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 Discussion on question by Khun Chang:…

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Quillo
May 20 15:14
Does this answer help? physics.stackexchange.com/a/265409/226902
 

 Discussion on answer by gandalf61: Is…

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Quillo
May 1, 2024 08:30
Speaking of models, apart from fractals of positive fractional dimension (physics.stackexchange.com/q/811926/226902), there may be mathematical objects that can be interpreted as having negative dimensions: physics.stackexchange.com/q/131256/226902 and links therein.
 

 Discussion between mma and FlatterMann

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Quillo
Jan 19, 2024 14:03
@mma great question, thank you for sharing the beautiful discussion in Beltrametti and Cassinelli's book. I do not know the answer, but I will think about it. Anyway, for the interested reader, the book is also great for the axiomatic approach to QM physics.stackexchange.com/a/363136/226902
 

 Discussion on question by Silly Goose…

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Quillo
Oct 9, 2023 01:24
@SillyGoose you may want to carefully check the linked question: you will see that it clearly parallels your derivation. Then, you may also check the derivation of $e^{-\beta E}$ given, e.g., here (it is just the standard textbook derivation, clearly explained): arxiv.org/abs/1903.02121
Quillo
Oct 9, 2023 01:24
@SillyGoose so why do you obtain $p=e^{-\beta H}$ if you re not interested in the canonical? In the microcanonical $p=\delta(E-H)$. Your derivation (involving a splitting of the whole system) is the derivation of the canonical in terms of the micro, as written in my first comment and explained in the linked question.
Quillo
Oct 9, 2023 01:24
Sorry, I can not follow the reasoning: do you have any references? Deriving the Gibbs weight from the microcanonical ensembke requires to split the whole system in two subsystems in thermal equilibrium. This question and the end of this answer can be helpful. Possible duplicate: Deriving the canonical ensemble from the microcanonical ensemble
 

 Discussion on question by Cham: Numer…

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Quillo
Aug 6, 2023 01:05
The log(p) plot is strange, it should go to minus infinity for the pressure approaching zero at the surface. Why do you use Method -> "StiffnessSwitching"? Did you try other methods? I think it's a numerical problem. My runge kutta 4 integrator for the TOV worked well. I was cutting at the iron layer because the last part of the star is completely irrelevant and tabulated realistic equatiosn of state never go so low in density anyway. However, the polytrope was working well with RK4.
Quillo
Aug 6, 2023 01:05
Hauahaha sorry Cham, silly mistake! The rest is however a reasonable piece of advice. Also, I agree with ProfRob that a linear-log plot would be great. The cutoff is arbitrary, you can justify it only on the basis of physical arguments and the iron layer is a possibility that is often used. When you have the log plot it would be amazing to see it for two different numerical methods (as you probably know you can select the method in the DSolve function).
Quillo
Aug 6, 2023 01:05
Cham, the point of my comment is that you should be able to find the radius at a given finite (small) density that, if converted to dimensional units, is close to the one of the Iron layer (i.e., the "cutoff" you are mentioning at the end of the question). Then I remember that for certain polytropic indexes, the solution is unbounded, maybe it's worth to try also with gamma smaller than 1 or bigger than 2: do you know what happens in those cases? (for example, the ideal Fermi gas should have $\gamma=5/3<1$... if I remember correctly, it is now 20 years since I last solved the TOV numerically).
Quillo
Aug 6, 2023 01:05
Which $\gamma$ are you using? Also, you may consider the radius at $\sim10^4g/cm^3$, where you should have $^{56}$Fe nuclei (they have the highest binding energy per nucleon) plus electrons. The outer crust consists of a lattice of iron-56. At $\sim10^4g/cm^3$ the atoms are fully ionized owing to the high density. In any case, try to see the difference in radius between the radius at $10^4g/cm^3$ and, say, $10^6g/cm^3$: it should be almost the same! Related: TOV equation and the equation of state, solving numerically
 

 Discussion on question by Jackson Wal…

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Quillo
Aug 5, 2023 05:13
I asked for clarification, hoping to understand the question and its context. You completely changed the question after my first comment, despite the posted answer is related to the first version. Regarding physics: yes, you can do that calculation proposed in the current version. It is the "entropy per particle" and is commonly used in thermodynamics (but again, this is not what you were asking in the first version). Of course, the entropy per particle of a fluid is not necessarily the entropy of a system consisting of a single particle.
Quillo
Aug 5, 2023 05:13
Is this question somehow related to this one? physics.stackexchange.com/q/774037/226902
 

 Discussion on question by Rohit Shekh…

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Quillo
May 11, 2023 16:21
Does this answer your question? Why is there a controversy on whether mass increases with speed?
 

