Why must partialS/partialα_n = -β_n where S is the complete integral?

[giraffe714]

TL;DR Summary

I don't understand why the derivative of the Jacobi complete integral with respect to the constant α must be another constant, and furthermore why that constant is negative.

As stated in the TLDR, I don't understand why the derivative of the Jacobi complete integral with respect to the constant α must be another constant, and furthermore why that constant is negative. The textbook I'm following, van Brunt's The Calculus of Variations proves it by taking:
$$ \frac{\partial}{\partial \alpha_1} (H + \frac{\partial S}{\partial t}) = \frac{\partial^2 S}{\partial \alpha_1 \partial t} + \sum_{k=1}^n \frac{\partial^2 S}{\partial \alpha_1 \partial q_k} \frac{\partial H}{\partial p_k} = 0 $$
and then just stating that
$$ \frac{\partial S}{\partial \alpha_1} = -\beta_1 $$
is satisfied identically but i cant figure out a) how those two equations are even related and b) from what I can tell, if $\frac{\partial S}{\partial \alpha_1} = -\beta_1$ where $\beta_1$ is constant, that means $\frac{\partial^2 S}{\partial \alpha_1 \partial t}$ must be zero? but if it's zero, then in the original equation
$$ \frac{\partial}{\partial \alpha_1} (H + \frac{\partial S}{\partial t}) = \frac{\partial^2 S}{\partial \alpha_1 \partial t} + \sum_{k=1}^n \frac{\partial^2 S}{\partial \alpha_1 \partial q_k} \frac{\partial H}{\partial p_k} = 0 $$
the term $\sum_{k=1}^n \frac{\partial^2 S}{\partial \alpha_1 \partial q_k} \frac{\partial H}{\partial p_k}$ must be zero and I just don't see why that's true?

 

[TSny]

The textbook I'm following, van Brunt's The Calculus of Variations proves it by taking:
$$ \frac{\partial}{\partial \alpha_1} (H + \frac{\partial S}{\partial t}) = \frac{\partial^2 S}{\partial \alpha_1 \partial t} + \sum_{k=1}^n \frac{\partial^2 S}{\partial \alpha_1 \partial q_k} \frac{\partial H}{\partial p_k} = 0 $$
and then just stating that
$$ \frac{\partial S}{\partial \alpha_1} = -\beta_1 $$

I happened to find online access to this text, so I will refer to specific pages and equations in the book.

The equation $\dfrac{\partial S}{\partial \alpha_1} = -\beta_1$ is just a specific example of the general equation $P_k = -\dfrac{\partial S}{\partial Q_k}$ found in (8.25) on page 173. Note that at the bottom of page 175 we have $Q_k = \alpha_k$ and $P_k = \beta_k$.

from what I can tell, if $\frac{\partial S}{\partial \alpha_1} = -\beta_1$ where $\beta_1$ is constant, that means $\frac{\partial^2 S}{\partial \alpha_1 \partial t}$ must be zero?

No, $\dfrac{\partial^2 S}{\partial \alpha_1 \partial t}$ does not have to be zero. This can be confusing. $S(t, q, \alpha)$ generally depends on $t$ both explicitly and also implicitly through the various $q_k(t)$. So, $\dfrac{\partial^2 S}{\partial \alpha_1 \partial t}$ will also generally depend on $t$ explicitly and implicitly through the $q_k(t)$. As time progresses, $t$ and $q(t)$ change in such a way that $\beta_1 = - \dfrac{\partial S}{\partial \alpha_1}$ remains constant during the time evolution of the system. Similarly for the other $\beta_k$.

However, $\dfrac{\partial^2 S}{\partial \alpha_1 \partial t}$ is generally not zero. This is because the notation $\dfrac{\partial^2 S}{\partial t \partial \alpha_1}$ is interpreted as $$\frac{\partial^2 S}{\partial t \partial \alpha_1} = \left. \frac{\partial}{\partial t} \left( \frac{\partial S}{\partial \alpha_1} \right)\right|_{q_k, \alpha_k}$$ where the $q_k$ are held constant while taking the partial derivative with respect to $t$. See footnote 7 on page 178.

