01:00 - 21:0021:00 - 22:00

1:49 AM
@Amit i would also mark that person as my sworn enemy

3 hours later…
4:50 AM
Is it possible to geometrically depict the Wigner rotation angle in a (2+1) dimensional spacetime diagram?

5:13 AM
when considering two "weakly interacting" einstein solids, we denote a "mesostate" as a state of the composite system specified by the energy in einstein solid one and einstein solid two. we compute the number of microstates for each mesostate by multiplying the microstates of the corresponding individual subsystem macrostates.

@SillyGoose i love miso soup

we can only compute the number of microstates in this way because we are assuming (weak interactions) that the accessible microstates of einstein solid one are independent of the macrostate of einstein solid two?
hm maybe im confused

@SillyGoose om nom

i suppose it's valid anyways
@Relativisticcucumber why doe

@SillyGoose cuz ymacf

1 hour later…
6:30 AM
@GiorgiLagidze hi that proof has a simple fix. assuming product terms dont exist, we get to $aL'+a^2/2 L''+... = \frac{\partial f }{\partial t}$. the rhs is independent of $x$ because $\frac{\partial f}{\partial x}=0$. So the lhs is also independent of $x$, which implies $L$ is at most linear in $x$ (if L had higher powers of $x$ then the lhs would be dependent on $x$)
this also fixes the conclusion of this proof so that $L$ that is linear in $x$ is allowed @GiorgiLagidze

Found a 1950's journal for pseudoscience stuff

6:54 AM
@RyderRude hey. problem with it is the original proof tries to show why L can't contain $x$ at all and what you just did still allows the possibility that L can be linear in $x$, it fixes the conclusion about $\frac{\partial L}{\partial x} = 0$, but as i said, it allows linearity in x which is not what proof aimed to prove

7:10 AM
@GiorgiLagidze yes. if you dont revise the conclusion of the original proof, then it just cant be fixed. the only fix i can see is to add isotropy as an assumption of that proof as well, so that you can later rule out the x dependent cases

exactly.
btw, what do you think about your proof ? @RyderRude hackmd.io/NeANi7kQRVm717rBJ6D66A

it is correct :P
it has three assumptions : homogeneity, isotropy and absence of product terms

hackmd.io/NeANi7kQRVm717rBJ6D66A can you check this ?

if we have a perturbed hamiltonian $H_0 + \lambda V$ with g-fold degeneracy for energy $E_d$ why can we not just solve the characteristic equation exactly as opposed to doing perturbation theory?

@GiorgiLagidze here you have concluded that a homogenous and isotropic lagrangian of two particles cannot depend on both x_1 and x_2
but this is not true

7:14 AM
yes, i mean in inertial frame, space is homogeneous and isotropic, so for 2 particles, that yields that L can't depend on x1 and x2, but this is wrong because for 2 particles, we know L depends on x
so where does my conclusion fail ?

homogeneity implies that L is of the form L(x_1-x_2)

why ? if it's L(x_1 + x_2), it's still homogeneous :P

hmm, no it is not. x1+x2 is not invariant under x1+a and x_2+a
x_1-x_2 is

it's not invariant the same way it's not for single particle
for single particle, we had: $L' = f(v) +g(x+a)$
and we saw that it yields the same E.O.M as $L = f(v) +g(x)$

lemme think. then i will reply

7:20 AM
@Sanjana what about the diagrams on Wiki is insufficient?

ok so the rhs of the EL eqn is dL(x1+x2)/dx1. This is not invariant under the transformation x_1=x1+a, x_2=x_2+a

@SillyGoose what do you think specifying the energy does if not specifying the accessible microstates (they're those with that energy)?
@Slereah I love the question marks; at least they're slightly uncertain about the order of all things :P

@ACuriousMind it is cutting edge research

@RyderRude hm... are you sure ? if L contains x1 + x2 in linear form, it's invariant

the edge in question being the edge of the trangle?

