Eulerian vs. Lagrangian description

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  • #1
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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
 

Answers and Replies

  • #2
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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
 
  • #3
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Yeah but why/how can you do that
 
  • #4
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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]
 
  • #5
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Anyone else?
 

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