"In the foregoing discussion the basic task in quantum dynamics is reduced to finding an observable that commutes with H and evaluating its eigenvalues."

does this sentence mean evaluating the hamiltonian's eigenvalues or the observable that commutes with H's eigenvalues?

I think I killed math chat with my textbook problems. I will burden the h-bar now. Given a surface $4x^2+4y^2+z^2=16$ I'm asked to find the volume. What would be the first thing you do?

I thought to arrange it so x is the independent variable so to find bounds of the other two $x=\sqrt{4-y^2-\frac{z^2}{4}}$

the expression $4-y^2-\frac{z^2}{4} > 0$

setting y=0, $\frac{z^2}{4} < 4$

$-4< z < 4$

with z=0, $-2<y<2$

@Obliv why are you looking for the bounds?

I would just recognize that this formula describes an ellipsoid

Oh wait.. yeah I'm supposed to find them in terms of other variables

like $\int_a^b\int_{h_1(x)}^{h_2(x}\int_{g_1(x,y)}^{g_2(x,y)}f(x,y,z)dzdydx$

I mean, you can do that

but again, this is an ellipsoid, there are easier ways to figure out the volume of an ellipsoid than to do a triple integral

@SillyGoose both, probably? you didn't give enough context here to say for sure, but in any case, does it matter? won't whatever discussion follows make this clear?

True, but I'm trying to apply a general method since I'm not sure what types of volumes i'll be finding tomorrow

I just noticed, how come mathjax commands use dollar signs?

did I just uncover the secrets of the mathjax illuminati

only Knuth can answer that

well in the example that follows that statement, they state the energy eigenvalues. but i am confused about if we only know the compatible observable's eigenvalues, how to obtain the energy eigenvalues

but it's just likely that $ was simply a convenient symbol not yet used for anything else in TeX and since scientists rarely need to type actual dollar signs it's an acceptable choice for a reserved symbol

also how do we go into the next step... where $B$ does not necessarily commute with the time evolution operator. I am confused because going to the next step seems to imply that $B$ does commute with the time evolution operator?

Yeah plus knuth was american, could have been a euro sign or something lol. Didn't know about this man, very important figure in CS.

brb

@SillyGoose didn't you just say that this observable commutes with $H$

@SillyGoose this is like when people say "ah im so ugly" so people will be like "oh my gosh no, youre BEAUTIFUL"

The observable $A$ with eigenstates $|a'\rangle$ commutes with $H$

okay, so the implication is that $\lvert a\rangle$ are joint eigenstates of $A$ and $H$ then

as for the "commutes with time evolution" - look carefully, these exponentials just numbers so of course they commute with $B$

@Obliv the more likely thing that excludes the Euro sign is that it didn't exist yet :P

also, even among existing symbols, you really only want to choose ASCII symbols as special ones

oh, so $t$ is a value here, not a paramter...

@Relativisticcucumber well that is nott my intentionxD

@SillyGoose e x p o s e d

i think i am wondering if $A$ and $H$ have joint eigenstates, how to relate their eigenvalues

wait a minute...

so the unitary translation operators. 1) they are not observables. 2) So do we always act on state kets with the unitary translation operators evaluated at particular values? For example, you would never do an arbitrary time translation of a state ket?

Let $\mathcal{U}$ be the unitary time-evolution operator. To be precise, would I write $\mathcal{U(t_0, t')} |_t |\alpha\rangle$ to represent an arbitrary translation by $t$?

re-quackted

you've been honked

honk

@SillyGoose If I understand what you're asking in 2), the time evolution is not single operator but a one parameter group i.e. there is an evolution operator for each value of $t$

yes! i think that revelation has helped me better understand the situation

The same goes for 1D space translations or plane rotations

i also just commented this on the same topic...to try and check my understanding...