7:23 AM
@GiorgiLagidze it is invariant for linear functions like L= f(v) + k(x_1+x_2) though
@GiorgiLagidze yes, but this is not the only solution for it to be homogeneous
@GiorgiLagidze for $L(x_1-x_2)$, you can have arbitrary non linear functions

Fun little journal but overall not that interesting though

so non linear functions like L(x_1-x_2) are allowed for homogeneity

The articles tend to be mostly the same

@ACuriousMind Those are not space- time diagrams. If I boost in one particular direction then the axes of the transformed tilt towards each other to form a non-orthogonal pair of axes in a spacetime diagram. For non-collinear boosts I need atleast a 3 dimensional figure...

I guess I'm not sure why you'd need a spacetime diagram when you can evidently see the rotation already in a space diagram :P

7:26 AM
for e.g. the gravitational potential is of the form ~ 1/(x_1-x_2) @GiorgiLagidze

all the articles are basically "I will finally mend the modern divide between science and religion!"

this is non linear but makes for a homogenous law

@RyderRude for L(x_1 - x_2), homogeneity doesn't imply that L must be linear in $x1-x2$ which I agree with and we can't do the logic as we did for single particle.

Apparently the really nutty quantum mysticism stuff appears in the 1960's according to this book : books.google.fr/books?redir_esc=y&hl=fr&id=Nl26tCsGE5UC
But it is hard finding it in full

@GiorgiLagidze yes. and most real world interacting two particle lagrangians are of this form

7:28 AM
but for L(x1 + x2), homogeneity implies that they must be linear and isotropy implies that they can't even be linear which means that L can't ever have the form of L(x1+x2)

@GiorgiLagidze yes. there is no homogenous and isotropic lagrangian of that form
unless it is constant

@Slereah I mean I'd say that's an accurate representation of like 95% of pseudoscience :P

@ACuriousMind Won't it be fun? :p I mean even in cases of length contraction I can just draw a big rod and a small rod and be done with it...but it's like we already know it's gonna happen. Given those space diagrams nobody can even find out the expression for the angle or even conclude from the first of those diagrams that there is going to be a rotation (well, one can if he does the math without the diagrams too)...

linear functions of (x1+x2) are allowed for homogeneity but ruled out for isotroy @GiorgiLagidze

The last conclusion that we just made is this: L can never ever have the form of L(x1+x2) in any circumstance. because even if you don't consider homogeneous/isotropy space, one can always look at the L from such space. So what if the textbook gives us the problem: "$L = L(x1+x2)$

7:30 AM
If I can draw a spacetime diagram and say "Look! That angle that line makes with that line...that is the Wigner rotation angle"---I thought it would be nice...

The Tao of Physics might be the big quantum mysticism book according to this

@GiorgiLagidze the conclusion we made is that L cant have that for when L is both homogeneous and isotropic
when L is just homogeneous, that form is allowed and shows up in physical situations

@Sanjana I don't think I've ever seen a 3d spacetime diagram that I didn't find confusing :P It's already hard to think about the 2d versions "properly" without being misled by Euclidean intuition

yes, but note that we can always find reference frame in which space is both homogeneous and isotropic and from that reference frame, how can L have such form ?

for e.g. it shows up in accelerated frames motion of two non interacting particles
@GiorgiLagidze L is frame dependent

7:32 AM
@Slereah Are you trying to find the origin of quantum woo so time travelers have a new target or what's going on here?

@ACuriousMind you have to construct it
Out of pipe cleaner and popsicle sticks

L has that form in accelerated frames but ceases to have that form when u shift to inertial frame

@ACuriousMind just doing an overall review of the history of GR including popular conceptions

@GiorgiLagidze L is frame independent only when u switch between inertial frames. otherwise, L is frame dependent

hm, i think you didn't get me. L = L(x1+x2) in accelerated frame. ok. now, I look at it from inertial frame. L is still L(x1+x2). no ?

7:34 AM
no
L changes its form when u shift from accelerated to inertial frame
it may be easier to see this with just one particle
whats L for a free particle in an inertial frame and an accelerated frame?

in inertial frame, mv^2/2

and in accelerated frame, the EL is a= c
so u get 1/2mv2 + mcx

@Slereah ah, and the cranks didn't do you the favour of separating the GR nonsense from the quantum nonsense cleanly? :P

It tends to be a bit mixed up together usually yes

@GiorgiLagidze this should make sense. the same particle follows a straight trajectory in an inertial frame but a curved one in an accelerated frame

