Can someone explain what does "decay" mean in nuclear physics ? I know about some nuclei decays into lighter nuclei to get stability but that lighter nuclei can be assumed to be a part of that decaying nuclei (together with some more nucleons) . So can we similarly say that an electron is a constituent of a W- boson because W- bosons decay into electrons and anti neutrino ?

Also do the quarks inside a proton experience coulomb force because of charges on them ?

@Ankit Yes, the quarks do experience the Coulomb force, but this is relatively small compared to the strong force so it doesn't play a large role in reactions of hadrons.

@Ankit A nucleus is a composite object made up from protons and neutrons, and in a nuclear reaction like uranium fission all that happens is the neutrons and protons separate and rearrange into different nuclei (and free neutrons).

But a fundamental particle like a W boson is not composite i.e. it is not made up from other particles. Instead it can transform into other particles. This happens because quantum field theory allows fundamental particles to be created and destroyed.

In effect what happens in the W⁻ ⟶ eν reaction is that the W particle is destroyed, and the energy released is used to create the electron and the antineutrino.

evening @JohnRennie :)

@antimony Hi :-)

@DanielUnderwood FWIW, John Baez was doing a blog-like thing called This Week's Finds in Mathematical Physics on Usenet, before the WWW existed and before the term "blog" was invented. In a sense, he was one of the inventors of blogging.

@PM2Ring blogging is not much new

People have been doing regular articles for a long time

Sometimes in websites, sometimes in newspapers!

also in newsletters, for old nerds

@Slereah Fair point. In that case Benjamin Franklin was a very early blogger. :)

Also Luther

Early blogging

The thing about blogs (and This Week's Finds) is combining personal "what's happening in my life" stuff with a technical column. That didn't really happen much before blogs.

Ben Franklin also extensively practiced sock-puppetry. He used a huge number of pseudonyms, to make it less obvious that a large number of the articles in his newspaper & almanac were written by him.

Gonzo journalism

web.archive.org/web/20100510034949/http://…

@PM2Ring yeah I actually saw some usenet stuff a while back (I think on his site) and went on a little dive

it seems that unmoderated lists these days are "interesting" and I didn't see much activity on moderated lists unfortunately

kind of makes me sad to miss the early eras of the internet before my time

I didn't have an internet connection back then, but I was active on FidoNet. I think things were a little more civilised back then. We still needed moderators, but they rarely needed to do much. On FidoNet, the technology itself had a moderating effect: arguments are less likely to spiral into a frenzy of anger when there's a day between replies. ;)

But I wish I'd been on Usenet when Terry Pratchett was active there.

Hey!

So there is this cactus diagram in Feynman diagrams

What process does the loop show?

it's a 3-loop correction to the free propagator

No I mean

Is this correct??

it's just a diagram

how could it be "incorrect"?

Can someone help me out.. I have been taught that electric field and magnetic fields are like two different faces of same coin . They are just the same thing viewed from different perspective. But if that's true : Why does one form a closed loop while the other can't ?

@ACuriousMind no I meant the momenta I have labelled

Because I don't understand whether it is a particle decaying into 3 particles or a particle decaying into two and annihiting which gives a 3rd

@Ankit I don't know about faces but it's like one can't live without other

@Korra it's a diagram that contributes to the propagator, there is no "decay" here

you have a particle entering from the left, doing stuff, then a particle just like it leaving at the right - why do you think this is a "process" at all?

@Ankit the "different faces of same coin" means that if you accept special relativity + Coulomb's law, then you get the rest of electromagnetism automatically, i.e. you can't have electric fields without magnetic fields because what is a pure electric field in one frame is an electromagnetic field in another frame

@ACuriousMind I don't understand. What do you mean by contributes to the propagator?

it's not supposed to mean that electric and magnetic fields have no differences

@Korra let's take a step back - why did you draw this diagram to begin with?

i.e. what are you trying to compute?

@ACuriousMind No I just found it in Schroeder and I was just wondering what it implied

A diagram alone doesn't mean anything

you need to know at least what theory it has been drawn for (Standard Model? Scalar $\phi^4$? Something else?) and what particles the lines represent to say anything

@ACuriousMind phi 4

@ACuriousMind so it's not an annihilation process?

@ACuriousMind shape of the line tells you the theory, obviously

is it wavy or curly???

@Korra why would it be?

