Relativistic change of mass and De-broglie's equationFrom De Broglie's equation λ = h m v We can see that λ depends inversely on mMy question is:If the m in the equation indicates the kinetic mass of the body then from this relation m = m 0 √ ( 1 − v 2 / c 2 ) m would increase with v. Thus the λ value should change doubly...first for the rise of v and second for the rise of mOr is the m in De Broglie's equation representing rest mass, meaning that I am altogether wrong.I am a layman in Physics and I stumbled on this question while understanding electron microscopy

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Relativistic change of mass and De-broglie&#039;s equationFrom De Broglie&#039;s equation  λ  =      h          m      v      We can see that   λ depends inversely on   mMy question is:If the   m in the equation indicates the kinetic mass of the body then from this relation  m  =            m      0                      √            (      1      −              v        2                    /                    c        2            )        m would increase with   v. Thus the   λ value should change doubly...first for the rise of   v and second for the rise of   mOr is the   m in De Broglie&#039;s equation representing rest mass, meaning that I am altogether wrong.I am a layman in Physics and I stumbled on this question while understanding electron microscopy

Calvin Oneill · Accepted Answer

In case you want a less mathematical answer:what you say is not wrong but a rather upside-down way of thinking of it.When de Broglie first came up with formula   λ  =      h    p   in 1923 he was motivated by the way quantities transform under Einstein&#039;s relativity theory.Einstein had previously established two principles:Energy is directly proportional to frequency.   (  E  =  h  f  )Energy and momentum   (  E  ,  p  ) naturally transform together when you change your reference frameSo the only way to make sense of a relativistic quantum matter wave is if you pair up   (  f  ,  λ  ) in a similar way, meaning that wavelength and momentum must be related by the same constant h.   (  p  =      h    λ    )So the complexity eluded to in your question does not come from quantum theory, but rather Einstein&#039;s original relativity theory: the relation between momentum and velocity is not a simply linear relation   (  p  =  m  v  ) but a more complex one where   v appears twice:  p  =            m      v              1      −                        v          2                          c          s                    [  m denotes rest mass here]

zuzogiecwu · Accepted Answer

Let us restrict our studies to one spatial dimension plus time, coordinates   (  t  ,  x  )  .In the frame where the particle is in rest, consider a field   ϕ that oscillates with a frequency proportional to the rest mass/rest energy:   ϕ  =  sin  ⁡      ω    0    t where       ω    0    =  2  π      f    0    =  2  π        m    0        c    2        /    h  =      m    0        c    2        /    ℏ  .Now we introduce a relativistic inertial frame   (      t    ′    ,      x    ′    ) in which the particle moves with speed   v in the positive   x direction. The Lorentz transformation between these frames is      x    ′    =  γ  (  x  +  t  v  )  ,        t    ′    =  γ  (  t  +  x  v      /        c    2    )  x  =  γ  (      x    ′    −      t    ′    v  )  ,    t  =  γ  (      t    ′    −      x    ′    v      /        c    2    )where   γ  =  1      /        1    −          v      2              /              c      2        .Thus,  ϕ  =  sin  ⁡                    m        0                    c        2              ℏ    t  =  sin  ⁡                    m        0                    c        2              ℏ    γ  (      t    ′    −      x    ′    v      /        c    2    )  =  sin  ⁡            γ              m        0                    c        2              ℏ    (      t    ′    −      x    ′    v      /        c    2    )  =  sin  ⁡            m              c        2              ℏ    (      t    ′    −      x    ′    v      /        c    2    )    =  sin  ⁡            m              c        2                    t        ′            −      m      v              x        ′              ℏ    =  sin  ⁡            E              t        ′            −      p              x        ′              ℏ    =  sin  ⁡  (  ω      t    ′    −  k      x    ′    )The wavelength is then given by  λ  =            2      π        k    =            2      π              p              /            ℏ        =            2      π            ℏ        p    =      h    p    .

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