I've been working on my dynamics homework when I've run into confusion with derivatives and could use some help regarding that. The question is as follows:

When varying a functional with respect to a variable, do we consider different derivatives of functions to be independent? For instance, what is the functional derivative of $\ddot{x}$ with respect to $\dot{x}$?

For instance, I have an equation $L = x + \dot{x}+\ddot{x}$. What is the partial derivative of this with respect to $\dot{x}$?

  • 2
    $\begingroup$ $\ddot x$ isn’t a functional so you can’t take its functional derivative with respect to anything. $\endgroup$ – G. Smith Aug 13 '20 at 0:09
  • $\begingroup$ The action is a functional $S = \int{\rm d}t L$. $\endgroup$ – David Aug 13 '20 at 0:24
  • $\begingroup$ @David can you clarify what you mean? I'm unsure how to apply that to the derivatives of x-dot $\endgroup$ – Chip Aug 13 '20 at 0:35
  • $\begingroup$ A functional is an object that depends on functions, but not their dependent variables. Example: Because the action $S$ is the integral of the Lagrangian over time, it only depends on the functions $x,\dot x$ etc, but not $t$. $\endgroup$ – David Aug 13 '20 at 0:38
  • $\begingroup$ @David oh were you responding to the person above you with your comment/ I wasn't sure if that was an answer because It didn't clarify it for me. I guess I'm asking for a general case for how this derivative with respect to x-dot would work as x-dot varies with time $\endgroup$ – Chip Aug 13 '20 at 0:42

This is a fairly common misconception. The action functional $S$ eats a function $q$ and spits out the following number:

$$S[q] := \int_{t_1}^{t_2} L\big(q(t),\dot q(t), t\big)\ dt$$

where the Lagrangian $L: \mathbb R^3 \rightarrow \mathbb R$ is just a function of three variables. One might have, for example,

$$L(a,b,c) = \frac{1}{2} mb^2 - \frac{1}{2} m\omega^2 a^2$$

where $m$ and $\omega$ are constants. In that case, the action would be

$$S[q] := \int_{t_1}^{t_2} L\big(q(t),\dot q(t),t\big) = \int_{t_1}^{t_2} \left(\frac{1}{2}m \dot q^2(t) - \frac{1}{2}m\omega^2 q^2(t) \right) dt$$

Given some $(a,b,c)$, we can linearize the Lagrangian to find its value at a nearby point $(a+\delta a,b+\delta b,c+\delta c)$:

$$L(a+\delta a,b+\delta b,c+\delta c) = L(a,b,c) + \left[ \big(\partial_1L\big)\cdot \delta a+ \big(\partial_2L\big)\cdot \delta b +\big(\partial_3L\big)\cdot \delta c\right]$$

where $\partial_nL$ is the derivative of $L$ with respect to its $n^{th}$ slot. Therefore, if we add a small $\eta$ to $q$, we get

$$S[q+\eta]=\int_{t_1}^{t_2} L\big(q(t)+\eta(t),\dot q(t) + \dot \eta(t),t\big) \ dt $$ $$\simeq \int_{t_1}^{t_2} L\big(q(t),\dot q(t),t\big)\ dt + \int_{t_1}^{t_2} \big[(\partial_1 L) \eta(t) + (\partial_2 L) \dot \eta(t) \big] dt$$

Integration by parts then gives

$$S[q+\eta]-S[q]\simeq \int_{t_1}^{t_2} \big[(\partial_1 L) - \frac{d}{dt}(\partial_2 L)\big]\eta \ dt$$

where we've used the fact that $\eta(t_1)=\eta(t_2)=0$. If we demand that this vanish for arbitrary (differentiable) $\eta$, we must have that

$$\frac{d}{dt}\left[\bigg(\partial_2 L\bigg)\big(q(t),\dot q(t),t\big)\right] = \bigg(\partial_1 L\bigg)\big(q(t),\dot q(t),t\big)$$

Notation being what it is, it is standard to say something like $$L(x,v,t) = \frac{1}{2}m v^2 + \frac{1}{2}m\omega^2x^2$$ $$\frac{\partial L}{\partial x} = m\omega^2 x$$ $$\frac{\partial L}{\partial v} = mv$$

But this is a bit misleading. What we've written as $\frac{\partial L}{\partial x}$ is really the derivative of $L$ with respect to its first slot, evaluated at the point $(x,v,t)$. The same is true when we write $L = L(q,\dot q,t)$; the fact that $q$ and $\dot q$ are related to each other by differentiation is irrelvant, because $q(t)$ and $\dot q(t)$ are the values (not functions!) we plug in to the first and second slots after we take the partial derivative of $L$.

