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Consider the Euler-Lagrange equation in multiple dimensions (which is actually a system of equations): $$ \frac{\partial L}{\partial q_i}(t, \mathbf{q}(t), \dot{\mathbf{q}}(t)) - \frac{d}{dt}\frac{\partial L}{\partial \dot{q}_i}(t, \mathbf{q}(t),\dot{\mathbf{q}}(t)) =0. $$ My question is the following:

Is the relationship $$\dot{\mathbf{q}}(t)=\frac{d}{dt}\mathbf{q}(t)$$ an assumption of the Euler-Lagrange equation, or is it a conclusion?

My argument -- where are the mistakes? I have argued that it is a conclusion before, see in this answer, I don't know how to prove or disprove that this is correct.

Specifically, when evaluating $\displaystyle\frac{\partial L}{\partial q_i}$ no terms of the form $\displaystyle\frac{\partial \dot{q}_i}{\partial q_i}$ show up, and when evaluating $\displaystyle\frac{\partial L}{\partial \dot{q}_i}$ no terms of the form $\displaystyle\frac{\partial q_i}{\partial \dot q_i}$ show up, which wouldn't make sense if $\dot{q}_i$ was being treated as a function, for example, if $q_i = \frac{1}{2}t^2$ then $\dot{q}_i = t$ so then $\displaystyle\frac{\partial q_i}{\partial \dot{q}_i} = t \not=0$.

However, neglecting such terms would/does make sense if the $\dot{q}_i$ are being treated as independent variables, rather than functions. The only problem then is that one can't say that one independent variable is the derivative of another independent variable.

BUT, along the solution curves, the independent variable denoted by an abuse of notation $\dot{q}_i$ could/would be/is an implicit function of $t$, call it $f_i(t)$, as would the independent variable $q_i$ be an implicit function of $t$ along the solution curve call it $g_i(t)$, and then a consequence of the Euler-Lagrange equation might be then that $f_i(t) = \frac{d}{dt}g_i(t)$.

This seems to be a natural way to think of the situation/setup if one thinks about differential equations as "phase flows" in "phase space", e.g. as in the first section of the first chapter of Vladimir Arnold's excellent book on ODE's.

The answers to this question seem to suggest this might be a possibility: Derivative with respect to $y'$ in the Euler-Lagrange differential equation; however they don't state whether or not the relation $\dot{\mathbf{q}}(t)=\frac{d}{dt}\mathbf{q}(t)$ is a consequence or not, although that should be implied since it could not be an assumption if $y'$ were an independent variable, as the answers suggest.

Note: Before dismissing this possibility out of hand, I would ask one to at least consider that partial derivative notation is notorious for obfuscating and making more confusing issues surrounding implicit differentiation and implicit functions, see for example here, here, here, here, here, here, here, here, here: "A problem with Jacobi's notation is that... substitution of variables can lead to absurd formulae or, at least, to formulae that require careful interpretation", and here.

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Chill2Macht
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  • @Bye_World Not really; I understand that it is thought of as a function of time only, but if it is a function, then how can you differentiate with respect to it? That would seem to be possible only if the generalized velocity was an independent variable. But if it is an independent variable, then you can't say that it has an "antiderivative". – Chill2Macht Sep 14 '16 at 23:15
  • @Bye_World But then if you use the chain rule those terms I mentioned which do disappear shouldn't disappear (I think). Or is one substituting into the independent variable "$y$" the generalized velocity, i.e. $\tilde{L}(t)= L(t, q(t), \dot{q}(t)$, where $L=L(t,x,y)$. I guess my question could also be "is it possible to formulate the EL equation in such a way that this relationship is a conclusion and not an assumption?" – Chill2Macht Sep 14 '16 at 23:16
  • Related and also helpful like the accepted answer https://math.stackexchange.com/a/697189/327486 – Chill2Macht Dec 31 '21 at 03:30

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Let $L: \Bbb R^3 \to \Bbb R: (x_1, x_2, x_3) \mapsto L(x_1, x_2, x_3)$ and $q: \Bbb R\to \Bbb R: t \mapsto q(t)$. Then denote by $\dot q$ the derivative of $q$. Let $\gamma: \Bbb R \to \Bbb R^3: t\mapsto (t,q(t),\dot q(t))$. Now consider the restriction of $L$ to $\gamma$:

$$(L\circ \gamma)(t):\Bbb R\to \Bbb R \\ t\mapsto L(t,q(t),\dot q(t))$$

This is the usual way of writing the Lagrangian.

Now the expressions $\frac{\partial L}{\partial q}$ and $\frac{\partial L}{\partial {\dot q}}$ in the EL equation are really just the partial derivatives of the above function $L$ wrt its second and third arguments: $$\frac{\partial L}{\partial q}:= \frac{\partial L}{\partial x_2} \qquad \frac{\partial L}{\partial \dot q}:= \frac{\partial L}{\partial x_3}$$

Thus the value of these partials does not depend on the fact that $\dot q = \frac{dq}{dt}$.

