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Mechanics by Landau

  • Thread starter Advent
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  • #1
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Hi!

Just reading the first book by Landau in the theoretical physics course, and I need some guidance about one point (cf. 4, Lagrangian for a free particle.) Notations: L and L' are Lagrangians refered to different inertial frames of reference. e is a element of velocity between L and L'.
It says " We have L' = L (v'^2) = L (v^2 + 2ve + e^2 ). Expanding this expression in powers of e and neglecting the terms above the first order, we obtain

L (v'^2) = L (v^2) + partial derivative of L respect to v^2 times 2ve​

So, if this is supossed to be a Taylor serie with n from zero to one, why appears the derivate of L with respect to v^2. Can someone, please, make all steps explicit?


Any help would be very appreciated!
 

Answers and Replies

  • #2
dx
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L is a function of v², and when you change to the new intertial frame v² becomes v² + 2v⋅ε to first order in ε because (v + ε)² = v² + ε² + 2v⋅ε . The change in L to first order in ε is therefore (∂L/∂v²)(2v⋅ε).
 
  • #3
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Thanks for the post.
I don't see why is not (∂L/∂v²)(v²+2v⋅ε)
 
  • #4
dx
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If you have a function f and you want the change in f from a to a + Δ, you have to multiply f'(a) by Δ, i.e. f(a + Δ) = f(a) + f'(a)Δ.

In our case, v² becomes v² + 2v⋅ε, so you multiply by 2v⋅ε.
 
  • #5
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Thanks for your help, clear now! Should I study calculus of variations in order to continue with the book?
 
  • #6
dx
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No it's not really necessary. The only place where it is used in this book is in section 2, where the Euler-Lagrange equations are derived, and a quite clear description of it is given there.
 

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