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joigus

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joigus last won the day on September 2

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249 Beacon of Hope

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About joigus

  • Rank
    Organism
  • Birthday 05/04/1965

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  • Interests
    Biology, Chemistry, Physics
  • College Major/Degree
    Physics
  • Favorite Area of Science
    Theoretical Physics
  • Biography
    I was born, then I started learning. I'm still learning.
  • Occupation
    teacher

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  1. My favourite positron generator is the atmosphere.
  2. Now that I think about it, the product of the first 23 is, as you say, 223092870. Now, the number of ways in which you can sum two (arbitrary) natural numbers to give N is just 111'546'435, because 223'092'870 is even. So the problem reduces to finding how many numbers there are between 1 and 111'546'435 that are not divided by 1, 2, 3, 5, ..., 23 (the 1st 23 primes). The only thing I can say is that's not an elementary problem. How did you get your conjecture @Tinacity? Counterexample: \( \pi \) is irrational; \( 1-\pi \) is irrational too. but \( \pi + 1- \pi = 1 \).
  3. She didn't specify, but I thought it was kind of implied... For arbitrary products of primes, certainly nothing like that can be proved with the present mathematics. But for 223092870, I just don't know.
  4. The maths for predicting that kind of thing is called number theory. I know very little about number theory. It studies connections between numbers. The result that you propose reminds me of some theorems by Fermat. Have you proved it, or is it just an intuition? Maybe @wtf or @Sensei, or @mathematic, or @taeto can help you. @studiot is encyclopedic. Maybe he can help you too. Number theory is not very interesting for physics, AFAIK. And physics and mathematical physics are my turf.
  5. joigus

    MathML test

    It's a mark-up language; a subset of XML: https://www.w3.org/Math/whatIsMathML.html Sorry what I posted was inline LateX, but you can read it as MathML by right-clicking.
  6. I will bee looking into this more closely later, and see if I can add anything significant, but first off I'd like to tell you about what seems to be a misconception in what you say in your question. Elementary particles are not like tinker-toy assemblies from which you can split the parts. They are instantiations of one basic thing, say the electron, of which you can obtain more and more copies by providing the necessary energy (2\( mc^2 \), with the electron mass.) By doing so, you produce particle-antiparticle pairs of the given kind. The universe before the big bang (following the standard inflationary model) had particle number zero. So, as others (Swansont, Bufofrog) say, no. This would not be the picture: Connecting with MigL's comments, all the fields must have to have been there (flavours, colours,...), but with zero expected value for the number of particles or antiparticles. This, in turn, connects with the very interesting question of "did the laws themselves appear at the moment previous to the bang, or were they already there?" Lee Smolin is one of the most notable proponents of this question. The electron-positron state that you talked about is called the positronium. It's very short lived. I don't think it lives long enough on average to produce relativistic velocities.
  7. For all it's worth, I'll say users @Ghideon and @MigL are right AFAIK.
  8. And yes, MigL is right. In order to get back to its original state you must turn it around 720º (4\( \pi \) radians). It was corroborated in experiments with external magnetic fields. Although that really may be confusing to you.
  9. An ordinary rotation is cycling around an axis. An electron is not like that. No matter how you look at it, it always rotates with angular momentum h/2 or -h/2 in any direction you look. So no, it doesn't act like a planet. Electrons do not transform into photons; the either emit photons of absorb them. Electrons might absorb photons if there are photons there to be absorbed. Electrons radiate when they are accelerated, or when they are in an excited state. I don't really understand what behaviour you are picturing the magnetic field to inhibit.
  10. This could be hard proof of tunneling for macroscopic objects...
  11. Because the theory predicts that charged particles with this angular momentum of spin will deflect in a magnetic field; and the deflection has been measured --Stern-Gerlach experiment. That was the epochal experiment that's still very much in use. There are others based on photon emission, etc. There are many, many other indirect checks. Ferromagnetism, for example.
  12. Thank you, Zapatos. I've found a little bit more of the geological history on Wikipedia: https://en.wikipedia.org/wiki/Kegon_Falls From https://www.visitnikko.jp/en/spots/kegon-falls/ The place is such a tourist magnet that it's kinda difficult to find something more science-oriented about it: Geology, fauna and flora, etc.
  13. Another real-estate agency's hopes have been shattered, as we speak.
  14. joigus

    Today I Learned

    I always thought they were laughing.
  15. I think you're confusing what equation (problem) you actually have, and what values of \( k \) make sense for your problem. You also seem to have some confusion about what values of \( k \) are valid for your problem. In your OP, you're proposing up to four different equations: \[x^{2}+1-1/k=0\] \[x^{2}+1-1/k=1\] \[x^{2}+1+1/k=0\] \[x^{2}+1+1/k=1\] So the first question I would ask you is: Which one is it? Maybe you want to solve all of them. The question about what values of \( k \) make sense is simpler. Only those with \( k\neq0 \) make sense for the equations you've written. Moving something or other to the right hand side has nothing to do with it, unless you do it multiplicatively. For example, if your equation were, \[k\left(x^{2}-k\right)=1\] you would have to be careful not to divide by \( k \) in case that parameter were \( 0 \). In general, when you have both unknowns (the thing you want to solve for) and parameters (which define an infinite family of possible equations, one for each possible value of the parameter), you must discuss the equation for every possible value of the parameter. And you must leave out those values of the parameter that don't give a sensible equation. I hope that helps.
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