- #1
happyparticle
- 431
- 20
- 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.
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.
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