# Lagrangians NOT of the form $T-U$

My Physics teacher was reluctant to define Lagrangian as Kinetic Energy minus Potential Energy because he said that there were cases where a system's Lagrangian did not take this form. Are you are aware of any such examples?

Update: Here I'm of course assuming that $T$ and $U$ stands for the kinetic and the potential energy, respectively. Also:

1. adding a total time derivative term to the Lagrangian, or

2. scaling the Lagrangian with a non-zero multiplicative constant

do not change the Euler-Lagrange equations, as Dilaton and dmckee points out in the comments. Needless to say that I'm not interested in such trivial modifications (1&2).

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How did he define it? –  DJBunk Jan 13 at 2:49
Different kinds of terms can be added to a $T-U$ Lagrangian, as long as they do not change the corresponding equations of motion. –  Dilaton Jan 13 at 10:38
Here is another example. –  Qmechanic Jan 13 at 12:41
In a classical mechanics course, the nonconstraint forces are conservative and $L$ is definitely always $T-V$. The whole point of Lagrangian mechanics is to get at a minimal set of differential equations for the system as easily as possible. Definitions like "it is some function that gets the right equations blah blah blah" are useless, because you can never solve any problems with them. And if a method doesn't make predictions, it's not physics. –  Chris White Jan 13 at 17:39
It should be clear that you can multiply the Lagrangian by any non-zero constant (even a dimensional one) and obtain another Lagrangian, but I don't really consider that an enlightening case. –  dmckee Jan 14 at 12:54
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For a relativistic free particle you would think that the Lagrangian would be like $$\tag{1} L ~=~ T ~=~ E-E_0~=~(\gamma -1)m_0c^2. \qquad(\leftarrow\text{Turns out to be wrong!})$$ This is not the case! Instead it is $$\tag{2} L ~=~ -\gamma^{-1}m_0c^2.$$ These two functions look like

and are not the same. This choice (2) of kinetic term gives a canonical momentum

$$p~:=~\frac{\partial L}{\partial v}~=~\gamma m_0v,$$

as it should be.

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The action integral can be in "Jacobi action" form, which looks like:

$$S = 2\int^{B}_{A}\sqrt{(E-V)T}\,\mathrm{d}t$$

where usually $E$ is constant, $V=V(x)$ is the potential energy, and $T=2m$ is the kinetic energy.

For more on this, see:

1. Brown, J. D. and J. W. York (1989). "Jacobi’s action and the recovery of time in general relativity". Physical Review D40, 3312–3318. doi:10.1103/PhysRevD.40.3312.
2. Lanczos, C. (1970). The Variational Principles of Mechanics. University of Toronto Press, Toronto.

There are many other versions for deriving the equations of motion from variational calculus, see:

1. Spivak, M. (2010). Physics for Mathematicians, Mechanics I. Publish or Perish.
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In one of his classical mechanics lectures (I believe the latest set), Leonard Susskind answered a similar question by saying (and I cannot directly quote because I don't have the video in front of me) that Lagrangians are simply functions that lead to the correct equations of motion. I will add that those equations of motion can be solved and the resulting behavior compared to Nature as a test of correctness. Susskind went on to day that there is no rule that a system's Lagrangian must be T - U and that there can be "cross terms" that describe certain interactions. He went further to say something that really stuck with me, and that is when we're learning calculus, we never ask, "From where do we get the functions that we're learning to analyze?" We basically make them up or guess them or deduce them from observed behaviors (in physics, anyway). That statement seemed rather profound to me.

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As far as I know, in classical mechanics $L$ is defined exactly as difference between kinetic and potential energy. Conversely, it's the Hamiltonian that not alway equals $T+U$, and should be defined as as Legendre transform of Lagrangian.

In more complicated models, as in field theory, the Lagrangian could be more complicated. This is because Lagrangians, as Hamiltonian operators in quantum mechanics, are not determined by a universal rule or by a theorem. They are chosen only because they work, i.e. because of an analogy with classical mechanics, or because they lead to physically verified Euler equations. In this case, there is no special reason for which a lagrangian should be separable in two distinct $U$ and $T$ terms.

