Question

From the Lagrangian

L=−12ϕ□ϕ−12m2ϕ2 we get the equation of motion □ϕ=−m2ϕ and from it we get that the oscillation of ϕ in time is ω=k2+m2−−−−−−−√.

My question is: following the same reasoning, what are (1) the equation of motion and (2) the ω
of the following Lagrangian?

L=−12ϕ□ϕ−12m2ϕ2−14λϕ4 (where m2<0 and λ>0).

Answer 1

The equation of motion is □ϕ = −m^2ϕ - λϕ^3. The angular frequency is ω = √(m^2 + λϕ^2/m).

Step-by-step explanation:

To find the equation of motion for the given Lagrangian L=−12ϕ□ϕ−12m2ϕ2−14λϕ4, we need to use the Euler-Lagrange equations. Applying these equations to the Lagrangian, we obtain the equation of motion:

□ϕ = −m^2ϕ - λϕ^3

The equation of motion is □ϕ = −m^2ϕ - λϕ^3. The angular frequency is ω = √(m^2 + λϕ^2/m).

To find the angular frequency (ω), we can solve the equation of motion as a harmonic oscillator by assuming a solution of the form ϕ = A e^(iωt). Plugging this into the equation of motion, we get the angular frequency:

ω = √(m^2 + λϕ^2/m)