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Why is the lagrangian extremized

by copernicus1
Tags: extremized, lagrangian
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copernicus1
#1
Jul29-13, 05:33 PM
P: 83
I've been reading a lot about path integrals lately, and I've found it fascinating to see at the quantum level how the extremal values of the lagrangian are basically the only ones that contribute when the action is large and therefore we get the classical path.

Something that continues to puzzle me, though, is why the lagrangian function in particular is extremized. Why should T-V be the quantity that gets extremized instead of some other function? Is there any consensus view on this question?

Thanks!
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WannabeNewton
#2
Jul29-13, 06:39 PM
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We have to first find the functions that extremize the action by solving Lagrange's equations. Nothing is extremized a priori.
BruceW
#3
Jul29-13, 06:54 PM
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yes, it is the action that is extremized, not the Lagrangian. I think what copernicus meant is along the lines of "why should the action have a particular form?" Or similarly "why should the Lagrangian have a particular form?". I think the answer is pretty much that this 'form' encodes the physical properties, or laws, of our system. It is similar to asking "why is F=ma?" or "Why is the electromagnetic force proportional to the charge it is acting on?" It is the physical law that we are postulating. We have to start somewhere, with some kind of principle. And our choice of principle is often guided by simplicity and common sense. For example isotropy, invariance to time reversal, e.t.c.


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