Lagrangian?

kiwiheretic

This is just a made up problem of a fake trajectory to help me understanding the Lagrangian

I am trying to compute the action of this graph if m = 1 then I assume that the Lagrangian is the integral of T-V = 1 - 0.5t^2 for 0 <= t < 1 and T = 1.5 - (.5 - 0.5t^2) for 1 < t <= 2.

This is assuming that T is the slope of the line squared and that V is mx and I'm basically absorbing g into the units for x as V is traditonally = mgh.

Does this look like I am approaching this in the right way? I am probably getting myself confused here as I am really not sure how to integrate this.

Pmb

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I am trying to compute the action of this graph if m = 1 then I assume that the Lagrangian is the integral of T-V ...
No. The Lagrangian, L, is T - V, not the integral of it. Where did you get the notion it was the integral of T - V?

kiwiheretic

I probably used a wrong word. I meant the action.

In Taylors book chapter 7 equation 7.8

$image=https://latex.codecogs.com/png.latex?%5Clarge%20S%20%3D%20%5Cint_%7Bt_1%7D%5E%7Bt%5E2%7D%20%5Cmathcal%7BL%7D%20dt&hash=8fa43f0f16ec8f955a726d6a30e290d2$

I really just made up a simple example to see if I could calculate the action to see if I was understanding it correctly.

The basic idea is that I compute T-V for all points $t_1 < t < t_2$ (or at least a sampling of them). If x = f(t) and $v = \dot{x}$ then given that $T = \frac{1}{2} m v^2$ then this should just be the slope of the line on the x-t graph at that point squared. V is of course the potential defined at every point x.

However I am noticing some peculiarities, namely that x=0 is a solution and therefore it's going to be interesting finding a stationary functional that's not that.

I'm currently working on a Jupyter notebook to see.

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