# A Fundamental Arguments For The Form Of The Lagrangian, L=T-U

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1. Sep 5, 2016

### Gabriel Golfetti

I am trying to establish a Rationalist approach to Physics as a side project, and have taken Hamilton's Principle as one of the few postulates in my work. I've developed the concept enough to arrive at the usual stuff, like the Euler-Lagrange equations, Newton's First Law and Nöther's Theorem, but I still haven't been able to describe interactions between systems.

What do I mean by that? Suppose that we have some sort of entity that we know follows a certain Lagrangian $\mathcal{L}_{free}=T$ while it's in free motion. Now we create conditions for this entity to interact with something else (whatever it is, it doesn't matter), and it starts to follow a new Lagrangian $\mathcal{L}$. In most applications of the Lagrangian, we set $$\mathcal{L}=T-U,$$ Where $U$ is a function that depends only on the nature of the interaction and can be deduced experimentally through the equations of motion. My question now is about why we can do this. Why can we guarantee the term that describes the interactions is additive instead of completely altering the Lagrangian? Keep in mind that I'm not assuming Newton's Second and Third Laws hold.

Here are two of my failed attempts at resolving this issue:
• Interaction as a perturbation
Suppose we can modulate the strength of our interaction through some constant $\alpha$. Then we can say that our Lagrangian is of the form $$\mathcal{L}=T-\alpha U,\,U\equiv-\frac{\partial\mathcal{L}}{\partial\alpha},$$ And if we take $\alpha$ to be the 'correct' value for our equations of motion and then absorb the constant into the $U$ term, we have our new Lagrangian.
The problem with this approach is that some interactions can't be modulated to the proper intensity, e.g. very quick isentropic processes or anything with hysteretic behavior, and therefore this model is at least flawed.​

• Interaction as a constraint
This time, we assume that we can write our interactions as some sort of constraint $f=0$, and from this we find stationary points for the free action: $$S_{free}=\int T\,\mathrm{d}t, \text{subject to } f=0.$$ We know from Variational Calculus that this is equivalent to finding stationary points to the functional $$S=S_{free}-\lambda f=\int T-U\,\mathrm{d}t, \text{where }U\equiv-\lambda\frac{df}{dt},$$ For some lagrangian multiplier $\lambda$. This new functional will now be called our action.
Now, the thing that bothers me here is that I am still assuming that the free action is still 'stationary' despite having no idea how the interactions may affect the Lagrangian.​

If any of you Lagrangian enthusiasts could lend me a hand in getting to a fundamental reason for this general form of the functional, I'd be very grateful. Thanks anyway.

2. Sep 5, 2016

### ShayanJ

Are you saying physics is not rational?

3. Sep 6, 2016

### Gabriel Golfetti

Haha, no. When I say Rationalist I mean it in the philosophical sense, i.e. that the Universe has an underlying logical structure, and any of its properties can be deduced with reason. In the old days, it was opposed to Empiricism, which basically argues that we can only learn about the Universe with experiment. Nowadays, we realize that both of these go hand in hand, and I'm trying to focus on Rationalism for my approach to Physics.