The typical first application of Lagrange's equation is showing that it implies Newton's law for a particle whose Lagrangian is $L=\frac{1}{2}mv^2-V(x)$. Plugging this Lagrangian into Lagrange's equation, we have that $$\frac{\partial}{\partial x}\left(\frac{1}{2}mv^2-V(x)\right)=\frac{d}{dt}\frac{\partial}{\partial v}\left(\frac{1}{2}mv^2-V(x)\right)$$

Typically at this point books will conclude that, therefore, we must have Newton's 2nd law


However, how can we be sure of the following?

$$\frac{\partial}{\partial x} \frac{1}{2}mv^2+\frac{d}{dt}\frac{\partial}{\partial v}V=0$$

The above result of Newton's law seems to depend on this, but I'm not sure how to go about showing that the last equation is true.

  • $\begingroup$ The main question (v2) Why are $x$ and $v$ independent variables? is a duplicate of physics.stackexchange.com/q/885/2451 and links therein. $\endgroup$ – Qmechanic Feb 9 '15 at 0:36
  • $\begingroup$ But I have a problem with that. Most people's answers mentioned that x and v are independent because you are free to set any initial x and v as initial conditions. But that only means that v is not uniquely determined once you know x; I doubt that it implies that the derivative of v with respect to x must be 0. Do you know what I mean? Sure, there is not a unique function of x that will describe v, but that doesn't necessarily imply that the derivative of v with respect to x must be 0, does it? $\endgroup$ – Physics Llama Feb 9 '15 at 1:10

Imagine your kinetic energy $T$ and your potential energy $V$. The kinetic energy, $T$, depends on your velocity, the potential energy, $V$, depends on your location. So if you want to know your Lagrangian $L=T-V$ then you need to know both.

So imagine a plane where the $x$-axis tells your position and the $y$-axis tells you your speed, the Lagrangian $L=T-V$ is a function of the plane. And you can imagine looking at how the Lagrangian changes as a function if you move around this plane in the $x$ direction or in the $y$ direction. These are exactly what $\partial L / \partial x$ and $\partial L / \partial v$ are meant to denote. They do not denote what happens in time or as the particle moves. it is about how the Lagrangian would have been different had the position or the speed been different.

In three dimensions, you can imagine a 6d space, with the first three directions telling you the three parts of the position vector, and the next three giving the three parts of the velocity. The Lagrangian $L=T-V$ is a function of the 6d space. And you can imagine looking at how it changes if you had been at a different $x$ or at a different $\dot x$. The "rate" of change for these different position/velocity values are exactly what $\partial L / \partial x$ and $\partial L / \partial \dot x$ are meant to denote.

You must think of the Lagrangian $L=L(x,y,z,\dot x, \dot y, \dot z)$ as a function of hypothetical values of position $(x,y,z)$ and velocity $(\dot x, \dot y, \dot z)$, and take partials as a function, i.e. holding all the other variables (inputs to the function) constant. It is not motion in real space it is a hypothetical change in what the values of position or velocity could have been, you want to see how the Lagrangian would have been different for different initial positions and different initial velocities.

edit to respond to information in the comments

The Lagrangian could be different if the initial position were different. The Lagrangian could be different if the initial velocity were different.

But if you changed only the initial velocity, then the potential energy $V=V(x)$ doesn't change so $\partial V /\partial v =0$.

Similarly if you changed only the initial position, then the kinetic energy $T=T(v)$ doesn't change so $\partial T /\partial x =0$.

Therefore when you check to see how the Lagrangian $L=T-V$ changes when you adjust the initial position you get $\partial L/\partial x=0-\partial V/\partial x$ and when you check to see how the Lagrangian $L=T-V$ changes when you adjust the initial velocity you get $\partial L/\partial v=\partial T/\partial v-0$

I don't think it is always helpful to think about derivatives of $x$ with respect to $v$ (or vice versa), just think about how the potential energy changes if your initial velocity changes or think about how the kinetic energy changes if your initial position changes.


In the context of Lagrange's equations, $\frac{\partial}{\partial x}$ acts on the Lagrangian $L(x,\dot{x})$, which is a function of $x$ and $\dot{x} = v$, which are independent variables. Thus $$\frac{\partial v}{\partial x} = 0.$$ Secondly, if the potential depends on velocity, then it (probably) isn't a good potential, and Newton's laws (probably) aren't going to hold.

Edit: To address why $v$ and $x$ are independent variables, the whole point of the least action principle is that at any point along the trajectory, the particle is free to have any position and velocity, because there is no equation of motion relating the two (yet). However, in demanding that the action be stationary with respect to these trajectories, we obtain an equation of motion that fixes them.

It is a bit like asking whether some function $f(s)$ is a maximum at $s=3$. Now, we just solve $f'(s) = 0$, keeping $s$ variable, and then plug in $s=3$ to verify. It would be a mistake to set $s=3$ right away and then try to vary $f(3)$. It is a mistake to do the same thing when finding solutions to equations of motion, but that mistake sneaks in more easily than you would think.

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    $\begingroup$ How could x and v be independent variables in general? As a contradiction to that statement, consider the simple harmonic oscillator. There is a dependence between the two variables, isn't there? $\endgroup$ – Physics Llama Feb 9 '15 at 0:33
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    $\begingroup$ @PhysicsLlama: Actually, now that I (really) think about it, $$\frac{\partial}{\partial x}v=\frac{\partial}{\partial x}\frac{\partial}{\partial t}x=\frac{\partial}{\partial t} 1=0$$ in general with or without the equations of motion. $\endgroup$ – Ryan Unger Feb 9 '15 at 3:13
  • $\begingroup$ @0celo7 At first I was extremely happy with that answer, but consider that we're not taking the derivative of speed with respect to position, it's v^2. So I think your argument might not work :( $\endgroup$ – Physics Llama Feb 9 '15 at 3:25
  • $\begingroup$ $$\frac{\partial}{\partial x}v^2=2v\frac{\partial}{\partial x}v=2v\cdot 0=0$$ $\endgroup$ – Ryan Unger Feb 9 '15 at 3:26
  • $\begingroup$ @lionelbrits the potential for a charged particle in an electromagnetic field depends on velocity. $\endgroup$ – Eric Angle Feb 9 '15 at 14:29

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