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Photon - h/(lamda * c)

  1. Mar 18, 2008 #1

    Rel

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    Hello

    I red in web side " The photon has mass " Is that real ?


    and what about


    m = h / lamda * c
     
  2. jcsd
  3. Mar 18, 2008 #2

    CompuChip

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    Hi, welcome to PF.

    There is indeed a certain energy assigned to the photon, which is related to its frequency [itex]\omega = 2\pi\nu = 2\pi c / \lambda[/itex] by
    [tex]E = \hbar \omega = \frac{h}{2 \pi} \omega = \frac{h c}{\lambda}[/tex]
    When we invoke the famous [itex]E = mc^2[/itex], the equivalence between energy and mass, we can indeed think of this energy as representing a mass
    [tex]m = \frac{E}{c^2} = \frac{h}{\lambda c}[/tex]

    The "rest" mass is usually defined by
    [tex](m_0c^2)^2 = E^2 - (\vec p c)^2 [/tex]
    where p is the (three)-momentum. If we take this momentum to be [itex]\vec p^2 = m^2 \vec v^2 = (m c)^2 = (h / \lambda)^2[/tex] then this formula gives
    [tex](m_0c^2)^2 = (h c / lambda)^2 - (h c / lambda)^2[/tex]
    so [itex]m_0 = 0[/itex]. It should, because no massive particle can travel faster than light.

    These considerations are important in processes such as pair production and annihilation, where conservation of 'mass' (= energy) must hold.
     
  4. Mar 23, 2008 #3

    Rel

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    Thank you compuchip

    You gived to me what I want
     
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