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De Broglie Waves

  1. Jan 1, 2012 #1
    According to Broigle,

    λ=h/p
    where
    p=momentum,
    h=planck constant, and
    λ=wavelength

    But that means,
    if λ increases, then p will increase
    p=mv
    and so v will increase along with the wavelength

    But what if the v is that of light,i.e, c?
     
  2. jcsd
  3. Jan 1, 2012 #2
    For starters, why increase? λ and p are inversely proportional. Or it's New Year's morning and I'm not thinking straight.
     
  4. Jan 1, 2012 #3
    danR is correct. If [tex]\lambda [/tex] decreases, then [tex]p[/tex] increases. But, this is not a cause and effect thing.

    That aside, if [tex]v = c[/tex], then [tex]p \ne mv[/tex].

    Best wishes
     
  5. Jan 1, 2012 #4
    when p increases ,i.e, mv (that of light)
    then either m or v (of light) should increase.....
    then what will be that would increase?
     
  6. Jan 2, 2012 #5
    There clearly is an inverse relationship between the two parameters of your interest. Also Lambda can't increase as your post suggests , in Q.M as you may know light comes in packets ' photons' which have specific energy levels.

    For a given wave , it's wavelength and speed can be related by : c = fλ
     
  7. Jan 2, 2012 #6
    This is true only for non-relativistic particles! You want [tex]v=c[/tex], which is relativistic. Most likely you are talking about light (electromagnetic radiation). In the relativistic case [tex]p=E/c[/tex], where [tex]c[/tex] is the constant speed of light. For light the photons have zero mass and when you increase [tex]E[/tex] the momentum [tex]p[/tex] increases but the speed of light [tex]c[/tex] remains constant.

    The point is that [tex]p=mv[/tex] does not apply when [tex]v=c[/tex].

    Best wishes
     
  8. Jan 3, 2012 #7
    Oh i see. thanks.
     
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