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?

  • $\begingroup$ How do you know that's what $E$ is? If you insert (3) into (1) and assume the usual expressions for $T$ and $p$ you should be able to get some intuition for what is going on here. $\endgroup$ – ZachMcDargh Mar 6 '18 at 19:55
  • $\begingroup$ so that gives me $E = p\dot{x} - T + V = p\dot{x} - \frac{p^2}{2m} + V$? I just feel so lost with what to look for... $\endgroup$ – MRT Mar 6 '18 at 21:19
  • The precise statement is the following:


  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.


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

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.

  • $\begingroup$ Yeah I have the Euler-Lagrange equation $$\frac{\mathrm{d}}{\mathrm{d}t}\bigg(\frac{\partial L}{\partial \dot{q}_i}\bigg) - \frac{\partial L}{\partial q_i} = 0$$ But I don't understand how to apply it? $\endgroup$ – MRT Mar 6 '18 at 18:35
  • $\begingroup$ Take $q=x$, then define $p$ canonically. $\endgroup$ – J.G. Mar 6 '18 at 18:44
  • $\begingroup$ what do you mean by canonically? $\endgroup$ – MRT Mar 6 '18 at 21:22
  • $\begingroup$ Use canonical momentum, not kinetic momentum, viz. $p=\partial_\dot{q} L$. $\endgroup$ – J.G. Mar 6 '18 at 21:23
  • $\begingroup$ Would that mean that $$p = \frac{\partial L}{\partial \dot{x}} \Rightarrow L = p\dot{x} \Rightarrow E = p\dot{x} - p\dot{x} = 0$$ so is conserved? Or would it be that $$\frac{\mathrm{d}}{\mathrm{d}t}\Big(p\Big) - 0 = 0 - 0 = 0$$ $\endgroup$ – MRT Mar 6 '18 at 21:37

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