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Eulerian vs. Lagrangian description

  1. May 30, 2010 #1
    1. The problem statement, all variables and given/known data
    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


    2. Relevant equations



    3. 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
     
  2. jcsd
  3. May 30, 2010 #2
    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
     
  4. May 30, 2010 #3
    Yeah but why/how can you do that
     
  5. May 30, 2010 #4
    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]
     
  6. May 30, 2010 #5
    Anyone else?
     
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