# Assumptions reg. Kinetic energy and Potential energy in the Lagrangian formulation

I have recently been introduced to Lagrangian mechanics. My previous exposure to Lagrangian math has been in the form of optimizing constrained functions using Lagrange multipliers.

I get the math behind the Euler-Lagrange equations. I understand the proof behind conservation of energy using these equations assuming time-translation invariance. I also believe I understand that symmetries will always result in some conserved quantity. No challenges there.

But to my untrained eye, it appears there are a few assumptions we make in the process, and I'm having some trouble understanding why these assumptions are true. Or, maybe my assumptions are wrong (in which case, I don't get the math after all)?

1. We appear to know that the principle of stationary action is true for the universe. E.g. there is an excellent answer here about why the Principle of Stationary action is true. I am convinced.

2. We define Kinetic energy of the system to be to be $$T = \sum f(\mathcal{P}_n(\dot{q}))$$ where $$\mathcal{P}_n$$ is a polynomial of some degree.

3. We define $$V(q)$$ to be the potential energy of the system.

4. We assume the system is time-translation invariant.

5. We define Lagrangian to be $$L(\dot{q}, q) = T(\dot{q})-V(q)$$.

Questions:

1. Why is T only a function of $$\dot {q}$$? How do we know for sure?

2. Why is V only a function of $$q$$? How do we know for sure?

I've been trying to understand why these assumptions are true for a few days now, and I find myself going in circles. Can someone give me an intuition (or references) for why these assumptions are true?

• I might recommend that you split this up into a few different questions; the format here is not amenable to answering several different questions in one post. Perhaps three posts, one dealing with questions 1 & 2, one with question 3, and one with question 4. Apr 17, 2020 at 12:02
• Also, concerning questions 1 & 2: check out the Lagrangian for a charged particle in a magnetic field, which is not of the form $T - V$ at all but generates the correct equations of motion. I'll try to write up a discussion of it as a full answer later. Apr 17, 2020 at 12:05
• Regarding 1, 2 and 4 c.f. sections 1 - 5 and sec. 25 of Landau's Mechanics. Apr 17, 2020 at 12:17
• I really don't know why this question was closed, but I again recommend you check the reference I gave above, also the first section (and chapter 2) of volume 2. Apr 17, 2020 at 14:00
• @MichaelSeifert I updated the question per your recommendation. Can this question be opened again? Apr 22, 2020 at 11:58

Those assumptions are not enforced by nature, but occasionally by mathematical convenience. A potential that only depends on position gives a conservative vector field, but we allow for vector potentials too, as pointed out in the comments.

As for $$T$$, the assumption $$T(\dot q)$$ is quite a strong one. A more physical one is to assume that $$T$$ is quadratic in $$\dot q$$, with the quadratic form allowed to depend on $$q$$. Surely, if you take the standard kinetic energy in Euclidean coordinates, you will get position variables in the mix as soon as you switch to polar coordinates, but you'd still have a quadratic form in $$\dot q$$.

Regarding question 3. I don't see why the universe is involved in the discussion. If you're hinting at the fact that one assumes the Lagrangian that describes the dynamics of the universe to be of the form suggested in the OP, then we have just seen that that's certainly not the case, even for trivial mechanical systems.

• I meant the Kinetic energy is defined as some polynomial of velocity. Updated the question accordingly. Apr 22, 2020 at 12:28

It seems prudent to point out some counterexamples:

• In general the Lagrangian $$L$$ does not need to be of the form $$T-U$$, cf. this Phys.SE post.

• In general, the Lagrangian $$L(q,\dot{q},t)$$ could depend explicitly on time $$t$$, e.g. if there are external forces/sources, c.f. this Phys.SE post.

• In general the potential $$U(q,\dot{q},t)$$ can depend on velocity $$\dot{q}$$, cf. e.g. my Phys.SE answer here.

• In general the kinetic term $$T(q,\dot{q},t)$$ can depend on position $$q$$. Consider e.g. the kinetic energy of a non-relativistic point particle in spherical coordinates.