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## Definition of the lagrangian function

According to Landau-Lifchitz course, the definition of the least action principle contains two substantial points.

• First it tells us that any mechanical system is fully characterized by one function that depends on generalized coordinates, on the first time derivative of the generalized coordinates and on the time. Such a function is called the lagrangian.

• The second point deals with the minimization problem itself. The motion of the system satisfies the following. Considering two distinct instants and the associated generalized coordinates that describe the position of the system at those two instants.

Between those two points the motion is made such as the integral of the lagrangian function between those two instants is minimum.

From there you can get the Lagrange's equation. Nothing is said about $L = T-U$.

## Expression of the lagrangian for a free particle

Considering a free material point we choose to describe the motion in a well specific kind of frame. A frame where the space can be considered homogeneous, isotropic and where the time uniform seems to be the wisest choice. Assuming such a frame exists (it is called a galilean frame of reference) what would be the form of the lagrangian?

Because the space is homogeneous the lagrangian can't contain any term involving the generalized coordinates. In other words the laws of motion cannot depend on where the system actually is. Because the time is also homogeneous, we get the same conclusion, the time cannot explicitly appear in the lagrangian.

The space is also isotropic, it means that the laws of motion cannot depend on the direction of the motion in the space. Then the lagrangian only depends on the norm of the speed and thus not on the direction of the speed vector. Then the lagrangian function only depends on the absolute value of the speed or on the square of the speed vector. $L = a v^2$.

If you put this form in the Lagrange equation's you'll get that $v^2$ is a constant independent of time. Then you obtain the first Newton's law. Pursuing the reasoning with the study of two galilean frames moving from one to another will end on L proportional to the square of speed.

## General expression of the lagrangian

Consider an isolated system constituted of several particles. You can discribe the interactions between all particles with a function that depends on the position of each particle only. You can call this function $-U$.

It's important to see why this function cannot depend on time. In classical mechanics we consider that the interaction propagates itself instantaneously from one particle to another. Then the time cannot explicitly appear int this -U function.

Hence the general form of the lagrangian function is $L = T-U$. Using the uniformity of time and Lagrange's equations you will be able to find that a certain quantity doesn't depend on time : $$E=\sum_i \dot{q_i} \frac{\partial L}{\partial \dot{q_i}} - L$$ Using the form $T-U$ of the lagrangian, the above relation and the Euler's homogeneous function theorem you will get : $$E = T + U$$ Now, and only now, you can say that the total energy of the motion is the sum of two distinct terms. The first one only depends on velocity and is called kinetic energy. The second term only depends on position and it is called potential energy.

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Good review, but not directly answer the question. I am not quite sure the last sentence "Now, and only now". For Euler's homogeneous, we definitely have E=T+U, but I am not sure for the reverse direction –  hwlau Jan 14 at 1:40
The point was to show the way to obtain $L=T-U$ as an expression of the lagrangian. This last form highly depends on what assumptions we made, and one in particular : we only consider conservative forces, i.e. the total energy is always conserved. –  ChocoPouce Jan 14 at 22:15

Just a pair of remarks. The second is the most interesting, in my view.

(1) The Lagrangian of a charged particle in an assigned electromagnetic field still has a Lagrangian ${\cal L}= T-U$, but here $U$ is not a standard position-depending function, since it generally depends also on $\dot{q}$ and $t$ as is well known (see Jackson's textbook, for instance).

The difference between the structure of $T$ and $U$ is now that the dependence of $U$ on $\dot{q}$ is of the first order instead of the second one as in $T$. Otherwise the determinism (normality'' of Eulero-Lagrange equations) could be violated. However one cannot think of $U$ as a potential energy. The same structure of $U=U(t,q, \dot{q})$ arises if one includes in ${\cal L}$ inertial forces when working in a generic non-inertial reference frame.

