Formal definition of virtual displacement I'm trying to read a book on classical mechanics, and I'm having a hard time trying to know what exactly is a virtual displacement, sometimes called a variation. In the Lagrangian "formalism" the differential forms $dx$ are changed by "virtual displacements" and noted by $\delta x$. As far as I understand the unique difference between $dx$ and $\delta x$ is that $\delta x$ doesn't depends on time, that is, $\delta x$ is a purely spatial differential form.
However in a book of classical mechanics I see that $\delta \dot q$ can be "integrated" respect to $dt$ to give $\delta q$, what makes no sense if $\delta q$ will be a differential form. So, I would like to know a precise and very formal definition of some expression like $\delta x$. I tried to find such formal (mathematically rigorous) definition in many books of mechanics or differential geometry but I could not find one, so I would appreciate some reference or explanation.
 A: Assume that the functional (a function/map/operator that has a function as input and a number as output) is given as
$$F[ρ]=\int_\Omega f\Bigl(x,ρ(x),∇ρ(x)\Bigr)\,dx.$$
For the variation in directional-derivative form you then get
\begin{align}
δF[ρ;ψ]&=\lim_{s\to0}\frac{F[ρ+sψ]-F[ρ]}{s}
\\[.3em]&=\int_\Omega \left[ \frac{∂f}{∂ρ}\Bigl(x,ρ(x),\nabla ρ(x)\Bigr)ψ(x)+\frac{∂f}{∂∇ρ}\Bigl(x,ρ(x),\nabla ρ(x)\Bigr)∇ψ(x)\right]\,dx.
\end{align}
Here $∂ρ$ and $∂∇ρ$ in the partial-derivative symbol just indicate the place in the arguments of $f$ that the partial derivative is taken for, there is no other special magic involved. If one writes $f(x,y,v)$ as the "canonical" arguments of $f$, the partial derivatives could also be written as $\frac{∂f}{∂y}$ and $\frac{∂f}{∂v}$.
One could use $δρ$ as function symbol instead of $ψ$, in a slight abuse of notation $∇δρ$ might then be written as $δ∇ρ$. Or you might declare $δρ(x)[ψ]=ψ(x)$ as the evaluation operator for the derivative direction function $ψ$.
The variation in differential-quotient form $\frac{δF[ρ]}{δρ(x)}$ can now be defined as some limit over the support of $ψ$ shrinking to the point $x$, or more systematically over the demand that
$$
δF[ρ;ψ]=\int_\Omega \frac{δF[ρ]}{δρ(x)}δρ(x)[ψ]\,dx+\int_{∂\Omega}...
$$
To that end one would have to get rid of the derivative terms $∇ψ$. That is achieved using some variation of the fundamental theorem of calculus, generally called Stokes theorem for differential forms (Greene, Gauss,...). This gives in interior points
$$
\frac{δF}{δρ(x)}[ρ]=\frac{∂f}{∂ρ}\Bigl(x,ρ(x),\nabla ρ(x)\Bigr)-∇^*\frac{∂f}{∂∇ρ}\Bigl(x,ρ(x),\nabla ρ(x)\Bigr).
$$
A: What you need is the definition of the functional differential. To complement the explanation from Wikipedia, you can also have a look at the answer I provided for this somewhat simpler question.
To summarize the key idea, consider the total derivative for an $n$-dimensional vector space (or a function with $n$ inputs)
$$
df(x, h) = \sum_{i=1}^n \frac{\partial f}{\partial x_i} h_i \Rightarrow
df(x, h) = \sum_{i=1}^n \frac{\partial f}{\partial x_i} dx_i(x, h)
\textit{, for $dx_i(x, h) = h_i$}
$$
and then take the same idea and map it to functionals (functions that take a function as input) to get
$$
\delta f[\rho,\psi] = \int \frac{\delta f}{\delta \rho}(x) \psi(x) dx \Rightarrow
\delta f[\rho,\psi] = \int \frac{\delta f}{\delta \rho}(x) \delta\rho(x) dx 
\textit{, for $\delta\rho[\rho, \psi](x) = \psi(x)$}
$$
A: I found an answer that fit my doubts. If we have a manifold $M\times \mathbb{R}$ and a coordinate system $\varphi :=(q_1(t),\ldots ,q_n(t),\dot q_1(t),\ldots ,\dot q_n(t),t)$ then we can set $\delta p:=i^*dp$ where $i:M\times \{t_0\}\to M\times \mathbb{R}$ for every coordinate $p$. This is what I assumed from first place. However in a book I see written something like
$$
\int_{t_1}^{t_2}\frac{\partial L}{\partial \dot q_k}\delta \dot q_k\,d t=-\int_{t_1}^{t_2}\delta q_k \frac{d}{d t}\left(\frac{\partial L}{\partial \dot q_k}\right)\,d t\tag1
$$
This confused me a lot, however if we think as $\delta p$ as above then we dont need to introduce an extraneous $d t$ in the integral, because we will have $\delta \dot q=\ddot q\,d t$ and we just can write
$$
\int_{t_1}^{t_2}\frac{\partial L}{\partial \dot q_k}\delta \dot q_k=-\int_{t_1}^{t_2}\frac{d}{d t}\left(\frac{\partial L}{\partial \dot q_k}\right)\delta q_k\tag2
$$
assuming that the $dt$ in the expression $\delta \dot q=\ddot q\,d t$ is the differential form on the cotangent bundle in the domain of $\psi :\mathbb{R}\to M\times \{t_0\}$, where $\psi :=(q_1(t),\ldots ,q_n(t),\dot q_1(t),\ldots ,\dot q_n(t),t_0)$. By now this is the best theoretical setting that I've found to understand the notion of "virtual displacement" given in books of classical dynamics.
This is almost the same idea shown here.
