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Planck scale dimensions

  1. Dec 27, 2009 #1

    Planck energy:
    [tex]E_P = m_P c^2 = \sqrt{\frac{\hbar c^5}{G}}[/tex]

    Gravitational radius:
    [tex]r_G = \frac{r_s}{2} = \frac{G m_P}{c^2}[/tex]

    Gravitational radius is equivalent to Compton wavelength:
    [tex]r_G = \overline{\lambda}_C[/tex]

    [tex]\frac{G m_P}{c^2} = \frac{\hbar}{m_P c}[/tex]

    Planck force is a constant in the Einstein field equation:
    [tex]F_P = \frac{E_P}{r_G} = m_P c^2 \left( \frac{c^2}{G m_P} \right) = \frac{c^4}{G} = \frac{8 \pi T_{\mu \nu}}{G_{\mu\nu}}[/tex]

    The maximum ratio of energy per gravitational length:
    [tex]\boxed{\frac{c^4}{G} = \frac{8 \pi T_{\mu \nu}}{G_{\mu\nu}}}[/tex]

    Are Planck scale dimensions the maximum limits in the Universe?

    http://en.wikipedia.org/wiki/Planck_force" [Broken]
    http://en.wikipedia.org/wiki/Planck_mass" [Broken]
    http://en.wikipedia.org/wiki/Planck_energy" [Broken]
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Dec 30, 2009 #2
    Where did you get that expression for the quotient of the Stress-enery-tensor and the Einstein-tensor from? As far as I know, it's not valid to divide two tensor-valued quantities.
  4. Dec 30, 2009 #3

    From Einstein's field equation:
    [tex]G_{\mu \nu} = \frac{8 \pi G}{c^4} T_{\mu \nu}[/tex]

    It is not possible to divide tensors. However, It is possible to divide the solutions after the tensors have been solved for a specific solution.

    For example:
    [tex]\boxed{\frac{c^4}{G} = \frac{8 \pi T_{r r}}{G_{r r}}} \; \; \; \mu = \nu[/tex]

    Are Planck scale dimensions the maximum limits in the Universe?
    Last edited: Dec 30, 2009
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