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Linear Sound Wave Equations

  1. Apr 1, 2015 #1
    Taken from my lecturer's notes, how did they make the jump from 8.5 to 8.6 and 8.7?

    sound1.png

    Even after differentiating (8.5) with time I get

    [tex] \rho_0 \frac{\partial^2 \vec u'}{\partial t^2} + \nabla \frac{\partial p '}{\partial t} = 0 [/tex]
    [tex] \frac{\partial^2 p'}{\partial t^2} + \rho_0 c^2 \nabla \cdot \frac{\partial \vec u'}{\partial t} = 0 [/tex]

    Is there a relation between ##\vec u## and ##p## I am missing?
     
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  3. Apr 1, 2015 #2

    vanhees71

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    Note that
    $$\vec{\nabla} \cdot \partial_t \vec{u}=\partial_t \vec{\nabla} \cdot \vec{u}.$$
     
  4. Apr 1, 2015 #3
    Is ##\nabla \cdot \vec u ## somehow related to pressure?
     
  5. Apr 1, 2015 #4

    vanhees71

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    It's somewhat related to density. Using the continuity equation, which expresses mass conservation (valid in non-relativistic physics from very basic principles)
    $$\partial_t \rho + \vec{\nabla} (\rho \vec{v})=0.$$
     
  6. Apr 1, 2015 #5
    [tex] \frac{\partial m}{\partial t} = - \int \rho \vec v \cdot d\vec S [/tex]
    [tex] \int \frac{\partial \rho}{\partial t} dV = -\int \rho \nabla \cdot \vec v dV [/tex]

    This implies that ##\frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \vec v) = 0 ##.
     
  7. Apr 1, 2015 #6

    boneh3ad

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    Well yes that equation is satisfied by default, as it is the continuity equation.

    You can relate the velocity and pressure through Euler's equation.
     
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