7:37 AM
true. @RyderRude let me think on it and will reply later :P thanks a lot

assuming we have more than 1 dimensions. but even in 1D, the EoM are very different

Relativity in pop culture tends to be much later than you'd think really overall
but I guess that's always true of science
takes a while for things to filter down to the public
Like basically nobody outside of scientists knew about relativity before the 1920's
although quite a lot of discussion about higher dimensions in the 19th century

Where is John Rennie nowadays? Is he okay?

but then again that was something that had existed since Gauss and the public was unaware of it until the 1870's, so same principle there

@GiorgiLagidze after some thinking, for two particles, homogeneity alone implies : $L= f(v1,v_2) + g(x_1-x_2) + c(x_1+x_2)$. isotropy revises this to : $f(v_1^2, v_2^2) + g((x_1-x_2)^2)$

7:58 AM
"According to Wikipedia (hmm. . . ), “F ” is for ferme (French for “closed”)
and σ for somme (French for “sum” or “union”). “G” is for gebiet (German for
“neighborhood”) and δ for durchschnitt (German for “intersection”)."

or rather it should be $cx_1 + dx_2$ . for accelerated frames, this is $m_1ax_1+m_2ax_2$

@RyderRude you didn't have to add $c$ function. adding constant $a$ to x1 and x2 is enough

@GiorgiLagidze yes. i meant c as a constant implicitly
it's not a function

@Slereah the public might have been a bit busy with a little world war :P

I mean pop science was still a thing
It was just focused on other fields
Mostly electricity and natural sciences
Also human sciences were pretty popular

8:06 AM
not a lot of obvious applications of relativity around in everyday life

electricity was mysterious back then. tesla used to do weird experiments with it, which is y it may have been part of pop sci
there was also experiments to revive dead people using electricity
x rays were also experimented a lot on people

Lots of racial science stuff, too

@RyderRude when they discovered that x-rays were able to see bones inside people, they put x-rays beside soon-to-be-dead-people. they were expecting that it would see the souls emerging :P

@GiorgiLagidze LOL. those poor people
@GiorgiLagidze there were related experiments about measuring the weight of a soul

yes. all of them resulted in nothing but kudos for them to trying it

8:15 AM
yes. admittedly, this "x-ray can see souls" sounds like an interesting experiment from 20th century perspective. same for "electricity can revive dead people"
these experiments happened because x rays and electricity were not understood
with modern scientific understanding, it is hard to imagine people try these wild random experiments with new substances

@RyderRude $L = f(v_1) + g(x_1 - x_2)$ is this lagrangian for inertial frame ?
$L' = f(v_2) + g(-x_1 + x_2)$ this definitely gives different e.o.m, and fails for isotropy
you could also argue that it must have complied with isotropy as well and to do that, it must have been independent of $x_1$ and $x_2$

@GiorgiLagidze yes. and it should be f(v1,v2)
@GiorgiLagidze you need to do (x_1-x_2)^2 to pass both homogeneity and isotropy

yep. so linear form of $x_1 - x_2$ is still for non-inertial frame as it turns out

yes. c(x_1-x_2) is not isotropic where c is a constant
@GiorgiLagidze no, it's not for non ineetial frames either
i dont know of any physical situation where this would be relevant
accelerated frames have m_1ax_1 + m_2ax_2
the signs of both terms are the same
the sign of the term comes from the sign of acceleration of the frame. so all particles have the same sign

8:42 AM
@GiorgiLagidze ok so this can show up in an inertial frame with an external uniform electric field. the two particles must be of equal but opposite charge
then you have the term qE(x1-x2)

yes, i think everything is clear now. just summarizing my thoughts to make sure that i feel confident

2 hours later…
10:58 AM
@Amit hi :)

11:15 AM
Me playing chess
Is this a good or bad win
I'm white
Where can I improve?

@MoreAnonymous i thought it was good
@MoreAnonymous but i thought you gave away a pawn in the beginning

11:39 AM
Yea ... I'm not perfect ... but no one else is either
How do I get gotham chess the review the game?

