Undergrad What does it mean for a Lagrangian to have "explicit" time dependence?

[aliens123]

Suppose I had a Lagrangian
$$L = q+ \dot{q}^2 + t.$$
This has explicit time dependence. Now consider another Lagrangian:
$$L = q+ \dot{q}^2 .$$
Which has no explicit time dependence. But after solving for the equations of motion, I get $$\dot{q} = t/2 + C.$$
So I could now write my Lagrangian as:
$$L = q+ (t/2 + C)^2 .$$
Now it has explicit time dependence.

 

[jedishrfu]

We have an old thread on the topic:

https://www.physicsforums.com/threa...me-dependence-in-lagrangian-mechanics.164414/

perhaps it can explain it best to you.

and this quora article that mentions the ∂L/∂t=0 means the Lagrangian is not explicitly dependent on time.

https://www.quora.com/In-Lagrangian...dence-refer-to-their-mathematical-definitions

 

[PeroK]

Suppose I had a Lagrangian
$$L = q+ \dot{q}^2 + t.$$
This has explicit time dependence. Now consider another Lagrangian:
$$L = q+ \dot{q}^2 .$$
Which has no explicit time dependence. But after solving for the equations of motion, I get $$\dot{q} = t/2 + C.$$
So I could now write my Lagrangian as:
$$L = q+ (t/2 + C)^2 .$$
Now it has explicit time dependence.

The Lagrangian is not an equation of motion. The Lagrangian is used to obtain the equations of motion. You cannot substitute an equation of motion back into the Lagrangian.

The foundation of the Lagrangian approach is to treat $q$, $\dot q$ and $t$ as independent variables and not as dynamic quantities that are related to each other.

 

[vanhees71]

The two Lagrangians are equivalent and thus describing the same physical system. The equations of motion are the same. It's very easy to show by explicitly evaluating the Euler-Lagrange equations that any two Lagrangians $L$ and $L'$ are equivalent iff
$$L'(q,\dot{q},t)=L(q,\dot{q},t)+\frac{\mathrm{d}}{\mathrm{d} t} \Omega(q,t),$$
for an arbitrary function $\Omega$.

That's equivalent to the demand that the variation $\delta S$ of the action is unchanged, which of course implies that the Euler-Lagrange equations are the same.