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Calculating the energy of an electron

  1. Apr 28, 2005 #1
    Let's say that an electron is having the speed of X m/s.

    So the energy is sometimes calculated like this:

    (1)[tex]E= \frac{mv^2}{2}[/tex]

    and sometimes like this

    (2)[tex]E = mc^2 + \frac{mv^2}{2}[/tex]

    When should I use 1 and when should I use 2?

    Thank you for any help.
     
  2. jcsd
  3. Apr 28, 2005 #2

    SpaceTiger

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    When we talk about energy, we usually talk about conserving it in a system. This means that the total energy at one time is equal to the total energy at another time. A particle's total energy is given by:

    [tex]E=\gamma mc^2[/tex]

    where [tex]\gamma[/tex] is the Lorentz factor:

    [tex]\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]

    In the limit of small v, the first equation goes to your second one:

    [tex]E=mc^2+\frac{1}{2}mv^2[/tex]

    Now, if we have some hypothetical set of particles and we want to conserve total energy, we can write:

    [tex]\sum_{i=1}^{N}E_{i,1}=E_x+\sum_{i=1}^{N}E_{i,2}[/tex]

    where Ex includes energy in radiation, fields, etc. This can be expanded to

    [tex]\sum_{i=1}^{N}(m_{i,1}c^2+\frac{1}{2}m_{i,1}v_{i,1}^2)=E_x+\sum_{i=1}^{N}(m_{i,2}c^2+\frac{1}{2}m_{i,2}v_{i,2}^2)[/tex]

    If, for all particles:

    [tex]m_{i,1}=m_{i,2}[/tex]

    then the equation simplifies to

    [tex]\sum_{i=1}^{N}\frac{1}{2}m_{i,1}v_{i,1}^2=E_x+\sum_{i=1}^{N}\frac{1}{2}m_{i,2}v_{i,2}^2[/tex]

    because the mc2 terms are the same on both sides.

    In words, as long as the components of your system are preserving their rest mass, then the mc2 term is not needed. In practice, this corresponds to cases in which the velocities are well below that of light. This is why classical theory works in the low-velocity limit.
     
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