Is the Variation of the Action Correct Under a Boost?

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SUMMARY

The variation of the action under a boost is calculated using the action integral ##S = \int L dt = \int \frac{1}{2} m v^{2} dt##. The solution involves determining the change in the Lagrangian ##\delta L##, leading to the expression ##\delta S = \int \frac{1}{2} m (2 v \delta v + v_{0} \cos \theta \delta v) dt##. The final result simplifies to ##\bigg( \int (mv + v_{0} \cos \theta) dt \bigg) \delta v##, confirming the correctness of the approach taken in the discussion.

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Homework Statement



Calculate the variation of the action ##S = \int L dt = \int \frac{1}{2} m v^{2} dt## under a boost ##\vec v \rightarrow \vec v + \vec v_{0}##.

Homework Equations



The Attempt at a Solution



##\delta S##
##= \int \delta L\ dt##
##= \int \frac{1}{2} m\ \delta(\vec v^{2})\ dt##
##= \int \frac{1}{2} m\ \delta(v^{2} + vv_{0}cos \theta + v_{0}^{2})\ dt##
##= \int \frac{1}{2} m\ (2 v\ \delta v + v_{0}\ cos \theta\ \delta v )\ dt##
##= \bigg( \int (mv + v_{0} cos \theta)\ dt\ \bigg)\ \delta v##

Am I correct so far?
 
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## \delta{v^2} = 2\mathbf{v}\cdot\delta{\mathbf{v}} ##
 

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