Why does Cauchy's EOM not contain D ( ρ u i ) D t but instead contains ρ D u i D t ?

Question

Why does Cauchy&#039;s EOM not contain             D      (      ρ              u        i            )              D      t       but instead contains   ρ            D              u        i                    D      t       ?

a2t2esdg · Accepted Answer

It&#039;s essentially a consequence of continuity. From Evan&#039;s response, what you essentially want to establish is      d          d      t            ∫    V    ρ      u          d    3    x  =      ∫    V    ρ            D              u                    D      t              d    3    xwhich you can do in the following way. From Reynold&#039;s transport theorem, we know that                    (1)                                          d                                              d                        t                                    ∫                      V            (            t            )                          ρ                  u                                  d          3                x        =                  ∫                      V            (            t            )                                    (                      D                          D              t                                (          ρ                      u                    )          +          (          ρ                      u                    )          ∇          ⋅                      u                    )                          d          3                x            note that the continuity equation is            D      ρ              D      t        +  ρ  ∇  ⋅      u    =  0so we can expand   D  (  ρ      u    )      /    D  t from (1) as            D      (      ρ              u            )              D      t        =  ρ            D              u                    D      t        +      u              D      ρ              D      t          =  ρ            D              u                    D      t        −      u    (  ρ  ∇  ⋅      u    )putting this into (1) yields      d          d      t            ∫    V    ρ      u          d    3    x  =      ∫    V    ρ            D              u                    D      t              d    3    xThe intuition here is that you&#039;re trying to track the rate of change of some quantity X multiplied by ρdV while simultaneously following dV. Well, ρdV represents a mass element, and mass is assumed to be conserved. So the if you expand this out in a product rule, the sum is only concerned about how X is changing, hence the rate of change of XρdV should just be DX/DT multiplied by ρdV.

atestiguoki · Accepted Answer

Look at the integral version of the balance of linear momentum first. It is in the form                     G            ˙        =      F  , where G is the linear momentum of a finite blob of continuum and F is the net force acting on the blob. For a blob occupying a current region       R  , the integral form is      D          D      t            ∫                  R              ρ      v    d  v  =      ∫          ∂              R                  t    d  a  +      ∫                  R              ρ      b    d  vwhere t are the supplied tractions, and b are body forces. You should be able to prove that      D          D      t            ∫                  R              ρ      v    d  v  =      ∫                  R              ρ            D              v                    D      t        d  von your way to deriving the local, or differential version of the balance law.

Why does Cauchy's EOM not contain D(ρu_i)/(Dt) but instead contains ρDu_i/(Dt)?

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