Jacobi's principle of least action

[dRic2]

Homework Statement

Consider the action integral of Jacobi's principle
$$A = \int_{\tau_1}^{\tau_2} \sqrt{2} \sqrt{E-V} \sqrt{ \Sigma a_{ik} q_i' q_k'} d \tau$$
($q_i' = \frac {dq_i}{d \tau} = \frac {dq_i}{dt} \frac {dt}{d\tau} = \dot q_i t'$).
Show that the $p_i$ associated with this integrand satisfy the following identity:
$$\frac {\Sigma b_{ik} p_i p_k} {2 (E-V)} = 1$$
where $b_{ik}$ are the coefficients of the matrix which is the reciprocal of the original matrix $a_{ik}$.

Relevant Equations

Symmetrical form of the action integral
$$A = \int_{t_1}^{\tau_2} \sum_i p_i q'_i d \tau$$

I'm studying a chapter on the parametric form of the canonical equation so basically the author says that time is no more an independent variable but it is expressed as a function of an other variable called $\tau$. In this way the canonical integral is reduced to the one I've written in "Homework Equations". The problem is that in this way I add a variable to my problem so I must also add a condition between the various $p_i$ in order to solve the integral. But I don't know how to derive that relation for the Jacobi's principle.PS: I know this is not strictly related to the exercise, but I never really understood Jacobi's "critique" of Lagrange's method. The book I'm reading said that he used this approach (expressing the time not as independent variable, but as a function of some other parameter) because he criticized that otherwise the process of varying between definite limit was not possible.

Thanks
Ric

 

[TSny]

Hello.

Just to be clear, are you asking how to obtain the identity $$\frac {\Sigma b_{ik} p_i p_k} {2 (E-V)} = 1$$ from $$A = \int_{\tau_1}^{\tau_2} \sqrt{2} \sqrt{E-V} \sqrt{ \Sigma a_{ik} q_i' q_k'} d \tau \,\,\, \rm ?$$

 

[dRic2]

Kind of. That integral can't be solved because $\tau$ is a variable arbitrary introduced. I think that relation between $p_i$ must be provided in order to solve it. This is how I interpreted the question. Sounds possible ?

 

[TSny]

My interpretation of the homework statement is that from the integrand of the integral in $$A = \int_{\tau_1}^{\tau_2} \sqrt{2} \sqrt{E-V} \sqrt{ \Sigma a_{ik} q_i' q_k'} d \tau$$ you can obtain expressions for the momenta $p_i$. Using these expressions, you can then prove the identity $$\frac {\Sigma b_{ik} p_i p_k} {2 (E-V)} = 1$$

I'm not sure what you mean when you say,

That integral can't be solved because $\tau$ is a variable arbitrary introduced.

By "solving" the integral, do you mean evaluating the integral? There is generally no need to explicitly evaluate the integral, as the integral is used only to obtain the differential equations for the actual path of motion via the variational principle $\Delta A = 0$. However, if you wanted to, it seems to me that you could (in principle) evaluate the integral for an arbitrary path $\left\{q_1(\tau), q_2(\tau), ..., q_N(\tau) \right\}$ connecting the initial and final points. The value of the integral will not depend on the particular choice of the parameter $\tau$. The integral depends only on the choice of path and the value of the total energy $E$.

 

[dRic2]

I'm not sure what you mean when you say,

Ok. The integral that can't be solve that I was referring to was $$\int_{\tau_1}^{\tau_2} \sum p_i q_i' d\tau$$
In order to solve the above integral I need a relation between $p_i$ that I can obtain from
$$A = \int_{\tau_1}^{\tau_2} \sqrt{ 2(E-V)} \sqrt{ \sum a_{ik} q_i' q_k' } d \tau$$
So, in my second post, I basically misunderstood my own statement
Thanks for pointing that out, now it is definitively clearer.

 

[dRic2]

So
$$p_i = \frac {\partial L}{ \partial q_i'} = \frac {\sqrt{2(E-V)} \sum_k a_{ik} q_k' } {2 \sqrt{\sum a_{ik} q_i' q_k'}}$$
Now I should find a way to exploit the fact that $b_{ik}$ is the inverse matrix of $a_{ik}$, am I on the right path ?

 

[TSny]

So
$$p_i = \frac {\partial L}{ \partial q_i'} = \frac {\sqrt{2(E-V)} \sum_k a_{ik} q_k' } {2 \sqrt{\sum a_{ik} q_i' q_k'}}$$
Now I should find a way to exploit the fact that $b_{ik}$ is the inverse matrix of $a_{ik}$, am I on the right path ?

OK, you're on the right path. But I think you missed a factor of 2 in the numerator, which is easy to do. Try to see that $$\frac {\partial}{ \partial q_j'} \sum_{ik} a_{ik} q_i' q_k'= {\color{brown}{2}} \sum_k a_{jk} q_k'$$

 

[dRic2]

(I'll drop the summation sign from now on)

$$p_i = \sqrt{2(E-V)} \frac {a_{ik}q_k'} {\sqrt{a_{ik}q_i'q_k'}}$$

$$b_{ij}p_i p_j = 2(E-V) \frac {b_{ij}a_{ik}q_k'a_{lj}q_l'} {\sqrt{a_{ik}q_i'q_k'} \sqrt{a_{lj}q_l'q_j'} }$$

(I was a little sloppy in choosing the indices of summation)

Now I exploit the fact that $B$ is the inverse of $A$, so I get $b_{ij}a_{ik} = \delta_{jk}$ and thus $\delta_{jk}a_{lj} = a_{lk}$. Finally:

$$ (...) = 2(E-V) \frac {a_{lk}q_k'q_l'}{a_{ik} q_i' q_k'} $$

where I use the fact that $i$, $k$, $j$ and $l$ are dummy indices to simplify the denominator. The numerator is identical to the denominator in the final equation so the exercise is done.

I can't believe I was stuck because I couldn't remember that $b_{ij}a_{ik} = \delta_{jk}$... for inverse matrices... I feel so stupid right now

BTW do you happen to know the argument between Jacobi ad Lagrange that I mentioned at the end of my original post ?

 

[TSny]

Looks good.

BTW do you happen to know the argument between Jacobi ad Lagrange that I mentioned at the end of my original post ?

I'm not familiar with this, other than the brief remarks in Lanczos' text near the bottom of page 136:
https://archive.org/details/VariationalPrinciplesOfMechanicsLanczos_201610/page/n155
(similar remark middle of page 134)

 

[dRic2]

That is the text I'm reading and the one where i got this exercise from ahahahah

 

[TSny]

Interesting. On what page of Lanczos is this exercise given?

 

[dRic2]

In the link you posted you can find it at pagE 188

 

[TSny]

Ah, thanks.

 

[TSny]

Some of the history is given in https://www.jstor.org/stable/27900374?seq=1#metadata_info_tab_contents of 1912, which you can download. I only glanced through it, but section v looks relevant.