Eulerian vs. Lagrangian description

  • Thread starter Thread starter coverband
  • Start date Start date
  • Tags Tags
    Lagrangian
Click For Summary

Homework Help Overview

The problem involves a particle's motion described by a position vector in terms of time, with the goal of finding its velocity in both Lagrangian and Eulerian forms. The discussion centers around the mathematical manipulation of these expressions.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the substitution of terms in the velocity vector derived from the position vector, questioning the validity and physical meaning of such substitutions.

Discussion Status

Some participants have offered insights into the mathematical identities that allow for substitutions, while others are seeking clarification on the reasoning behind these operations. The discussion is ongoing with various interpretations being explored.

Contextual Notes

There appears to be some confusion regarding the physical implications of the mathematical operations involved, as well as the assumptions made in the substitutions. No explicit consensus has been reached on these points.

coverband
Messages
170
Reaction score
1

Homework Statement


A particle moves so that [tex]\vec{x} \equiv [x_0 exp(2t^2), y_0 exp(-t^2), z_0 exp(-t^2)][/tex]. Find the velocity of the particle in terms of [tex]x_0[/tex] and t (the LAgrangian description) and show it can be written as [tex]\vec{u} \equiv (4xt, -2yt, -2zt),[/tex] the Eulerian description


Homework Equations





The Attempt at a Solution


[tex]\vec {u} = [4tx_0 exp(2t^2), -2ty_0 exp(-t^2), -2tz_0 exp(-t^2)][/tex]

The exp's go to 1 !?

Thanks
 
Physics news on Phys.org
No, i suppose it's just that in your velocity vecotr you can substitute the term [tex]x_0exp(2t^2)[/tex] with x - "first part" from your [tex]\vec{x}[/tex] vector and so on
 
Yeah but why/how can you do that
 
Well, i guess because they're equal. I see no physical meaning of such operation, it's just the mathematical identity that allows you to subsitute it.
Let's consider a single function [tex]f(x)=exp(2x)[/tex]. Obviously, [tex]\frac{df}{dx}=2exp(2x)[/tex]. And you can write [tex]\frac{df}{dx}=2f(x)[/tex], which is true for all x, but doesn't really mean anything.

E:and so, in your example, you just get [tex]x_x=x_0exp(2t^2)[/tex], [tex]u_x=4tx_0exp(2t^2)=4tx[/tex]
 
Anyone else?
 

Similar threads

Replies
1
Views
2K
  • · Replies 59 ·
2
Replies
59
Views
13K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
5
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
13
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
5
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K