# Lagrangian 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?

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How did he define it? – DJBunk Jan 13 at 2:49
He struggled to define it concisely. He merely said that it was an abstract function of generalized coordinates, velocity and time that produced "sensible" equations of motion when Euler-Lagrange equations were calculated. I don't know if he was trying to prepare us for the generalization of the Principle of Least Action to other subject areas or whether he was referring to systems still in the realm of Classical Mechanics. – ZAC Jan 13 at 2:52
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
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 $$L = T = (\gamma -1)mc^2$$ This is not the case, it is $$L = -\gamma^{-1}mc^2$$ These two functions look like

and are not the same. This choice of kinetic term gives a canonical momentum of $p=\gamma mv$.

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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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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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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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## 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 doesn't 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

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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