Question about Lagrange multipliers

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  • #1
dRic2
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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
 

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  • #2
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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?
 
  • #3
dRic2
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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 ?
 
  • #4
wrobel
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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##
 
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  • #5
dRic2
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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
 
  • #6
wrobel
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Why do not you take a regular textbook in analysis? V. Zorich Mathematical Analysis for example
 
  • #7
dRic2
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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"
 
  • #8
wrobel
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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.
 
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  • #9
dRic2
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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)
 
  • #10
wrobel
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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
 
  • #11
dRic2
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You can always define a new variable ##\lambda' = \frac 1 {\lambda}## any get the same result
 
  • #12
wrobel
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You can always define a new variable λ′=1λ\lambda' = \frac 1 {\lambda} any get the same result
what about ##\lambda=0##? :)
 
  • #13
dRic2
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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.
 

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