joigus

I've been known to act like a cretin, but I'm certainly no Cretan.

Parity is always well defined for any particle or system of particles. OTOH, parity is an involution. This means that if you apply it twice, the state goes back to itself. This, in turn, implies that quantum mechanical particle states can be either even or odd under parity, and nothing else. Given that parity is not conserved in Nature (not even parity, but parity as such), the final state could have but even or odd character under parity, but both are OK. I cannot be sure of what your mental framework is, or your level of acquaintance with the principles of quantum mechanics, but remember that several particles (think decay products) are not a sum of individual particle states, but a product. I have a feeling that's what's bothering you. Is it?

It seems that only @Ken Fabian mentioned comedy. Here's an (allegedly) underrated classic: https://en.wikipedia.org/wiki/Quark_(TV_series)

Here's the quote: https://books.google.es/books?id=jJzGNl9K5SIC&pg=PA277&lpg=PA277&dq=murray+gell-mann+"close+to+a+proof"&source=bl&ots=Wsndi_9ZiJ&sig=ACfU3U1TVvvcpwOmBDk1GxpNphUqSMmSdQ&hl=en&sa=X&ved=2ahUKEwjNxeDZ-pL0AhVoBGMBHWRADHMQ6AF6BAgCEAM#v=onepage&q=murray gell-mann "close to a proof"&f=false https://www.worldscientific.com/worldscibooks/10.1142/7101 Article: Particle Theory, from S-Matrix to Quarks The quote is: (My emphasis.)

Simple and enlightening.

I would say these are problems for so-called numerical relativity; that is, numerical simulations with Einstein's equations, and parametrizing the collision data. I'd think impact parameter and velocities of approach, masses, and so on. I've found a reports-like paper that could be what you're looking for. https://arxiv.org/pdf/1909.06085.pdf I hope it lives up to your expectations. Haven't had time but to skim through it a little bit.

I'm proud of you, young man. I wish I could give you 5 reputation points for this. You didn't just believe me, and went on and performed the experiment. I could have been lying to you.

Ah. Seems like by groping in this relative darkness, you've come up with a very interesting question. I'll let Andrew Wiles do the talking, because I think he's got something to say that's related to your question: [Quoted from Did earlier thoughts inspire Grothendieck? by Frans Oort, who refers to the BBC documentary by S. Singh and John Lynch: Fermat’s Last Theorem. Horizon, BBC 1996.] [With thanks to Thomas Riepe] Taken from: https://micromath.wordpress.com/2011/11/06/andrew-wiles-on-doing-mathematics/ I'd read this quote before as referred to number theory, which Collatz's conjecture is really about. The idea is that you could be microns away from the switch and you wouldn't know. So I'm very skeptic of any relevance of any statement on how close you are to a proof of something. Seems like something that cannot be graded. But that's my feel of it. I have a heartfelt love for pure mathematics, but I don't have the training of a pure mathematician. Taking up from Wiles' metaphore, I'd ask: Is there a way to judge whether the switch is within your reach before you've touched it? I know I'm waxing metaphorical only, but I think you perfectly understand what I mean. I remember the late Murray Gell'Mann making similar comments on some 'mathematical people' saying they were 'close to a proof'.

Interesting... By definition, any conjecture that stands unproven is beyond the reach of present day mathematics. So Erdos is being kind of tautological, whether on purpose or not, I don't know. And as to the second sentence, you're comparing two unponderables: 1) The time it will take humans to solve Collatz's conjecture 2) The time it will take humans to get nuclear fusion to be practical ('practical' = ?) I'd say there's no answer to your question ('I wonder...'). Unless 'there's no answer to your question' is considered to be an answer to your question'. And that, assuming you're asking a question --maybe you're just projecting a wondering.

So how do the EM fields generated by the apparatus exactly cancel the interference pattern? A little bit of maths would help someone like me. I like your line of thinking, but a line of thinking doesn't make a theory. The idea that electrons are somehow both particles and waves, as embodied by standard quantum mechanics, is far superior in that it correctly predicts all properties that we know about so-called elementary particles. Plus, Your theory --once it's formulated in a mathematical fashion-- should be able to explain such other phenomena as those implied in the Aharonov-Bohm effect (a 360º rotation of the electron completely 'inverts' its state). Now, that is weird in terms of EM alone; EM fields do not explain it. One double-slit doesn't make quantum mechanics. QM goes much deeper. It ushers in a whole new way of thinking about physics.

Sorry, I meant "aluminum foil". I'm an old person, you see. Plus my vocabulary is somewhat out of whack due to age, language acquisition, and random influences. The important thing is doing the experiment. It's a very simple experiment, anybody can do it; and it illustrates perfectly Faraday's cage to shield from EM waves that Swansont was talking about.

Wrap your mobile phone in tin foil and tell a friend to call you. See what happens.

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.

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).

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.

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.

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.

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.

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...

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.

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.

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.

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?

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".

That's a bunch of commited students though... Commited to do Seppuku if you fail.