Linearization Euler's equation

[happyparticle]

Homework Statement

First order linearized Euler's equation for a small perturbation

Relevant Equations

##mna (\frac{\partial}{\partial t} + \frac{\vec{v}}{a} \cdot \nabla ) \vec{u} = - \nabla P - mn \nabla \phi##

I'm trying to linearize (first order) the Euler's equation for a small perturbation ##\delta##

Starting with ##mna (\frac{\partial}{\partial t} + \frac{\vec{v}}{a} \cdot \nabla ) \vec{u} = - \nabla P - mn \nabla \phi## (1)

##\vec{u} = aH\vec{x(t)} + \vec{v(x,t)}##
Where a is the scale factor and H the Hubble parameter.

Also,
##n = \bar{n}(t) (1 + \delta(\vec{x},t))## (2)
## P(\vec{x},t) = \bar{P}{t} + \delta P(\vec{x},t)## (3)
## \phi(\vec{x},t) = \bar{\phi}{t} + \delta \phi(\vec{x},t)## (4)

Plugging (2), (3), and (4) in (1) and after some algebra, I got:

##m \bar{n} a \dot{a} \vec{v} + m \bar{n} a \dot{\vec{v}} + m \bar{n} a H \vec{v} + m \bar{n} \delta a \dot{a} \vec{v} + m \bar{n} \delta a \dot{ \vec{v}} + m \bar{n} \delta a H \vec{v} = - \nabla \delta P - m \bar{n} \nabla \delta \phi - m \bar{n} \delta \nabla \delta \phi ##

The correct answer should be:

##m \bar{n} a (\dot{\vec{v}} + H \vec{v}) = - \nabla \delta P - m \bar{n} \nabla \delta \phi##

Thus, I'm wondering if, for example, this part ##a \dot{a}## is a second order one?

I'm not totally sure how to linearize an equation.

 

[ergospherical]

These can be fiddly. The first thing to get clear is the distinction between the proper coordinates ##\mathbf{r}## and comoving coordinates ##\mathbf{x}##, with ##\mathbf{r} = a \mathbf{x}##. This implies two important relationships:
\begin{align*}
\nabla_{\mathbf{r}} &= a^{-1} \nabla_{\mathbf{x}} \\ \\
\frac{\partial}{\partial t} \bigg{|}_{\mathbf{r}} &=\frac{\partial}{\partial t} \bigg{|}_{\mathbf{x}} + \frac{\partial \mathbf{x}}{\partial t} \bigg{|}_{\mathbf{r}} \cdot \nabla_{\mathbf{x}} = \frac{\partial}{\partial t} \bigg{|}_{\mathbf{x}} - H \mathbf{x} \cdot \nabla_{\mathbf{x}}
\end{align*}The Euler equation in its usual form -- expressed in terms of ##\nabla_{\mathbf{r}}## and ##(\partial/\partial t)\big{|}_{\mathbf{r}}## -- is the following:$$\left(\frac{\partial}{\partial t} \bigg{|}_{\mathbf{r}} + \mathbf{u} \cdot \nabla_{\mathbf{r}} \right) \mathbf{u} = - \frac{1}{m\bar{n}} \nabla_{\mathbf{r}} P - \nabla_{\mathbf{r}} \phi$$You can plug in your linearisations ##P = \bar{P} + \delta P, \ \mathbf{u} = H \mathbf{r} + \mathbf{v}, \ \phi = \bar{\phi} + \delta \phi##. After subtracting off the zero order equation, check that you obtain:$$\frac{\partial \mathbf{v}}{\partial t} \bigg{|}_{\mathbf{r}} + (H\mathbf{r} \cdot \nabla_{\mathbf{r}}) \mathbf{v} + (\mathbf{v} \cdot \nabla_{\mathbf{r}}) H\mathbf{r} = - \frac{1}{m\bar{n}} \nabla_{\mathbf{r}} P - \nabla_{\mathbf{r}} \phi$$Can you re-write this equation in terms of ##\nabla_{\mathbf{x}}## and ##(\partial/\partial t)\big{|}_{\mathbf{x}}## using the previous expressions? That should give you your answer (after dropping the subscript ##\nabla_{\mathbf{x}} \equiv \nabla## and ##(\partial/\partial t)\big{|}_{\mathbf{x}} \equiv \partial/\partial t##).

 

[happyparticle]

Thank you!
It works. I understand why it didn't work.

First, I had replaced ##n## with ##\bar{n}(1+\delta)##. I would like to know why you replaced ##n## with ##\bar{n}##.

Moreover, it seems like you put ##\frac{\partial}{\partial t} H\vec{r} = 0##. I'm not sure understand why.

 

[ergospherical]

It's true that ##n = \bar{n}(1+\delta)## but the linearization of the pressure force is ##\frac{1}{\bar{n}} \nabla_{\mathbf{r}} \delta P##

It's not true that ##\frac{\partial}{\partial t} \big{|}_{\mathrm{r}} (H \mathbf{r}) = 0## -- instead, this term is part of the 0th order equation. So it is killed when you subtract that off.

 

[happyparticle]

Maybe I don't understand, but there is no other similar term to subtract with. Also, does ##\frac{\partial}{\partial t} \big{|}_{\mathrm{r}} ## means r fixed? So we consider r as a constant?

 

[ergospherical]

The zeroth order equation (i.e. with no perturbations) is:
$$\frac{\partial}{\partial t} \bigg{|}_{\mathbf{r}} (H\mathbf{r}) + H\mathbf{r} \cdot \nabla_{\mathbf{r}} (H\mathbf{r}) = - \frac{1}{m\bar{n}} \nabla_{\mathbf{r}} \bar{P} -\nabla_{\mathbf{r}} \bar{\phi}$$
When you plug in the linearisations, e.g. ##H\mathbf{r} \rightarrow H\mathbf{r} + \mathbf{v}##, etc., you get something like$$\frac{\partial}{\partial t}\bigg{|}_{\mathbf{r}}(H\mathbf{r} + \mathbf{v}) \quad + \quad (\dots) \quad = \quad (\dots) \quad$$where I haven't typed out all the other terms. You get the first order linearization of the equations by subtracting the zeroth order equation (and also keeping only terms that are no higher than first order). The ##(\partial/\partial t)\big{|}_{\mathbf{r}}(H\mathbf{r})##'s cancel and you are left with only ##(\partial \mathbf{v}/\partial t) \big{|}_{\mathbf{r}}##

For the second question, yes, the ##\big{|}_{\mathbf{r}}## notation is the standard partial derivative notation of keep that variable fixed in the differentiation. (I.e. in this case, you take the time derivative at fixed proper coordinates.)

 

[happyparticle]

Thanks a lot! Things are so much clearer now.