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Finding rate of relativistic work, special relativity.

  1. May 4, 2012 #1
    Hello all,
    1. The problem statement, all variables and given/known data
    Given [tex] \frac{dE}{dt}=\frac{d(m(u))}{dt}\cdot u[/tex]
    show that [tex] \frac{dE}{dt}=\frac{m_0}{(1-u^2)^{\frac{3}{2}}}u\frac{du}{dt} [/tex]
    where u is velocity, m(u) is relativistic mass, and [itex] m_0 [/itex] is rest mass.

    2. Relevant equations
    [tex] m(u)=\frac{m_0}{\sqrt{1-u^2}}[/tex]

    3. The attempt at a solution
    Substituting in m(u) gives [tex] \frac{dE}{dt} = \frac{d}{dt}(\frac{m_0}{\sqrt{1-u^2}})\cdot u [/tex]
    Using the chain rule
    [tex] = \frac{d}{du}(\frac{m_0}{\sqrt{1-u^2}})\frac{du}{dt}\cdot u [/tex]
    [tex] = \frac{m_0 u}{(1-u^2)^{3/2}}\frac{du}{dt} \cdot u [/tex]
    From here I get into trouble. My first thought was since u should be parallel to [itex] \frac{du}{dt} [/itex] the dot product just becomes multiplication, but then there's an extra u factor. Any help from here would be greatly appreciated.
  2. jcsd
  3. May 4, 2012 #2
    The problem statement is ambiguous. The quantity in parenthesis should be written as mu, and not m(u). More precisely, it should be moγu.
  4. May 4, 2012 #3
    I don't think the relativistic mass should be a vector, so that dot must be just an ordinary product between two scalars. I do agree that the problem seems a little strange.
  5. May 6, 2012 #4
    That makes much more sense, thanks for the help guys
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