Some notes. I've been playing around a bit with "Classical Mechanics Lite" to get a better understanding of exactly what's happening in terms of conservation laws, symmetries, and the seemingly derivative-like relationship between linear momentum $\mathbf p = m \mathbf v$ and kinetic energy $T = \frac{1}{2} m \dot{\mathbf x} \cdot \dot{\mathbf x}$. The reason I'm calling this "Classical Mechanics Lite" is because I'm working under the following assumptions:

  • The coordinates $x_i$ are Cartesian coordinates (not generalized coordinates).
  • Since there's no coordinate curviture, I'm not worrying about distinguishing covariant and contravariant indices, as one would do in usual Classical Mechanics.
  • Kinetic energy $T$ is strictly a function of the velocity variables $\dot x_i$ (again, because there is no curvature in the coordinates).
  • Potential energy $U$ is strictly a function of the position variables $x_i$ (I'm not worrying about complicated potentials like the one for the Lorentz force).

Here's what I've found so far.

  • The $i$th component of net force $m\ddot x_i$ can be written in terms of kinetic energy $T= \frac{1}{2} m \dot{\mathbf x} \cdot \dot{\mathbf x}$ in two ways. My sense is that each of these representations yields a conservation law under the proper conditions. $$\displaystyle m\ddot x_i = \frac{d}{dt} \left( \frac{\partial T}{\partial \dot x_i} \right) \label{1}\tag{1}$$ $$\displaystyle m\ddot x_i = \frac{\partial}{\partial \dot x_i} \left( \frac{dT}{dt} \right) \label{2}\tag{2}$$
  • Assuming we're only worried about conservative forces, then by definition of potential energy $U$, we have $m \ddot{\mathbf x} = - \mathbf \nabla U$, so $$ m\ddot x_i = -\frac{\partial U}{\partial x_i} \label{3}\tag{3}.$$
  • Combining equations $\ref 1$ and $\ref 3$ gives conservation of momentum $p_i = \partial T / \partial \dot x_i$ when $-\partial U / \partial x_i = 0$, as expected. This explains the derivative relationship between momentum and kinetic energy. $$0 = - \frac{\partial U}{\partial x_i} = \frac{d}{dt} \left( \frac{\partial T}{\partial \dot x_i} \right) \Rightarrow \boxed{p_i = \frac{\partial T}{\partial \dot x_i} \text{ is conserved}}$$
  • Combining equations $\ref 2$ and $\ref 3$ almost give conservation of energy in general. $$-\frac{\partial U}{\partial x_i} = \frac{\partial}{\partial \dot x_i} \left( \frac{dT}{dt} \right)$$ $$0 = \frac{\partial}{\partial \dot x_i} \left( \frac{dT}{dt} \right) + \frac{\partial U}{\partial x_i}$$ $$0 = \frac{\partial}{\partial \dot x_i} \left( \frac{dT}{dt} \right) + \frac{\partial}{\partial \dot x_i} \left( \frac{\partial U}{\partial x_k} \dot x_k \right)$$ $$0 = \frac{\partial}{\partial \dot x_i} \left( \frac{dT}{dt} \right) + \frac{\partial}{\partial \dot x_i} \left( \frac{dU}{dt} \right)$$ $$0 = \frac{\partial}{\partial \dot x_i} \left[ \frac{d}{dt} (T+U) \right] \label{4}\tag{4}$$ $$\text{(this is where I get stuck)}$$

My question. Beyond equation $\ref 4$, I'm expecting to find $\frac{d}{dt} (T+U) = 0$, which would imply that $T+U$ is conserved. However, this cannot be directly deduced, even if taken to be true for all $i$s, since it could be possible to find a function $f$ such that $$\frac{d}{dt}(T+U) = f(x,\ddot x,\dddot x, \ldots),$$ where $x$ represnts all of the $x_i$s, $\ddot x$ represents all of the $\ddot x_i$s, etc, and still have the relationship in equation $\ref 4$ satisfied. What additional assumption (ideally rooted in some sort of physical principle or intuition) would need to be taken in order to deduce conservation of energy $T+U$? I feel like just plainly assuming that $f = 0$ is too big of a jump, and isn't based in any sort of physical principle or intuition.

