# The speed of sound of the inflaton field

1. Jan 12, 2010

### chronnox

I've been reading about inflation and i encountered that one can always define the sound's speed as

$$c_s^2 \equiv \frac{\partial_X P}{\partial_X \rho}$$

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

$$P+\rho=2 X$$ and

$$\rho-P= 2 V$$

taking variations of these last to equations i obtain

$$\delta P = - \delta \rho + 2 \delta X$$ (1) and

$$\delta P = \delta \rho - 2 \partial_\phi V \delta \phi$$ (2)

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

2. Jan 12, 2010

### cristo

Staff Emeritus
The problem is that the literature often uses $c_{\rm{s}}^2$ to mean two different things, sometimes simultaneously. Looking at things from a thermodynamic perspective, one can write $P=P(\rho,S)$, and then perturb to give
$$\delta P=\frac{\partial P}{\partial\rho}\delta \rho +\tau \delta S$$
where $\frac{\partial P}{\partial\rho}$ 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 $P=P(X,\phi)$. Then, the adiabatic sound speed can be written as
$$c_{\rm{s}}^2=\frac{\partial P}{\partial\rho}=\frac{\partial_X P +\partial_\phi P}{\partial_X\rho+\partial_\phi\rho}$$. 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
$$\tilde{c_{\rm{s}}}^2=\frac{\partial_X P}{\partial_X\rho}$$. 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.