In this situation crucially the Hamiltonian is time independent => there exist stationary states => arbitrary state ket can be expressed as combination of stationary states => unitary time-evolution operator (UTEO) acting on the arbitrary state ket can be turned into a non-hamiltonian dependent exponential via taylor expanding the UTEO and using the eigenvalue relation for energy => the UTEOs are now solely dependent on a parameter t, so can be pulled out.

Thus, the UTEO does NOT commute with observables in general, but the circumstances of this situation allow us to pull out the UTEO.

also, every day i inch infinitesimally closer to applying group theory in physics

I don't understand the problem with commutation here. For time independent hamiltonians the commutation of time evolution operator solely depends on $H$

@uhoh That sounds quite difficult! FWIW, I tend to agree with the Skyfield commenter who said "I don't think there's anything wrong with a sharp discontinuity at the limb of the Sun, because there is a real, physical discontinuity there - the Sun isn't really transparent to anything, apart from neutrinos, and we can't measure accurate directions for them."

One of the questions you linked has this answer: astronomy.stackexchange.com/a/32734/16685 which has a link to sns.ias.edu/~jnb/SNdata/Export/BP2004/bp2004stdmodel.dat I couldn't find more recent data, or the equations it's based on. Anyway, here are some plots created using that data. I copied the data to Github so that SageCell can read it.

Here's the plotting script:

sagecell.sagemath.org/…

Here's the same plot, with the radius scaled to use the Sun radius from the data, 695980 km, which is a bit larger than the modern IAU nominal photosphere radius of 695700 km. I scaled the mass by the Sun's Schwarzschild radius, 2.953250077 km.

Eg, the mass within 100000 km has a Schwarzschild radius of 0.5 km

I assume it's a polytrope function. Or a minor modification of a polytrope. But we don't know the parameters. en.wikipedia.org/wiki/Polytrope

We can get a rough idea of the gravitational deflection by using the shell theorem, and ignoring all the mass above a given radius. Of course, we really need to do a tedious integration which accounts for that mass. The actual radius is a large multiple of the Schwarzschild radius, so we can use the simple gravitational deflection equation $\theta = \frac{2r_s}b$ which I posted in astronomy.stackexchange.com/a/49874/16685

Note that the maximum deflection is roughly double the deflection of a ray that grazes the Sun's surface, 1.75 arcsecs.

@PM2Ring "We can get a rough idea of the gravitational deflection by using the shell theorem" Are you really sure? The shell theorem applies to Newtonian gravity (potentials, forces) but is the math behind weak gravitational lensing of a ray of light really using that?

The 2r_s/b applies to a ray passing an external mass source, not through a mass distribution. I'm not asking if you think it should be okay, I'm asking if you are sure this is correct (to 1st order)

Did you find a source somewhere confirming that you can simply integrate (rho/b)d^3x over a mass distribution to get a deflection of a ray passing through the source?

@uhoh Sorry for the delay. I was catching up on the transcript, so I didn't see your ping.

There's a version of the shell theorem in GR, connected to Birkhoff's theorem. en.wikipedia.org/wiki/…

@uhoh No, I'm not really sure. That's why I said it's a rough idea.

However, the spacetime curvature is quite small, so a semi-Newtonian approach should be roughly correct.

@PM2Ring You need to use this formalism en.wikipedia.org/wiki/… and do an "integral over the gravitational potential Phi along the line of sight" So perhaps yes you can use the shell theorem to get Phi so based on your link, I think yes the shell theorem applies

@PM2Ring perhaps in the end, that all reduces down to what you already did :-)

You can always try projets.lam.fr/projects/lenstool/wiki to check your results!

and look for a copy of link.springer.com/book/10.1007/978-3-662-03758-4

this is nice (starting at slide #8) events.asiaa.sinica.edu.tw/school/20140210/talk/…

I have written a couple of different programs that compute photon trajectories around black holes. The most recent one uses arbitrary ptecision arithmetic & elliptic integrals, and I'm quite confident it gives accurate results. I used it for the deflection diagram in that Astronomy.SE answer. The earlier one uses numerical integration, and needs a lot of time steps to get reasonable accuracy for trajectories that loop around the BH multiple times.

