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Reynolds Transport Theorem derivation question

  1. Aug 9, 2011 #1
    Hi,

    I'm struggling to understand one step in the derivation of the Reynold's Transport Theorem. I get as far as:

    \begin{equation}
    \frac{d}{dt} \int_{V(t)} f(\mathbf x, t) \, dV = \frac {d} {dt} \int_{V_0} f(\mathbf X,t) J(\mathbf X,t) \, dV = \int_{V_0} J \frac {\partial f} {\partial t} + f \frac {\partial J} {\partial t} \, dV
    \end{equation}

    From here, I want to write the RHS as:

    \begin{equation}
    \int_{V_0} \frac {\partial f} {\partial t} J + f \, ({\nabla \cdot} {\mathbf v}) J \,\, dV = \int_{V(t)} \frac {\partial f} {\partial t} + f \, ({\nabla \cdot} {\mathbf v}) \,\, dV
    \end{equation}

    However, all of my sources inexplicably replace the \begin{equation} \frac {\partial f} {\partial t} \end{equation} with a \begin{equation} \frac {Df} {Dt} \end{equation} Could anyone explain their reasoning or my error?

    Thanks!
     
  2. jcsd
  3. Aug 9, 2011 #2
    \begin{equation} \frac {Df} {Dt} \end{equation} is just the material derivative (the derivative with respect to time while keeping the initial position constant).
     
  4. Aug 9, 2011 #3
    But isn't the material derivative given by the total derivative wrt time, \begin{equation} \frac {df} {dt} = \frac {\partial f} {\partial t} + (\mathbf v \cdot \nabla) f \end{equation} whereas the integrand over the control volume just contains a partial derivative?
     
  5. Aug 9, 2011 #4
    Yes, it is if f is a scalar field. But in the Reynolds transport theorem, f is a vector field. For a vector field V, the material derivative is defined as follows:

    \begin{equation} \frac {Df} {Dt} = \frac {\partial V} {\partial t} + (\mathbf V \cdot \nabla) V \end{equation}
     
  6. Aug 10, 2011 #5
    Afraid I still don't follow. Why is the material derivative involved at all? In the ordinary course of physics we never swap a partial derivative for a total derivative, and as I understand it, the material derivative gives:
    \begin{equation}
    \frac {df(\mathbf x (\mathbf X,t),t)} {dt}
    \end{equation}
    which is not the integrand. I'm fairly certain the transport theorem is equally valid for scalars and vectors, so that shouldn't have made a difference either.
     
  7. Aug 10, 2011 #6
    You are taking the material derivative because the volume is a material fluid (a function of initial position and time where there is no flux of mass) as opposed to a control volume which is just a function of position with a fixed volume. Basically, you can still use the transport theorem on a control fluid, you just need to take the partial derivative instead of the material derivative. The same thing goes for a scalar field; the theorem is still applicable, you just need to consider that when deriving it.
     
  8. Aug 13, 2011 #7
    Thanks for pointing out that distinction, it clears things up.
     
  9. Aug 13, 2011 #8
    Glad to be of assistance.
     
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