"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 $x=\sqrt{4-y^2-\frac{z^2}{4}}$
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?
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?
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 $t$?
@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 $t$
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."
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?
@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 :-)
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.
@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.
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@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
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
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.
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
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
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 $\int_{t_\text{min}}^{t_\text{max}} \rho(t)\mathrm{d}t$
@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
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].
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.
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
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
@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
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
well, first you should write code that actually implements the intended algorithm for your specific cdf (whose normalization constant depends on $t_\text{min},t_\text{max}$)
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
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
I joined the site 8 years ago now and I'd be curious how things have changed over time. Have we been gaining traffic? Losing traffic? Are there particular seasons or times of year that users flock to the site?
Nothing important, just a curiosity. I think that with 25k reputation I could view this...