v} \times \vec{B})$ not for EM wave.
 
æther
 
@RyanUnger $ds^2=\frac{dxdy}{xy}$ is a pushforward of the Minkowski metric: $ds^2=dxdy$, so I guess it implies flatness?
 
5:10 PM
@schn to make it more egg shaped, you need to prefix the (x - x_o) term with a factor; to rotate it, you'll need to add in a $b(x - x_o)(y - y_o)$ cross term
 
a pushforward? No, just use log/exp coordinates
that will give you mink in double null
dx/x = du
dy/y = dv
 
@geocalc33 it is surprisingly hard to tell whether a metric is equivalent to another metric just by looking at it. The only reliable way to tell a metric is the same as the Minkowski metric is to compute the Riemann tensor and show it is zero.
 
@schn Personally, I'd recommend the 3D GeoGebra grapher for this - you can play with the sliders, etc.
 
@NiharKarve Nice, thanks for the reply. Will try.
 
@BioPhysicist ah ok :)
 
5:13 PM
@123 I don't know what you mean by that. Forces act on bodies, not waves. The magnetic force on a body with charge $q$ and velocity $v$ in a magnetic field $B$ is $qv\times B$.
whether that magnetic field belongs to an EM wave or comes from something else is completely irrelevant
 
@NiharKarve By cross term, do you mean adding $b(x - x_o)(y - y_o)$ to the exponential term, so $e^{...}+b(x - x_o)(y - y_o)$?
 
Yes
 
@JohnRennie alright I better compute the Riemann tensor :)
 
When the duplicate flag you raised gets accepted, does it automatically add a comment with "Does this answer your question..."?
 
@geocalc33 Mathematica is your friend here. Doing it manually is a lot of work.
 
123
5:18 PM
@ACuriousMind i meant to say if Direction of magnetic force and electric force of EM wave is same it apply force on particle in same direction. How do we know it is due to electric force or magnetic force. This is the problem
 
@NiharKarve do you mean if you vote to close as a duplicate?
 
@JohnRennie yeah
 
Yes, when you vote to close as a duplicate the SE automatically adds the comment.
 
Ok, I was worried I involuntarily wrote that and had no recollection of it
 
123
@JohnRennie pls look at glance on my question also. It also required your contribution. I want to understand the phenomenon.
 
5:21 PM
@123 it's too late in the day for me to do physics i'm afraid. Ask me tomorrow morning.
 
123
@JohnRennie OoooooKay....
 
@RyanUnger I think I see your point about using log/exp coordinates. But what if you don't use that coordinate change?
 
@NiharKarve I've tried Geogebra and I can see it somehow being rotated when adding the cross term, but in WolframAlpha the level curves look a bit strange. Consider first adding a prefix to $(x-x_0)$.
@NiharKarve Then adding a cross term...
Some level curves grapher would be useful here.
 
Er, the cross term needs to be inside the exponential (sorry if I was a bit unclear) - like this: $f(x, y) = e^{a(x-x0)^2 + b(x-x0)(y-y0) + c(y-y0)^2}$
 
5:53 PM
@geocalc33 then you compute the Gauss curvature
it's easy because the metric is manifestly conformally flat
but like there's a log or exp involved in that formula and then it'll boil down to something trivial
 
6:12 PM
@NiharKarve When rotating the function, it has a tendency to explode at the ends. See: geogebra.org/3d/ryhuqf5q
@NiharKarve Or maybe it is only undefined there.
 
 
2 hours later…
8:29 PM
@Slereah you should preorder amazon.com/…
 
Should I?
 
yeah
 

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