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If the Lagrangian does not depend explicitly on time, then the quantity $E$ given by $$E := p\dot{x} - L \tag{1}$$ is conserved.

I'm really confused. Normally the total energy is given by $$E = T + V.\tag{2}$$ Our definition of the $\textbf{Lagrangian}$ is $$L(x,\dot{x}, t) = T - V\tag{3}$$ with $T$ being the kinetic energy and $V$ being the potential energy. So I think to rearrange to get $$L = p\dot{x} - E = p\dot{x} - T - V.\tag{4}$$ But I don't know what the kinetic and potential are?

Qmechanic
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MRT
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2 Answers2

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I'm not the special case $L=T-V$ with $T=\frac{p^2}{2m},\,p=m\dot{x}$, you can show $p\dot{x}=2T$. The expression you've been given for $E$ is always conserved; you can prove this using an Euler-Lagrange equation.

J.G.
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  • The precise statement is the following:

Given

  1. the definition of Lagrangian momentum $p_i:=\frac{\partial L}{\partial v^i}$,

  2. the definition of Lagrangian energy $h:=\sum_{i=1}^n p_i v^i-L $, and

  3. the Euler-Lagrange eqs. $\frac{dp_i}{dt}\approx \frac{\partial L}{\partial q^i}$,

  4. the Lagrangian $L(q,v)$ does not depend explicitly on time $t$,

then the Lagrangian energy is conserved $\frac{dh}{dt}\approx 0$.

  • For the perspective of Noether's theorem, see e.g. this related Phys.SE question.

Note that:

  • A Lagrangian $L$ is not necessarily on the form $T-V$, cf. this Phys.SE post.

  • Mechanical energy $T+V$ is not necessarily the same as Lagrangian energy.

  • OP's Eqs. (2)-(4) do not necessarily hold.

References:

  1. H. Goldstein, Classical Mechanics; Chapter 2.
Qmechanic
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