Canonical equations from action integral

Question

I'm going through Lanczos Variational Principles of Classical Mechanics and on page 169 it says that we can form the action integral

$$A=\int_{t_1}^{t_2} \left[\sum p_i\dot q_i-H(q_1,...,q_n;p_1,...,p_n;t)\right]dt$$

from which we can get do the variation to get

$$\delta A=0=\frac{dp_i}{dt}+\frac{\partial H}{\partial q_i}$$

$$\delta A=0=-\dot q_i + \frac{\partial H}{\partial p_i}$$

Now this is straightforward for the $\delta p$ variation, $$p_i\delta\dot q_i + \dot q_i\delta p_i -\frac{\partial H}{\partial q_i}\delta q_i - \frac{\partial H}{\partial p_i}\delta p_i - \frac{\partial H}{\partial t} \delta t = 0$$

$$\rightarrow \dot q_i\delta p_i - \frac{\partial H}{\partial p_i}\delta p_i = 0$$

but for the $\delta q$ variation I'm finding I need to do something tricky looking and potentially incorrect: $$p_i\delta\dot q_i + \dot q_i\delta p_i -\frac{\partial H}{\partial q_i}\delta q_i - \frac{\partial H}{\partial p_i}\delta p_i - \frac{\partial H}{\partial t} \delta t = 0$$

$$\rightarrow \delta(\frac{d}{dt} q_i) p_i -\frac{\partial H}{\partial q_i}\delta q_i = 0$$ and using the fact that d and $\delta$ commute $$\rightarrow \frac{d}{dt}\delta(q_i) p_i -\frac{\partial H}{\partial q_i}\delta q_i = 0$$ $$\rightarrow \frac{d}{dt}p_i \delta q_i -\frac{\partial H}{\partial q_i}\delta q_i = 0$$

I can recover the equation as given, except I'm off by a minus sign! So, I'm wondering if the trick I used is valid, and why I'm getting a minus sign off the equation. Any help would be greatly appreciated!

Answer 1

Your error is when going from $\delta(\frac{d}{dt} q_i) p_i$ to $\frac{d}{dt}p_i \delta q_i$. In doing so you must change sign: $$ \delta \left( \frac{d}{dt} q_i \right) p_i = \frac{d}{dt}(\delta q_i) p_i = \frac{d}{dt}(\delta q_i \, p_i) - \delta q_i \, \frac{d}{dt}p_i $$ Now, $\frac{d}{dt}(\delta q_i \, p_i)$ vanishes on integration if $\delta q_i$ is taken to vanish at the end points, so we end with $- \delta q_i \, \frac{d}{dt}p_i.$

Answer 2

You are treating $\delta$ like a differential. $\delta$ is a variation, not a differential. I feel like the only way I can tell you what went wrong in your manipulations, is to show you the correct way of dealing with a variation.

Consider an functional , which feeds one smooth vector function $\mathbf{f}$ and spits out an integral (which is called the action) $$ S[\mathbf{f}] = \int_{a}^{b} F(\mathbf{f}(t),\dot{\mathbf{f}}(t),t) dt $$ here $\mathbf{f}$ furthermore needs to satisfy $\mathbf{f}(a)=\mathbf{x}$, $\mathbf{f}(b)=\mathbf{y}$ with $\mathbf{x},\mathbf{y}$ fixed and given. The function $F$ is say another smooth function.

What your book is trying to tell you is the function $\mathbf{f}$ making the action extremum, satisfies those equations. This is what one does. Let $\mathbf{g}$ be the (unknown) extremum solution, $\epsilon >0$ a small number and $\mathbf{h}$ a function with $\mathbf{h}(a)=\mathbf{h}(b)=0$. This is a perturbation of the extremum solution. To put things in perspective, by $\delta \mathbf{f}$ we mean $\mathbf{f}-\mathbf{g}=\epsilon \mathbf{h}$. This is what a variation means.

Now $$ F(\mathbf{f}(t),\dot{\mathbf{f}}(t),t)- F(\mathbf{g}(t),\dot{\mathbf{g}}(t),t) = F(\mathbf{g}+\epsilon\mathbf{h},\dot{\mathbf{g}}+\epsilon \dot{\mathbf{h}},t) - F(\mathbf{g},\dot{\mathbf{g}},t) = \epsilon\left[ \frac{\partial F}{\partial \mathbf{f}}\cdot \mathbf{h}+ \frac{\partial F}{\partial \dot{\mathbf{f}}}\cdot \dot{\mathbf{h}} \right] $$ the equality being up to $\epsilon^2$ order. Therefore $$ \delta S := S[\mathbf{f}]- S[\mathbf{g}]=\epsilon \int_a^b \left[ \frac{\partial F}{\partial \mathbf{f}}\cdot \mathbf{h}+ \frac{\partial F}{\partial \dot{\mathbf{f}}}\cdot \dot{\mathbf{h}} \right]dt $$ Doing an integration by parts over the second summand (and using the fact that $\mathbf{h}(a)=\mathbf{h}(b)=0$ $$ \delta S =\epsilon \int_a^{b} \left[ \frac{\partial F}{\partial \mathbf{f}}{\color{red}-\frac{d}{dt}} \frac{\partial F}{\partial \dot{\mathbf{f}}} \right]\cdot \mathbf{h} dt $$ Now if $S$ is extremum at $\mathbf{g}$, then $\delta S =0$ for all choices of $\mathbf{h}$. This forces $$ \boxed{ \frac{\partial F}{\partial \mathbf{f}}-\frac{d}{dt} \frac{\partial F}{\partial \dot{\mathbf{f}}}=0} $$ which is called the Euler-Lagrange equation. You can either use, the ideas in this derivation to find Hamiltonian equations of motion from scratch, or you can just use the Euler-Lagrange equation to your action. I hope by now it is clear how unsafe what you were doing originally was.