12:05 PM
@RyderRude since we concluced that L can't contain x, how do we conclude that it can't contain $t$ for free particle ? our logic was that time must be homogeneous, hence, the followings: $L = f(v) + g(t)$ and $L' = f(v) + g(t+a)$ should yield the same e.o.m which they do if you use EL which seems that even if L contains $t$, we still got time homogeneity. :P

@GiorgiLagidze yes, but g(t) is just a total derivative, so you can just throw it away. it doesnt change dynamics unlike linear V(x)

yes, so what my argument is that even if L depends on $t$, we still got time homogeneity, so from this logic, we can't imply that "L can't contain t"

Or any good chess reviewer?

yes, L can contain $t$, but what we should say is that you can always choose an L which doesnt contain $t$ if you have time homogeneity

same thing would apply to $x$ as well. you can always choose an L without $x$ and you would have space homogeneity
:D

12:11 PM
no, if you throw away x from 1/2mv2+x, the EoM becomes different
if u throw away t from 1/2mv2 + t, the EoM is the same
because functions of only t are always a total derivative term

my point is time homogeneity doesn't imply that L can't contain t

then it's true. total derivative terms of time can be there

then why does landau say again that time homogeneity implies L can't contain t
another mistake ?

because Landau doesnt want to go into technical details. he is giving you an intuitive motivation :P

i think i got better proof why L can't contain t :P

12:19 PM
how can you prove something that's wrong? :P
you just yourself gave the counter example

$\frac{d}{dt}(\displaystyle\sum_i \dot q \frac{\partial L}{\partial \dot q} - L) = -\frac{\partial L}{\partial t}$
This is energy conservation principle
if L depends on t, rhs will be non-zero
which means energy is not conserved
lhs is energy :P
and for freely moving particle, energy must be conserved
which means L can't contain t
WDYT ?

hmm
there should be something wrong here

I don't think so

because 1/2mv2+ g(t) does give you the dynamics of a free particle
the rhs is indeed non-zero for this

it gives you dynamics, but tells you nothing about energy conservation
if you use EL, ok, you got E.O.M the same
but that doesn't mean energy must be conserved along the path
I think this is better proof than using time homogeneity

12:24 PM
ok so the rhs actually cancels out anyway
calculate lhs for 1/2mv2+g(t)

@GiorgiLagidze Energy is not special, just like with any other Noether charge you need to use the proper formula when the symmetry is a quasi-symmetry

just plug this lagrangian into your eqn, you will get energy conservation
@GiorgiLagidze we get d/dt (mv2-1/2mv2-g(t))=-d/dt g(t)

$\frac{d}{dt}(\dot q m\dot q - \frac{1}{2}m\dot q^2 - g(t)) = -\frac{\partial g(t)}{\partial t}$

yes so d/dt(g(t)) cancles out
giving u d/dt (1/2mv2)=0

hm. true, but where is the mistake in my argument ? it's proved that $\dot q \frac{\partial L}{\partial \dot q} - L$ is the energy we know

12:27 PM
@GiorgiLagidze where is that proved?

like ACuriousMind said, this formula needs to be modified to get the energy in case of quasi symmetries in time

Since you can add arbitrary total time derivatives to the Lagrangian this is not always the energy
it's just usually what we mean by energy, but properly energy is the Noether charge of time translation, not that specific expression (and again that expression gets modified when time translation is a quasi-symmetry)

@ACuriousMind landau shows it

landau must have made some assumptions in the proof that it is an exact symmetry

12:31 PM
@GiorgiLagidze the text explicitly says it's assuming time-independence of the Lagrangian

so it's an exact symmetry. no total derivatives

@ACuriousMind yes, he made $\frac{\partial L}{\partial t} = 0$, but in my case, we don't make it 0
i guess what you're saying is if we don't make it 0, left side won't be energy anymore

@GiorgiLagidze so what is the problem? You're breaking the assumptions of the text, so why should this still be energy and/or conserved?

again, you cant prove something that is wrong. you just yourself gave the counter example with a t dependence :P
but then you proceeded to provide a proof that it must be t independent despite knowing your counter example
Landau says it must be t independent because Landau assumes exact symmetries. the point is to give an intuitive argument
he doesnt want to get caught in irrelevant technicalities like an addition of g(t), which does not help the reader

12:49 PM
@ACuriousMind he mentions time independence because he is proving conservation law of energy :)
even if L is dependent on time, the following formula - chat.stackexchange.com/transcript/message/64481720#64481720 is still energy E = K + U

why would you think so?
22 mins ago, by ACuriousMind
it's just usually what we mean by energy, but properly energy is the Noether charge of time translation, not that specific expression (and again that expression gets modified when time translation is a quasi-symmetry)

so you agree with this formula right - chat.stackexchange.com/transcript/message/64481720#64481720 ?

