0
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ Does the Euler-Lagrange equation coincide with the desired equation of motion? If so, what else do you need? $\endgroup$ – Emilio Pisanty May 2 '17 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. $\endgroup$ – Gabriel Sandoval May 2 '17 at 1:21
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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)

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Gabriel Sandoval May 2 '17 at 1:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.