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Everything posted by joigus
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Wrap your mobile phone in tin foil and tell a friend to call you. See what happens.
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Suggestion: By telling everybody here what you have set out to do with your work. Then, try to outline, Physical principles to be applied. Mathematical formulas to be proven/tested, etc., and how they comply with aforementioned principles. Calculations. Conclusions. Discussion. I hope that helps.
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Strong nuclear force goes to zero when distances become very small (asymptotic freedom); electromagnetic forces go to zero at very large distances (inverse-square-of-distance for electrostatic forces, and inverse-of-distance for radiation). I'm not sure that's what you're looking for though. You may be asking about something like shielding. EM can be shielded, as Swansont said. Strong force cannot be shielded at any distance range where it's sizeable, due to its complicated polarity. Gravity cannot be shielded either, due to not having any polarity at all. Weak force is not at all about shielding. It's a very different kind of force having to do with decay (plus neutrinos cannot be efficiently shielded).
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This is kind of the argument I was thinking about. Wouldn't that wall have lots of turbulence, with the rotors somehow working against each other? I would like to hear more about this.
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What does it give you for the #photon/#nuclei ratio? Why is it nearly 1010? Why are there about 1090 photons in the universe? Those are the questions you should be asking yourself in the context of your model. I meant the visible universe --within the Hubble radius.
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You should be your number one critic. Isn't your model isotropic just because you assume isotropy from the get go? That's called begging the question, and I think that's what you're doing. I'm perhaps the wrong person to discuss these matters with. I tend to think almost everything one thinks is wrong. Including myself.
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Isotropy (sameness in all directions.) If the universe starts out at Planck scale, that's much, much smaller than atomic scale in time and length, and far, far bigger than atomic mass in energy scale. Keep in mind that Planck's time and length are ridiculously small, while Planck's mass is more like the size of an amoeba or a neuron, or something like that... Funny, isn't it? So disparate! It's kept me wondering for more than 30 years. It still does.
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Three strikes, you're out! ------ 47 minutes ago, stephaneww meant to say: My model/egality explains the homogeneity isotropy of the visible universe no? do you agree? ----- No. I don't see that either. How? This is also obvious, may I point out...
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No, sorry. I don't see that coming out of your idea. The present universe is not homogeneous. You should aim at explaining the present degree of inhomogeneity --deviations from homogeneity.
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OK, from scratch. You're using the fact that the Hubble time is a certain number of times the Planck time. That much makes sense. So, \[ \frac{m_{P}}{2}+\stackrel{\left(T_{H}/t_{P}\textrm{ times}\right)}{\cdots}+\frac{m_{P}}{2}=\frac{T_{H}}{t_{P}}\frac{m_{P}}{2} \] Now, the Planck scale is defined by, \[ t_{P}=\sqrt{\frac{c^{5}}{\hbar G}} \] \[ r_{P}=\sqrt{\frac{\hbar G}{c^{3}}} \] \[ m_{P}=\sqrt{\frac{\hbar c}{G}} \] Now, you seem to be saying that, \[ M_{H}=\frac{m_{P}}{2}+\stackrel{\left(T_{H}/t_{P}\textrm{ times}\right)}{\cdots}+\frac{m_{P}}{2} \] Forget about dimensionless numbers. We're safer here if we do dimensions. Then, \[ M_{H}=\frac{1}{2}\frac{t_{H}}{t_{P}}m_{P}=\frac{1}{2}t_{H}\sqrt{\frac{\hbar c}{G}}\sqrt{\frac{c^{5}}{\hbar G}}=\frac{1}{2}t_{H}\frac{c^{3}}{G} \] Is that it? Is that what you're saying? Dimensions check now. Sorry I made a mistake last night: \[ \frac{1}{2}t_{H}\frac{c^{3}}{G}=T\frac{L^{3}T^{-3}}{LL^{2}T^{-2}M^{-1}}=M \] Now, you may well be saying something else, in which case I would be very interested to know what it is. PS: I agree with Swansont that saying that the universe at the time of the big bang was the size of an atom is a gross mis-estimation in the light of the current cosmological model.
