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E=mc^2 Derivation

  1. Jan 4, 2012 #1
    Hi

    Is this derivation of E = mc^2 correct? ... I have some doubt at the red place ...

    [itex]\large{ F = \frac{dp}{dt} = \frac{d}{dt}(mv) }[/itex]

    [itex]\large{ F = v\frac{dm}{dt} + m\frac{dv}{dt} }[/itex]

    Let this force cause a displacement dx

    [itex]\large{ dW = F \cdot dx }[/itex]

    Assuming body was initially at rest and this work is converted into kinetic energy and increase it by dK

    [itex]\large{ dK = F\cdot dx }[/itex]

    [itex]\large{ dK = v\frac{dm}{dt}dx + m\frac{dv}{dt}dx }[/itex]

    [itex]\large{ dK = mvdv + v^2dm }[/itex] --- Equation 1


    Now using eqn

    [itex]\large{ m = \frac{m_o}{ \Large{ \sqrt{1-\frac{v^2}{c^2}} } } }[/itex]

    Squaring both sides,

    [itex]\large{ m^2= \frac{{m_o}^2}{1-\frac{v^2}{c^2}} }[/itex]

    [itex]\large{ m^2c^2 - m^2v^2 = {m_o}^2c^2 }[/itex]

    differentiating the expression

    [itex]\large{ 2mc^2dm - 2mv^2dm - 2vm^2dv = 0 }[/itex]

    [itex]\large{ c^2dm = mvdv - v^2dm }[/itex]

    Using this in eqn 1

    [itex]\large{ dK = c^2dm }[/itex]

    Integrating

    [itex]\large{ K = \int_0^K{dK} = \int_{m_o}^{m}{c^2dm} }[/itex] < --- HERE

    [itex]\large{ K = c^2(m - m_o) }[/itex]

    Total energy of body,

    [itex]\large{ E = K + m_o c^2 }[/itex]

    [itex]\large{ E = c^2(m - m_o) + m_o c^2 }[/itex]

    [itex]\large{ E = mc^2 = \frac{m_o c^2}{ \Large{ \sqrt{1-\frac{v^2}{c^2}} } } }[/itex]

    Also [itex]\large{ K = E - m_o c^2 = (m - m_o)c^2 }[/itex]

    [itex]\large{ \Delta E = \Delta m c^2 }[/itex]
     
  2. jcsd
  3. Jan 4, 2012 #2

    BruceW

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    I think this step isn't right. Maybe you got the negative sign in the wrong place? Up until this bit, I'd say it was all correct.
     
  4. Jan 4, 2012 #3
    Oops ... it was a typo
    It would be ...

    [itex]\large{ c^2dm = mvdv + v^2dm }[/itex]

    What i was confused was that, we write E = mc2 and not moc2 because when moving mass increase, and that is a way of saying (if i'm not wrong) that KE of body adds to its mass.
    So is it ok for me to take limits as mo → m ?
    I mean, when i'm treating KE as separate and not a part of addition in mass, should i really take these limits? ... Well i know my these sentences are also against some steps of my derivation above ... but ,,, ummm,,, i dunno ,,, please help me get rid of my confusion ...

    [itex]\large{ K = \int_0^K{dK} = \int_{m_o}^{m}{c^2dm} }[/itex]
     
  5. Jan 4, 2012 #4

    jedishrfu

    Staff: Mentor

  6. Jan 4, 2012 #5

    BruceW

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    Yes, I see it was simply a typo, and the rest of the derivation is correct :) (for 1-d motion, of course).

    m is the relativistic mass. m0 is the rest mass. So you can say that the KE of a body adds to its relativistic mass, but not to its invariant (rest) mass.

    And yes, those are the correct limits of integration, because when KE is zero, m=m0 (in other words, the only energy it has is rest energy).
     
  7. Jan 4, 2012 #6

    BruceW

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  8. Jan 4, 2012 #7
    So you mean that, the integrated eqn of K accounts for change in mass, ... m-mo mass is added to the particle when it moves ...
    Well is makes sense now ... I don't know why i was thinking all that before ,,, thank you for your help BruceW !! :biggrin:
     
  9. Jan 4, 2012 #8

    BruceW

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    No worries. Practice makes perfect!
     
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