 Discussion on question by Juvia Locks…

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Quillo
May 10, 2023 04:23
Does this answer your question? Why are stars, planets and larger moons (approximately) spherical in shape (like, the Sun, the Moon, the Earth, and other planets)?
 

 Discussion on question by Mikayla Eck…

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Quillo
Apr 3, 2023 06:48
@Mikayla Eckel Cifrese please, try to streamline the question to the essential point, it's really hard to follow.
 

 Discussion on question by Fawzy Hegab…

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Quillo
Apr 2, 2023 07:56
@FlatterMann you may think about Copenhagen but you write unclear things like "A wave function is a description of a quantum mechanical ensemble". My point is warning users against the possibile misunderstanding coming from the fact there there is an ensemble interpretation not to be confused with Copenhagen. Everything may be clear in your mind, but if you write things like that (potentially misleading and without quoting authoritative references) then you have to be open to clarify what you mean. Have a nice day, I am not going to continue anymore.
Quillo
Apr 2, 2023 07:56
@FlatterMann the widely accepted probabilistic ideas by Born (that are part of the "orthodox" interpretation for decades) and the less known "ensemble interpretation" are different things. Check Sakuray, Prugovecki, or any other good book (they talk about "ensembles" only referring to the "density matrix", not pure states). Clear and authoritative reference: "The ensemble interpretation and context-dependence in quantum systems" by Home & Whitaker, 1986 or "Ensemble interpretations of quantum mechanics. A modern perspective": doi.org/10.1016/0370-1573(92)90088-H
Quillo
Apr 2, 2023 07:56
@FlatterMann: 1. You say that "A wave function is a description of a quantum mechanical ensemble" is Copenhagen (confusing, as it sounds more like the "ensemble interpretation"). 2. Ensemble interpretations of quantum theory state that the wave function (or pure state) describes an ensemble of identically prepared systems. This is in clear contrast to the “orthodox” (read "Copenhagen”) interpretation(s): the pure state provides a description of an individual system. 3. Wiki article is good and consistent with authoritative references, e.g. doi.org/10.1016/0370-1573(92)90088-H
Quillo
Apr 2, 2023 07:56
@FlatterMann I think you're mixing concepts here, the Wiki article I am talking about is well made and clear. Give it a shot.
Quillo
Apr 2, 2023 07:56
@FlatterMann according to Wiki (see link I posted) the fact that a pure state is a way to describe an ensemble is not the "Copenhagen" but the "Ensemble" interpretation (this is how I interpreted your initial comment, but maybe I misunderstood it).
Quillo
Apr 2, 2023 07:56
@RogerVadim thank you, it's an interesting answer (I 100%agree with it).. but I do not see a relation with the idea that the wavefunction itself describes an ensemble (a pure state describes an ensemble). However, this was useful to clarify the matter: en.wikipedia.org/wiki/Ensemble_interpretation
Quillo
Apr 2, 2023 07:56
@FlatterMann do you have a reference for this kind of interpretation?
 

 Discussion on answer by John Doty: Do…

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Quillo
Feb 20, 2023 17:08
@anukarma receiving downvotes is not a shame, it can mean a number of things. First of all, I can mean that people find that the answer is not particularly useful or justified, or maybe needlessly pretentious (and, as a matter of fact, I see that you have accepted the answer that expands my previous comment on $E = (p^2+m)^{1/2}$). Since you seem new to this website, please do not think that downvoting is something related to shame (more on downvoting on: physics.meta.stackexchange.com/search?q=downvoting).
Quillo
Feb 20, 2023 17:08
@JohnDoty Ok, as I said not wrong but, ultimately, it's your personal taste. For me, not very illuminating, as I do not understand what you're trying to "illuminate".
Quillo
Feb 20, 2023 17:08
Sorry, I do not follow you. A box of massless photons has energy. Mass is always "rest mass" in almost all modern books and papers, which, I believe, it's a good way to kill any possible semantic misunderstanding (after all, mass is just a parameter in the Lagrangian, some systems have it, others no).
Quillo
Feb 20, 2023 17:08
I agree with @SebastianRiese , it's a "stretch". I agree that it's not wrong but I also fail to see how this should clarify things. I would rather say that in relativity $E = (p^2+m)^{1/2}$ rather than $E=p^2/2m$, so for $m=0$ you get $E=|p|$. How is this an "obfuscation"?
 

