The lagrangian is always phrased as $L(t,q,\dot{q})$.

If you magically knew the equations $q(t)$ and $\dot{q}(t)$, could the Lagrangian ever be written only as a function of time?

Take freefall for example. $$y(t)=y_0 + v_0t -(1/2)gt^2$$ $$\dot{y}(t)=v_0 -gt$$

Can the Lagrangian now be written as: $$L=KE-PE=(1/2)m\dot{y}^2-mgy=(1/2)m(v_0 -9.8t)^2-mg(y_0 + v_0t -4.9t^2)$$

Now the Lagrangian is written only as a function of time, and we can put in a time and find out what the Lagrangian is at any point in the motion. Is this legitimate? Forgive me if this is simple.

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    $\begingroup$ To me it seems like you could do so - but of what use would it be? What you are trying to achieve is linked to wether you find it "legitimate", so maybe for us to unser the question, you could elaborate what you mean by "legimimate". $\endgroup$ Commented Aug 6, 2020 at 17:16
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    $\begingroup$ I was just wondering if my logic and math were accurate. It is probably not useful since if you already have q(t) you can track anything about the system you would like. $\endgroup$ Commented Aug 6, 2020 at 17:23
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    $\begingroup$ Related: physics.stackexchange.com/q/307794/50583 $\endgroup$
    – ACuriousMind
    Commented Aug 7, 2020 at 8:41
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    $\begingroup$ What is the use of the Lagrangian? Its use is that we can derive the equations of motion (and the integrals of motion) from it by studying how the action integral reacts to a variation of $q$, $\dot q$. If you fix $q$, $\dot q$ there's nothing to vary. So, what would the use of this "Lagrangian" be? Literally, you could evaluate the Lagrangian as you describe -- but it would be a useless object. It is interesting to ponder how and why this is different from the usual way we deal with functions, where we are interested in evaluating them but not so much in their functional form. $\endgroup$
    – tobi_s
    Commented Aug 7, 2020 at 13:30
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    $\begingroup$ I very much disagree there JonathanZ. I direct your attention to this glorious contraption. twitter.com/i/status/755139996399595520 $\endgroup$ Commented Aug 8, 2020 at 4:40

2 Answers 2

  1. Using the EL equations (or inserting their solutions) into the action functional typically destroys the stationary action principle. For examples, see this Phys.SE post.

  2. Concerning OP's title question: If a Lagrangian $L(t)$ only depends on time $t$, then it can be written as a total time derivative $L=dF/dt$, and that means it is equivalent to the trivial Lagrangian $L=0$, i.e. the EL equations are trivial.


When doing this, it removes one of the great powers Lagrangian mechanics: the Euler-Lagrange equations, which are

$$\frac{d}{dt} \frac{\partial L}{\partial \dot{q}} = \frac{\partial l}{\partial q}$$

Notice in the equations, we have partial derivatives. So if we made $L=L(t)$, then the Euler Lagrange equations would give you $0=0$ (a beautiful fact, but nonetheless not very helpful). Plus, the Euler-Lagrange equations are meant to give you equations of motion, so one probably wouldn't be able to substitute equations for $q(t)$ or $\dot{q}(t)$ anyways.

Edit: There is actually a more fundamental problem when you do this (and thank you so much to Qmechanic and ZeroTheHero for bringing this to my attention). If you simply plug back the motion into your Lagrangian and then try to apply the Euler-Lagrange equations again, you will more than likely get the wrong equation of motion. For example take the simple spring:

$$L = \frac{m\dot{x}^2}{2} - \frac{kx^2}{2} \implies m\ddot{x}=-kx \implies x=\sin(\sqrt{k/m}\cdot t)$$ (just one possible solution)

where we used the EL eqs. However, if we plug this back into our Lagrangian, we will not get the right equations of motion:

$$L = \frac{m\dot{x}^2}{2} - \frac{k\sin^2(\sqrt{k/m}\cdot t)}{2} \implies m\ddot{x} = 0$$(!!!)

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    $\begingroup$ One cannot solve for $\dot q$ and plug it back into the Lagrangian as the resulting EOM will not be correct. Thus there is everything wrong with doing this. See also the post by @Qmechanic below. $\endgroup$ Commented Aug 6, 2020 at 18:22

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