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Comments to the question (v1):

I) In the Lagrangian $L(q(t),\dot{q}(t),t)$, one must distinguish between implicit time dependence via the variables $q(t)$ and $\dot{q}(t)$, and explicit time dependence.$^1$

However, the implicit time dependence in the Lagrangian $L$ only makes sense in the context of a fixed (but arbitrary, possibly virtual) path $$\tag{1} [t_i,t_f]~\stackrel{q}{\longrightarrow}~\mathbb{R}^n.$$ The implicit time dependence would typically be different for another path.

II) In fact a (possibly virtual) path (1) is technically speaking not the input for a Lagrangian $L$. Rather the Lagrangian $$\tag{2} \mathbb{R}^n\times\mathbb{R}^n\times [t_i,t_f]~\stackrel{L}{\longrightarrow}~\mathbb{R}$$ is a function $$\tag{3} (q,v,t)~\mapsto~ L(q,v,t) $$ (as opposed to a functional) that only depends on

1. an instant $t\in[t_i,t_f]$,

2. an instantaneous position$^2$ $q\in\mathbb{R}^n$, and

3. an instantaneous velocity $v\in\mathbb{R}^n$;

not the past, nor the future.

Notice that we here use the symbol $v$ rather than the notation $\dot{q}\equiv \frac{dq}{dt}$. This is because the ability to differentiate $\frac{dq}{dt}$ would imply that we know (at least a segment of) a path (1) rather than just information about an instantaneous state $(q,v,t)$ of the system.

III) In contrast, the action $$\tag{4} S[q] ~:=~ \int_{t_i}^{t_f}dt \ L(q(t),\dot{q}(t),t)$$ is a functional (as opposed to a function) that depends on a (possibly virtual) path (1).

For more details, such as, e.g., an explanation how calculus of variations works, why $q$ and $v$ are independent variables in the Lagrangian (3) but dependent variables in the action (4), etc.; see e.g. this related Phys.SE post and links therein.

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$^1$ By the way, if the Lagrangian $L(q,v)$ has no explicit time dependence, then the energy

$$ \tag{5} h(q,v)~:=~v^i \frac{\partial L(q,v)}{\partial v^i}-L(q,v) $$ is conserved, cf. Noether's theorem.

$^2$ Here $q\in\mathbb{R}^n$ denotes an $n$-tuple, as opposed to eq. (1) where $q$ denotes a path/curve.

I) In the Lagrangian $L(q(t),\dot{q}(t),t)$, one must distinguish between implicit time dependence via the variables $q(t)$ and $\dot{q}(t)$, and explicit time dependence.$^1$

However, the implicit time dependence in the Lagrangian $L$ does only make sense in the context of a fixed (but arbitrary, possibly virtual) path $$\tag{1} [t_i,t_f]~\stackrel{q}{\longrightarrow}~\mathbb{R}^n.$$ The implicit time dependence would typically be different for another path.

II) In fact a (possibly virtual) path (1) is technically speaking not the input for a Lagrangian $L$. Rather the Lagrangian $$\tag{2} \mathbb{R}^n\times\mathbb{R}^n\times [t_i,t_f]~\stackrel{L}{\longrightarrow}~\mathbb{R}$$ is a function $$\tag{3} (q,v,t)~\mapsto~ L(q,v,t) $$ (as opposed to a functional) that only depends on

1. an instant $t\in[t_i,t_f]$,

2. an instantaneous position$^2$ $q\in\mathbb{R}^n$, and

3. an instantaneous velocity $v\in\mathbb{R}^n$;

not the past, nor the future.

Notice that we here use the symbol $v$ rather than the notation $\dot{q}\equiv \frac{dq}{dt}$. This is because the ability to differentiate $\frac{dq}{dt}$ would imply that we know (at least a segment of) a path (1) rather than just information about an instantaneous state $(q,v,t)$ of the system.

III) In contrast, the action $$\tag{4} S[q] ~:=~ \int_{t_i}^{t_f}dt \ L(q(t),\dot{q}(t),t)$$ is a functional (as opposed to a function) that depends on a (possibly virtual) path (1).

For more details, such as, e.g., an explanation how calculus of variations works, why $q$ and $v$ are independent variables in the Lagrangian (3) but dependent variables in the action (4), etc.; see e.g. this related Phys.SE post and links therein.

--

$^1$ By the way, if the Lagrangian $L(q,v)$ has no explicit time dependence, then the energy

$$ \tag{5} h(q,v)~:=~v^i \frac{\partial L(q,v)}{\partial v^i}-L(q,v) $$ is conserved, cf. Noether's theorem.

