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2 votes

Worldsheet action in the presence of background fields in complex coordinates

The integration measure is always a top-form. This means that $d^2\sigma$ is really (proportional to) a 2-form $\mathrm{d}\sigma\wedge\mathrm{d}\tau$. $d^2z$ is really (proportional to) a 2-form $\...
Qmechanic's user avatar
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2 votes

Why the choice of Configuration Space in Hamilton's Principle is $(q, \dot{q}, t)$?

Well, the function $f$ in the action integral is always identical to the Lagrangian in Hamilton's principle: $$\delta I=\delta\int_1^2L(q,\dot q,t)\;dt=0,\;\;\text{For Some Path in Configuration space}...
Albertus Magnus's user avatar
0 votes
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Is there a proof that a physical system with a *stationary* action principle cannot always be modelled by a *least* action principle?

On the question whether it is possible to go from Newton's laws to Hamilton's stationary action: Yes, that is possible. (As pointed out by physics stackexchange contributor Kevin Zhou: in physics you ...
Cleonis's user avatar
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0 votes

Is there a proof that a physical system with a *stationary* action principle cannot always be modelled by a *least* action principle?

Yes it is typically the case for field theories. A simple construction involves many harmonic oscillators with large frequencies. When there are an infinite number, the frequencies can be unbounded ...
LPZ's user avatar
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Is there a proof that a physical system with a *stationary* action principle cannot always be modelled by a *least* action principle?

If you divide the action integral by the fixed time between the two fixed points in space, the variation principle states, that all paths, invariant with respect to the time mean of the diffenence ...
Roland F's user avatar
-1 votes

Directly integrating the Lagrangian for a simple harmonic oscillator

The action is the difference of time averaged kinetic and potential energy, that is, zero.
my2cts's user avatar
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0 votes

Directly integrating the Lagrangian for a simple harmonic oscillator

Indeed you have the option of actually performing the integration as stated in (1). $$ S = \int\left(\frac{1}{2}m\dot{x}^2 - \frac{1}{2}kx^2\right)dt.\tag{1} $$ That is to say: The only way to ...
Cleonis's user avatar
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1 vote

Directly integrating the Lagrangian for a simple harmonic oscillator

This question can be cast in a convenient way by using the initial conditions $$ x_a=x(t_a) \quad x_b=x(t_b), $$ and taking $$ x(t)=A(x_a,x_b)\cos(\omega t)+B(x_a,x_b)\sin(\omega t). $$ Then, you can ...
Jon's user avatar
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3 votes
Accepted

Directly integrating the Lagrangian for a simple harmonic oscillator

Well, if we know the classical solution $q_{\rm cl}:[t_i,t_f] \to \mathbb{R}$ (which we do for the harmonic oscillator), we can plug it into the action functional $S[q]$ and obtain the on-shell action ...
Qmechanic's user avatar
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