In Itzykson & Zuber p.3 the Hamilton equations are derived. They start by defining the action by the Hamiltonian -

$$Ldt=pdq-Hdt\Rightarrow I=\intop_{t_1}^{t_2}\left[pdq-Hdt \right].\tag{1.11}$$ Then they say the change in $I$ is the integral of the change - $$\delta I=\intop_{t_1}^{t_2}\left[\delta p\left(\dot{q}-\frac{\partial H}{\partial p}\right)+p\frac{d}{dt}\delta q-\frac{\partial H}{\partial q}\delta q \right]dt,\tag{1.11a}$$

and after "integrating by parts the $p\frac{d}{dt}\delta q$ term" they get
$$\frac{\delta I}{\delta p\left(t\right)}=\dot{q}-\frac{\partial H}{\partial p},\qquad-\frac{\delta I}{\delta q\left(t\right)}=\dot{p}+\frac{\partial H}{\partial q}.\tag{1.11b}$$

My questions are:

  1. It seems the get the derivatives by dividing with $\delta (p(t)), \delta (q(t))$ while ignoring the integral. How is this justified?

  2. How was the integral by parts calculated? what I get is $(p\delta q)|_{q_1,\delta q_1}^{q_2,\delta q_2}-\intop_{t_1}^{t_2}\dot{p}\delta qdt$ which is not $\dot{p}$, unless I can ignore the integral again.

Latter on they say that if theres a new lagrangian $$L=L+qF(t)\tag{1.13a}$$ then

$$\frac{d}{dF}I=\intop_{t_1}^{t_2}dt\frac{\partial L'}{\partial F}=\intop_{t_1}^{t_2}dtq=Q(t),$$ where $Q(t)$ is the real trajectory, cf. eq. (1.14).

  1. How is it possible that by integrating $q$ over time we get $Q(t)$? how can the answer even depend on time? shoulnd the anser be the total distance traveled?

You should read up on functional derivatives.

  1. This is exactly the definition of the functional derivative as given on Wikipedia: If $\delta I = \int (\cdots)\, \delta q(t) \,\mathrm dt$, then the $(\cdots)$-part is by definition the functional derivative $\frac{\delta I}{\delta q(t)}$.
    Compare this to the definition of the gradient: $$ F(\vec x + \delta \vec x) - F(\vec x) = \nabla F(\vec x) \cdot \delta \vec x + \mathcal O(|\delta \vec x|^2) . $$ In the functional case, the vector $\vec x$ is replaced by the function $q(t)$ and the dot product by the integral.

  2. After the integration by parts, you have $$ \delta I = \int \left[ \delta p\, (\dot q - \frac{\partial H}{\partial p} ) - \delta q\, (\dot p + \frac{\partial H}{\partial q}) \right] \mathrm dt , $$ because $\delta q$ is defined to be zero at times $t_1$ and $t_2$.

  3. Your formula is wrong, it is not $\frac{\mathrm d I}{\mathrm d F}$ but $$ \frac{\delta I}{\delta F(t)} = \int \mathrm d\tau \frac{\delta L'(\tau)}{\delta F(t)} = \int \mathrm d\tau q(\tau) \delta(t-\tau) = q(t) . $$ The object $\frac{\delta L'(\tau)}{\delta F(t)}$ is something slightly new again, it obeys the rule $\frac{\delta q(\tau)}{\delta q(t)} = \delta(t-\tau)$, compare with $\frac{\partial x_i}{\partial x_j} = \delta_{ij}$ in the more familiar case. Try to understand the fact that, if $I[q] = \int L[q(t)] \,\mathrm dt$, $$ \frac{\delta I}{\delta q(t)} = \int \frac{\delta L[q(\tau)]}{\delta q(t)} \mathrm d\tau .$$


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