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In Tong’s preliminaries for QFT, he makes the following step while showing Noether’s theorem:

$(\frac{\partial L}{\partial q}\delta q + \frac{\partial L}{\partial \dot q}\delta \dot q) = \left(\frac{\partial L}{\partial q} - \frac{d}{dt}\left(\frac{\partial L}{\partial \dot q} \right)\right)\delta q + \frac{\partial L}{\partial \dot q}\delta q$

How do you get algebraically from the left to the right side there?

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The equality you asked about is wrong, and I will show why. Since $$\frac{\partial L}{\partial \dot{q}}\delta \dot{q} =\frac{\partial L}{\partial\dot{q}}\left ( \frac{d}{dt}\delta q \right )$$ we can write using the product rule: $$\frac{d}{dt}\left ( \frac{\partial L}{\partial \dot{q}}\delta q \right ) = \frac{d}{dt}\left ( \frac{\partial L}{\partial \dot{q}} \right )\delta q+\frac{\partial L}{\partial \dot{q}}\delta \dot{q}$$ and so $$\frac{\partial L}{\partial \dot{q}}\delta \dot{q} = \frac{d}{dt}\left ( \frac{\partial L}{\partial \dot{q}}\delta q \right )-\frac{d}{dt}\left ( \frac{\partial L}{\partial \dot{q}} \right )\delta q$$ and so the equality would be: $$\frac{\partial L}{\partial q}\delta q + \frac{\partial L}{\partial \dot{q}}\delta \dot{q} = \left ( \frac{\partial L}{\partial q} - \frac{d}{dt}\left ( \frac{\partial L}{\partial \dot{q}} \right )\right )\delta q + \frac{d}{dt}\left ( \frac{\partial L}{\partial \dot{q}} \delta q\right )$$ which is the same as your equality, except it has a time derivative on the last term of the RHS. Notice that the first two terms in the RHS vanish by the Euler-Lagrange equation.

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  • $\begingroup$ You’re right, I misrepresented Tong! And your answer still made me understand the intended reasoning, well done! $\endgroup$
    – Urdatorn
    Dec 13 '20 at 21:20

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