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.
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?
@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
@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
@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
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?
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
@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...
@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.
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)
@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)...
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: "$L = L(x1+x2)$
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...
@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 ?
@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 : $f(v_1^2, v_2^2) + g((x_1-x_2)^2)$
"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”)."
@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
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
@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
@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
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)
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)
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.
@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
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
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 wonder if there is anyone that has worked with AdS/CFT. Specifically, with things like entanglement wedge reconstruction and subregion-subalgebra duality.
@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 ?
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. 2. In this article : http://nicadd.niu.edu/~dhiman/courses/phys684_10/lectures/process_rates.pdf 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
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 ?
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
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
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
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
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 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