I don't know what else to say

as long as you don't assume it's energy right ?
just call it some quantity, but chat.stackexchange.com/transcript/message/64481720#64481720 left hand side, let's call it some quantity Q(assume it's not energy)
you can now say - $\frac{\partial L}{\partial \dot q} = \frac{\partial T}{\partial \dot q}$
so - $Q = \displaystyle\sum_i \dot q \frac{\partial L}{\partial \dot q} - L = \displaystyle\sum_i \dot q \frac{\partial T}{\partial \dot q} - L$.
Since $T$ is a quadratic function of velocities($\dot q$), plugging in $k\dot q$ instead of $\dot q$ would result in $k^2T(\dot q^2)$ where $n=2$. Hence it's homogeneous function. So the following also holds true - $\dot q\frac{\partial T}{\partial \dot q} = 2T$. Substituting this into $Q$, we get: $O = 2T - L = 2T - (T - U) = T + U$ which means our quantity(O) is total energy(E) of the system.
and by $T$, I mean Kinetic energy

I don't really know what you're trying to do here

12:55 PM
well, that's what Landau does :)
shows that Q = T + U

sure but he's assuming the Lagrangian is time-independent to begin with
I don't know what you're arguing about or trying to show here

even if Lagrangian is time dependent, the quantity is still Q = T+U that's what i show

of course if you assume that $L = T(\dot{q}) + U(q)$ you're also assuming that the Lagrangian time-independent
a time-dependent Lagrangian does not have the form $T+U$
I really don't know what the point here is supposed to be

i think i was wrong. for time dependent lagrangian, L is not T - U right ?

@GiorgiLagidze this proof is assuming L=T-U
yes, in your own counter example L= T-U +g(t)

12:59 PM
yeap, T-U assumes it doesn't contain $t$ explicitly
which means my original proof was incorrect as left side is not energy :D
@RyderRude all I wanted to achieve is show why L can't depend on $t$ the same way as we showed that L can't depend on $x$

@GiorgiLagidze you can't show something that's wrong :P
even homogeneity+isotropy does not rule out total derivatives like g(t)

yeah :)

@GiorgiLagidze what you can show is this : take the EL eqn of L= f(v) + V(x,t), and demand that the EL eqn doesnt depend on $t$. You will prove that dV/dx must be independent of $t$ @GiorgiLagidze
so this rules out any products of x and t due to time homogeneity

for V(x,t) sure
for g(t), no

V(x)+g(t) is not a product of x and t, so it's not a counter example to this theorem

1:05 PM
but get your point

@GiorgiLagidze yes, you cant prove that that must not be there in L

for why it can't depend on $\vec v$, it seems easy right. you have $L = f(v)$ and $L' = f(-v)$ and e.o.m results are different
unless L contains even power of $v$

honestly I think you're spending far too much time on silly arguments about the free Lagrangian :P

yes

the reason I spend such huge time is not because only for lagrangian :P

1:09 PM
in the real-world, all your Lagrangians will depend on both $\dot{q}$ and $q$, and if you're really unlucky they'll also depend on $t$ but mostly not

honestly, due to Landau's such complicated one page analysis, I learned a ton
which definitely will be useful
i didn't even know galillean relativity and didn't know reference frame logic
I learn by doing and listening to you guys. For example, when I read taylor series on Landau, i didn't even know at first taylor series, so i learned that as well

Landau isn't really written for an audience that just learned what a Taylor series is; you would probably learn much better if you tried to learn from sources written for your level

Ofc, I agree @ACuriousMind and as I said, this free lagrangian is the last piece I'm reading from landau. and i am almost done <3

nvm then

1:16 PM
hmm ok. how are you ? @user726941

Fine thanks, how are you?

i am fine too :)

Dunno how many variants of this coV thing there are, but I keep catching a different cold every other week

i read that sometimes it's good to let the fever burn out. the immune system fights off the thing on its own

Yeah, that's what I do.