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I'm all ears. I don't know how it affects the fact that you can plug in any dependence a(t) for the expansion parameter in Hubble's law, so there is no direct proportionality between the life of the universe and the amount of matter that remains inside the Hubble horizon. And yes, the sum is finite, but you still have to explain the coefficients. I don't have a lot of time now, but I see you're using dimensionless time in one factor and dimensionful quantities in another. Please, clarify your notations and we may have a meaningful conversation. The expression I derived is correct if you use dimensionful quantities throughout and the sum is understood literally the way you wrote it down. I may be able to give you more details later. Now it seems like you''re implying something like, \[m_{H}=\sum_{n=1}^{Nt_{H}/t_{P}}n\frac{m_{P}}{2}\] for some discrete index n and some upper limit N, which went unspecified. If you drop the attitude that you're teasing out bits of cosmological wisdom and piecewise letting them fall on me like clues of a riddle, it will help along the conversation immensely.
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It's joigus. Hello. Sorry, I don't speak French. And I didn't make any mistakes. This is a divergent series. =(1+2+...+)mp/2 If you're implying a different summation rule (Borel summation), or re-ordering, you should say so. What summation criterion are you using? No. My expression checks for units of mass. I bothered to check. It it were as you say, it wouldn't. And it does. I didn't write mP*tP anywhere. Check it out. It's a finite sum, by the way, so it should be no problem. ?? I mean, no ordering criterion or Borel transform should affect it. Are we on the same page?
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I don't think your formula is correct. What you're essentially doing (if I understand you correctly) is adding up \( \frac{m_{P}}{2} \) a certain number of times, \[ \frac{m_{P}}{2}+\cdots+\frac{m_{P}}{2} \] and positing that this amounts to Hubble's mass*. Let's call that \( m_{H} \) for brevity if you will. How many times are you adding up half a Planck mass? As many as there are Planck times in a Hubble time. So, what you're saying amounts to, \[ m_{H}=\frac{1}{2}\frac{t_{H}}{t_{P}}m_{P}=\frac{1}{2}t_{H}\sqrt{\frac{\hbar c}{G}}\sqrt{\frac{\hbar G}{c^{5}}}=\frac{1}{2}t_{H}\frac{\hbar}{c^{2}} \] In other words, the "Hubble mass" is proportional to the Hubble time. For starters, I don't think there's any reasonable definition of a Hubble mass that can be related to the expansion parameter. The amount of mass that's trapped within a Hubble radius is the amount that happens to be there due to the whole expansion history of the universe, and should in no way be assumed proportional to time. Keep in mind that the expansion parameter, in the FLRW models can have in principle any time-dependence that you want to postulate. It can be accelerated, decelerated, oscillating, etc. *I'm assuming that by "Hubble mass" you mean the mass of the universe within a Hubble radius. What I'm trying to tell you is that the amount of mass in there is nothing to do with the Hubble "mechanism".
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That's a bunch of commited students though... Commited to do Seppuku if you fail.
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Forever? Trying to play @dimreepr here.
