ajb Posted December 24, 2014 Posted December 24, 2014 (edited) Has anyone see a Lagrangian of the following type before? If so where? Consider [math]\mathbb{R}^{2}[/math] equipped with local coordinates [math](x,y)[/math], then then Lagrangian I am interested in is of the form [math]\mathcal{L} = \dot{x}\dot{y} - y^{2}[/math] where I have set any necessary dimensional constants to 1. I won't say exactly where I have encountered Lagrangians like this quite yet, but I will later once I have finished the work. Right now I was wondering if anyone has seen something similar in any context? Technically it is of mechanical type but not what I would call standard. Edited December 24, 2014 by ajb
ALex7JA Posted January 14, 2015 Posted January 14, 2015 Yes. I know this. It's used in weapons technology. The code is a bit dubious in it's nature that you speak of though. Thoroughness.
ajb Posted January 15, 2015 Author Posted January 15, 2015 Yes. I know this. It's used in weapons technology. The code is a bit dubious in it's nature that you speak of though. Thoroughness. I was not expecting that... Anyway you can show that this Lagrangian is classically equivalent to [math]L' = \frac{1}{2}(\ddot{x})^{2}[/math] and we have a higher order Lagrangian. What at first seems strange to me is that y is like an auxiliary field, but it is dynamical. Up to a numerical factor and a sign y is the acceleration of x. I was wondering if some Lagrangian like that in terms of x and y naturally appears in some applications, maybe not as higher order Lagrangians are not too common and I would say not very well studied.
elfmotat Posted January 15, 2015 Posted January 15, 2015 The most unusual thing about it is that the equations of motion are third-order. The closest thing I've seen would be a classical self-force, like the Abraham-Lorentz Force (and all the problems associated with it), which could be expressed in a Lagrangian like the following: [math]L= (\dot{x})^2-k(\ddot{x})^2[/math]
ajb Posted January 16, 2015 Author Posted January 16, 2015 There are a few higher order Lagrangians discussed in the literature, these arise due to simplifications of the model and for sure cannot be fundamental as they break Newton's laws. However, it is not like the literature is full such examples.
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