$^2$ Here $q\in\mathbb{R}^n$ denotes an $n$-tuple, as opposed to eq. (1) where $q$ denotes a path/curve.

Comments to the question (v1):

I) In the Lagrangian $L(q(t),\dot{q}(t),t)$, one must distinguish between implicit time dependence via the variables $q(t)$ and $\dot{q}(t)$, and explicit time dependence.$^1$

However, the implicit time dependence in the Lagrangian $L$ only makes sense in the context of a fixed (but arbitrary, possibly virtual) path $$\tag{1} [t_i,t_f]~\stackrel{q}{\longrightarrow}~\mathbb{R}^n.$$ The implicit time dependence would typically be different for another path.

II) In fact a (possibly virtual) path (1) is technically speaking not the input for a Lagrangian $L$. Rather the Lagrangian $$\tag{2} \mathbb{R}^n\times\mathbb{R}^n\times [t_i,t_f]~\stackrel{L}{\longrightarrow}~\mathbb{R}$$ is a function $$\tag{3} (q,v,t)~\mapsto~ L(q,v,t) $$ (as opposed to a functional) that only depends on

1. an instant $t\in[t_i,t_f]$,

2. an instantaneous position$^2$ $q\in\mathbb{R}^n$, and

3. an instantaneous velocity $v\in\mathbb{R}^n$;

not the past, nor the future.

Notice that we here use the symbol $v$ rather than the notation $\dot{q}\equiv \frac{dq}{dt}$. This is because the ability to differentiate $\frac{dq}{dt}$ would imply that we know (at least a segment of) a path (1) rather than just information about an instantaneous state $(q,v,t)$ of the system.

III) In contrast, the action $$\tag{4} S[q] ~:=~ \int_{t_i}^{t_f}dt \ L(q(t),\dot{q}(t),t)$$ is a functional (as opposed to a function) that depends on a (possibly virtual) path (1).

For more details, such as, e.g., an explanation how calculus of variations works, why $q$ and $v$ are independent variables in the Lagrangian (3) but dependent variables in the action (4), etc.; see e.g. this related Phys.SE post and links therein.

--

$^1$ By the way, if the Lagrangian $L(q,v)$ has no explicit time dependence, then the energy

$$ \tag{5} h(q,v)~:=~v^i \frac{\partial L(q,v)}{\partial v^i}-L(q,v) $$ is conserved, cf. Noether's theorem.

$^2$ Here $q\in\mathbb{R}^n$ denotes an $n$-tuple, as opposed to eq. (1) where $q$ denotes a path/curve.

Comments to the question (v1):

I) In the Lagrangian $L(q(t),\dot{q}(t),t)$, one must distinguish between implicit time dependence via the variables $q(t)$ and $\dot{q}(t)$, and explicit time dependence.$^1$

However, the implicit time dependence in the Lagrangian $L$ only makes sense in the context of a fixed (but arbitrary, possibly virtual) path $$\tag{1} [t_i,t_f]~\stackrel{q}{\longrightarrow}~\mathbb{R}^n.$$ The implicit time dependence would typically be different for another path.

II) In fact a (possibly virtual) path (1) is technically speaking not the input for a Lagrangian $L$. Rather the Lagrangian $$\tag{2} \mathbb{R}^n\times\mathbb{R}^n\times [t_i,t_f]~\stackrel{L}{\longrightarrow}~\mathbb{R}$$ is a function $$\tag{3} (q,v,t)~\mapsto~ L(q,v,t) $$ (as opposed to a functional) that only depends on

1. an instant $t\in[t_i,t_f]$,

2. an instantaneous position$^2$ $q\in\mathbb{R}^n$, and

3. an instantaneous velocity $v\in\mathbb{R}^n$;

not the past, nor the future.

Notice that we here use the symbol $v$ rather than the notation $\dot{q}\equiv \frac{dq}{dt}$. This is because the ability to differentiate $\frac{dq}{dt}$ would imply that we know (at least a segment of) a path (1) rather than just information about an instantaneous state $(q,v,t)$ of the system.

III) In contrast, the action $$\tag{4} S[q] ~:=~ \int_{t_i}^{t_f}dt \ L(q(t),\dot{q}(t),t)$$ is a functional (as opposed to a function) that depends on a (possibly virtual) path (1).

For more details, such as, e.g., an explanation how calculus of variations works, why $q$ and $v$ are independent variables in the Lagrangian (3) but dependent variables in the action (4), etc.; see e.g. this related Phys.SE post and links therein.

