# Tag Info

### 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 ...
• 167k
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 ...
• 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 ...
• 38.5k
1 vote

I am not sure if this is your problem ?. with the non holonomic constraint equations $$\left[ \begin {array}{c} {\dot x}-f \left( x \right) \\ {\dot y}-g \left( y \right) \\ {\dot m}-h \left( m,x,y \... • 8,379 1 vote Accepted ### How does the boundary term matter in scalar field and in more general cases? Note first of all that it is usually important to specify appropriate boundary conditions (BCs) to render a variational principle well-posed, i.e. to ensure that the functional/variational derivative (... • 167k 1 vote ### Variational operator confusion One can in principle vary infinitesimally the S_0[X,e] action (2) simultaneously wrt. both the e and X variables. The coefficient function in front of \delta X will then give the Euler-... • 167k 1 vote ### Confusion with the variational operator \delta and finding variations Yes, that happened. I guess you meant$$ \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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