 Discussion on answer by John Doty: Ca…

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Quillo
Jan 13, 2023 21:32
@Anixx still an amplifier :)
Quillo
Jan 13, 2023 21:32
@Anixx no you can not just use a single photon/graviton, there is always some amplification apparatus at a certain point.
Quillo
Jan 13, 2023 21:32
@Anixx that's exactly my point. You see: a graviton/photon ALONE can not collapse the wave function (after all they can not give rise to any non-unitary, non-local, discontinuous change). I agree with JohnDoty that you (incorrectly) put the cut where the interaction is still reversible, at the graviton level.
Quillo
Jan 13, 2023 21:32
@Anixx The point of my criticism is this: in the Copenhagen interpretation "collapse" is postulated to be a feature of interaction with "classical" systems. Practically, it is a black-box for an irreversible interaction with a classical system (i.e. a multitude of degrees of freedom that are not accounted for in the unitary evolution: irreversible and classical here go hand in hand). A graviton (or photon) is not a classical environment.
Quillo
Jan 13, 2023 21:32
@Anixx Sorry if I am perplexed, but what does it mean "graviton is used for measurement"? Is there a special graviton used for measurements different from gravitons that lead to interaction used in unitary evolution? (The same question applies to photons).
Quillo
Jan 13, 2023 21:32
@Anixx I am confused: you state that there is no collapse "in theory" (i.e. everything unitary at the microscopic level... difficult to disagree) but then you state in your answer that a graviton can cause the collapse...
 

 Discussion on question by Charles Hud…

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Quillo
Oct 2, 2022 21:02
We can never observe the ground state because nothing happens there (stationary state). If you let the ground state interact with a "measurement device" you perturb it (i.e. a new term in the hamiltonian pops out, to model the interaction with the "detector", so the ground state is not the ground state anymore). We always interact with excitations. In QFT the excitations are the "particles" (in condensed matter, the "quasiparticles").
 

 Discussion on question by Dill: Flowe…

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Quillo
Sep 16, 2022 05:50
@Dill, ok what a stupid unit system! In a nutritional context, the "large" cal unit is used almost exclusively. It is generally written "calorie" with lowercase "c" and symbol "cal", although the capitalized form "Calorie" (with symbol "Cal") or the name "kilocalorie" (with symbol "kcal") are also used. So, I think you are right. Anyway, try to do the edit by adding your calculation.
Quillo
Sep 16, 2022 05:50
@Dill, are you sure that you are not considering the calories released by the combustion of "petroleum" (colloquially called "oil")? world-nuclear.org/information-library/facts-and-figures/… Anyway, your question is interesting and it's a pity that has been closed. You can edit it so it can be reopened. I suggest you to add details and links to your information and to put the focus on the physics and energetics going on inside the seed: intechopen.com/chapters/21496
Quillo
Sep 16, 2022 05:50
@Dill, oil is about $40 \,J/g$ (i.e. vegetable oil is almost 9 calories per gram), meaning that a $0.5mg$ seed can contain at most $0.02 J$.
 

 Discussion on question by Jordi: Why …

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Quillo
Jul 2, 2022 18:41
Mass is just a term $AA$, so why not adding also $AAAA$? You do not add what is not needed (Occam's razor).
 

 Discussion on question by SergioMarqu…

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Quillo
May 1, 2022 19:11
arxiv.org/abs/physics/9810017 relativistic motion of an arbitrary point of an accelerated rigid rod is discussed for the case when velocity and acceleration are directed along the rod's length.
 

 Discussion on question by Libertarian…

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Quillo
Apr 20, 2022 14:48
I am not an expert, but maybe this issue is more clear when one considers the Milne-Cartan formulation of non-relativistic mechanics. See section 3 of this paper by Brandon Carter.
 

 Discussion on question by Mohammad Ja…

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mrc ntn
Jul 31, 2020 23:47
I have to say that I didn't check the profile yesterday, now I see your point, thank you.
mrc ntn
Jul 31, 2020 23:47
I think that there is no need to be so aggressive in the comments. It seems to me that the question is interesting and legit. Then OK, you can disagree that it is a simpler version of the TP, or about the initial comment on the "acceleration"... but who cares, the question still makes sense.
 

 Discussion on answer by Stratiev: Why…

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mrc ntn
Jul 8, 2020 16:01
@LokusPokus probably knows these basic facts. Maybe his question is more subtle. In Minkowsky (far from the galaxy cluster) the geodesics make an angle with the coordinate lines. If we use this criterion (constancy of the angle) also in the distorted region, then we obtain the path B. Why, instead, the geodesics look like A and the "constancy of the angle" does not work? (of course the "not-so-satisfactory" answer is: because you have to solve the geodesics equation, which is not equivalent to use the "constancy of the angle" prescription).