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A Plankian Derivation

  1. Apr 11, 2009 #1
    1. The problem statement, all variables and given/known data

    Just need to know if i have derived this correctly.

    2. Relevant equations



    3. The attempt at a solution


    Taken the dimensions of the square of the mass-energy formula, and applying the equation to both sides (even though approximated) the equation of [tex]\frac{\lambda}{hf}[/tex] which in this case is also squared, which yields [tex]h^2c^2[/tex] Strictly using Natural Units, is then state:

    [tex]h^2c^2(E^2) \approx \lambda^2 E^2(M^2c^4)[/tex]

    Now by simple derivation, divide both sides by the wavelength [tex]\lambda[/tex], so that

    [tex]\frac{h^2}{\lambda}\frac{c^2}{\lambda}\frac{(E^2)}{\lambda} \approx \frac{\lambda^2}{\lambda}\frac{E^2}{\lambda}(\frac{M^2c^4}{\lambda})[/tex]

    which gives numerically:

    [tex]E^2(\frac{M^2c^4}{\lambda})=\lambda E^2[/tex]

    since [tex]E=\frac{hc}{\lambda}[/tex] and also because [tex]\frac{\lambda^2}{\lambda}\frac{E^2}{\lambda}[/tex] reduces to [tex]\lambda E^2[/tex].

    And finally, using the same equations, instead of dividing the derivation by the wavelength, i divided it by [tex]2\pi[/tex], and the result was in this expression:

    [tex]\hbar^2c^2\frac{E}{2 \pi}[/tex]

    Cheers in advance!
     
    Last edited: Apr 11, 2009
  2. jcsd
  3. Apr 12, 2009 #2

    CompuChip

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    What have you derived? And from what?
     
  4. Apr 13, 2009 #3
    equation 1 in the OP requires to know that

    [tex]hc=\lambda E[/tex]

    and to know we are using the sqaured value of [tex]E[/tex] in the generalized form [tex]E=Mc^2[/tex].

    Knowing this, then equation 1 is simply the plugging in of these values. The rest of the commands of the derivation after this to equation 3, the equation yields [tex]\lambda E^2[/tex], simply a relation to sqaured value of [tex]E[/tex] and it's wavelength [tex]\lambda[/tex].

    The final derivation took a different course. Using equation 1 again, the division of [tex]2\pi[/tex] gives:

    [tex]\sqrt{\hbar^2 c^2(\frac{E}{2\pi})[/tex]

    (knowing that [tex]\frac{h^2}{2\pi}=\hbar[/tex])

    which became an expression which leads to an equivalance between the kinetic energy

    [tex]\hbar c(\frac{KE}{2\pi})=\sqrt{\hbar^2 c^2(\frac{E}{2\pi})}[/tex]

    If the equation was now squared on both sides, i came to the derivational expression of:

    [tex]\hbar^2c^2 \frac{E}{2\pi}[/tex]
     
    Last edited: Apr 13, 2009
  5. Apr 13, 2009 #4
    I would also like approval of the following derivation, where i have derived the planck time and found relationships between the derivations to solve for the quantization of planck charge:

    I define to begin with, the gravitational constant:

    [tex]\frac{M - c(\hbar)}{M} = G[/tex]

    By rearrangement we can have:

    [tex]M - c \hbar = GM[/tex] (1)

    This is obviously quite a large value, if not quantized. Equation (1) can be rearranged also:

    [tex]c \hbar = M(G+1)[/tex]

    We are simply returning the [tex]M[/tex] to the right hand side, but expressed in brackets that will distribute it. Therergo, we have:

    [tex]c \hbar = GM^2[/tex]

    Now multiply [tex]c^2[/tex] to both sides:

    [tex]c^2 \dot c \hbar = GMc^2[/tex]

    so that

    If [tex]G(Mc^2)[/tex] is true, then it is the same as [tex]G(E)[/tex]. Therefore, the following must also be true:

    [tex]G(E)= \hbar c^3= GMc^2 = \sqrt{h^2 f^2 G^2}[/tex]

    because [tex]E=hf[/tex].

    Knowing that [tex]f^2 \cdot G^2h^2 = h^2f^2 \cdot G^2[/tex] then now take away [tex](f)[/tex] from both sides and rearrange:

    [tex]G^2h^2 = h(G+1)[/tex]

    Now divide both sides by the quantization of [tex]c^5[/tex], and you have

    [tex]\sqrt{G^2h^2/c^5}= t_{pl}[/tex]

    Which is exactly the Planck Time.

    Now moving on to charge relationships, Knowing the previous work, one can derive this set of relationships:

    [tex]\frac{M - \hbar c}{M} = \frac{M - q^2}{M} :\ [\sqrt{\hbar c} = q][/tex]

    Where [tex]q[/tex] is Plancks Charge, we see that the square root of [tex]\hbar[/tex] is the precise value of the quantization of the charge. Since the fine structure constant is so much larger than the gravitational constant, [tex]\alpha > \alpha_g[/tex], where [tex]\alpha_g[/tex] here denotes the gravitational coupling constant, it seems interesting to note that:

    [tex]\frac{G\sqrt{M^2}}{\hbar c} = \frac{GM}{\sqrt{M(G+1)}}=\frac{GM}{q}[/tex]

    which would be a relation to the charge again, and this leads to my final set of relations:

    [tex]\frac{GM}{\sqrt{M(G+1)}}=\frac{G \sqrt{\alpha_g}}{q}[/tex]

    Does this all seem right?

    (I had to edit for the expression of the planck time. I certainly didn't know in latex the expression P_t lead to... is it the number for silver... anyway, redited)
     
  6. Apr 14, 2009 #5
    No takers?
     
  7. Apr 14, 2009 #6
    I bolded this part becuse i've realized i've made some mistakes. So it's totallly invalid now. But the planck charge derivation seems correct.
     
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