Question about Lagrange multipliers

[dRic2]

I'm having some trouble understanding the following proof ($a_{ik}$ and $b_{ik}$ are constants)

Our problem is to find the stationary value of $a_{ik}q_iq_k$ under the auxiliary condition the we move on the surface $b_{ik}q_iq_j = 1$.

We drop the auxiliary condition and minimize $a_{ik}q_iq_k - \frac 1 {\lambda} b_{ik}q_iq_k$.

Shouldn't it be $a_{ik}q_iq_k - \frac 1 {\lambda} (b_{ik}q_iq_k-1)$ ?

(Summation convention is used)

Thanks Ric

 

[Paul Colby]

One thing I don't understand is Lagrange multipliers. However, one thing I do follow is that adding a constant to a Lagrangian doesn't change the equations obtained by the variation. Isn't the term you've added, $-\frac{1}{\lambda}$, held constant in the variation?

 

[dRic2]

Yes I see you point and in fact I am a bit puzzled myself, but then what would be the difference between the condition $b_{ik}q_iq_k = 1$ and some other condition $b_{ik}q_iq_k = c$ or even $b_{ik}q_iq_k = 0$ since I can always neglect the constant term ?

 

[wrobel]

You can not neglect constant. Loosely speaking, you have a condition $g(x)=c$ and your task is to find a critical point of a function $f=f(x)$ on the manifold $\{x\in\mathbb{R}^m\mid g(x)=c\}$. To do that you must solve the following system
$$g(x)=c,\quad \frac{\partial f}{\partial x^i}=\lambda \frac{\partial g}{\partial x^i},\quad i=1,\ldots,m$$
you have m+1 equations and m+1 unknowns: $\lambda,x^1,\ldots,x^m$

 

[dRic2]

I know, that's why I'm asking. I don't understand the argument given in that proof. I have to specify that I do not need to evaluate the minimum of that function, but the values of $q_i$ such that the function has minimum, maybe in that sense the constant can be neglected... I don't know

 

[wrobel]

Why do not you take a regular textbook in analysis? V. Zorich Mathematical Analysis for example

 

[dRic2]

Maybe I was not expressing myself very well. I was reading a book. The book is not about math nor about Lagrange multipliers, it is about classical mechanics. While reading the book I ran into the argument posted in the first post which I don't understand.

PS: maybe the confusing part is the word "proof". I didn't mean that the proof is about Lagrange multipliers. The first line should read "I'm having some trouble understanding the following statement"

 

[wrobel]

I also do not understand that argument and I don't care because I understand what Lagrange multipliers are and I can solve the problem independently on that book.

 

[dRic2]

I can solve the problem independently on that book.

Ok, so let's try to solve it. Do you agree with me that if $x_i = q_i$, $f(x_i) = a_{ik}q_iq_k$, $g(x_i) = b_{ik}q_iq_k$ and $c = 1$ then the system becomes:
$$b_{ik}q_i = \lambda a_{ik} q_i$$
plus the condition $b_{ik}q_i q_k = 1$. (Note that I used $\frac 1 {\lambda}$ as a multiplier to stick to the convention introduced by the author of the book)

 

[wrobel]

Note that I used 1λ\frac 1 {\lambda} as a multiplier to stick to the convention introduced by the author of the book)

it is not a matter of convention, $\lambda$ must be situated as I said

 

[dRic2]

You can always define a new variable $\lambda' = \frac 1 {\lambda}$ any get the same result

 

[wrobel]

You can always define a new variable λ′=1λ\lambda' = \frac 1 {\lambda} any get the same result

what about $\lambda=0$? :)

 

[dRic2]

Ok you are right.

But in the end I solved my problem. In this particular case the formal procedure and the one used by the author are the same, he probably used this "trick" to get a faster result.

 

[wrobel]

something in addition the Lagrange multipliers
https://www.physicsforums.com/goto/post?id=6325380