# Time dependent lagrangian

1. Aug 3, 2011

### Petr Mugver

Suppose I have a mechanical system with l + m degrees of freedom and an associated lagrangian

$L(\alpha,\beta,\dot{\alpha},\dot{\beta},t)$

where $\alpha\in\mathbb{R}^l$ and $\beta\in\mathbb{R}^m$.
Now suppose I have a known $\mathbb{R}^l$-valued function f(t) and define a new lagrangian

$M(\beta,\dot{\beta},t)=L(f(t),\beta,\dot{f}(t), \dot{\beta},t)$

Do the equations that derive from M correctly describe the motion of the initial mechanical system, where the first l degrees of freedom are constrained to the motion f(t) (by means of an external force)?

2. Aug 3, 2011

### BruceW

$$\frac{\partial L}{\partial q} = \frac{d}{dt} \frac{\partial L}{\partial \dot{q} }$$
where q is any one of the coordinates of $\alpha$ or $\beta$ (i.e. there are l+m separate equations).
From this, I would assume that $M(\beta , \dot{\beta} , t)$ would be correct, because this simply gives the m equations, which correspond to $\beta$.
(I know I've not done a rigorous proof or anything, but this seems to make sense to me).

3. Aug 4, 2011

### Petr Mugver

Uhm, please don't let me write the formula, but when you take the derivative with respect to t of the momentum dM/dv, don't you get extra terms due to f(t) and df(t)/dt?

4. Aug 4, 2011

### Bill_K

Yes, this is correct. In fact it's done all the time: write down T and V as if the particles were entirely free, and then impose the constraints on them. The Lagrangian M you get from this will describe the motion of the constrained system.

What you cannot do is the other way around: trying to substitute f(t) into the equations of motion you originally derived from L.

5. Aug 4, 2011

### Petr Mugver

Yes, it sounds so obvious. I feel stupid now... :)