Newtonian fluid mechanics: Navier-Stokes equation

Click For Summary

Homework Help Overview

The discussion revolves around understanding the Navier-Stokes equation in the context of Newtonian fluid mechanics. The original poster expresses confusion regarding specific terms and concepts presented in their textbook, particularly about the right side of the equation and the introduction of additional terms that were not clearly explained.

Discussion Character

  • Exploratory, Conceptual clarification

Approaches and Questions Raised

  • Participants are attempting to clarify the distinction between material derivatives and regular derivatives, with questions about how to identify when a derivative is material. There is also a focus on understanding the implications of the terms presented in the Navier-Stokes equation and their relevance to fluid acceleration.

Discussion Status

The discussion is ongoing, with participants sharing their understanding and seeking clarification on specific concepts. Some guidance has been offered regarding the terminology used in different textbooks, but there is no explicit consensus on the confusion surrounding the material derivative.

Contextual Notes

There is mention of varying terminology in textbooks, which may contribute to the confusion regarding the definitions and applications of derivatives in fluid mechanics.

Feodalherren
Messages
604
Reaction score
6

Homework Statement


woot.png


Homework Equations


Navier-Stokes

The Attempt at a Solution


Not really trying to solve a problem, trying to understand what is going on in my textbook. So look at the stuff in red first. I see where all that is coming from, it's clear to me. However, the stuff in green indicates that in the example case the right side of the equation should be zero, instead they throw in the stuff in blue out of the blue. What exactly happened here that they totally missed to explain?!
 
Physics news on Phys.org
Are you familiar with the material derivative, d/dt? It is defined as:
$$\frac{d}{dt}=\frac{\partial}{\partial t}+u\frac{\partial}{\partial x}+v\frac{\partial}{\partial y}+w\frac{\partial}{\partial z}$$
The material derivative of u is equal to the x-component of the fluid acceleration.
$$\frac{du}{dt}=a_x=\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}+w\frac{\partial u}{\partial z}$$

Chet
 
  • Like
Likes   Reactions: Feodalherren
Okay yeah I remember that from class. Thanks.

I'm still slightly confused though. How am I supposed to know if it's a material derivative of just a regular derivative? What exactly makes it a material derivative? If I take the expression "at face value" then u does NOT depend on t and it equal zero.
Thanks again!
 
Feodalherren said:
Okay yeah I remember that from class. Thanks.

I'm still slightly confused though. How am I supposed to know if it's a material derivative of just a regular derivative? What exactly makes it a material derivative? If I take the expression "at face value" then u does NOT depend on t and it equal zero.
Thanks again!
It's all a matter of the terminology your textbook or professor uses. Some fluids textbooks use d/dt and others use D/Dt. Of course, if you are familiar with the NS equations, you know to look for that.

Chet
 
  • Like
Likes   Reactions: Feodalherren

Similar threads

  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 5 ·
Replies
5
Views
5K
  • · Replies 11 ·
Replies
11
Views
3K
Replies
9
Views
3K
  • · Replies 14 ·
Replies
14
Views
8K
  • · Replies 35 ·
2
Replies
35
Views
13K
  • · Replies 7 ·
Replies
7
Views
3K