I guess it wouldn't be too hard to modify either program to handle variable mass. But I'm not very motivated to do that. ;)

Okay, so your arbitrary precision will give precise results, but have you duplicated a published result to confirm that your underlying math is correct? precise ≠ accurate/coorect.

Anyway around a black hole ≠ through the Sun.

projets.lam.fr/attachments/download/347/lenstool_dummies.pdf starting on page 4

gotta go, later!

@uhoh The underlying maths is just a well-known elliptic integral. You can read about it in Divergent reflections around the photon sphere of a black hole, which I've linked in various places, including physics.stackexchange.com/a/680961/123208 Also see, hepweb.ucsd.edu/ph110b/110b_notes/node81.html

@uhoh Yes, I realise that. ;)

@uhoh And I know that 2r_s/b is definitely wrong. :) It's certainly adequate when b >> r_s, but the simple Padé approx is better, and only a tiny bit more work to calculate.

@antimony No, it's a fairly new feature. See meta.stackexchange.com/q/382019/334566

> Not only can you create multiple saved lists, but now you can also leave specific private notes for each saved question or answer. Use this as you see fit – collect thoughts, remind yourself why you saved a post in the first place, or draft responses. These notes are private to you, as are all of your lists and saves.

ER=EPR is only true for anti deSitter spaces, correct?

@Asklepian Last time I checked, ER = EPR was just a conjecture, not something proven for any specific case

@Asklepian yes, though I believe there have been attempts to make sense of it in de Sitter as well.

@PM2Ring hey, how are you. I read those two articles you posted here about the planetary motion.

@ACuriousMind Hello ACM, how are you as well? I think I'm here with a much better explanation to my question about the elliptical orbits.( :-( after a sleepless night)

I'm posting that image again to maintain the flow.

By radii I meant instantaneous radii there, which is variable with time. As You can observe here, whenever the object is tending towards those 2 points (1,2) the planet is changing its direction of velocity very rapidly as compared to the other points on its elliptical path. If it is so, then the acceleration must be of greater magnitude when the planet is tending towards that region.

And as you know that the centripetal acceleration is equal to $a=frac{v^2}{r}$ and if we say that the $r$ here is the instantaneous radius And if this radius is getting shorter with time then the centripetal acceleration must be increased (As I elaborated this in other words above)

if the acceleration is increasing when the planet is moving towards these two points then the net force acting on that planet must also be increased that also looks good when the planet is moving towards the point 2, as here the Gravitational force is increasing as it's getting closer to the sun.

But wait, Now the contradiction starts here, if we look at the point 1 which is far away from the sun then how is it possible to get the same behaviour of motion of that planet which is just a repetition of the case when it's closer to the sun? As the Gravitational force is too small at that point as compared to when it's closer to the sun.

Then from where is it getting such a great amount of acceleration at point 1, that it's changing it's direction of velocity much rapidly (same as when the planet is tending towards point 2)?

@TejasDahake the two points do not have the same behaviour - the velocity of the planet at the two points is different

the direction of the velocity is the same up to a sign, but drawing the ellipse doesn't show you how fast the planet is at that point - PM2Ring already posted the vis viva equations that tells you the speed

the curvature doesn't tell you anything about the speed

from the drawing alone, you can just tell the direction of velocity (tangent to the ellipse at that point)

I'm not talking about the speed here, I'm taking about the direction of velocity which is changing very rapidly

When the planet is moving towards those two extreme points

so? at 2, the planet is faster than at 1. Turns out that it is exactly that much slower at 1 that the smaller gravitational force at that larger distance has the same effect on the trajectory as the larger gravitational force at 2 has on the faster velocity there

Here you are telling about the symmetrical logic behind this, but how would you explain if you have to deal with this thing physically?

