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Speed of sound in an isentropic ideal gas

  1. Apr 9, 2015 #1
    1. The problem statement, all variables and given/known data
    [tex] \rho_0, c_0 [/tex] is the mean density, the mean speed of sound in the ideal gas.
    Is the following correct?
    [tex] c(\rho)=c_0\left(\frac{\rho}{\rho_0}\right)^{\frac{\kappa-1}{2}} [/tex]

    2. Relevant equations
    [tex] p = const * \rho^\kappa, c=\sqrt{\frac{\partial p}{\partial \rho}} [/tex]

    3. The attempt at a solution
    [tex] c=\sqrt{\frac{\partial p}{\partial \rho}} = \sqrt{const*\kappa*\rho^{\kappa-1}}=const*\rho^{\frac{\kappa-1}{2}} [/tex]
    With c(\rho_0)=c_0, I get:
    [tex] c(\rho)=c_0\left(\frac{\rho}{\rho_0}\right)^{\frac{\kappa-1}{2}} [/tex]

    Can I then say, that the refractive index is:
    [tex] n(\rho)=\frac{c_0}{c(\rho)}=\left(\frac{\rho}{\rho_0}\right)^{\frac{1-\kappa}{2}} [/tex]
    Hence, the ratio of 2 refractive indexes, like it is needed in the refraction law, is independent of [tex] \rho_0 [/tex]?
    Is there a mistake in the reasoning? Thank you very much for your help!
     
  2. jcsd
  3. Apr 11, 2015 #2

    MarcusAgrippa

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    Gold Member

    I assume that your equation of state is [itex] p = A \rho^\kappa [/itex]. Then
    [itex] c^2 = \displaystyle \frac{dp}{d\rho} = \kappa A \rho^{\kappa-1} [/itex]
    You then appear to want to eliminate A - is that right? If so, use the equation of state again to write
    [itex] p_0 = A \rho_0^\kappa \ \ \Rightarrow \ \ A = \frac{p_0}{{\rho_0}^\kappa} [/itex]
    so that
    [itex] c^2 = \displaystyle \frac{dp}{d\rho} = \kappa \frac{p_0}{{\rho_0}^\kappa} \rho^{\kappa-1} [/itex]
    I am not sure where you want to go from here.

    The following calculation is more common:
    [itex] c^2 = \displaystyle \frac{dp}{d\rho} = \gamma \frac{p}{\rho} [/itex]
    This is how this result is usually stated.

    You seem to want to introduce [itex] c_0 [/itex], though I can't see why. Presumably you define
    [itex] {c_0}^2 = \displaystyle \left.\frac{dp}{d\rho}\right|_0 = \gamma \frac{p_0}{\rho_0} [/itex]
    in which case
    [itex] c^2 = {c_0}^2 \displaystyle \frac{p}{p_0} \frac{\rho_0}{\rho} [/itex]
    or
    [itex] c = {c_0} \displaystyle \sqrt{\frac{p}{p_0} \frac{\rho_0}{\rho}} = c_0 \left( \frac{\rho}{\rho_0} \right)^{(\kappa -1)/2} [/itex]
    as you claim.

    Finally, you want to speak about a refractive index, but I don't know what that means in this context. You are looking at the propagation of sound waves in an homogeneous isotropic medium which, in its undisturbed state it at pressure [itex] p_0 [/itex] and density [itex] \rho_0 [/itex]. You have no interfaces where the wave passes from one medium into another, so you won't encounter any refraction. Your two sound speeds are for the same medium (?) and so cannot be used define a refractive index.

    In fact, I don't know how to interpret your sound speed c. The usual interpretation is that [itex] c_0 [/itex] is the sound speed for small amplitude waves in the original medium. The speed c is calculated with different values of [itex] p, \rho [/itex] which are not constant, but change sinusoidally about mean values [itex] p_0, \rho_0 [/itex] as a wave propagates through the medium. So what does it represent physically? The speed of a wave travelling through your wave? That makes no sense. Does your wave (large amplitude, perhaps) change its speed as the medium compresses and rarefies? Will this lead to shock waves?

    If you want to define a refractive index for sound waves, how would you do it? Sound waves do not propagate through a vacuum, so you cannot take the sound speed in vacuum and divide it by the sound speed in your medium. I suppose you could use the sound speed in some suitably defined standard medium (air at STP?) and then measure all sound speeds relative to that - but I have never seen this done.
     
    Last edited: Apr 11, 2015
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