Why do we extremize the Lagrangian in the Hamilton principle instead of energy?

[anbhadane]

I know that by extremizing lagrangian we get equations of motions. But what if we extremize the energy? I am just little bit of confused, any help is appreciated.

 

[hilbert2]

The Lagrangian is a quantity describing a whole trajectory between instants of time $t_1$ and $t_2$. The energy is a property of a single instant $t$. The extremal problem gives the same trajectory as Newton's II law.

 

[anuttarasammyak]

The principle of least action says

\delta S=\delta \int_{t_1}^{t_2}L(q,\dot{q},t)dt=0
with $q(t_1),q(t_2)$fixed. $L(q,\dot{q},t)$ is defined on the paths. It gives Lagrange equation of motion.

By its Legendre transformation Hamiltonian H, i.e. energy expressed by $p_i$ and $q_i$ is

H(p,q,t)=\Sigma p_i\dot{q_i}−L

So the principle of least action is stated as

\delta S=\delta \int_{t_1}^{t_2}[\Sigma p_i\dot{q_i}−H]dt=0

This gives Hamilton equation of motion.

 

[weirdoguy]

I know that by extremizing lagrangian

We extremize the action, not lagrangian.

 

[vanhees71]

There's also an action principle using the Hamiltonian,
$$H(q,p,t)=\dot{q}^{k} p_k-L,$$
where
$$p_k=\frac{\partial L}{\partial \dot{q}^k}$$
is the canonical momentum. The action reads
$$S=\int_{t_1}^{t_2} \mathrm{d} t (p_k \dot{q}^k-H),$$
and the Hamilton principle says $S$ is to be minimized under variations of the phase-space trajectories $(q^k(t),p_k(t))$ with the $p_k$ unrestricted and the $q^k$ at fixed boundaries $q^k(t_1)$ and $q^k(t_2)$.

The formalism works, because it is equivalent to Newton's equations of motion for systems with no constraints and with D'Alembert's principle for systems with constraints. The reformulation in terms of a variational principle is just an important mathematical tool to analyze the equations of motion in more general detail. Of particular importance is Noether's theorem and the fact that you can formulate everything in terms of Poisson brackets, because in this way classical mechanics looks pretty similar to quantum mechanics in the operator formalism, which is the way how Dirac formulated his version of quantum mechanics, which reveals its structure most clearly of all formalism.

Quantum mechanics, written in terms of Feynman's path integrals, most easily explains "why" the action principle in classical mechanics holds, it's an approximation for the quantum dynamics, if the situation is such that the action becomes very large compared to $\hbar$. Then the path integral for the propagator can be approximated by the stationary-phase approximation.

Another way to come to the same result is to use Schrödinger's wave-mechanics formulation and use the WKB approximation, which is a formal expansion in orders of $\hbar$, using the ansatz $\psi(t,\vec{x})=\exp(\mathrm{i} S/\hbar)$. In leading order you get the Hamilton-Jacobi partial differential equation which is equivalent to the Euler-Lagrange or Hamilton canonical equations of motion, and $S$ is the classical action in this leading-order approximation.

 

[anbhadane]

The Lagrangian is a quantity describing a whole trajectory between instants of time and . The energy is a property of a single instant .

so, basically we find first path and it automatically satisfies the minimum energy requirement?

 

[anbhadane]

We extremize the action, not lagrangian.

sorry, I was saying action with energy as function.

 

[anuttarasammyak]

so, basically we find first path and it automatically satisfies the minimum energy requirement?

Not the minimum energy as you say but the stationary action would be satisfied. As you know Lagrangean is kinetic energy $\mathbf{-}$ potential energy. How did you come to the idea that not Lagrangean but energy, i.e. kinetic energy $\mathbf{+}$ potential energy, should play some role in action principle ?

 

[anbhadane]

How did you come to the idea that not Lagrangean but energy, i.e. kinetic energy potential energy, should play some role in action principle ?

I know in action we use lagrangian which is T - V, but i am saying instead of T-V, can we use T+V? anyway it's function too.

 

[PeroK]

I know in action we use lagrangian which is T - V, but i am saying instead of T-V, can we use T+V? anyway it's function too.

Can you see why minimising or maximising $T + V$ would not work? Imagine an object in a gravitational field.

 

[anuttarasammyak]

I know in action we use lagrangian which is T - V, but i am saying instead of T-V, can we use T+V?

Lagrangian L = T - V = 2T - (T+V) = 2T - H as post #3 says. This expression of Lagrangian, i.e. integrand for action, using energy H ( and T ) might be of your interest.

 

[anbhadane]

Imagine an object in a gravitational field

Thank you. Now I got it.

 

[anbhadane]

Lagrangian L = T - V = 2T - (T+V) = 2T - H as post #3 says

Anyway I am now clear with my doubt. 2T - H is another form of L so basically it's the same as L. I was interested in only T + V. Thank you for your valuable responses.

 

[vanhees71]

I know in action we use lagrangian which is T - V, but i am saying instead of T-V, can we use T+V? anyway it's function too.

The answer is that Hamilton's variational principle in configuration space (i.e., the Lagrangian version of the principle) works with the Lagrangian $L(q,\dot{q},t)=T-V$, i.e., it gives the correct equations of motion known from Newton's Laws.

It's also not simply a minimum as a function of $q$ and $\dot{q}$ but what's minimized (or rather extermized) is the action functional
$A[q]=\int_{t_1}^{t_2} \mathrm{d} t L(q,\dot{q},t),$
i.e., you look for the curve $q(t)$ in configuration space with fixed endpoints $q_1=q(t_1)$ and $q_2=q(t_2)$ which makes the action stationary. This leads to the Euler-Lagrange equations,
$$p=\frac{\partial L}{\partial \dot{q}}, \quad \dot{p}=\frac{\partial L}{\partial q}.$$