Wednesday, March 20, 2013

Higgs mechanism

what fraction of your body weight is due to the Higgs mechanism? in what way is the remaining mass being produced?

bonus question: can you show that parity transformation $\vec{r}\rightarrow -\vec{r}$ introduces a factor $(-1)^\ell$ in the spherical harmonics $Y_{\ell m}(\theta,\phi)$?

4 comments:

  1. Mass of quarks in proton and neutron = 10 MeV
    Mass of proton / neutron = 940 MeV
    Mass of electrons et al is negligible.

    Thus, about 1 %

    The rest comes from the binding energy of the proton / neutron.

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    1. thanks for the fast reply! CQW completely agrees :)

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  2. maths bonus question: parity transformation in spherical coordinates means $(\theta,\phi)\rightarrow(\pi-\theta,\phi+\pi)$. for the spherical harmonics $Y_{\ell m}$, \begin{equation} Y_{\ell m} = \sqrt{\frac{2\ell+1}{4\pi}}\sqrt{\frac{(\ell-m)!}{(\ell+m)!}} P_{\ell m}(\cos\theta)\exp(-\mathrm{i}m\phi) \end{equation}
    and the associated Legende-polynomials $P_{\ell m}(\mu)$
    \begin{equation}
    P_{\ell m}(\mu) \propto \frac{\mathrm{d}^{\ell+m}}{\mathrm{d}\mu^{\ell+m}}(\mu^2-1)^\ell \end{equation}
    with $\mu=\cos\theta$ on can quickly show that parity inversion of the exponential gives $(-1)^{-m}$ because
    \begin{equation}
    \exp(-\mathrm{i}m(\phi+\pi)) = \exp(-\mathrm{i}\pi)^m\exp(-\mathrm{i}m\phi)=(-1)^{-m}\exp(-\mathrm{i}m\phi)
    \end{equation}
    and the cosine transforms $\mu\rightarrow -\mu$, such that the associated Legendre polynomials transform according to
    \begin{equation} P_{\ell m}(-\mu) = (-1)^{\ell+m}P_{\ell m}(\mu).
    \end{equation}
    collecting these results yields a factor of $(-1)^\ell$ for the spherical harmonics.

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    1. and I'd like to add that if the spherical harmonics $Y_{\ell m}$ are defined with $\exp(+\mathrm{i}m\phi)$, you'd obtain a factor $(-1)^{\ell+2m}$ under parity inversion, which is of course $=(-1)^\ell$.

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