3 votes

Proof of principle of stationary action when the Lagrangian is not $L=T-V$

In many systems (in particular outside the topic of classical mechanics) the principle of stationary action is taken as a first principle/axiom, i.e. it has no proof per se. The choice of action is ...
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  • 167k
2 votes
Accepted

Why does the trajectory of a relativistic particle "minimises its Minkowski distance"?

It is straightforward to derive that OP's action (1) is $$S~=~ - m_0c ~ \Delta s, $$ where $$\Delta s~=~c\Delta\tau $$ is the spacetime distance $$\Delta s ~=~\int \!ds$$ in the $(+,-,-,-)$ sign ...
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  • 167k
1 vote

Why does the trajectory of a relativistic particle "minimises its Minkowski distance"?

Let's just justify this physically without making an explicit appeal to the Lagrangian math. Assume the Minkowski metric and $c=1$ units. We already know, if the theory is going to make any sense at ...
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1 vote

How to derive the Lagrangian for a system of first order equations of motion?

I am not sure if this is your problem ?. with the non holonomic constraint equations $$\left[ \begin {array}{c} {\dot x}-f \left( x \right) \\ {\dot y}-g \left( y \right) \\ {\dot m}-h \left( m,x,y \...
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1 vote
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How does the boundary term matter in scalar field and in more general cases?

Note first of all that it is usually important to specify appropriate boundary conditions (BCs) to render a variational principle well-posed, i.e. to ensure that the functional/variational derivative (...
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  • 167k
1 vote

Variational operator confusion

One can in principle vary infinitesimally the $S_0[X,e]$ action (2) simultaneously wrt. both the $e$ and $X$ variables. The coefficient function in front of $\delta X$ will then give the Euler-...
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1 vote

Confusion with the variational operator $\delta$ and finding variations

Yes, that happened. I guess you meant $$ \delta f = \sum_i \frac{\partial f}{\partial x_i} \delta x_i $$ on your third equation. Also you've implicity fixed inital $t_0$ and final $t_1$, so that your ...
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  • 179

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