--

$^1$ By the way, if the Lagrangian $L(q,v)$ has no explicit time dependence, then the energy

$$ \tag{5} h(q,v)~:=~v^i \frac{\partial L(q,v)}{\partial v^i}-L(q,v) $$ is conserved, cf. Noether's theorem.

$^2$ Here $q\in\mathbb{R}^n$ denotes an $n$-tuple, as opposed to eq. (1) where $q$ denotes a path/curve.

Comments to the question (v1):

I) One must distinguish in the Lagrangian $L(q(t),\dot{q}(t),t)$ between implicit time dependence via the variables $q(t)$ and $\dot{q}(t)$, and explicit time dependence.$^1$

However, the implicit time dependence in the Lagrangian $L$ only makes sense in the context of a fixed (but arbitrary, possibly virtual) path $$\tag{1} q:[t_i,t_f]\to \mathbb{R}^n.$$ The implicit time dependence would typically be different for another path.

II) In fact a (possibly virtual) path (1) is technically speaking not the input for a Lagrangian $L$. Rather the Lagrangian $$\tag{2} L:\mathbb{R}^n\times\mathbb{R}^n\times [t_i,t_f]~\to~\mathbb{R}$$ is a function $$\tag{3} (q,v,t)~\mapsto~ L(q,v,t) $$ (as opposed to a functional) that only depends on

1. an instant $t\in[t_i,t_f]$,

2. an instantaneous position $q\in\mathbb{R}^n$, and

3. an instantaneous velocity $v\in\mathbb{R}^n$;

not the past, nor the future.

Notice that we here use the symbol $v$ rather than the notation $\dot{q}\equiv \frac{dq}{dt}$. This is because the ability to differentiate $\frac{dq}{dt}$ would imply that we know (at least a segment of) a path (1) rather than just an instantaneous position.

III) In contrast, the action $$\tag{4} S[q] ~:=~ \int_{t_i}^{t_f}dt \ L(q(t),\dot{q}(t),t)$$ is a functional (as opposed to a function) that depends on a (possibly virtual) path (1), i.e. both the past and the future.

For more details, and how calculus of variations work, see also e.g. this related Phys.SE post and links therein.

--

$^1$ By the way, if the Lagrangian $L(q,v)$ has no explicit time dependence, then the energy is conserved, cf. Noether's theorem.

Comments to the question (v1):

I) In the Lagrangian $L(q(t),\dot{q}(t),t)$, one must distinguish between implicit time dependence via the variables $q(t)$ and $\dot{q}(t)$, and explicit time dependence.$^1$

However, the implicit time dependence in the Lagrangian $L$ only makes sense in the context of a fixed (but arbitrary, possibly virtual) path $$\tag{1} [t_i,t_f]~\stackrel{q}{\longrightarrow}~\mathbb{R}^n.$$ The implicit time dependence would typically be different for another path.

II) In fact a (possibly virtual) path (1) is technically speaking not the input for a Lagrangian $L$. Rather the Lagrangian $$\tag{2} \mathbb{R}^n\times\mathbb{R}^n\times [t_i,t_f]~\stackrel{L}{\longrightarrow}~\mathbb{R}$$ is a function $$\tag{3} (q,v,t)~\mapsto~ L(q,v,t) $$ (as opposed to a functional) that only depends on

1. an instant $t\in[t_i,t_f]$,

2. an instantaneous position$^2$ $q\in\mathbb{R}^n$, and

3. an instantaneous velocity $v\in\mathbb{R}^n$;

not the past, nor the future.

Notice that we here use the symbol $v$ rather than the notation $\dot{q}\equiv \frac{dq}{dt}$. This is because the ability to differentiate $\frac{dq}{dt}$ would imply that we know (at least a segment of) a path (1) rather than just information about an instantaneous state $(q,v,t)$ of the system.

III) In contrast, the action $$\tag{4} S[q] ~:=~ \int_{t_i}^{t_f}dt \ L(q(t),\dot{q}(t),t)$$ is a functional (as opposed to a function) that depends on a (possibly virtual) path (1).

For more details, such as, e.g., an explanation how calculus of variations works, why $q$ and $v$ are independent variables in the Lagrangian (3) but dependent variables in the action (4), etc.; see e.g. this related Phys.SE post and links therein.

--

$^1$ By the way, if the Lagrangian $L(q,v)$ has no explicit time dependence, then the energy

$$ \tag{5} h(q,v)~:=~v^i \frac{\partial L(q,v)}{\partial v^i}-L(q,v) $$ is conserved, cf. Noether's theorem.

$^2$ Here $q\in\mathbb{R}^n$ denotes an $n$-tuple, as opposed to eq. (1) where $q$ denotes a path/curve.