Translational invariance
For any isolated system (without external forces) of particles interacting only with each other, all the inter-particle forces should remain unchanged if the whole system is
translated (offset) en masse by some displacement $\myv{\epsilon}$. In particular, we'd expect
$$U(\myv{r}_1+\myv{\epsilon},....\myv{r}_n+\myv{\epsilon})-U(\myv{r}_1,...\myv{r}_n) = 0$$
The kinetic energies also should remain unaffected: $d\myv{\epsilon}/dt=0$ and so the velocities are not changed by a translation of the system: $$\frac{d}{dt}(\myv{r}+\myv{\epsilon}) = \frac{d}{dt}\myv{r} +\frac{d}{dt}\myv{\epsilon}=\frac{d}{dt}\myv{r}.$$
Thus, the Lagrangian should remain the same on translation by an arbitrary distance. Right? Let's work out a consequence of this...
If $\myv{\epsilon} = \epsilon\uv{x}$, then the overall change in the Lagrangian, which must sum to zero is only formally dependent on the $x$-coordinates of the $n$ particles in the system. In what follows, $i$ is a particle index... $$0 = \sum_{i}^n \frac{\del {\cal L}}{\del x_i} = \sum_i \frac{d}{dt}\frac{\del {\cal L}}{\del \dot{x}_i}.$$
Now, $\frac{\del {\cal L}}{\del \dot{x}_i}=(\myv{p}_i)_x$ is the momentum of particle $i$ in $x$-direction. Our sum, which is still zero, can now be written $$0=\sum_i \frac{d}{dt}(\myv{p}_i)_x = \frac{d}{dt}\sum (\myv{p}_i)_x = \frac{d}{dt}P_x$$
$P_x$ is the $x$-component of the total momentum. We could just as easily take the displacement in the $y$- or $z$-directions. So, total momentum is conserved.
This is Newton's law of inertia--that objects (more generally, the center of mass of any isolated system) will continue moving at a constant speed if not acted on by an outside force. It appears here as a consequence (many names for this:)
- of the symmetry of the Lagrangian with respect to translations in space.
- of the invariance of the Lagrangian with respect to translations in space.
- because physics works the same in Goshen and Berlin.
(Linear) momentum conservation is a consequence of the homogeneity of space.
In a similar vein, we can say that the angular momentum of a physical system is conserved when the Lagrangian ('Physics') is the same in all directions, or as a consequence of the isotropy of space.
Derivatives vs Partial Derivatives
A partial derivative of a function of several variables is the change in the function as one variable is changed while all the others are held constant.
The derivative of a function is the change in the function as one variable is changed.
Consider for example, $g(t) = t^2-3$, and $f(g,t) = 5g$. What are...
- $\frac{\del f}{\del g} = $
- $\frac{\del f}{\del t} = $
- $\frac{df}{dt}=$
This particular $f(g,t)$ has no explicit time dependence. (But it might still depend on $t$...).
Time invariance
In general, we write the Lagrangian as a function ${\cal L} = {\cal L}(q_1...q_m, \dot{q}_1...\dot{q}_m, t)$ of $m$ generalized coordinates. If the system consists of $n$ particles, then $m=3n$.
As time moves forward, the Lagrangian changes according to...(using the chain rule for functions of many variables) $$\begineq\frac{d}{dt}{\cal L} &=& \sum_\alpha^m \frac{\del {\cal L}}{\del q_\alpha}\frac{d}{dt}q_\alpha + \sum_\alpha^m \frac{\del {\cal L}}{\del \dot{q}_\alpha}\frac{d}{dt}\dot{q}_\alpha + \frac{\del {\cal L}}{\del t}\frac{d}{dt}t\\ &=&\sum_\alpha^m \frac{\del {\cal L}}{\del q_\alpha}\dot{q}_\alpha + \sum_\alpha^m \frac{\del {\cal L}}{\del \dot{q}_\alpha}\ddot{q}_\alpha + \frac{\del {\cal L}}{\del t},\endeq$$
where we're summing over a coordinate index $\alpha$ (not a particle index).
The partial derivative $\frac{\del {\cal L}}{\del t}$ is the change in the Lagrangian due to any explicit dependence of the Lagrangian on time.
But each of $q_{\alpha}=q_{\alpha}(t)$ and $\dot{q}_{\alpha} = \dot{q}_{\alpha}(t)$ is itself a function of time.
So $\frac{d}{dt}{\cal L}$ captures the rate of change of the Lagrangian due to both its explicit time dependence, as well as the implicit time dependence of the positions and velocities. *
We have called $p_{\alpha} = \frac{\del {\cal L}}{\del \dot{q}_{\alpha}}$ the generalized momentum. By the Euler-Lagrange equation, $$\frac{\del {\cal L}}{\del q_{\alpha}} = \frac{d}{dt}\frac{\del {\cal L}}{\del \dot{q}_{\alpha}} = \frac{d}{dt}p_{\alpha} = \dot{p}_{\alpha}.$$
Using these equations for $p_{\alpha}$ and $\dot{p}_{\alpha}$, we can re-write the time derivative, equation (4) as $$\begineq \frac{d}{dt}{\cal L} &=& \sum \dot{p}_{\alpha}\dot{q}_{\alpha} + \sum p_{\alpha}\ddot{q}_{\alpha} + \frac{\del {\cal L}}{\del t}\\ &=&\frac{d}{dt}\sum p_{\alpha}\dot{q}_{\alpha} + \frac{\del {\cal L}}{\del t}.\endeq$$
Or, re-arranging lightly: $$-\frac{\del {\cal L}}{\del t} = \frac{d}{dt}\left( \sum p_{\alpha}\dot{q}_{\alpha}-{\cal L}\right).$$
The equation above says that when there is no explicit time dependence of the Lagrangian (a condition that has been true of almost all of the Lagrangians we've dealt with so far), that the quantity in brackets called the "Hamiltonian" ${\cal H}$ is conserved:
$${\cal H} \equiv \sum_{\alpha} p_{\alpha} \dot{q}_{\alpha} - {\cal L}.$$
For one dimensional motion, there is only one term in the sum, and the Hamiltonian is $$\begineq{\cal H} &=& p \dot{x} - {\cal L}=mv\,v-(T-U(x))\\ &=&mv^2-\frac12mv^2+U=\frac12mv^2+U\\ &=&E.\endeq$$
${\cal H}$ is conserved when the Lagrangian does not depend explicitly on the time. When this is the case, the Hamiltonian is identical with the energy of the system. We can state this in more exalted terms...
Energy conservation occurs because time is homogeneous.
An example of a Lagrangian which *does* depend explicitly on time: Imagine that you are trying to write the Lagrangian for a charged particle in between the plates of a capacitor. Imagine further that the voltage difference on the plates varies in time $\Rightarrow$ the electric field between the plates varies in time $\Rightarrow$ the force on a particle, $\myv F(t)=q\myv E(t)$ varies in time.