(2) Consider a classical particle on the real line immersed in a liquid giving rise to a friction force $-\gamma v$, with $\gamma>0$ constant. We can also suppose that there is a positional force with potential energy $U=U(x)$. $m>0$ is the mass of the particle and we use its coordinate $x$ as Lagrangian coordinate. This system is not invariant under time reversal, however there is a Lagrangian for this system:

$${\cal L}(t,x, \dot{x}) = e^{\gamma t/m} \left(\frac{1}{2}m \dot{x}^2 -U(x)\right)\:.$$

Indeed, it immediately produces the correct Newtonian equation:

$$m \frac{d^2x}{dx^2} = -\frac{dU}{dx} - \gamma \frac{dx}{dt}\:.$$

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To derive the field equations of general relativity (in vacuum), the Lagrangian density is simply the Ricci scalar, which measures deviations from flat space-time. This is a good example of a Lagrangian that has no real "energy" interpretation: in vacuum there is clearly no energy in classical mechanics!

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The point is rather subtle for a non-mathematically literate physicist, since the distinction is quite technical. According to Arnold (see References), we give the following definitions.

Definition. Let $M$ be a differentiable manifold, $TM$ its tangent bundle and $L : TM \to M$ a differentiable application. An application $\gamma : \mathbb R \to TM$ is a motion in a lagrangian system with configuration manifold TM and lagrangian function $L$ if and only if $\gamma$ is extremal for the functional

$$\Phi(\gamma) = \int_{t_0}^{t_1} \mbox{dt} \, L(\dot{\gamma}).$$ $\dot{\gamma}$ is said velocity vector, $\dot{\gamma}(t) \in TM_{\gamma(t)}.$

Local coordinates $q_1, \dots, q_n$ of the point $\gamma(t)$ evolve according to the Euler-Lagrange equation

$$\frac{\partial L}{\partial q} = \frac{\mbox{d}}{\mbox{d}{t}}\frac{\partial L}{\partial \dot{q}} .$$

Now, suppose $M$ is a riemannian manifold, i.e. a couple $(M, g)$, with $M$ differentiable manifold and $g$ a positive-definite quadratic form, usually indicated as $\langle \cdot , \cdot \rangle$. In this case, and only in this case, we can define a kinetic energy as is usually meant:

Definition Let $M$ be a riemannian manifold. A quadratic form $K = \frac{1}{2}\langle v, v \rangle$, where $v \in TM_x$, defined on all tangent bundles is called kinetic energy. We say that $U$ is a potential energy if and only if $U : M \to \mathbb R$ is a differentiable function.

Definition. A lagrangian system on a riemannian manifold is said natural if and only if $L = K - U$, for some $K$ and $U$ previously defined.

In classical mechanics, one deals with riemannian manifolds all the time (apart from "pathological" situations), so he does not care about the distinction. If fact, in basic courses that a problem never arises. But it should be pointed out (by teachers, I mean) that Minkowski space $\mathcal M^4$ of special relativity is not a riemannian manifold, actually it is a pseudo-riemannian one (the metric is not positive-definite), so the definition of lagrangian must be taken with care. It is clear that the situation in general relativity is even more "dramatic" and defining a lagrangian is a non-trival problem.

The most widely known example of such a lagrangian is, I think, that of a free particle in special relativity: $L = -mc^2 \sqrt{ 1 - \frac{v^2}{c^2} }$. (See Goldstein)

References. V.I. Arnold, Mathematical methods of classical and celestial mechanics, chapter IV.ù H. Goldstein, C. Poole, J. Safko, Classical mechanics, 3d edition, Par. 7.9.

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I always considered the canonical example to be the lagrangian for a point charge (with charge $q$ and mass $m$) in an external EM field: $L(\vec{x},\vec{v}) = \frac{1}{2} m v^2 -q \phi + q \vec{A}(\vec{x}) \cdot \vec{v}$, where $\phi$ is the scalar potential for the electric field, and $\vec{A}$ is the vector potential for the magnetic field.