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The relationship between the quantum and the classical
joigus replied to geordief's topic in Modern and Theoretical Physics
I think the answers that you got here are more than enough to go by. But let me try and add a very strong inkling that the quantum must be more fundamental. There is a well-known result of formal (mathematical) quantum mechanics called Ehrenfest's theorem, which tells you that whatever evolving quantum configuration reproduces classical physics at least for the expected values. So classical mechanics is somehow in the guts of quantum mechanics. But there is no way that you can get the quantum with all its peculiarities from classical mechanics. The closest you can get to that is very vague, but it does exist. It's called the Hamilton-Jacobi equation. Some people like to say that, had William Rowan Hamilton spent one more sleepless night, he may have surmised something like quantum mechanics was plausible. But I think that's an overstatement. How would he have figured out the need of a fundamental constant like \( \hbar \)? -
And the spoken one:
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Consider me a ramshackled old monument* that needs your attention from time to time. And is very thankful for it. The uses of the present continuous that I can remember are, 1) Pointing out, reminding, or plainly informing the listener about the activity the speaker is involved in at the time of speaking: I can't talk now, I'm working 2) Pointing out, reminding, or plainly informing the listener about a future activity the speaker already --at the time of speaking-- has made arrangements for. It's impossible for me to attend, I'm seeing the doctor at that time 3) Pointing out, reminding, or plainly informing the listener about an annoying habit they indulge in: You're always talking back to me To which I would add: 4) Quite deliberately using function 2) for emphasis, when the speaker is only signaling their firm intention: I'm not talking to you anymore! And that much is what I remember, although there may be other, literal of figurative. But English is very rich indeed, and what it may lack here and there in grammatical structures, it makes up for in creativity. I've always been a fan of, I must be going** *Maybe not a monument, and just a piece of work. **Top that! A modal with "be going"
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LOL. Thanks. Yes, I took a quick look at the Wikipedia page, but I couldn't figure out, was he a terrible teacher? To @StringJunky: Going over your example I realised it's perfectly OK to use perfect future tense in some cases. What threw me off the tracks in the particular example was the combination of the action (making cakes, rather a short-term action), the time adverb "once", and the use of the perfect future. The overall effect of this accumulation of elements results, to me, in a very unnatural sentence. Again: "Once I make another cake, I will have possessed three cakes" I'd rather say: "Once I make another cake, I will have three cakes" What I mean is, frequency adverbs, as well as time adverbs, strongly constrict the tenses. E.g., "I often visit old monuments" Is OK. But, "I'm often visiting old monuments" is awful, from the point of view of good English grammar. Perfect? This largely depends on the state of the apple.
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It doesn't make sense to me, which doesn't mean it doesn't make sense. What does it mean to add half Planck's mass from one to the ratio Hubble's time/Planck's time? I can't make heads or tails of it. Where did you get it from? Mmmmm. I think I have an intuition of what you're trying to do there... Think Hubble-to-Planck units. Also, what's the purpose? I'll get back to you. In the meantime, perhaps someone can answer.
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Thank you. I know. I possess an almost inexhaustible patience, but I don't own it. I can own a car, or a house though. They are always on my mind. I will have owned them all by the end of 2022. But I will have never possessed them. I was once trying to explain this to a student who insisted that gerunds are nouns.
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You're right. It makes perfect sense. It's just that I rarely ever hear or read that use of verbs denoting possession. And I'm wondering why that is. I've searched on Google, and I get about 3'900'000 occurrences of "I will have owned," and 393'000 of "I will have possessed." It is perhaps relevant to say that most of them --and the first pages of them-- are from grammar sites, and not real language in use. But let me add that that doesn't necessarily mean there's anything wrong with them. Perhaps it's something people are less likely to say for some reason. To take an extreme example, "my dog funded the project" is a perfect from the grammatical and syntactical point of view, though it's unlikely that we would ever hear or read that anywhere. I can't say I'm familiar with his work... Perhaps for good reasons...
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English verbs... Who needs them. The answer is everyone! I like to think about English verbs in different degrees of "malleability". They go all the way from modal verbs --which I like to think of as verbal particles, rather than "real" verbs--, going through stative verbs --reflecting status, rather than "fixed for all time", at least in my understanding--, down to ordinary action verbs. But the most important difficulty with this categorisation --with any categorisation perhaps-- is that these qualities change with time and with how people --with special focus on native speakers-- actually use these things. Language changes, and if people somehow agree that it's OK to say "I'm loving it" --never mind McDonald's"--, then it's OK to say "I'm loving it."
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The Schrödinger's cat thought experiment proves there is no God
joigus replied to VenusPrincess's topic in Religion
I'm sure there's a gif of the gaps too. There's a gif for everything!