Now let's look at $\frac{d}{dt}\frac{\partial L}{\partial \dot q}$. This is really the derivative of the restriction of $\frac{\partial L}{\partial x_3}$ to the path $\gamma$

$$\frac{d}{dt}\frac{\partial L}{\partial \dot q} := \left[\frac{d}{dt}\left(\frac{\partial L}{\partial x_3}\circ \gamma\right)\right](t)$$

So the EL equation, written in this new notation is:

$$\left(\frac{\partial L}{\partial x_2}\circ \gamma\right)(t) - \left[\frac{d}{dt}\left(\frac{\partial L}{\partial x_3}\circ \gamma\right)\right](t) =0$$

or written slightly nicer (but also slightly less clear):

$$\left.\frac{\partial L}{\partial x_2}\right|_{\gamma(t)} - \frac{d}{dt}\left(\left.\frac{\partial L}{\partial x_3}\right|_{\gamma(t)}\right) = 0$$

Or equivalently, one could interpret the EL equation as saying that this particular function is the null function on $\Bbb R$:

$$\frac{\partial L}{\partial x_2}\circ \gamma - \frac{d}{dt}\left(\frac{\partial L}{\partial x_3}\circ \gamma\right) = 0$$

  • "Maybe this is an abuse of notation and one should define new variables $\eta = q_i(t)$ and $\xi=\dot{q}_i(t)$..." So then by assumption, the inputs to $L$ are assumed to lie in the image of $\mathbb{R}$ under $q_i$ on the $\eta$-line and in the image of $\mathbb{R}$ under $\dot{q}_i$ on the $\xi$ line? My confusion stems from the fact that I am not sure if $q_i$ and $\dot{q}_i$ are given, or if they are implicit functions derived from how the the values of the points of solution curves on the $\eta$/$\xi$ lines respectively relate to their value on the $t$ line – Chill2Macht Sep 15 '16 at 08:30
  • Is the Euler-Lag equation an ODE for a system of univariate functions in $t$, these functions being derived by the composition of a multivariable function $L(t, \eta, \xi)$ with $q(t)$ and $\dot{q}(t)$? (Because in this case the relationship between $q$ and $\dot{q}$ would have to be an assumption) Or is it a PDE for the function $L(t, \eta, \xi)$, whose solution curves lead to implicit functions for both $\eta$ and $\xi$ in terms of $t$, which we denote $q(t)$ and $q_i(t)$ respectively? (because then the relationship between the two functions would have to be a consequence of the EL equation) – Chill2Macht Sep 15 '16 at 08:33
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    $q_i$ is not given. The EL equation is a PDE whose solution is the $q_i$ (or $q_i$'s) that extremizes the classical action (the integral of $L$ wrt $t$). –  Sep 15 '16 at 11:21
  • "whose solution is the $q_i$ that extremizes the classical action" so the $q_i$ is an implicit function of $\eta$ in terms of $t$ given by the solution of the PDE? In particular as a consequence writing $L(t, q(t), ...)$ instead of $L(t, \eta, ...)$ is an abuse of notation? – Chill2Macht Sep 15 '16 at 11:26
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    I don't know that I'd call $q_i(t)$ an implicit function of $\eta$. There is an explicit relationship between them (equality). $\eta$ is an implicit function of $t$, however. –  Sep 15 '16 at 11:32
  • But isn't $\eta$ an independent variable? How can it be a function of anything? Isn't that like saying $\mathbb{R}$ is a function of some other set? – Chill2Macht Sep 15 '16 at 13:03
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    When we say $y = f(x)$ and $z=g(x,y)$ we're treating $y$ like a variable within $g$, but its values are always in the image of $f$. So we treat $y$ as both another name for the function $f$ (outside $g$) and as an independent variable (inside $g$). When we bring in partial derivatives the lines between variable and function become even more blurred. –  Sep 15 '16 at 14:33
  • I get than and agree with that, but isn't that then essentially abuse of notation? I don't see how it's not. It might be a convenient/practical abuse of notation, perhaps so much so that it is better to use than to avoid the abuse of notation, but it still seems like an abuse of notation. For example as explained in this video: https://www.youtube.com/watch?v=F1E27SFuI1I. Right now my concern is less about using practical notation (indeed I have already used the equation successfully before) than having the correct theoretical understanding of the EL equation. – Chill2Macht Sep 15 '16 at 14:47
  • Like basically I want to know what would be the completely correct way to write out the Euler-Lagrange equation if I was the most obnoxious pedant in human history, and that seems to preclude treating the same quantity as both a function and an independent variable in different contexts. Again, it doesn't seem necessary to blur the lines between variables and functions when using partial derivatives, just convenient. I want to understand how to "unblur" them in the specific context of the Euler-Lagrange equation to be 100% sure that I am blurring them "correctly" when I do blur them. – Chill2Macht Sep 15 '16 at 14:50
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    @William Did you ever see my edited response? I think this should be correct. –  Sep 24 '16 at 15:33
  • I'm sorry about that; you were right, I hadn't seen it -- I appreciate the thought and effort you put into this; it makes a lot more sense to me now. This is exactly the type of answer I was hoping for – Chill2Macht Sep 24 '16 at 15:39