Time dependence of the Lagrangian of a free particle?

Question

I am working through Landau's book on Classical Mechanics. I understand the logic and physics of isotropy and homogeneity of space-time behind the derivation of the Lagrangian for a free particle, but I am confused regarding its time dependence. When we calculate the action as the integral of the Lagrangian for a wiggly trajectory, the velocity is obviously dependent on time and so should be the Lagrangian. Of course, we extremize the action to find the true trajectory of motion. But this time dependence of Lagrangian for wiggly trajectories is very confusing to me because this indicates to me that Lagrangian is dependent on time for a free particle. Where am I making the mistake here?

Also, when we have a particle in a position dependent potential, the velocity (and kinetic energy) is again dependent on time for any trajectory we choose and even for the true trajectory. But then again we write the velocity as independent of time in the Lagrangian. Why is that so?

Answer 1

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.

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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.

Answer 2

Think about an action principle as an abstract mapping of trajectories $\mathbf r = (\vec r(t), t_0, t_1)$ to some number $S(\mathbf r)$ which no longer is explicitly time-dependent.

Now, sometimes, the action can be computed from another function. The other function just has a bunch of parameters, which we can call $\{\alpha_i\}, ~ \{\beta_i\}, ~ \tau$. From the perspective of this function, those are just numbers.

We say that this function is a Lagrangian if we compute the action from it by doing a time integral:

$ S(\mathbf r) = \int_{t_0}^{t_1} dt~ L(\{\alpha_i = r_i(t)\}, ~ \{\beta_i = \dot r_i(t)\},~ \tau = t) $

In other words, the function doesn't "know about" physics just yet; the physics is imbued by the least-action principle. The Lagrangian is just a function which takes a bunch of variables and produces a number. Sure, if you "know" that this time integral will be done you can apply that function to a real particle's trajectory and then you'd find that it was time-dependent, but by itself, it's just a function, used by the action principle for all trajectories, to compute their action.

So when we skip writing the parameters as $\{\alpha_i\}, ~ \{\beta_i\}, ~ \tau$ and instead write them as $\{r_i\}, ~ \{v_i\}, ~ t$ that is just us reusing these symbols as names for the numbers that we're eventually going to put into the function. This "punning" (as the programming language Haskell calls it) is a little confusing at first but it helps us remember what was what when we later go to substitute the values in.

Answer 3

You say " When we calculate the action as the integral of the Lagrangian for a wiggly trajectory, the velocity is obviously dependent on time and so is the Lagrangian".

How exactly is the velocity dependent on time? Before applying the least action principle and find the trajectory of the object, we have a Lagrangian dependent on velocity (through the kinetic energy term), and position and (eventually) time (through the potential energy). We have no idea how the velocity depends on time. There is a continuum of forms of dependences, because there is a continuum of forms of trajectories that the object may follow in principle. This is why, before minimizing the action, we take in the Lagrangian the velocity as a variable in itself.

We don't know the trajectory before minimizing the action, s.t. we have no relationship between velocity and time.