What do you mean "deal with this thing physically"?

we way we know this is the correct physical behavior is because these ellipses are the solutions to the Kepler problem

That is correct, but when kepler found this thing true, then he must have thought so too, that why are these planets behaving like this

like...that's just what you get when you solve the equations of motion

I'm not sure what sort of "why" you're looking for here

it's just what solving $F=ma$ tells you happens in this case

By "deal with this thing physically", I meant that how would you explain this to me with the help of physics behind this. The reason you explained here was on the basis of mathematical result i think.

I reject the idea that a mathematical derivation is not a physical explanation :P

I want intuitive feel I meant

@ACuriousMind Another reason you'll find about this is :-

I'm not sure what you expect me to say/do here

I'm perfectly content with the answer that ellipses are the solution to the Kepler problem

if you want a different answer, you have to ask someone else :P

Okay I will. But the video I shared is telling about some different reason for the planetary orbits to be elliptical.

What is your point of view on that?

I'm not watching a video right now

Okay but whenever you'll watch that, please notify me.

Also, could you please notify me if this is for real or just more fake news :P

Is there a pdf for the exponential decay in which the maximum and the minimal lifetime of a particle is included? I know the typical exponential decay $N(t)=N_0e^{- frac t \tau}$

But the minimal and maximal lifetime that a particle can have are not present in this pdf

description

why would there be a minimal/maximal lifetime?

Because if decay is involved

a particle,depending on conditions

has a finite lifetime before decaying into other particles

example: muon decay

no it doesn't

Well that's the theoretical example I am reading right at this moment

the "lifetime" in the typical muon exercise is just the mean lifetime

yeah

there's no minimum/maximum lifetime

But how do I incorporate this :

The theoretical lifetime of the particles is 𝜏= 2µs, which roughly corresponds to the lifetime of the muon. Due to the overlapping of the detector signals, lifetimes smaller than 𝑡min= 1µs cannot be reliably measured. The measurement electronics are only active up to the point in time 𝑡max= 10 µs.

More than the solution I want to understand how the formula works

I am asked to write a pdf, including t,tau,t_min and t_max

But other then writing "..formula.." for t_min<t<t_max or 0 otherwise

I have no real clue as to how are the values relevant for the pdf

that's not about minimum or maximum lifetimes of a particles, that's talking about the minimum or maximum values for the lifetime your particular setup can measure

Yes that

but that wouldn't affect the pdf, would it?

you have some pdf $\rho(t)$ depending on $\tau$ that tells you how likely it is for the particle to decay at time $t$. The probability to measure decay with your setup is just $\int_{t_\text{min}}^{t_\text{max}} \rho(t)\mathrm{d}t$

Yes

But I am asked to write the pdf function

Which is weird

How can I do that, without having data to analize (in theory) or other informations

I mean you should know what the pdf is when you say "the theoretical lifetime of the particles is $\tau$..."

where does this "theoretical lifetime" come from if not from a pdf?

@ACuriousMind It's somehow given as info. No explanation as to how we got that

And i am asked to find the exponential decay pdf

but

1) I have no deep knowledge in decay, other than basis stuff. 2)If this is what it is asked and I can't do it. More then doing it, I want to understand how you can find a pdf? You need some sort of feedback to write that

@imbAF so if you know it's an exponential decay with lifetime $\tau$, what's the problem?

the problem is

that the pdf has as variables (t,tau,t_min,t_max)

I know this

$N(t)=N_0e^{- frac t \tau}$

I don't know what that means. What exactly does the exercise tell you to do?

One sec

The aim is to measure the lifetimes of particles that are stopped in an absorber after passing through a detector. The decays are again registered by the same detector via the decay products. The theoretical lifetime of the particles is 𝜏= 2µs,

which roughly corresponds to the lifetime of the muon. Due to the overlapping of the detector signals, lifetimes smaller than 𝑡min= 1µs cannot be reliably measured. The measurement electronics are only active up to the point in time 𝑡max= 10 µs. The number of decays registered in this way is only very small with 𝑁=50 registered events, so that in an unbinned maximum likelihood fit (i.e. all data points are considered in the likelihood function,

not just the entries in bins of a histogram) an exponential function should be adjusted to the lifetimes measured in the interval [𝑡min,𝑡max].