As an example, look at the expression for $\beta_2$ on page 181 for the geometrical optics example: $$\beta_2 =\frac{\alpha_2 A}{\mu^2} - t$$ where $A$ is the function of $q_1(t)$ given on page 180. $\beta_2$ is a constant of the motion. But $$\frac{\partial^2 S}{\partial t \partial \alpha_2} = \frac{\partial}{\partial t}(-\beta_2) = - \frac{\partial }{\partial t} \left(\frac{\alpha_2 A}{\mu^2} - t \right ) = 1$$ since $A$ is fixed while taking the partial with respect to $t$.

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As a side note, I get a slightly different result for $\beta_1$ and $\beta_2$ for the geometrical optics example. I get $$\beta_1 = \dfrac{2 \alpha_1 A}{\mu^2} - q_2$$ $$\beta_2 = \dfrac{2 \alpha_2 A}{\mu^2} - t.$$ These have a factor of 2 in the fractions that do not appear in the expressions in the book. But I could have made a mistake. My results for $q_1(t)$ and $q_2(t)$ are $$q_1 = \frac{\mu^2}{4\alpha_2^2}(\beta_2+t)^2 + \frac{\alpha_1^2+\alpha_2^2}{\mu^2}$$ $$q_2 = \frac{\alpha_1}{\alpha_2}t + \frac{\alpha_1}{\alpha_2}\beta_2 - \beta_1$$

 

[giraffe714]

. S(t,q,α) generally depends on t both explicitly and also implicitly through the various qk(t).

That actually makes sense, thank you. I think I was just missing that S is a function of q_k which are also functions of t so that clears it up. One question though, just to make sure: the fact that $P_k = \beta_k = const.$ doesn't *follow* from $Q_k = \alpha_k = const.$, instead they're just *chosen* to be this way - and further there is also no actual guarantee that an S that satisfies this can be found or exists?

But ∂2S∂t∂α2=∂∂t(−β2)=−∂∂t(α2Aμ2−t)=1 since A is fixed while taking the partial with respect to t.

And just to be perfectly clear - when we're taking this partial derivative, we disregard the q_k(t) dependence, which is why it doesn't have to be zero. If this is true then I assume the total time derivate of S would be constant, correct?

 

[TSny]

One question though, just to make sure: the fact that $P_k = \beta_k = const.$ doesn't *follow* from $Q_k = \alpha_k = const.$, instead they're just *chosen* to be this way

I don't think the $P_k$'s are "chosen" to be constants. See the first part of section 8.4.1 which explains why both the $P_k$'s and the $Q_k$'s are constants. They are constants because of Hamilton's equations $$\dot Q_k = -\frac{\partial \hat H}{\partial P_k}$$ $$\dot P_k = -\frac{\partial \hat H}{\partial Q_k}$$ with the requirement that $\hat H = 0$. Thus $\dot Q_k = 0$ and $\dot P_k = 0$.

- and further there is also no actual guarantee that an S that satisfies this can be found or exists?

$S$ exits as long as a solution to the Hamilton-Jacobi equation (8.31) exists.

And just to be perfectly clear - when we're taking this partial derivative, we disregard the q_k(t) dependence, which is why it doesn't have to be zero.

Yes, that's right.

If this is true then I assume the total time derivate of S would be constant, correct?

The total time derivative of $S(q_k, \alpha_k, t)$ is not a constant. The total time derivative of $\dfrac{\partial S}{\partial \alpha_k}$ is zero. See (8.38). And this corresponds to $\dfrac{\partial S}{\partial \alpha_k} = -\beta_k$ being constant.

The time derivative of $S$ is interesting: $$\frac{dS}{dt} = \sum_k \frac{\partial S}{\partial q_k} \dot q_k + \frac{\partial S}{\partial t} = \sum_k p_k \dot q_k - H(q_k, p_k, t) = L$$ where $L$ is the Lagrangian of the system in the original coordinates $q_k$. Here, we used $\frac{\partial S}{\partial q_k} = p_k$ and $\frac{\partial S}{\partial t} = -H$.