Euler equations for ideal fluids, approximations

Click For Summary
The Euler equations for ideal compressible flow are presented with small-value expansions for density and pressure. The user questions why the right-hand side of the second equation simplifies to \(\rho_0 \nabla \cdot v\) instead of including \(\delta \rho\). It is clarified that \(\delta \rho\) is negligible compared to \(\rho_0\), justifying the approximation. This simplification is crucial for analyzing fluid dynamics under the assumption of small perturbations. Understanding these approximations is essential for solving problems related to ideal fluid flow.
Niles
Messages
1,834
Reaction score
0

Homework Statement

The Euler equations for ideal compressible flow are given by
<br /> \partial_t v + (v\cdot \nabla)v = g-\frac{1}{\rho}\nabla p \\<br /> \partial_t \rho + \nabla \cdot(\rho v) = 0<br />
In my book these are written in terms of the small-value expansions \rho = \rho_0 + \delta \rho, p = p_0 + \delta p and the equations become
<br /> \partial_t v = -\frac{1}{\rho_0}\nabla \delta \rho \\<br /> \partial_t (\delta \rho) = -\rho_0 \nabla \cdot v<br />

In the second equation, I don't understand why the RHS becomes \rho_0 \nabla \cdot v instead of (\rho_0+\delta \rho) \nabla \cdot v?

Thanks in advance.
 
Physics news on Phys.org
\delta \rho \nabla \cdot v is negligible compared to \rho_0 \nabla \cdot v because \delta \rho &lt;&lt; \rho_0
 

Similar threads

  • · Replies 32 ·
2
Replies
32
Views
3K
Replies
5
Views
3K
Replies
7
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 26 ·
Replies
26
Views
3K
  • · Replies 0 ·
Replies
0
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K