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Variations wrt. Time Derivatives

Physics Asked by Chip on December 26, 2020

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 $dot{x}$?

3 Answers

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 $frac{delta ddot{x}(t)}{delta x(t^{prime})}=delta^{primeprime}(t!-!t^{prime})$ 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.

Answered by Qmechanic on December 26, 2020

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

$$S[q] := int_{t_1}^{t_2} Lbig(q(t),dot q(t), tbig) dt$$

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

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

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

$$S[q] := int_{t_1}^{t_2} Lbig(q(t),dot q(t),tbig) = int_{t_1}^{t_2} left(frac{1}{2}m dot q^2(t) - frac{1}{2}momega^2 q^2(t) right) dt$$


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

$$L(a+delta a,b+delta b,c+delta c) = L(a,b,c) + left[ big(partial_1Lbig)cdot delta a+ big(partial_2Lbig)cdot delta b +big(partial_3Lbig)cdot delta cright]$$

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

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

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

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


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

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 $big(partial_2 Lbig)(x,dot x,ddot x)$, 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.

Answered by J. Murray on December 26, 2020

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 $$frac{partial L(dot{x},x)}{partialdot{x}}equivleft.frac{partial L(y,x)}{partial y}right|_{y=dot{x}},$$ 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 $partial/partialdot{x}$ 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 $partial L/partialdot{x}$ of the $L=x+dot{x}+ddot{x}$ in the question is just 1—as are $partial L/partial x$ and $partial L/partialddot{x}$

Answered by Buzz on December 26, 2020

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