Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Is this ok?

  1. Dec 2, 2011 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution

    Seems pretty strightforward:

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

    multiply both sides by [tex]c^4[/tex]

    [tex]\hbar c^5 = Gm^2c^4[/tex]

    knowing that [tex]m^2c^4[/tex] is [tex]E^2=(\hbar \omega)^2[/tex] then

    [tex]\hbar c^5 = G\hbar^2 \omega^2[/tex]

    dividing [tex]\hbar^2[/tex] off both sides


    then this is the same as


    because [tex]\omega^2 = \frac{k}{m}[/tex] so multiplying [tex]m[/tex] on both sides gives


    Then finally multiplying [tex]\hbar[/tex] on both sides gives you

    [tex]\frac{c^5}{\hbar}m \hbar=Gk\hbar[/tex]

    Has there been anything wrong in this so far?
    Last edited: Dec 2, 2011
  2. jcsd
  3. Dec 2, 2011 #2
    Basically it was suggested I had my dimensions wrong, but the derivation is so simple, I couldn't see anything wrong with. My question was simple enough I think.. is there anything wrong with it? Either it is wrong and I am doing this with the wrong dimensions, or it is right and there was no problem to begin with..

    which is it, please answer!
  4. Dec 2, 2011 #3


    User Avatar
    Science Advisor
    Homework Helper

    What does this derivation do ? What's it good for ?
  5. Dec 2, 2011 #4
    I haven't fully decided on this. There are two main features from the final equation written here in the OP.

    [tex]\frac{c^5}{\hbar}M\hbar = G(k\hbar)[/tex]

    which in another form is simply

    [tex]\frac{c^5}{\hbar}S = G(k\hbar)[/tex]

    This equation has the features that the left handside is in fact just a rescalling of the spin [tex]S[/tex] and the right handside has the feature of momentum of any quanta given as [tex](\hbar k)[/tex]. So for a specific wave equation, the right hand side may have an implication for

    [tex]\psi(x,t) = Aexp(ikx - i \hbar k2t / 2m)[/tex]

    because this has a definite momentum, p = ħk. But because this has the form of spin on the left handside, it can be futher written as

    [tex]\frac{c^5}{\hbar} \frac{\hbar}{2} \sigma_i = G(\hbar k)[/tex]

    Here [tex]\sigma_i[/tex] is the Pauli Spin Matrices, where the subscript represents any of the three dimensions of space [tex]\sigma_i = (\sigma_x, \sigma_y, \sigma_z)[/tex] and of course, if you use all three dimensions, this directly effects the momentum component to make [tex](\vec{\hbar k})[/tex].

    Of course, I have thought about other applications.
    Last edited: Dec 2, 2011
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook