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E = mc^2

  1. Dec 24, 2004 #1
    The energy of any piece of matter is given by
    E = mc^2 + Relative Energy

    Usuall the relative energy term is 1/2mv^2

    thus for v << c

    E = mc^2 + 1/2mv^2, but what if this particle was 100K As opposed to 0K

    does temperature play a role in the energy of an object using this definition?

    If so wouldn't a better definition of energy be given by

    E = Inertial Energy + Relative Energy

    if it was of this form, it would almost be like the Basic Kinematic Equation
    which states that the acceleration between two systems can be found by finding the relative acceleration added to any corrective terms due to various types of gyroscopic motion. Thus, energy would be the same way. Meaning no object as absolute energy, it is all relative depending on differences in velocity, temperature, and potentials between 2 observers. Does anyone agree?
     
  2. jcsd
  3. Dec 25, 2004 #2
    Actually if memory serves correctly the proper equation for it is:

    [tex]E_0 = m_0 c^2[/tex]
    [tex]E_k = \gamma m_0 c^2[/tex]
    [tex]\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}[/tex]

    [tex]E_t = E_0 + E_k[/tex]

    To get really general:

    [tex]E^2 = p^2c^2 + m^2_0 c^4[/tex]
     
  4. Dec 25, 2004 #3
    I'm not quite sure what you're getting at here.

    [tex] E = m_{0}c^2 + K = mc^2 [/tex]

    where [tex]m_{0} [/tex] is the rest mass of the object. [tex]m_{0}c^2[/tex] would be the rest (inertial) energy of the particle and K would be the relative energy. If you shift your frame of reference to that of the particle, it's kinetic energy is 0 and it reduces to it's inertial energy. That seems to me to be what you're saying.

    As for temperature, temperature is a measure of the {average} kinetic energy of the particles in an object. Temperature doesn't affect the kinetic energy, it *is* the kinetic energy. We don't usually talk about the temperature of a single particle because we measure it's kinetic energy directly.
     
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