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What goes wrong here?

  1. Mar 5, 2006 #1
    I'm just an undergrad who's interested in theoretical physics so please be gentle :biggrin:

    I've always had the question in mind that does an electron possess a "rest frequency" that you could derive from it's rest energy... and since I didn't find any info on this on the net I decided to give a go to play with some equations.

    Pardon me for not learning how to make the eqs with the code thingy:

    E = mc^2
    E = hf
    so
    hf = mc^2
    and
    f = mc^2 / h

    This would be the "rest frequency" of the electron (or some other particle). I don't know if it is legimate to derive the (assumed) frequency this way. Now if we include the velocity of the particle we get

    f = mc^2 / h√(1 - v^2 / c^2)

    and because ,\ (lambda) = v / f, we get

    ,\ = v / mc^2 / h√(1 - v^2 / c^2)
    = vh√(1 - v^2 / c^2) / mc^2

    Now this is really funny: if the wavelenght of the particle is derived with this formula, the result is 100 times smaller than the result given by the de Broglie formula. [ ,\ = (h / p)√(1 - v^2 / c^2) ]

    For example, if the speed of the electron is 0,1c, the formula I "proposed" gives the result

    ,\ = 2,41414856553*10^-13 m

    and the de Broglie formula

    ,\ = 2,41414856553*10^-11 m

    What goes wrong with my assumptions / formulas?
    Does the electron even possess such a thing as "rest frequency"?

    Btw. the other thing I find funny is that you can write the formula of the kinetic energy of the electron in the form

    E_k = h(f - f_o), where f = the frequency of the electron at v and f_o = the rest frequency of the electron

    Sorry if I ask/do stupid things, but I just want to know :smile:
     
  2. jcsd
  3. Mar 5, 2006 #2

    ZapperZ

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    Staff Emeritus
    Science Advisor
    Education Advisor

    Let me ask you one thing for you to consider if what you did here is valid.

    I am holding an object of mass m above the ground at a height h. The gravitational potential energy is mgh (of course). Do you think it is valid for me to equate mgh = hf? Just because I can form a mathematical equality, does it make sense physically?

    Zz.
     
  4. Mar 5, 2006 #3
    OK... never mind then, you can close this thread

    Oh well, you learn new stuff everyday :tongue:
     
  5. Mar 6, 2006 #4
    [itex]E = hf[/itex] is a relation for photons which are massless.

    [itex]E = m_0 c^2[/itex] is a relation for particles of rest mass [itex]m_0[/itex].

    So as ZapperZ pointed out above, you cannot equate the two, because they mean different things.
     
  6. Mar 6, 2006 #5

    jtbell

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    Staff: Mentor

    You're mixing up two different v's here. In the first equation, v is the speed of the particle. In the second equation, v is the speed of the de Broglie wave associated with the particle. These two speeds are not equal.

    [tex]v_{wave} = f \lambda = \frac{E}{h} \cdot \frac{h}{p} = \frac{E}{p} = \frac{\gamma m_0 c^2}{\gamma m_0 v} = \frac{c^2}{v}[/tex]

    where [itex]v[/itex] is the speed of the particle, and [itex]\gamma = \sqrt{1 - v^2/c^2}[/itex].
     
  7. Mar 9, 2006 #6
    Your question is not stupid at all. Take the inverse of your rest frequency f = mc^2/h, multiply it by c, and get c/f = h/mc - the Compton wavelength of a particle with mass m.

    What goes wrong here is that you (like many others) keep forgetting that the quantum formalism is nothing but an algorithm for calculating the probabilities of possible measurement outcomes on the basis of actual outcomes. What goes wrong here is that you think of mass, energy, and frequency (all being the same but measured in different units) in classical terms. This puts the cart in front of the horse. If you want to understand these concepts, you must understand them as mathematical aspects of the quantum mechanical probality algorithm.
     
  8. Mar 9, 2006 #7
    Thanks for the replies everyone!

    koantum, your post was an eye opener. Thanks.

    It's funny that you pointed out that you get the Compton wavelenght out of the eq. Didn't figure that out.

    Can't wait a few years to get to the uni to actually study this stuff.
     
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