In classical mechanics the Lagrangian of a system is a function of all its coordinates and their first-time derivative. For a system described by n coordinates it is of the form:

\( L ( { q }_{ 1 },{ \dot { q } }_{ 1 },{ q }_{ 2 },{ \dot { q } }_{ 2 },...,{ q }_{ n },{ \dot { q } }_{ n })= T-V \),

when T(kinetic energy) and V(potential energy) can be properly defined. Else it will be constructed in some other way but will still be a function of the \(q's\) and \(\dot { q }'s\) (and possibly \(t\) if the time dependance is explicit).

If well defined (by a clever choice of coordinates), the Lagrangian can be very useful to find the equations of motion of rather complex systems (see. *Euler-Lagrange equation*). It will yield a system of n second-order differential equations that can be solved numerically or, sometimes, analytically.

In a sense, the Lagrangian formulation of mechanics is less empirical than Newton's, there is a more theoretical motivation for it.

**On which of the fundamental laws of nature does the Lagrangian formulation of mechanics rely?**

Additional thoughts, details and technicalities you might have are most welcome as I'm currently learning about this subject; so please don't hesitate to share some of your knowledge.

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TopNewestActually, Lagrangian mechanics and Newtonian mechanics are mathematically equivalent, as one can be derived from the other. Lagrangian mechanics is based on the assumption that the integral of the Lagrangian along the true trajectory is a minimum.

For conservative forces like gravity, we can always identify \(\mathcal{L} \equiv T-U\), and then some story can be proposed that kinetic and potential energy minimize their discrepancy over the course of the trajectory a particle takes.

However, for non-conservative forces, it is no longer true that \(\mathcal{L}\equiv T-U\). For such forces, perhaps the best thing we can say is that \(\mathcal{L}\) is a thing that generates the correct dynamics when passed into Lagrange's equation of motion.

The great thing about Lagrangians (and Hamiltonians) is that they are scalar (numerical quantities), not vectors, and so don't require wizardly tricks in different coordinate systems. They can easily be re-written in any generalized set of coordinates (no vector calculus required to transform co-ordinates). It is in general very clear to see when a given quantity is conserved.

For example, if we know that the system's potential is independent of an angle \(\psi\), then we know immediately that the momentum about the given axis is conserved.

This is because \(\mathcal{L} \equiv T(\ldots, \dot{\psi}, \ldots) - V\), where \(V\) is not a function of \(\psi\), and therefore (by Lagrange's equation),

\[\frac{d}{dt}\frac{d\mathcal{L}}{d\dot{\psi}} = \frac{d\mathcal{L}}{d\psi} = \frac{dV}{d\psi} = 0\]

Thus

\[\frac{d}{dt}\frac{d\mathcal{L}}{d\dot{\psi}} = \frac{d}{dt}mr^2\dot{\psi} = 0\]

which shows the angular momentum is conserved. – Josh Silverman Staff · 2 years ago

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Calculus of variations is widely used in Engineering, especially in Optimal Control. – Abhishek Sinha · 2 years ago

This is called variational formulation of Newton's law. The general mathematical technique ofLog in to reply

Of course, the general framework for classical mechanics better be consistent by now! I only meant to say the Lagrangian formulation is sometimes more convenient than Newton's, though the two are equivalent. This is partly because, as you pointed out, changes of coordinates are easier to work out.

You gave the answer I was waiting for: Lagrangian mechanics is based on

the principle of least action: the orbit of the system is the one that minimizes a quantity called the action which is defined as:\(A=\int _{ { t }_{ 0 } }^{ t }{ L({ q }_{ 1 },...,{ q }_{ n },{ \dot { q } }_{ 1 },...,{ \dot { q } }_{ n })dt } \) .

I find this principle very powerful and beautiful. It seems deep and universal somehow.

As you shown, this formalism is also useful to work out conserved quantities, in particular the momentum conjugate to a cyclic coordinate is conserved. More generally if the Lagrangian is invariant under some infinetesimal change of coordinate, say:

\(\delta { q }_{ n }={ f }_{ n }({ q }_{ 1 },...,{ q }_{ n })\delta\) ,

then there is a conserved quantity given by:

\( \large Q =\sum { { p }_{ n }{ f }_{ n }({ q }_{ 1 },...,{ q }_{ n }) } \) , where \( \large { p }_{ n }=\frac { \partial L }{ \partial { \dot { q } }_{ n } }\)

In your example, the Lagrangian is invariant under the transformation \( \psi = \psi+ \delta \), we have \({ f }_{ \psi }=1\)

the conserved quantity associated with that symmetry is just

\(\large { p }_{ \psi }=\frac { \partial L }{ \partial \dot { \psi } } \) – Robin Plantey · 2 years ago

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You should try Landau and Lifshitz's Classical Mechanics Textbook. It's a classic on the subject of Langragians and starts from the base up. – Bhaskar Kumawat · 1 year, 10 months ago

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David Morin 's book on classical mechanics has a lot of material on lagrangian mechanics... You should try that book first... – Sarthak Behera · 2 years, 1 month ago

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– Rajdeep Dhingra · 2 years ago

U read this book ? And you are just 14 years old !!Log in to reply

– Sarthak Behera · 2 years ago

You too are 14 years it seems...Log in to reply

– Rajdeep Dhingra · 2 years ago

Yes, But U didn't answer my question.Log in to reply

– Sarthak Behera · 2 years ago

I have just gone through the book...Log in to reply

– Robin Plantey · 2 years, 1 month ago

Thank you for the suggestion. Actually you're right, a classic book such as this one would nicely complement the courses I've been taking...and those I'm going to take! Rest assured I'll look into it.Log in to reply