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I was informed by @hft that by combining the total time derivation:

$$\frac{dL}{dt} = \frac{\partial L}{\partial x}\dot{x} + \frac{\partial L}{\partial \dot{x}}\ddot{x} + \frac{\partial L}{\partial t}$$

and:

$$\frac{\partial L}{\partial t}=0$$

We can get:

$$\frac{d}{dt}\bigg(\frac{\partial L}{\partial\dot{x}}\dot{x}-L\bigg)=0$$

which shows conservation of energy. However, I don't fully understand how one can combine these two equations and get the equation shown above. Could someone please show this process step by step? Thanks.

Qmechanic
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Jungwoon Song
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2 Answers2

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There are two ways of seeing that the two equations can be combined to yield the desired result...

  1. The first one is simply to substitute $$\frac{d}{dt}\bigg(\frac{\partial L}{\partial \dot{x}}\dot{x}\bigg)-\frac{dL}{dt}= \frac{d}{dt}\bigg(\frac{\partial L}{\partial \dot{x}}\dot{x}\bigg)- \frac{\partial L}{\partial x}\dot{x}- \frac{\partial L}{\partial \dot{x}}\ddot{x}\\= \frac{d}{dt}\bigg(\frac{\partial L}{\partial \dot{x}}\dot{x}\bigg)- \bigg\{\bigg[\frac{\partial L}{\partial x}-\frac{d}{dt}\bigg(\frac{\partial L}{\partial \dot{x}}\bigg)\bigg]\dot{x}\bigg\}- \frac{d}{dt}\bigg(\frac{\partial L}{\partial \dot{x}}\dot{x}\bigg)$$ where in the second step I used that $\frac{\partial L}{\partial t}=0$ and in the second step I used the product rule! It is obvious now that the first and the third term vanish, whereas the second term can be identified as the equations of motion, and hence also vanish!

  2. The second way of combining the two equations such that the desired result emerges is by following the standard derivation for the conserved charge (namely the energy!)... The latter goes as follows: $$\frac{dL}{dt}=\frac{d}{dt}\bigg(\frac{\partial L}{\partial\dot{x}}\dot{x}\bigg)$$ given that the equations of motion hold and that the Lagrangian is not explicitly dependent on time (/ the partial time derivative vanishes). The equation above can be re-written as $$\frac{d}{dt}\bigg(\frac{\partial L}{\partial \dot{x}}\dot{x}-L\bigg)=0$$ which shows that the conserved charge is the quantity in the parenthesis, which for obvious reasons I denote as $E$ $$E=\frac{\partial L}{\partial \dot{x}}\dot{x}-L$$

I hope this helps! If there are any questions, please comment.

schris38
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You have to use the Euler -Lagrange equation also $\left( \frac{\partial L}{\partial x}-\frac{d}{dt}\frac {\partial L}{\partial \dot x}= 0 \right)$

user330563
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