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A particle is submitted to a time dependent force $$F(x,t)=\dfrac{k}{x^2}e^{-t/\tau}$$

Which is the Lagrangian of the particle?

I think that the force is derived from the potential $V$ and this potential has not explicit dependence of $\dot x$. So i can write
$$ \dfrac{d}{dt}\dfrac{\partial \mathcal L}{\partial \dot x} = m \ddot x$$

$$\mathcal L = T-\int \dfrac{\partial \mathcal L}{\partial x} dx$$ Then the lagrangian is $$\mathcal L = \dfrac{m}{2}\dot x^2 + \dfrac{k}{x}e^{-t/\tau}$$

Am i right?

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    $\begingroup$ Does the Euler-Lagrange equation coincide with the desired equation of motion? If so, what else do you need? $\endgroup$ Commented May 2, 2017 at 1:08
  • $\begingroup$ the statement does not say too much. I think that $F=\nabla V$ but have the time implicit. I just need the lagrangian. $\endgroup$ Commented May 2, 2017 at 1:21

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The only meaningful criterion is whether the Euler-Lagrange equation matches your desired equation of motion or not (regardless of time dependences or whatever). Since you have a candidate lagrangian already, it is then a straightforward calculation to see what Euler-Lagrange equation it predicts. If that gives the equation of motion you wanted, then you're good to go.

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In general, the Lagrangian is as simple as K-U. U is just minus the antiderivative of your force over x so you are correct although the term you wrote in the integral wasn't F (though I think that was just a slip)

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  • $\begingroup$ why that is not $F$, if the potential has not $\dot x$ explicitly then $\nabla V = -\dfrac{\partial \mathcal L}{\partial x}$. My problem is the time dependence. $\endgroup$ Commented May 2, 2017 at 1:27

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