  • $\begingroup$ Yeah you already have your answer, but if (2) is true, and if (3) is true, then all you can say is that the two right hand sides are equal. It's like if I say 3 = x and 3 = y, then x = y. Now prove x - y is 0. But you already proved that. Proving x = y is equivalent to proving x - y is 0. Or proving x - y is 0 is equivalent to proving x = y. You can't prove anything about, say z, from what we started out with. Likewise, from Newton's law, you can't prove anything about energy conservation. Newton's laws don't imply energy conservation. You have to define the idea of work. Then once you have $\endgroup$
    – DWade64
    Commented Jan 23, 2019 at 20:01
  • $\begingroup$ work, you can say Let W stand for work. Hey, W - W is 0! That's essentially energy conservation which you know (which we forced upon ourselves by 1) defining work in the first place and 2) defining potential energy to be the exact negative so we can force energy conservation of the form W -W is 0 or T + U is 0). But from Newton, we aren't led to this idea of work unless we want to $\endgroup$
    – DWade64
    Commented Jan 23, 2019 at 20:05
  • $\begingroup$ Actually nevermind. The analogy in my first comment doesn't fully apply to your situation. If you believe in your starting points ((1) and (3) are true), and if you believe everything you did was logical, and if the only conclusion you can draw from (4) is that the square brackets, d/dt (T +U), has to be zero (the right hand side can't be zero any other way), then you would have concluded that energy conservation has to be true because your starting points and everything else you did was true. This is a strange was to prove energy conservation, but it's what you'd have to conclude I think $\endgroup$
    – DWade64
    Commented Jan 23, 2019 at 22:26

2 Answers 2


Maybe I'm not seeing the problem, I will gladly delete this if that is the case, but you do not need to assume anything additional. If you use your Eq. (3) this is what you get

$$ \frac{{\rm d}}{{\rm d}t}(T + U) = m \ddot{x}_k \dot{x_k} + \frac{\partial U}{\partial x_k}\dot{x}_k = \left[ m\ddot{x}_k - m\ddot{x}_k \right]\dot{x}_k = 0 $$

So Eq. (4) is just a trivial $0 = 0$, which is completely consistent

  • $\begingroup$ There, I think, is the missing step! My last line was basically the $\partial / \partial \dot x$ version of all of this, which (after reflection with @knzhou 's comments), is probably adding a level of complexity / borderline invalidity to the thought process. This indeed completes the argumentation that $(d/dt)(T+U) = 0$. $\endgroup$ Commented Jan 23, 2019 at 19:33
  • $\begingroup$ This proof finale also eliminates introducing equation (2) entirely which will leave everyone happy. I think this the best option. $\endgroup$ Commented Jan 23, 2019 at 19:38
  • $\begingroup$ @TrevorKafka Great! I'm glad it was of help $\endgroup$
    – caverac
    Commented Jan 23, 2019 at 21:08

Unfortunately, you've made a fatal error at the very first step! It's wrong to interchange the $d/dt$ and $\partial / \partial \dot{x}_i$. This quickly leads to absurdities, such as in this question where the OP concludes the Euler-Lagrange equation is trivial. This point isn't explained well in most textbooks; for a related question see here. I'll explain why briefly, then show a derivation like yours that works.

In your setup, $T$ is a function of the two independent variables $q$ and $\dot{q}$, with no explicit dependence on time. To take a time derivative, you must restrict to a specific path $$q(t) = \gamma(t), \quad \dot{q}(t) = \frac{d\gamma(t)}{dt}.$$ In other words, the time derivative really means $$\frac{dT}{dt} \equiv \frac{d}{dt} T\left( \gamma(t), \frac{d\gamma(t)}{dt} \right).$$ Then $dT/dt$ is solely a function of $t$ and the path $\gamma$. It is no longer a function of $q$ and $\dot{q}$, so taking the partial derivative with respect to $\dot{q}$ doesn't even make any sense. The manipulations leading up to your equation (4) similarly don't make sense.

A big part of the confusion is that usually the path $\gamma(t)$ is written as $q(t)$, so a function of a path $q(t)$ looks unfortunately similar to a function of the variables $q$ and $\dot{q}$. Compounding the confusion, the Lagrangian can also have direct dependence on $t$ (in addition to $q$ and $\dot{q}$), but this still isn't the same as depending on the path $q(t)$. It's a big mess that's confused many students.

