Inertia and Relativity

Discussion in 'Alternative Theories' started by hansda, Dec 22, 2017.

  1. NotEinstein Valued Senior Member

    Messages:
    1,986
    Why are you trying to follow any link from there? I was just reminding you what we were talking about, so you can post those calculations you made that lead to your conclusion in post #110. Remember: you are the one making the claims here, so it's you who has to provide the evidence (calculation in this case).
     
  2. Google AdSense Guest Advertisement



    to hide all adverts.
  3. hansda Valued Senior Member

    Messages:
    2,424
    Well, follow this link https://en.wikipedia.org/wiki/Electron_magnetic_moment . You can get value of \(L \) from this link.

    As per my equations \(E=mc^2=hf=Iw^2k_2=Lwk_2 \). Consider \( L=Iw=mr^2kw\) or \( rw=\frac{L}{mrk}\). Considering values of m and r for electron, you can check the value for \(rw \).
     
  4. Google AdSense Guest Advertisement



    to hide all adverts.
  5. NotEinstein Valued Senior Member

    Messages:
    1,986
    And when you do that, what value do you get for the velocity with which the electron rotates?
     
  6. Google AdSense Guest Advertisement



    to hide all adverts.
  7. hansda Valued Senior Member

    Messages:
    2,424
    I have given you my equation. You can do the calculation.
     
  8. NotEinstein Valued Senior Member

    Messages:
    1,986
    No, I shall not. You are the one making the claim; you have the burden of proof. Or are you now admitting you didn't actually do this calculation, and your statement in post #110 was unfounded?
     
  9. hansda Valued Senior Member

    Messages:
    2,424
    https://en.wikipedia.org/wiki/Electron_magnetic_moment ; From this site consider "Spin magnetic dipole moment" section. We can consider \(L=S=\frac{\hbar}{2} \).

    From my equations I have observed \(mr \) is constant and \(mr=\frac{4\hbar}{c} \).

    Considering our earlier equation for \(rw \), we can write \(rw=\frac{L}{mrk}=\frac{\hbar}{2}\frac{c}{4\hbar}\frac{1}{k}=\frac{c}{8k} \).

    So tangential speed of spinning electron \(v_t=rw=\frac{c}{8k} \). From this we can see that \(v_t<c \)
     
  10. NotEinstein Valued Senior Member

    Messages:
    1,986
    Unless \(k\leq\frac{1}{8}\); what was \(k\) again?
     
  11. hansda Valued Senior Member

    Messages:
    2,424
    \(I=mr^2k \)
     
  12. NotEinstein Valued Senior Member

    Messages:
    1,986
    And what is its value (for an electron)?
     
  13. hansda Valued Senior Member

    Messages:
    2,424
  14. NotEinstein Valued Senior Member

    Messages:
    1,986
  15. hansda Valued Senior Member

    Messages:
    2,424
  16. NotEinstein Valued Senior Member

    Messages:
    1,986
    (Next time, could you please provide a better link, or more direction where exactly to find what you are referring to?)

    What value did you use for radius? Remember, the electron's radius has been measured to have an upper bound of \(10^-7\) fm!
     
  17. hansda Valued Senior Member

    Messages:
    2,424
    That is matching with the equations. \(m_pr_p=m_er_e \) or \(\frac{m_p}{m_e}=\frac{r_e}{r_p}\simeq 1836 \)
     
  18. hansda Valued Senior Member

    Messages:
    2,424
    Here actually I used the \(mr \) value, which is constant. From my equations also this can be observed.

    There may be a measurement error.
     
  19. NotEinstein Valued Senior Member

    Messages:
    1,986
    That doesn't answer the question. Why is the effective torque arm radius of an electron is equal to its physical radius? Please proof this assertion that you are making.
     
  20. NotEinstein Valued Senior Member

    Messages:
    1,986
    Is \(mr\) constant for all types of fundamental/non-composite particles?

    Unfortunately for you, there is not.
     
  21. hansda Valued Senior Member

    Messages:
    2,424
    That is the radius measured from spin axis.
     
  22. hansda Valued Senior Member

    Messages:
    2,424
    From my equations, it is becoming so.
     
  23. NotEinstein Valued Senior Member

    Messages:
    1,986

Share This Page