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What is the correct term for the energy represented by E=mc2

  1. Sep 12, 2015 #1
    Is E=mc2 the potential energy of a mass m? Is it the maximum energy such a mass can have? What is the correct term used to denote E in this context?

    For example, if an object is travelling at 10% of the speed of light, could one say it has a kinetic energy of 0.5 x m x 0.1² = 1/200th of its "maximum energy"?
     
  2. jcsd
  3. Sep 12, 2015 #2

    Dale

    Staff: Mentor

    Hi Keermalec, welcome to PF!
    Actually, it is the minimum energy. It is the energy that the mass has when it is at rest (or at least in a reference frame where its net momentum is 0).

    I usually call it "mass energy" or "rest energy".

    It would have a kinetic energy of (approximately) 1/200th of its rest energy. So its total energy would be 201/200th of its rest energy.
     
  4. Sep 12, 2015 #3

    jtbell

    User Avatar

    Staff: Mentor

    I usually use the symbol ##E_0## for rest energy to distinguish it from total energy E: ##E_0 = mc^2##.
     
  5. Sep 13, 2015 #4

    vanhees71

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    Science Advisor
    2016 Award

    For a massive classical particle the energy-momentum vector is
    $$p^{\mu}=m \gamma \begin{pmatrix} c \\ \vec{v} \end{pmatrix} \quad \text{with} \quad \gamma=\frac{1}{\sqrt{1-\vec{v}^2/c^2}}.$$
    The energy is
    $$E=p^0 c=\frac{m c^2}{\sqrt{1-\vec{v}^2/c^2}}=c \sqrt{m^2 c^2+\vec{p}^2}.$$
    Here, ##m##, is a Lorentz scalar, the socalled invariant mass of the particle. This is the notion of mass used almost exclusively in any modern textbook treating special relativity and in the physics community. The reason is that there is simply no need for another symbol for ##E/c^2##.

    Sometimes one also introduces the rest energy, ##E_0=m c^2## and the kinetic energy (e.g., when talking about the beam energy of a particle accelerator in fixed-target experiment),
    $$E_{\text{kin}}=E-E_0=c \sqrt{m^2c^2+\vec{p}^2}-m c^2.$$
     
  6. Sep 21, 2015 #5
    Thanks Dale, JTBell and Vanhees, that is exactly the answer I was looking for.
     
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