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Qmechanic ♦
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In-variance Invariance of Lagrangian under active coordinate transformation

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Qmechanic ♦
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This question is related to symmetry properties of the Lagrangian and conservation laws. Let us consider a one-dimensional case of a particle of mass $m$ moving along the $x$ axis such that the Lagrangian is given by $L = \frac{1}{2} m \dot{x}^2 $. Now, if there is an active coordinate transformation such that the physical location of the point mass changes from $x$ to $x + \delta x$, then this must happen in some interval of time ($t,t+\delta t$). The particle was at location $x$ at time instant $t$ and is at location $x +\delta x$ at time $t + \delta t$. Since $\dot{x}$ is in general a function of time, the Lagrangian has an implicit time dependence. Then how are we sure that the Lagrangian does not change under an active coordinate transformation  ?

This question is related to symmetry properties of the Lagrangian and conservation laws. Let us consider a one-dimensional case of a particle of mass $m$ moving along the $x$ axis such that the Lagrangian is given by $L = \frac{1}{2} m \dot{x}^2 $. Now, if there is an active coordinate transformation such that the physical location of the point mass changes from $x$ to $x + \delta x$, then this must happen in some interval of time ($t,t+\delta t$). The particle was at location $x$ at time instant $t$ and is at location $x +\delta x$ at time $t + \delta t$. Since $\dot{x}$ is in general a function of time, the Lagrangian has an implicit time dependence. Then how are we sure that the Lagrangian does not change under an active coordinate transformation  ?

This question is related to symmetry properties of the Lagrangian and conservation laws. Let us consider a one-dimensional case of a particle of mass $m$ moving along the $x$ axis such that the Lagrangian is given by $L = \frac{1}{2} m \dot{x}^2 $. Now, if there is an active coordinate transformation such that the physical location of the point mass changes from $x$ to $x + \delta x$, then this must happen in some interval of time ($t,t+\delta t$). The particle was at location $x$ at time instant $t$ and is at location $x +\delta x$ at time $t + \delta t$. Since $\dot{x}$ is in general a function of time, the Lagrangian has an implicit time dependence. Then how are we sure that the Lagrangian does not change under an active coordinate transformation?

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Tirthankar
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In-variance of Lagrangian under active coordinate transformation

This question is related to symmetry properties of the Lagrangian and conservation laws. Let us consider a one-dimensional case of a particle of mass $m$ moving along the $x$ axis such that the Lagrangian is given by $L = \frac{1}{2} m \dot{x}^2 $. Now, if there is an active coordinate transformation such that the physical location of the point mass changes from $x$ to $x + \delta x$, then this must happen in some interval of time ($t,t+\delta t$). The particle was at location $x$ at time instant $t$ and is at location $x +\delta x$ at time $t + \delta t$. Since $\dot{x}$ is in general a function of time, the Lagrangian has an implicit time dependence. Then how are we sure that the Lagrangian does not change under an active coordinate transformation ?