This is what I have as an info

aha

And I am asked to write a pdf for the exp decay

that you're doing a likelihood fit is already information you left out

But I am not sure

@ACuriousMind is it vital in theory?

I mean isn't that some technical stuff

they're just saying that the likelihood fit requires a pdf that is normalized on the interval $[t_\text{min},t_\text{max}]$

ahaa

so I got to find N_0

in the formula

after integrating

in that interval

yes, and you will find that $N$ depends on $\tau,t_\text{min},t_\text{max}$

I see

I thought they simply wanted a theoretical expression of the pdf

I never accounted for normality here

I think they expected you to understand that the likelihood fit needs a pdf normalized on the interval you're doing it on :P

Probably

I decided to take this course that has to do with python and statistics

and analyzing graphs etc, and extracting data and info from a graph

And, as a total newbe,

@ACuriousMind things like this

are bound to happen

but thx

can anyone help me how defense or some other secret satellites are different from regular GSAT

Can you make threads related to python here? Codes etc, that are related to physics?

by here= physics.stack...

@imbAF No, see "Implementation details of computational tasks" here

Ah so it's off topic

ok

But it says computational physics

that means, it;s one of the topics no?

that means questions e.g. about numerical algorithms can be on-topic as they are applied in physics

as the section I pointed to says:

> While computational physics is on topic, we are not a programming site. If your question is about implementing computational code - in particular, if it's about writing, compiling, debugging or optimizing code, or about a specific language or library - then it is off topic. It may be suitable for Computational Science or Stack Overflow, however.

Yes I read

but IDK if my question falls in that list :P

you said "related to python" and that says "[...]or about a specific language [...] - then it's off-topic"

more than that

I simply need create a function which generates random variables with exp. distribution within a certain interval (t_min and t_max), which is something that I know how to do, until the moment when tau, is one of the arguments in the function. so fct(N,tau,tmin,tmax)

And I wanted to ask why do I need tau, when I can easily produce random numbers of exp. distri within a range pretty easily, if this tau ain't involved

...what is the "exponential distribution" here that you're drawing from when you don't have a $\tau$?

the $\tau$ is part of the exponential distribution $\mathrm{e}^{-t/\tau}$, you can't just "generate a number from an exponential distribution" without specifying it

you produce random uniform numbers within a range

and combine that with np.exp

no, that's not right

see stackoverflow.com/q/2106503/3929857 for how to generate exponentially distributed numbers from a uniform distribution

or u could use this: stackoverflow.com/questions/69172510/…

see, essentially all questions you might have are on SO already, there is no need to ask them at physics.SE

Yeah, nearly there, if it wasn't for that tau thing in the function. But I am not going to ask in physics.SE so no worries

the $\tau$ is the $\lambda$ in the first thread I linked

I'll check it

I've always wondered why most people write $E$ instead of $\mathcal{H}$ in the dispersion relation in relativity

And then this "energy operator" pops up

why would you write $H$? people don't usually do relativity in the Hamiltonian formalism

If you're expressing it wrt the momenta you should use the hamiltonian

Unless you mean the momenta in the lagrangian formalism

no, not really

the Hamiltonian is specifically a function of $x$ and $p$ where $p$ is off-shell independent from $x$

it is not always equal to "energy", and $p$ is not always kinematic momentum

Yes of course

unless you are in the Hamiltonian formalism where you are using these $x$ and $p$ as your dynamical variables, you shouldn't call something the Hamiltonian

even if it may be on-shell numerically to it

I mean that $\sqrt{p^2c^2+m^2c^4}$ is the Hamiltonian corresponding to the free particle lagrangian

no it isn't :)

Oh

the Hamiltonian in relativity is zero due to reparametrization invariance

relativity is necessarily a constrained Hamiltonian theory

Oh well sorry, I mean the non covariant one

$L=mc^2/\gamma$

ew

I follow L&L and call that the Hamiltonian :P

@ACuriousMind lol

is a linear combination of spherical bessel and neumann functions the general solution to the radial schrödinger equation?

oops nevermind

@ACuriousMind Are you familiar with python?

