Calculus of Variations - Virtual displacements

Question

I am currently reading "The Variational Principles of Mechanics - Cornelius Lanczos", in which the author talks about the variation of a function $F(q_1, q_2, \dots q_n)$ where $q_1, q_2, \dots q_n$ are the generalized coordinates

$$F=F(q_1, q_2, \dots q_n)$$

$$ \delta F=\frac{\partial F}{\partial q_1}\delta q_1+\frac{\partial F}{\partial q_n}\delta q_n+\dots+\frac{\partial F}{\partial q_n}\delta q_n \tag{1} $$ next he writes, $\delta q_1=\epsilon a_1, \delta q_2=\epsilon a_2, \dots ,\delta q_n=\epsilon a_n$

where $a_1, a_2 \dots , a_n$ are the direction cosines, and $\epsilon$ is a small variation. I find it wrong to use the same $\epsilon$ for all $\delta q_i $'s as it seems to be inconsistent with dimensions

For example if we are dealing with spherical coordinates $(r, \theta, \phi)$, according to the above the individual variations become

$\delta r=\epsilon \hat{r}$, So I expect $\epsilon$ to have a dimension of $r$ (unit vectors are dimensionless)

$ \delta \theta = \epsilon \hat{\theta},$ now $\epsilon$ has a dimension of $\theta$?

substituting these in $\text{eq}(1)$ makes it worse,

Further I like to think (I may be wrong) of the $\delta F$ as $\nabla F$, since both seem to have the same form, but as we know $\nabla$ is different for different coordinates, but $\text{eq}(1)$ seems to use $\delta$ as we use for Cartesian coordinates

Am I missing out something?

P.S

This still isn't a very great problem since the main goal is to find the stationary value which anyways leads to a conclusion $\frac{\partial F}{\partial q_k}=0$

Answer 1

I agree with you that his choice of notation is questionable.

Regarding $\epsilon$, I believe it can only make sense if it is chosen to be dimensionless; this is the only way for putting it in front of every variable to work.

Next regarding the gradient analogy, I believe you are misinterpreting.

When you have a function of multiple variables say $f(x,y,z)$, the total variation reads:

\begin{equation} df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy + \frac{\partial f}{\partial z} dz \end{equation}

This result is independent of how you define the gradient and holds whatever the variables mean.

To define the gradient from the above formula, you need to come up with a notion of length measure and a dot product in the space of variables you are considering and then define $\nabla f$ as the vector field such that:

\begin{equation} df = \nabla f \cdot d\vec{l} \end{equation}

where $d\vec{l}$ is an infinitesimal vector element that reads differently in different coordinate systems.

• In cartesian coordinates it would read $d\vec{l} = dx \: \hat{u}_x + dy \: \hat{u}_y + dz \: \hat{u}_z$

• In cylindrical coordinates it would read instead $d\vec{l} = dr \: \hat{u}_r + rd\theta \: \hat{u}_{\theta} + dz \: \hat{u}_z $

Thus leading to coordinate-system-dependent gradient expressions.

In any case, the variational formula you wrote is exactly analogous to the first variation formula I wrote and not to a gradient.