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Relativistic mass

  1. Mar 4, 2009 #1
    Let me see if I got this right. A photon has no rest mass, and it travels at the speed of light. Knowing the wavelength of the photon will give us the momentum using the De Broglie relation. Now can we say that the momentum of the photon is = relativistic mass of the photon X the speed of light and from that know the relativistic mass, whatever that is?

    My other question, take an electron instead. If we know the wavenumber of the electron how do we calculate the speed of the electron.

    thanx for your reply

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  2. jcsd
  3. Mar 4, 2009 #2

    turin

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    If you want to, but you will most likely be misunderstood. Relativistic mass usually refers to sqrt{ E^2 - p^2 }. Many people even use the term to refer simply to E, but I find that a little strange.

    This question is a bit ambiguous. If you treat the electron as a plane wave, then you can determine the wave speed as v=omega/k. However, the result may surprise you, and this is probably not what you had in mind. If you treat the electron as a wavepacket (a localized bundle of electron-ness), so that it is more like the classical notion of a particle, then the speed of the packet is, to lowest order approximation, the group velocity. This depends on the nature of the wave packet.
     
  4. Mar 4, 2009 #3
    But that means that we have to know E and p of the photon, and those are given by De Broglie and Einstein E = h-bar x omega, right? Is it this relativistic mass which is reponsible for the bending of light in gravitational field?
    ¨
    I just started reading quantum mechanics on my own, preparing for next semester, so I hope u bare with me. My confusion is this. When we apply De Broglie to any other particle then the photon do we account for relativistic effects and how do we do that. For instance an electron accelerated to a speed which give relativitic effect, we only know the rest mass. If the electron is then diffracted and the wavelength is measured, can we deduce anything about the velocity of the electron. Thanx for your patience
     
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