1:21 PM
great!
some people think of fever as a bad thing because they dont know what it is
vaccines are truly genius though.

Sweating it out works for me, most of the time.

i too do that sometimes

These "colds" all attack my noise.

our bodies literally have this security system with officers and fbi and stuff
the immune system department is like. a huge city. it's on kurzesagt
@user726941 what do you mean

No fever, just uncontrollable sneezing.

1:27 PM
this is the video maybe
@user726941 oh. i have trouble standing up sometimes. i dont get sneezes

With extreme watering of the nose and eyes for me.

that must be terrible.

9 mins ago, by user 726941
These "colds" all attack my noise.
noise
👃 nose, sorry
That video opens with viruses flying up the nose.

1 hour later…
2:47 PM
Yo @RyderRude

@Amit long time

Yus, I took some time off to focus on the important things in life
Like video games and fantasy novels

oh, great :)

🙃🙃

are you learning more physics rn ?

2:51 PM
No
I think I forgot who Newton was

oh, same
i havent read novels in long

But there is something I gotta check. I heard that Faraday apparently had in writing the idea behind the final correction Maxwell did to the eqns?? That's curious

Faraday wasnt much familiar with the mathematics of theories. it would be very impressive if he did
@Amit what was the final correction btw
i remember seeing in a documentary that Faraday was thrilled when Maxwell computed the speed of electromagnetic waves and found that it was the speed of light

I dont recall now, it's that additional term in the curl of B that creates a time varying electric field I think
Yes since Faraday wasn't "educated" he just wrote this stuff in words rather than equations

3:14 PM
yes, there is this term. it was called "displacement current'

3:30 PM
Yus

2 hours later…
5:01 PM
I wonder if there is anyone that has worked with AdS/CFT. Specifically, with things like entanglement wedge reconstruction and subregion-subalgebra duality.

5:15 PM
@ACuriousMind Hi, I have a brief question. If you remember our discussion about the cross section. In Wikipedia it says :"The cross section is a measure of probability that a scattering event will occur". A measure of probability -was what I struggled to understand. Is it accurate to say that the probability density function is also a measure of probability ?

5:45 PM
Sep 16 at 12:36, by ACuriousMind
@imbAF "measure of probability" is not a technical term
Sep 16 at 12:37, by ACuriousMind
they just mean that it's a quantity that is usually used to obtain the probabilities they're talking about

I don't really understand how these 2 answers, answer my question\
@ACuriousMind can't the same be said about the probability density function? A quantity that is usally used to obtain the probabilities ?

sure

So it's a way of quantifying the probability
of decay
I have 2 more questions:
1. For a particle decay with one channel, first of all can I write i.e $\Gamma=0.34s^{-1}$. If yes, how would you interpret that value.
The 2nd part I don't get it :
Since the dimension of Γ is the inverse of time, in our system of natural units, it
has the same dimension as mass (or energy). When the mass of an elementary
particle is measured, the total rate shows up as the irreducible “width” of the

I don't really know what's unclear here

", the total rate shows up as the irreducible “width” of the
shape of the distribution."

5:59 PM
for 1, the probability every second that the particle will have decayed in that second is 34%

@ACuriousMind Ok good to confirm my own intepretation
Shape of the distribution? Of which? How is whatever distribution is considered related to the mass of the particle?

they're talking about the Breit-Wigner distribution

Ok thanks I will read it
Because I want to understand what is it saying
One more thing before I continue

@imbAF For 1, I would interpret that value as you either have Al-24 (unlikely) or As-85. 😉

Al-24?

6:06 PM
as I said, unlikely

what is that?
For a particle decay with two channels, if $\Gamma=0.34s^{-1}$ and $\Gamma_1=0.18s^{-1}$ and $\Gamma_2=0.16s^{-1}$. These $\Gamma_1$ and $\Gamma_2$ are already probabilities of decays to each channel. So why in the article it says that by doing the branching function, that is what gives us the probability of decay for a channel. In this case the probability of decay to channel 1 would be $0.18/0.34$. If this is the probability of decay to chanel one, that what is $0.18s^{-1}$ by itself ?

@PM2Ring do you have any opinion about the case of J.Assange

@imbAF where do you think the article says that?

