Lagrange equation: when exactly does it apply?

  • Thread starter Nikitin
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Hi! Does the Lagrange equation ONLY apply when the constraints are holonomic? What about the constraining forces acting on the system (i.e. normal force, or other perpendicular forces), do they make a system holonomic?

What about the Lagrange equation with the general force on the right hand side. I read in Goldstein that it can be, for instance, a non-conservative frictional force. Why? Where did that come from?
 

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BTW, I am talking about the Euler-Lagrange equation. This one, $$ \sum_j \frac{\partial L }{\partial q_j} - \frac{d}{d t} \frac{\partial L }{\partial \dot{q_j}} = 0$$ in case there was any confusion.

But what is up with the modified equation, ##\frac{\partial L }{\partial q_j} - \frac{d}{d t} \frac{\partial L }{\partial \dot{q_j}} = Q_j## ? When does this apply to a system, and for which generalized forces ##Q_j##s? It was not derived in Goldstein's book, just given.
 
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Another question, if somebody wants to answer: does ##\frac{\partial T}{\partial q_j}##, where ##T## is the kinetic energy of the system, always equal zero? Or do there exist situations where the kinetic energy has an explicit dependence on position?

It might seem like a strange question because kinetic energy is defined using total velocity, but I ask because one form of Lagrange's equation is ##\frac{d}{dt} \frac{\partial T}{\partial \dot{q_j}} - \frac{\partial T}{\partial q_j} = Q_j##.
 
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Another question, if somebody wants to answer: does ##\frac{\partial T}{\partial q_j}##, where ##T## is the kinetic energy of the system, always equal zero? Or do there exist situations where the kinetic energy has an explicit dependence on position?

It certainly can, in spherical coordinates (or polar) you have position dependence in the kinetic term.
 
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Check http://physics.clarku.edu/courses/201/sreading/AJP73_March2005_265-272.pdf [Broken] paper out. Does that help answer your questions?
 
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I hate to answer your question this way, but if you reread Goldstein chapter 1 and 2 enough, you will answer your questions. This was true for me.
 

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