Here's a different way to do it, which stays true to the spirit of your derivation. By the chain rule, $$\frac{dE}{dt} = \frac{\partial T}{\partial \dot{q}} \frac{d \dot{q}}{dt} + \frac{\partial U}{\partial q} \frac{dq}{dt}.$$ Now, applying the equation of motion $$\frac{d}{dt} \frac{\partial T}{\partial \dot{q}} = - \frac{\partial U}{\partial q}$$ we have $$\frac{dE}{dt} = \ddot{q} \frac{\partial T}{\partial \dot{q}} - \dot{q} \frac{d}{dt} \frac{\partial T}{\partial \dot{q}}.$$ However, because $T$ is only a function of $\dot{q}$, so is $\partial T / \partial \dot{q}$, so by the chain rule again, $$\frac{dE}{dt} = \ddot{q} \frac{\partial T}{\partial \dot{q}} - \dot{q} \frac{d\dot{q}}{dt} \frac{\partial}{\partial \dot{q}} \frac{\partial T}{\partial \dot{q}} = \ddot{q} \left(1 - \dot{q} \frac{\partial}{\partial \dot{q}} \right) \frac{\partial T}{\partial \dot{q}}.$$ However, because $T$ is quadratic in $\dot{q}$, $\partial T / \partial \dot{q}$ is linear in $\dot{q}$. That means that differentiating with respect to $\dot{q}$ and then multiplying by $\dot{q}$ again does nothing, so $dE/dt = 0$ as desired.

Of course, as the existing answer says, it would be much faster to get this result if we just used the fact that $T$ was quadratic from the beginning, but I figured you wanted a derivation that used some of the partial derivative machinery.

  • $\begingroup$ I definitely agree that commuting the derivatives is not true in general, but I do think it is true specifically for the potential energy function $T = (1/2) m \dot{\mathbf x} \cdot \dot{\mathbf x}$ in Cartesian coordinates. Equation 2 can be derived from the Work-Energy theorem: $\mathbf F \cdot d\mathbf x = dT$. Divide both sides by $dt$ to get $F \cdot \dot{\mathbf x}= dT/dt$ and then take the partial derivatives of both sides with respect to $\dot x_i$ to get $F_i = (\partial/\partial \dot x_i)(dT/dt)$. $\endgroup$ Commented Jan 23, 2019 at 19:15
  • $\begingroup$ Also, that original post you linked was mine back from September. ;-) I've gained a much greater understanding of partial and total derivatives now, but of course I still could be mistaken! :) $\endgroup$ Commented Jan 23, 2019 at 19:18
  • $\begingroup$ @TrevorKafka It doesn't make sense because the arguments don't line up. "$dT/dt$" only makes sense if you are thinking of a specific path $x(t)$. So it doesn't depend on $\dot{x}$ anymore, and taking the partial derivative with respect to $\dot{x}$ makes no sense. $\endgroup$
    – knzhou
    Commented Jan 23, 2019 at 19:19
  • 1
    $\begingroup$ Now, you might think, you don't need a path. If $T = m \dot{q}^2/2$, then $dT/dt$ simply means $m \dot{q} \ddot{q}$. That's fine, but in the subsequent $\partial / \partial \dot{q}$ you're considering $T$ as a function of three independent variables: $q$, $\dot{q}$, and $\ddot{q}$. That is still a mathematically valid idea, but it's not how Lagrangian mechanics is set up. In Lagrangian mechanics, the independent variables are $q$ and $\dot{q}$ from the start. If you have $L(q, \dot{q}, \ddot{q})$, it's not even clear how to define the Hamiltonian, for starters. $\endgroup$
    – knzhou
    Commented Jan 23, 2019 at 19:21
  • 1
    $\begingroup$ @TrevorKafka No, and this gets back to probably the most common confusion people have about Lagrangian mechanics. For a specific path, it's impossible to vary $\dot{x}$ and $x$ independently. If you change one, you automatically change the other. You can only vary $x$ and $\dot{x}$ independently before you take a specific path. (This is also explained more or less the same way in the top answer to your older question, with more detail.) $\endgroup$
    – knzhou
    Commented Jan 23, 2019 at 19:34

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