Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The speed of sound of the inflaton field

  1. Jan 12, 2010 #1
    I've been reading about inflation and i encountered that one can always define the sound's speed as

    [tex]c_s^2 \equiv \frac{\partial_X P}{\partial_X \rho}[/tex]

    where [tex]X \equiv \frac{1}{2} g^{ab} \partial_a \phi \partial_b \phi[/tex]. In the case of a canonical scalar field [tex]P=X-V[/tex] and [tex]\rho=X+V[/tex], so [tex]c_s^2=1[/tex]. That is what is obtained by definition. But i can always consider [tex]P[/tex] and [tex]\rho[/tex] as a function of [tex]P=(X,\phi)[/tex] and [tex]\rho=(X,\phi)[/tex] so

    [tex]P+\rho=2 X[/tex] and

    [tex]\rho-P= 2 V[/tex]

    taking variations of these last to equations i obtain

    [tex]\delta P = - \delta \rho + 2 \delta X[/tex] (1) and

    [tex]\delta P = \delta \rho - 2 \partial_\phi V \delta \phi[/tex] (2)

    Recalling that in general [tex]P=(\rho,S)[/tex] then [tex]\delta P = c_s^2 \delta \rho + \tau \delta S[/tex]. Thus if i read the coefficient of [tex]\delta \rho[/tex] of eq. (1) one obtains that [tex]c_s^2 = -1[/tex] and [tex]\tau \delta S = 2 \delta X[/tex], but if i read the coefficient of eq. (2) one obtains [tex]c_s^2 = 1[/tex] and [tex]\tau \delta S = - 2 \partial_\phi V \delta \phi[/tex], according to the definition the correct reading would be the one done by (2) but is there another explanation of why reading the coefficient [tex]c_s^2[/tex] from (2) is the correct way, or is there a motivation for the first definition for [tex]c_s^2[/tex]?
  2. jcsd
  3. Jan 12, 2010 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    The problem is that the literature often uses [itex]c_{\rm{s}}^2[/itex] to mean two different things, sometimes simultaneously. Looking at things from a thermodynamic perspective, one can write [itex]P=P(\rho,S)[/itex], and then perturb to give
    [tex]\delta P=\frac{\partial P}{\partial\rho}\delta \rho +\tau \delta S[/tex]
    where [itex]\frac{\partial P}{\partial\rho}[/itex] is then identified as the adiabatic sound speed-- i.e. the speed with which perturbations travel through the background.

    Now, for a scalar field we can parametrise as [itex]P=P(X,\phi)[/itex]. Then, the adiabatic sound speed can be written as
    [tex]c_{\rm{s}}^2=\frac{\partial P}{\partial\rho}=\frac{\partial_X P +\partial_\phi P}{\partial_X\rho+\partial_\phi\rho}[/tex]. By writing things like this, it should be apparent that this is not the same as the first expression you quote. It turns out that, for a scalar field, the speed of propagation is not the adiabatic sound speed, but in fact a different speed (say, the "effective sound speed"), which is defined as
    [tex]\tilde{c_{\rm{s}}}^2=\frac{\partial_X P}{\partial_X\rho}[/tex]. If you like, you can show this by calculating the Klein-Gordon equation for the perturbation of the field and looking at the term in front of the spatial derivative.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook