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What is the equation of motion for a single scalar field, which has a Lagrangian density in which the potential explicitly depends on time? For example: $$U(\phi,t)=\frac{1}{2}\phi^2 - \frac{1}{3} e^{t/T}\phi^3 + \frac{1}{8}\phi^4$$ where $T$ is a constant.

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1 Answer 1

It's just a Klein-Gordon equation with a RHS. Explicitly, the E-L equation for a scalar field is $$\partial_{\mu} {\partial {\mathcal L} \over \partial \phi_{,\mu}} - {\partial {\mathcal L} \over \partial \phi} = 0$$ so for your potential we have $$\square \phi + \phi - e^{t/T} \phi^2 + {1 \over 2} \phi^3 = 0$$

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Can u derive these equation from the first principle of variation of action.My point is that why coefficient in cubic phi term not differentiated with respect to time? –  Rakesh Jul 9 '11 at 7:00
@Rakesh Here's the derivation: physics.stackexchange.com/questions/1685/… I hope it also clarifies to you why one does not differentiate that term; the reason is no different from classical mechanics though: just take a Lagrangian $L = T - V(t)$ and see for yourself... –  Marek Jul 9 '11 at 17:31
@Rakesh: For a slightly more interesting model with explicit time dependence, consider moving the factor $e^{t/T}$ to instead multiplying the kinetic term... –  Qmechanic Jul 9 '11 at 19:13

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