# Variations wrt. Time Derivatives

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 $$\dot{x}$$?

• $\ddot x$ isn’t a functional so you can’t take its functional derivative with respect to anything. – G. Smith Aug 13 '20 at 0:09
• The action is a functional $S = \int{\rm d}t L$. – David Aug 13 '20 at 0:24
• @David can you clarify what you mean? I'm unsure how to apply that to the derivatives of x-dot – Chip Aug 13 '20 at 0:35
• 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 $t$. – David Aug 13 '20 at 0:38
• @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 – 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 $$\partial L/\partial\ddot{x}$$

• 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? – Chip Aug 13 '20 at 0:57