An "annihilation" for me involves particles going into the process that don't come out again, and in particular usually a particle and its antiparticle

finding the area under the integral curve

It depends on what ur integrating with respect to @geocalc33

Hi everyone! Does anyone study physics as if it is their hobby? Not some student in a university to score marks or is working professionally

@cOnnectOrTR12 there aren't actually many h bar regulars who are currently physics students/researchers

I meant in a general sense. Does somebody study because they love it.

@ACuriousMind so complete electric fields in one frame becomes magnetic field in the other . But shouldn't both have the same properties then ?

@Ankit no, it's not that the same vector field that's electric in one frame just becomes the magnetic field in another frame

you have to actually do the Lorentz transformation on the field

Hello. If we have a system, and we consturct the lagrange function of this system and then insert our Additional condiitons into the lagrange Function.. For example then we will recieve a lagrange function that is only dependend on two variables. Do we say then that the system have only two degrees of freedom of motion?

What if the lagrangian we constructed without inserting the additional conditions contain three variables (the conditions for example eliminate one) do we say the system has three degrees of freedom?

@ACuriousMind okay.. so does that transformation account for those differences in their properties ?

An example: $L = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2)$ is a free particle Lagrangian with two degrees of freedom. If we want to constrain it to the surface of a circle we can add a constraint using a Lagrange multiplier, by adding $\frac{1}{2}\lambda(x^2 + y^2 - R^2)$ to the Lagrangian, if we now choose polar coordinates the constraint vanishes and there is only one degree of freedom, $\theta$

@ACuriousMind can you tell me how crucial are the numericals in understanding some concept in physics?

@geocalc33 finding the curve whose which satisfies the differential equation which says it and it's derivatives are related in some way

In addition to the question ia sked, i am also intrigued to know, if the time is explicit in the lagrange function thus L(x1....xi,t) do we say that the freedom degrees are in this case i + 1 ? or do we not consider it ?

Time is treated as a parameter that the functions $x_1,x_2,...$ depend on as explicit functions via $x_1 = x_1(t)$, if $L$ depends explicitly on $t$ we don't interpret it as an additional variable we treat it as a parameter

In special relativity that changes but we then constrain it to be related to the other coordinates, but we can also treat it as a parameter that the others depend on if we want to

is the Big Rip model compatible with multiverse theory?

@bolbteppa Can you check my previous question on the degrees of freedom.

I just commented on the degrees of freedom in my post before with an example of a free particle vs. one constrained to a circle and losing a degree of freedom in doing this

If one universe exponentially gets larger how could it still be separate from other universes?

there is no "multiverse theory"

you're gonna have to ask about a specific theory

@bolbteppa what if i have a lagrangian dependent on for example $L( \phi, \theta)$ then i would say my lagrangian depends on phi and theta.. which are two degrees of freedom, however if insert a condition for example $ \phi = \omega * t$ where as $ \omega = const $ then i will now have one degree of freedom. Did i understand this correctly.

The time point is right, but the degrees of freedom point is not, in the example above I had two variables $(x,y)$ but we seen when we went to polar coordinates that only one variable really exists, $\theta$ (because $R$ is constant, in general it would be a variable, e.g. if you change the free particle Lagrangian I started with to polar coordinates, but when I add the constraint one disappears)

Yes

But why does changing the coordinate system change the degrees of freedom if we do not add constraints? or is that not what you meant?

@Slereah can you tell?

Because if you did not add the conidition you would still get two... nvm i got it i understand now.. Thanks @bolbteppa

@cOnnectOrTR12 i think you can understand most of physics through description, but if you want to test theory you need to understand how to do math :P

It doesn't, when I add the constraint $\frac{\lambda}{2}(x^2+y^2-R^2)$, $\lambda$ is a new 'variable', the 'equation of motion' for $\lambda$ just enforces $x^2+y^2-R^2 = 0$ so I can then solve this for $y = \sqrt{R^2-x^2}$ and re-insert this into the original Cartesian coordinate Lagrangian and see it's all in terms of $x$. But you need to examine it and realize this, naively at the beginning it looks like $x$ and $y$ are two variables

Yup i got what you mean.

This obviously gets very tricky quick and you need to be careful with 'holonomic constraints' vs other types etc

@RonaldVilliers what theory?