@imbAF somewhat? It's been, uh, around a decade since I last actually used it

I wanted to ask you about something. Your confirmation could help me

with this eg.bucknell.edu/~xmeng/Course/CS6337/Note/master/node50.html

you can generate random number with exp. distr. between 0 and 1

So What i did was:

b=-tau*np.log(1-x/tmax)

producing a single random variable at a time, and checking whether it is within the interval I want (t_min-t_max)

and accepting them if that is the case

you can use backticks ` to format code in chat

rng = np.random.default_rng( 42 )

aaa

Normally I cannot use the inverse transformation methode for other than values within 1 and 0 because np.log(1-x)

so x has to be within that interval

that's why i do this b=-tau*np.log(1-x/tmax)

kind of normalization using the largest value

I don't think you understood how this is supposed to work - why are you drawing $x$ from $[t_\text{min},t_\text{max}]$?

I am using this:

eg.bucknell.edu/~xmeng/Course/CS6337/Note/master/node50.html

it's a methode

yes, I know

it talks about drawing $y$ uniformly from $(0,1)$, and their $y$ is your $x$

@ACuriousMind because I need to generate random variables with exp. distribution, that represent average lifetime for muons

and it is said that the device cannot measure less then 1 or more then 10

so I need to "create" a sample with values within this range

yes, I know, but your x isn't the random variable that needs to lie between $t_\text{min}$ and $t_\text{max}$

it's b

yes

x is a variable with uniform distribution

size =1, so i get one value at the time

@imbAF yes but if you actually read the page you linked to (which by the way is exactly the same method as the SO answer I linked earlier), it needs to be uniform on (0,1) because you're equating it to a cumulative distribution function

I see

the idea is that if $N(t)$ is the cdf for $t$, then for $x$ uniform on $(0,1)$ you get a $t$ distributed according to the pdf underlying $N$ by doing $t = N^{-1}(x)$

the code you wrote does not implement this algorithm at all

@ACuriousMind nicely explained. Yes that's what it does

I am simply trying to produce the values within the interval,as it is asked

but I cannot

I am spinning around, left and right, and can't seem to find a solution

The idea is, that because we don't have a data sheet with values about the muon average time, we need to create one

and the values should be within 1 and 10

well, first you should write code that actually implements the intended algorithm for your specific cdf (whose normalization constant depends on $t_\text{min},t_\text{max}$)

In that link, the upper boundary of the integral for the cdf, is arbitrary, meaning it has no fixed value

only the lower boundary has one

it might turn out that you still need to do something special to constrain the values of $t$ this produces to actually lie in the desired interval, but currently that really isn't the problem you should be focusing on :P

@imbAF yes, that's just how cdfs work

then how do I get the normalization constant, when calculating the CDF?

I do not understand the question - just compute the cdf. Note that instead of your pdf being zero for $x<0$ as in the example on that page, your pdf is instead zero outside of $[t_\text{min},t_\text{max}]$

@ACuriousMind It is the 2nd step actually. I need to produce the correct range of values, because later on, on a 3rd state, I'll do 3000 Monte carlo simulations, and in each one, I'll be recording 50 values, which need to be correct

it seems to me that you are trying to put an algorithm into code that you haven't yet fully understood. That is almost always a bad idea - first make sure you actually understand how one would carry out an algorithm by hand, then try to code it

word for word

this is what I am asked

Write a function that generates 50 exponentially distributed random numbers in the sensitive detector interval [𝑡min,𝑡max].