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When deriving Newton’s law from Euler-Lagrange equation for a particle, the Lagrangian is defined as the kinetic energy minus the potential energy, but the problem is that the kinetic energy is defined as the work needed to raise the velocity of the particle to the current velocity and when deriving the kinetic energy expression we use Newton’s equation in the derivation since the work done equals force times distance and the force is substituted by m*a. My point is that we used Newton’s law to derive the kinetic energy from the work definition to find the Lagrangian and feed it to Euler-Lagrange equations to get back Newton’s law, so I don’t think that we “derived” Newton’s law or is something wrong?

Qmechanic
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Ali Khalil
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4 Answers4

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Lagrangian is defined as the kinetic energy minus the potential energy

You are wrong. Lagrangian has nothing to do with kinetic and potential energy. For example Einstein-Hilbert action is $S=\frac{1}{2k}\int R\sqrt{-g}\,d^4x$. There is no fast-hand way to see in there something like "potential" and "kinetic" energy.

There is only one true definition of Lagrangian. It is what makes our system to move along the path $S=\int L d\Omega=\max$. In every good book it is emphasized that this is simple coincidence of $L$ to have relation with energy. It is not random coincidence, but quite wonderful. But it is coincidence.

In mechanics it just happens that for certain type of simplest systems the underlying Lagrangian is connected to energies. Being that system slightly more complex, for example within nonlinear coordinates, you won't have same $\frac12 mV^2$ but some bilinear form of your coordinate space.

You need to understand the connections of things, so they present for you the most obvious and simplest systems where you can put a finger and say that "this is energy". It is energy exactly because energy is invariant of your system, same as the invariants of your system are what comes from variation of action.

sanaris
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  • I'm puzzled by this. Suppose I have a 1d harmonic oscillator. The action does NOT have this form: if anything it's not $\int d^4x$. Moreover what is the sense of a metric in a 1d configuration space? Wouldn't it be trivial? And what is $R$? Even more simply, suppose I have a free particle. What would $R$ be? Not saying you're incorrect just wondering how this "comes down" to ordinary physics. – ZeroTheHero Apr 06 '20 at 15:37
  • @ZeroTheHero there is good links for Feynman lectures and more https://en.wikipedia.org/w/index.php?title=Einstein-Hilbert_action&redirect=no – sanaris Apr 06 '20 at 15:50
  • @ZeroTheHero perhaps it is useful to point out, in the context of field theory, the difference of the Lagrangian(which is a FUNCTIONAL of $\phi(\cdot,t)$), and the Lagrangian density(which is a FUNCTION of $\phi(x,t)$ ), the latter is what goes in with $d^4x$(the 4 parameters that enter the theory). In point mechanics, there is no density-the only parameter is $t$, everything else being dynamical variables. There is no need for a Jacobian, there is no need for a $\sqrt{g}$, etc. Ofcourse, even in ordinary QM, the only parameter is $t$, and the above argument carries. Or am I missing something? – GRrocks Apr 06 '20 at 16:06
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    @sanaris, within the regime of the question, i.e. Classical Mechanics, $\mathcal{L}=T-V$ is a completely correct formula. Its not just for 'simple' systems i.e. it works completely for the double pendulum which isn't a 'simple' system. And I wouldn't refer to it as a mere coincidence, its not. And where does your 'true definition' $S=\int \mathcal{L} d\Omega = \text{max}$ come from? What is max? And why does the action equate to it? – SamuraiMelon Apr 06 '20 at 16:25
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    @GRrocks right I realize this but I'm not sure from reading the OP's question that the question deals with fields. It seems to deal with point particles. – ZeroTheHero Apr 06 '20 at 16:48
  • @ZeroTheHero I think I might agree with you; the premise is point mechanics here. – GRrocks Apr 07 '20 at 06:13
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1) The Lagrangian is simply a function of generalised coordinates and velocities(see also @sanaris above), which when put into an action and extremised gives you the time evolution of the coordinates(and thus, the velocities via differentiation). You do not know the trajectory, and thus the velocity, and thus the kinetic energy, before you have actually solved for $q(t)$. A priori, it is just some coordinate, with nothing to do with actual motion.

2) In fact, you can only interpret $\frac{1}{2}m\dot{q}^2$ as kinetic energy of the particle if $\dot{q}$ IS indeed the velocity-which happens only if you've already solved for $q(t)$. Only then can you define $\dot{q}(t)\equiv v$. This ofcourse is equivalent to having solved for $q(t)$ by Newton's equations, and then you can call it kinetic energy.

3) In a nutshell, you can assign meanings such as 'energy' to terms in the Lagrangian only AFTER you've solved the Euler Lagrange equations, and plugged in the actual $q(t)$ which the particle follows. Since his is identical to solving for $q(t)$ from Newton's equations, so all is consistent. Things like 'energy' only make sense on-shell, e.g. the Hamiltonian $H(q,p)$ is total energy ONLY WHEN $q,p$ solve Hamilton's equations. Otherwise it's a generic function.

GRrocks
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    yeah I think also this captures the subtle point about considering $\dot{q}$ and $q$ as independent coordinates in the Lagrange equations of motion. – ZeroTheHero Apr 06 '20 at 16:54
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You are right. The Newtonian and Lagrangian formulations are equivalent. Either can be used as a starting point for mechanics. Which you think more fundamental is largely a matter of personal choice.

My own preference is to formulate Newton's laws from conservation of momentum (the third law contains the physical content, the second just a definition of force). Conservation of momentum appears as a fundamental principle in general relativity, and can actually be proven for particle interactions in relativistic quantum theory. Others find the Lagrangian formulation appealing.

Charles Francis
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    but isn’t the principle of least action is more fundamental and leads to Lagrangian formalism? – Ali Khalil Apr 06 '20 at 08:23
  • Some people think so, but my own view is that action is just a formula in mathematics, not a fundamental physical principle. – Charles Francis Apr 06 '20 at 08:33
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    interesting viewpoint, what made you think like this? – Ali Khalil Apr 06 '20 at 08:49
  • @CharlesFrancis, as far as I understand a physical principle is supposed to explain the behavior of systems and principle of least action describes the dynamics very accurately. As you said it is largely a matter of taste, but could you please elaborate more on why conservation of momentum is more appealing as a fundamental law? I believe there are very good points that I have not observed. – exp ikx Apr 06 '20 at 10:03
  • This isn't the right place for a discussion. I have a chat room, https://chat.stackexchange.com/rooms/106297/particles-or-fields which would be better. – Charles Francis Apr 06 '20 at 10:43
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In Lagrangian mechanics, we take the kinetic & potential terms as axiomatic, i.e. we don't use Newton's second law to justify $K=\frac12m\dot{x}^2$, we just claim $K=\frac12m\dot{x}^2$. Newtonian, Lagrangian and Hamiltonian mechanics (and a few other options) are equivalent, but they assume different things.

J.G.
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