Do people ever use non-holonomic constraints

They seem very reasonable but I've never seen them in use

Probably too practical a tool for me to read

I'm just not sure, I keep meaning to get into this but I'm just hoping to come across some super easy explanation that makes it all trivial :(

scientists in movies don't do enough bibliography

That's like half the job

@RonaldVilliers I'm integrating under a real analytic integral curve w.r.t. the independent var.

If i integrate each integral curve like this, im mapping each integral curve to a real scalar

@gls got any suggestions for quantum simulators? there's the list on Quantiki here but it's a bit overwhelming

(also Quirk seems neat)

@Semiclassical it is, isn't it? The main developer is active on qc.SE btw, if you have any questions in that direction

@Semiclassical you've gotta be a bit more specific here. To do what exactly?

personally, I've never really found the need to use any "dedicated quantum simulator". Most of the time "quantum simulation" amounts to matrix multiplication anyway (at least, what I believe you mean with "quantum simulation" here), or solving some first-order differential equation if you work with dynamical systems. Not that there aren't cases in which this isn't trivial but... working in python, I found qutip to have pretty much all I needed

@geocalc33 In the example of it being a force field that your D.E describes, your integral curve is the force at every point on the curve, and your integral w.r.t time of that integral curve gives the "impulse". If you integrate w.r.t its position along some axis, you get what is called "work" done by the curve along that axis.

I might be wrong but I believe that's what the usage is in physics.

@glS yeah, that was a bit too broad. i'm looking at student resources/tools for a QC course

for creating reasonably simple circuits Quirk seems top-notch

esp. for the kinds of problems one sees in, say, Nielsen and Chuang

it doesn't, no

you can generalize the method, but it won't be as simple as that

see en.wikipedia.org/wiki/…

oh, second order

yeah, that's gonna be unpleasant

for instance, mathematica refuses to solve it

Nice

(for arbitrary c)

what if i knew exactly what c is

then maaaaybe. for instance, when $c=t$ you get Airy functions: wolframalpha.com/input/?i=dsolve+x%27%27%2Bt+x%3D0

hmmmm

@Semiclassical to play around for that, yes, definitely. You can also have fun building the tools you need yourself, which might be also even more fun/better for learning (assuming you have the time). I always found Mathematica to be best tool for learning and playing around like that

i'm so spoiled by Mathematica access

@MadSpaces doesn't look likely: wolframalpha.com/input/?i=dsolve+x%27%27%2Bsin%28t%29+x%3D0

analytically, anyways

Mathematica's symbolic solution doesn't seem any better

Can i write you in private chat in german

i'd rather not. there's really not much more for me to say

Well alrighty then.lol

@Semiclassical with MMA it's not too hard to do "circuit builders" that you can dynamically interact with, e.g. i.stack.imgur.com/osjEV.gif

oh, that's cute

my main use of MMA for quantum has simply been to rely on KroneckerProduct and PauliMatrix :P

and do linear algebra

@Semiclassical what do you mean?

as in, to compute unitary matrices

and do matrix multiplication

nothing sophisticated

but matrix multiplication doesn't need to be sophisticated

@Semiclassical I mean... almost all calculations in (this brand of) QM boil down to some form of that anyway. Being able to do more is cool in MMA mostly because it allows to visualise things in dynamical and flexible ways

yeah

otherwise MMA is the proverbial crane crushing a fly

you can find the code for the above on GitHub here by the way. I haven't touched it in a fairly long time though, so I can't promise it even works anymore. I just did that to play around when I wanted to visualise a few things about gates etc

neat

yeah, that's pretty short

If a given transformation produces a lagrangiagian of the form $ L_{new}= L_{old} + \dot r * r $ then it is not symmetrical transformation? ( Because as far as i know, the lagrangian needs to differ by a total time derivation, and i dont think there is a function F which produces the given right term by a derivation, .. right?

Well.. maybe if we set $F = r^2/2$.. does that work and then its okay to do this transformation? ... what about one where we get instead of $ \dot r * r $ we get $ \dot r *t $ ?

$\dot{r}r$ doesn't look like a total derivative

so probably not

what about $ d/dt F $ $F= r^2 / 2 $ this produces $ \dot r *r $!

True

Then that does sound like it would be a symmetry!

What about the second example i gave, where we have instead $ \dot r * t $ i can not seem to integrate this function.. even with parts