I have an equation $L=x+\dot x+\ddot x$. What is the partial derivative of this with respect to $\dot x$?

$L$ is a function, not a functional; it doesn't know how to take derivatives. It is a map which eats numerical values and spits out a numerical value. The only way to have a Lagrangian like that is to define $L(a,b,c)=a+b+c$, and then plug $x$ into the first slot, $\dot x$ into the second slot, and $\ddot x$ into the third slot$^\dagger$. If you do this, then $\big(\partial_2 L\big)(x,\dot x,\ddot x)$, which we we would usually write as $\frac{\partial L}{\partial \dot x}$ in an abuse of notation, would be equal to $1$.

$^\dagger$ Note that an action functional which involves second derivatives is mathematically problematic under most circumstances.


Here is the cheat sheet:

  • On one hand, in a partial differentiation the variable $x,\dot{x},\ddot{x},\ldots$ are independent. E.g. the partial derivatives $\frac{\partial \dot{x}}{\partial x}=0$ and $\frac{\partial x}{\partial \dot{x}}=0$ are zero.

  • On the other hand, in a functional differentiation the variable $x,\dot{x},\ddot{x},\ldots$ are dependent. E.g. the functional derivative $\frac{\delta \dot{x}(t)}{\delta x(t^{\prime})}=\delta^{\prime}(t\!-\!t^{\prime})$ and $\frac{\delta \ddot{x}(t)}{\delta x(t^{\prime})}=\delta^{\prime\prime}(t\!-\!t^{\prime})$ are derivatives of the Dirac delta distribution, while $\frac{\delta x(t)}{\delta \dot{x}(t^{\prime})}$ or $\frac{\delta \ddot{x}(t)}{\delta \dot{x}(t^{\prime})}$ are ill-defined/meaningless.

This is explained further in e.g. this and this related Phys.SE posts.


I am going to guess that you are working in the context of a Lagrangian, trying to derive the equation of motion. The Lagrangian for one-dimensional motion along a single direction $x$ is a function $L(\dot{x},x)$. (It can also be a function of $t$, but that’s not directly relevant for this discussion.) To derive the Euler-Lagrange equations, you need to take partial derivatives of $L$ with respect to both $x$ and $\dot{x}$, and this is done by considering $x$ and $\dot{x}$ to be completely independent variables.

Instead of $L(\dot{x},x)$, you can think of the Lagrangian as a function of two completely separate variables $L(y,x)$. Then the partial derivate with respect to $\dot{x}$ is essentially defined to be $$\frac{\partial L(\dot{x},x)}{\partial\dot{x}}\equiv\left.\frac{\partial L(y,x)}{\partial y}\right|_{y=\dot{x}},$$ taking the derivative of $L$ with respect to its first argument and evaluating it at that argument equal to $\dot{x}$.

When dealing with more complicated theories (involving not point particles but fields), defined in terms of an action $S=\int dt\,L$, the relationship between $\partial/\partial x$ and $\partial/\partial\dot{x}$ can be made more apparent, but at the at the level you are talking about, what I have described is probably the best way to think about things. This means, by the way, that the partial derivative $\partial L/\partial\dot{x}$ of the $L=x+\dot{x}+\ddot{x}$ in the question is just 1—as are $\partial L/\partial x$ and $\partial L/\partial\ddot{x}$

  • $\begingroup$ oh this makes sense! You are right - I am analyzing a double pendulum and part of the differential equation uses a partial with respect to x-dot. so as an addition, in my case x-dot varies with time - so would the partial with respect to x-dot not be x-dot-dot as the differentiation is not related to time, but in fact the x-dot term? $\endgroup$ – Chip Aug 13 '20 at 0:57

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