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Math Help - Differentiation

  1. #1
    Member roshanhero's Avatar
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    Differentiation

    I am trying to prove that the group velocity is equal to the particle velocity by relativistic purpose
    The relativistic energy is given by the relation
    E^2 = p^2c^2 + m_{0}^2c^4
    But I don't know how to differentiate it with respect to momentum "p"
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  2. #2
    Super Member Aryth's Avatar
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    Well, first you get E by itself:

    E = \sqrt{p^2c^2 + m_0^2c^4}

    Now, we take the derivative with respect to p on both sides:

    \frac{dE}{dp} = \frac{d}{dp} \sqrt{p^2c^2 + m_0^2c^4}

    First we use the chain rule setting u = p^2c^2 + m_0^2c^4

    \frac{dE}{dp} = \frac{1}{2\sqrt{p^2c^2 + m_0^2c^4}}\frac{d}{dp}(p^2c^2 + m_0^2c^4)

    Now, you differentiate what is in the parentheses:

    \frac{dE}{dp} = \frac{1}{2\sqrt{p^2c^2 + m_0^2c^4}}(2pc^2)

    The two's cancel, and a c factors out of the radical in the denominator, which simplifies the derivative to:

    \frac{dE}{dp} = \frac{pc}{\sqrt{p^2 + m_0^2c^2}}

    And that's the derivative, maybe you can continue with this.
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