To summarize my question first,
Given a classical mechanics Lagrangian, $L=L(x(t), \dot{x}(t); t)$,
Why the conservation law for time $t$-translational invariant system, under time variance $\delta t$, has $\delta L\vert_{\text{time trans}} \neq 0$ in general? But why the conservation law for space $x$-translational invariant system, under space variance $\delta x$, has $\delta L\vert_{\text{space trans}} = 0$ in general?
Why for deriving the conservation laws for time $t$-translational invariant system, we directly apply $\frac{\partial L}{\partial t}=0$? But why for deriving the conservation laws for space $x$-translational invariant system, we do not directly apply $\frac{\partial L}{\partial x}=0$ but use $\delta \dot{x}=\delta \ddot{x}=\dots=0$ to derive $\frac{\partial L}{\partial x}=0$?
PLEASE try to answer my question in a fundamental level, instead of the technical level. But please be explicit in your answer.
p.s. I tried to derive the Conservation Laws myself based on my own logic, but you can compare with Chapter 7.9 of Thornton-Marion Edition 5.
Details
1. Energy conservation:
If the system is time translational invariant, given a classical mechanics Lagrangian, $L(x(t), \dot{x}(t); t)$, I follow @Richard Myers previous advice, let us think about the process that we redefine a new $L'$ such that we assign $L(x(t+\delta t), \dot{x}(t+\delta t); t+\delta t)$ at the time $t+\delta t$ to the new $L'$ at the time $t$: $$ L'(x(t), \dot{x}(t); t) =L(x(t+\delta t), \dot{x}(t+\delta t); t+\delta t) \equiv L(x(t), \dot{x}(t); t) + \frac{\partial L}{\partial x} \delta x + \frac{\partial L}{\partial \dot{x}} \delta \dot{x} +\frac{\partial L}{\partial t} \delta t $$ $$ =L(x(t), \dot{x}(t); t) + (\frac{\partial L}{\partial x}\dot{x} + \frac{\partial L}{\partial \dot{x}}\ddot{x} +\frac{\partial L}{\partial t}) \delta t $$ If the system is time translational invariant, we interpret that the $L$ has no explicit time-dependent but only implicit time dependent (but why?) so $$\boxed{\frac{\partial L}{\partial t}=0},$$ we plug in to get $$ \Rightarrow \boxed{\delta L\vert_{\text{time translation}} \equiv L'(x(t), \dot{x}(t); t)-L(x(t), \dot{x}(t); t) =(\frac{\partial L}{\partial x}\dot{x} + \frac{\partial L}{\partial \dot{x}}\ddot{x} ) \delta t} $$ My question is that WHY this quantity is nonzero in general? $$ \boxed{\delta L\vert_{\text{time translation}} \neq 0} $$ so we cannot claim (why not?) $(\frac{\partial L}{\partial x}\dot{x} + \frac{\partial L}{\partial \dot{x}}\ddot{x})=0$, but we use the Lagrangian Eq Of Motion (EOM) (for time implicit dependent $L$) to write $\frac{\partial L}{\partial x}= \frac{d}{dt}\frac{\partial L}{\partial \dot{x}}$,so $$ \Rightarrow {\delta L\vert_{\text{time translation}} =(\frac{d}{dt}\frac{\partial L}{\partial \dot{x}}\dot{x} + \frac{\partial L}{\partial \dot{x}}\ddot{x} ) \delta t=\frac{d}{dt} (\frac{\partial L}{\partial \dot{x}}\dot{x} ) \delta t} $$ Then my understanding is that the total time derivative of $L$ becomes $$ \Rightarrow\boxed{\frac{dL}{dt} = \frac{\delta L\vert_{\text{time translation}}}{\delta t} =(\frac{d}{dt}\frac{\partial L}{\partial \dot{x}}\dot{x} + \frac{\partial L}{\partial \dot{x}}\ddot{x} ) \delta t=\frac{d}{dt} (\frac{\partial L}{\partial \dot{x}}\dot{x} )\neq 0} $$ $$ \Rightarrow\boxed{ \frac{d}{dt} (\frac{\partial L}{\partial \dot{x}}\dot{x} -L ) =\frac{d}{dt} H= 0 } $$ where $H$ is the Hamiltonian.
2. Momentum conservation:
In contrast, as we will see below, we have instead $$ \boxed{\delta L\vert_{\text{space translation}} = 0} $$
To be more explicit, let us think about the process that we redefine a new $L'$ such that we assign $L(x+\delta x, \dot{x}+\delta \dot{x} ; t)$ at the space $x+\delta x$ to the new $L''$ at the time $x$: $$ L''(x(t), \dot{x}(t); t) \equiv L(x+\delta x, \dot{x}+\delta \dot{x} ; t) =L(x(t), \dot{x}(t); t) + \frac{\partial L}{\partial x} \delta x + \frac{\partial L}{\partial \dot{x}} \delta \dot{x} $$ $$ \Rightarrow {\delta L\vert_{\text{space translation}} \equiv L'(x(t), \dot{x}(t); t)-L(x(t), \dot{x}(t); t) =( \frac{\partial L}{\partial x} \delta x + \frac{\partial L}{\partial \dot{x}} \delta \dot{x})} $$ It turns out that $\delta L\vert_{\text{space translation}}=0$ in general, but WHY? Also, if the system is space translational invariant, we do NOT interpret directly that the $L$ has $\frac{\partial L}{\partial x}=0$. BUT we (Thornton-Marion) claim that $$ \text{$\delta x$ has no explicit or implicit $t$ dependence} $$ So $$\delta\dot{x}=0$$ and we use $\delta L\vert_{\text{space translation}}=0$ and the EOM $\frac{\partial L}{\partial x}= \frac{d}{dt}\frac{\partial L}{\partial \dot{x}}$ to derive for any $\delta x$, $$ \boxed{\delta L\vert_{\text{space translation}}=0 \Rightarrow \frac{\partial L}{\partial x} =0 \Rightarrow \frac{d}{dt}\frac{\partial L}{\partial \dot{x}} = \frac{d}{dt} p=0} $$ which means the momentum conservation.
My questions are on the top.
PLEASE try to answer my question in a fundamental level, instead of the technical level. But please be explicit in your answer.
