How to justify drawing out the virtual differential $\delta$ to the front of the integral in the Hamilton principle?

Question

My textbook derives the equivalence of Hamilton principle to d'Alembert's principle as such:

\begin{align} 0&=\int_{t_1}^{t_2}\left(\sum_i(m_i\ddot{\vec{r}}-\vec K_i)\cdot\delta\vec r_i\right)dt=\\ &\ \ \vdots\\ =\int_{t_1}^{t_2}\delta(T-V)dt&=\delta\int_{t_1}^{t_2}(T-V)dt, \end{align} where the virtual displacements $\delta\vec r_i$ are differential functions of time.

The last equation is justified by saying:

".. we have exploited that for virtual displacements the times are not varied at all ($\delta t=0$) so that we could draw, for instance, the variation $\delta$ to the front of the integral."

Now I just can't figure out how "for virtual displacements the times are not varied at all" implies that we could draw $\delta$ to the front. At first I would imagine it means that virtual displacements are not dependent on time, therefore $\delta(T-V)$ is not influenced by the $dt$ in the integral and you can do whatever you want with $\delta(T-V)$. But the book states in the beginning of the derivation that

"the virtual displacements $\delta\vec r_i$ are differential functions of time",

and $T$ and $V$ are functions of $\delta\vec r_i$. So that contradicts my hypothesis.

Answer 1

I will first discuss why (I think) the virtual displacement $\delta\vec r$ is described as 'a function of time'.

The objective is to find the trajectory as a function of time.

Let's take a simple case as example: the case where the true trajectory, plotted as a function of time, is a parabola. Let's take the following curve:

y-coordinate: vertical axis
time coordinate: horizontal axis

$$ y = -t^2 + 1 \tag{1} $$

So that curve crosses the horizontal axis at the points [-1,0] and [1,0]. We make those points the start point and end point.

In that specific case we can implement the variation with a multiplication factor $\epsilon$:

$$ y = \epsilon (-t^2 + 1) \tag{2} $$

That way of implementing the variation ensures that at the points [-1,0] and [1,0] the variation is zero.

For the value '1' of the multiplication factor $\epsilon$ the variation coincides with the true trajectory. Variation is applied by varying $\epsilon$ away from that value '1'.

Change of the multiplication factor and the value of the time coordinate are independent of each other. (2) is a function of time because you need to apply the variation over an extended interval of time. Here that interval is from point [-1,0] to point [1,0].

That concludes my discussion of why (I think) the virtual displacement is described as a function of time.

When you change the multiplcation factor $\epsilon$ you change the height of the variation.

In a diagraam with cartesian coordinates the effect of the multiplication factor $\epsilon$ is at right angles to the direction of the time axis.

Integration is a linear operation: a multiplication factor can be drawn to the front of the integral.

Of course, the function that you are integrating may well be a quadratic function. or what have you. That means that for values of $\epsilon$ significantly away from 1 there will be an error. But that error at large variation is of no consequence. It is sufficient if in the region of infinitisemal variation the function can be treated as linear, and that is the case.

For any function you can do a Taylor series expansion. The first term of that expansion is linear. In the limit of infinitisemally small variation the linear term determines the outcome.

Answer 2

1. A virtual displacement is indeed frozen in time $t$, i.e. the virtual displacement is parametrized by some curve parameter, say $s$, different from time $t$, cf. e.g. this, this & this Phys.SE posts.

2. In particular, we may move the infinitesimal virtual displacement $\delta$ past the integral measure $\mathrm{d}t$ and the integration limits $t_1$ and $t_2$, i.e. back and forth between the integrand $L$ and the integral $S$.

3. However, this does not mean that the virtual displacement couldn't depend on other parameters as well, such as the generalized coordinates $q^j$ and time $t$, i.e. the virtual displacement doesn't necessarily have to be the same everywhere in the configuration space over time.

Answer 3

Just to offer a third perspective, when we are doing all of this stuff, prefixing a quantity with $\delta$ just means, “the change in that quantity due to changing from the path $\mathbf r(t)$ to the path $\mathbf r(t) +\delta \mathbf r(t).$” (These being vectors over a whole configuration space, so if you have four particles in 3D, $\mathbf r$ and $\delta\mathbf r$ are smooth functions from $[0,T]\to\mathbb R^{12}$ for example.)

That is, if you could calculate a kinetic energy with respect to the path parameter over the first path, $K(t),$ and for the second path $\hat K(t)$, then $\delta K(t)$ is literally just defined as $\delta K(t)=\hat K(t)-K(t).$ Similarly for $U(t).$

Now $\delta S$ is kinda different! Because it has no time parameter. So it has to be defined analogously as $$\delta S = S[\mathbf r +\delta \mathbf r] – S[\mathbf r].$$ But that's fine and it fits the same pattern. And so if $S =\int_0^T\mathrm dt~L(t), $ then by the pure linearity of the integral sign you immediately get$$\delta S = \int_0^T\mathrm dt~ \big(\hat L(t)-L(t)\big)= \int_0^T\mathrm dt~ \delta L(t).$$

This is also why it commutes with all sorts of derivatives, they distribute linearly over subtraction, that subtraction is all we mean when we write down $\delta.$ So later when we integrate by parts or so, I don't want you to get confused about $\delta \dot{\mathbf r}$ vs. ${\mathrm d\phantom t\over\mathrm dt}\big(\delta\mathbf r\big). $ It doesn't matter that it's a derivative over $t$, those derivatives are linear, it distributes over the subtraction just fine.

The path perturbation operator has no deeper meaning, that is just what it means, the difference between the value you get on the first path and the value you get on the other path. So all of these formulas that sound really mysterious are actually very trivial. Eventually you will assume that the two paths are infinitesimally close together, and derive some equations of motion, and those approximations frequently do not commute with other related quantities. You worry about those, don't worry about $\delta.$