I found this problem in Landau-Lifschitz vol.1 (Mechanics)
A simple pendulum of mass $m$, length $l$ whose point of support moves uniformly on a vertical circle with constant frequency $\gamma$.
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The problem is, of course, to find the Lagrangian. So I turned to Cartesian coordinates $x$ and $y$ in order to construct the problem in terms of $\phi$ and the rotation terms, i.e.
$$ x = a \cos{\gamma t} + l \sin{\phi}$$
$$ y = -a \sin{\gamma t} + l \cos{\phi}$$
I take the time derivative of each, square it, add it together, construct the kinetic and potential term, nothing fancy… (but here's the procedure anyway for the sake of completeness, maybe that's where the mistake is):
$$\dot{x} = -a \gamma \sin(\gamma t) + l \dot{\phi} \cos{\phi}$$
$$\dot{y} = -a \gamma \cos(\gamma t) – l \dot{\phi} \sin{\phi}$$
$$ \dot{x}^2 = a^2 \gamma^2 \sin^2(\gamma t) + l^2 \dot{\phi}^2 \cos^2{\phi} – 2al \gamma \dot{\phi} \sin{(\gamma t)} \cos{\phi}$$
$$ \dot{y}^2 = a^2 \gamma^2 \cos^2(\gamma t) + l^2 \dot{\phi}^2 \sin^2{\phi} + 2al \gamma \dot{\phi} \cos{(\gamma t)} \sin{\phi}$$
$$\dot{x}^2 + \dot{y}^2 = a^2 \gamma^2 + l^2 \dot{\phi}^2 + 2al\gamma\dot{\phi}\sin(\phi -\gamma t) $$
$$\text{also} \hspace{2cm} V=-mgy=-mg(-a \sin{\gamma t} + l \cos{\phi})$$
and I get this Lagrangian:
$$\mathcal{L} = \frac{1}{2}m(a^2 \gamma^2 + l^2 \dot{\phi}^2 + 2al\gamma\dot{\phi}\sin(\phi -\gamma t)) + mg(-a \sin(\gamma t) + l\cos{\phi})$$
Then I drop the terms which are total time derivatives, namely these:
$$ \frac{1}{2} ma^2 \gamma^2$$
$$-mga \sin(\gamma t)$$
That leaves me with this Lagrangian:
$$\mathcal{L}_{Me} = \frac{1}{2} m l^2 \dot{\phi}^2 + mal\gamma\dot{\phi}\sin(\phi -\gamma t) + mgl\cos{\phi}$$
But Landau gives this one:
$$\mathcal{L}_{Landau} = \frac{1}{2} m l^2 \dot{\phi}^2 + \color{red}{mal\gamma^2 \sin(\phi -\gamma t) }+ mgl\cos{\phi}$$
Am I missing something? Where did that red term come from?
Best Answer
Let's compare Landau's Lagrangian and the one given by $\mathcal L_{Me}$ in your question:
$$\mathcal{L}_{Landau}-\mathcal{L}_{Me}=mal\gamma^2\sin(\phi-\gamma t)-alm\gamma\dot\phi\sin(\phi-\gamma t)=\\ =mal\gamma\left((\gamma-\dot\phi)\sin(\phi-\gamma t)\right)=\\ =mal\gamma \frac{\text{d}}{\text{d}t}\cos(\phi-\gamma t)$$
Now the difference is obviously a total time derivative, thus you haven't made a mistake, just undersimplified the Lagrangian.