# Time dependence of the Lagrangian of a free particle?

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?

• Thanks for the response. But I would also like some insight into my second question. Why do we treat velocity as independent of time even in the case where we have an interaction ? Time is homogeneous but we know that for a true trajectory (or motion) velocity is changing with time in this case. – singularity Mar 3 '15 at 20:33

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.

• Very interesting ! I am reading the link you have provided here as well. But one question that arises here is how would you differentiate between implicit time dependence and explicit dependence in the case where Lagrangian actually is explicitly time dependent ? How would we make the difference precise ? I think I will try to work out the math here. Thank you. – singularity Mar 3 '15 at 21:00
• @stringcosmologyapplicant To satisfy the minimum action we must find a relation between the variables of the Lagrangian. So, we admit that the position and velocity depend on time. But before we find this relation, we know nothing of a particular dependence of the velocity on time. – Sofia Mar 3 '15 at 21:21

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.

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.

• Thanks for the response. I am sorry but your last sentence is a bit confusing to me. Are you saying that arbitrary trajectories may have time dependence of velocity but since we have chosen a Lagrangian which has time-independent velocity, the final equation of motion would have no relationship between time and velocity. – singularity Mar 3 '15 at 20:45
• @stringcosmologyapplicant No, no. I am saying that before applying the least action principle and find the trajectory of the object, we have a Lagrangian dependent on velocity I the kinetic energy term, and position and (eventually) time in the potential energy. We have no idea how the velocity depends on time. There is a continuum of forms of dependence, because there is a continuum of forms of trajectories available. This is why, in the Lagrangian we take the velocity as a variable in itself. (I continue) – Sofia Mar 3 '15 at 21:06
• @stringcosmologyapplicant Only when we learn what is the trajectory, by choosing the path that minimizes the action, we learn that the velocity depends on time in a certain way. But, as long as we don't define a trajectory, i.e. $\vec r(t), \vec {\dot r(t)}$ what we know about the velocity? Do we know on what it depends? – Sofia Mar 3 '15 at 21:08
• So it seems it doesn't make sense to talk about time dependence before applying the least action principle, since we do not really know the actual time dependence and the actual trajectory. – singularity Mar 3 '15 at 21:36
• @stringcosmologyapplicant yes. Before imposing the condition of minimal action, the velocity is treated as an independent variable because at a given time $\tau$ between $t_1$ and $t_2$ and at an whatsoever place in space, $\vec r$, the velocity $\vec v$ can be 5m/s, 0.32m/s, -200m/s, God knows what and in which direction. It behaves as an independent variable. Only by imposing the condition of minimum action, will we find that it has to depend on time in a certain way. And the condition of minimum action leads to the Euler-Lagrange eqs., which lead us explicitly to the form of dependence. – Sofia Mar 3 '15 at 21:55