In such cases, we are often interested in the branching fractions, i.e. the probabilities of the decay by individual modes. The branching fraction of mode i
is

I do not see the claim you made there
and it should be very obvious what the branching ratio is actually doing, I mean it's just a quotient

6:11 PM
branching fraction (which is $\frac{\Gamma_i}{\Gamma}$) is the probability of decay for a moe
mode
I mean, yeah, it showcases which of the channels is more probable of occuring

you can figure out what the branching ratio means as a probability just by applying the general laws of probability

but that has nothing to do with the probability of decay, the constant in the proportionality of decay with time

at some point you need to start drawing some inferences yourself instead of expecting texts to spell out every little detail or to be absolutely pedantic about their language

In my example there is a 50+% chance that the first channel is "chosen" but the rate of decay in that channel is 0.18
So, I believe I understand what the branching fraction showcases, and it has nothing to do with the probability of decay, instead, it has to do with the probability of a certain probability of decay occuring

it gives you the probability of a certain branch occurring assuming that the particle decays. It's just Bayes' theorem for A being a subset of B.

6:16 PM
Yes
But a branch occurring has nothing to do with the decay rate directly
I believe I tried to explain what I think of it above
I can't do a better description of it

6:35 PM
@Amit keeping focused on the important things in life is in itself important :P

7:24 PM
@ACuriousMind In the Breit Wigner Probability distribution, if the x axis is that of the energy, what is the y axis ?

...it's a probability distribution. What do you think the y-axis is for a probability distribution???

Aha
Note that for values of E off the maximum at M such that |E2 − M2| = MΓ, (hence |E − M| = Γ/2 for M ≫ Γ),. How does it come up with Γ/2 ?

@imbAF $E^2 = (M + \Gamma/2)^2 = M^2 + M\Gamma + \Gamma^2/4 \approx M^2 + M\Gamma \implies E^2 - M^2 = M\Gamma$.

7:40 PM
$E^2 = (M + \Gamma/2)^2$ where does this come from?

@imbAF it's literally what the article says with $\lvert E- M\rvert = \Gamma/2$? I just took off the absolute value bars

I mean that's my question
where does that come from
In the article it says "hence "

You can just read my logic from right to left

But if you remove the absolute value to the right, don't you have minus and plus options. You simply ignore the negative value
?

does it matter? did you try and see what you get when you choose the other sign?

7:47 PM
I did not

so do it!

Yes

I think you need to be in general a bit more willing to try and work stuff out instead of expecting a text to explain every little detail; the more advanced the topics you study get, the less of these minutiae they will typically explain

Yeah I need to do that
but at the same time, because I know I will fail
I opt to directly get the correct stuff
That way I conserve time
i.e you did that approx. in your derivation
I wouldn't

@imbAF the article said $M\gg \Gamma$
so we need to use that somewhere, and the usual effect of such comparisons is that you can neglect the powers of the smaller thing; this should be familiar to you as we do such approximations all the time in physics

7:54 PM
Yes it is familiar

let me stress I didn't just perfectly recall that derivation, I tried out a few things (in my head but there's no harm in doing this in writing) and then found the right one
your reaction to going down the wrong path shouldn't be "oh no I'm never going to figure this out" but "Hm, that didn't work, let's try something else"

And If my understanding is correct
E,M and Gamma all belong to the resonance/to the unstable particle that decays
To me it looks like a trick or a forced thing to compare the mass with the decay rate, just because we are using natural units

why would it be a "trick"?

I find it weird that just a notation convention can force such comparisons which I believe

in non-natural units you just get a bunch of constants everywhere

7:58 PM
has no physical interpretation

i.e. you'd have that they are proportional via a bunch of constant instead of equal

Ok

there's no meaningful different between those two situations, the non-natural units are just more annoying to write down

I see
The thing is the article has no graph

you can literally Google image search "Breit-Wigner distribution" and find hundreds of examples

8:01 PM
And I don't understand what is happening. You have energy, and a pdf, and you are trying to find the probability of creation of the resonance, but the energy is that of the resonance

how is that a problem

Well simply said, I don't understand what we get from it
So as energy increases, a peak is created around some energy value
but the energy of what increases?
Ah whatever, I can't really express my confusion, since I don't understand what is happening here

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