Alternative notion of kinetic energy in Newtonian and Lagrangian mechanics?

In classical mechanics, kinetic energy, potential energy, and work are defined by the equations $$T = \frac{1}{2}m\vec{v}\,^{2},\, \qquad \vec{F} = -\nabla U, \qquad\text{ and }\qquad W = \int_{\text{path }P} \vec{F}\cdot d\vec{r},$$ respectively. Two of the most important consequences of these equations are conservation of energy ($$dE/dt = 0$$ where $$E = T+U$$) and the work-energy theorem ($$W = T_{2}-T_{1}$$).

However, all three equations involve an inner-product in some way or another. Thus, without changing coordinates, we can consider alternative notions of kinetic energy, potential energy, and work given by $$T = \frac{1}{2}m\vec{v}\,^{T}g\vec{v},\, \qquad \vec{F} = -\nabla_{g} U, \qquad\text{ and }\qquad W = \int_{\text{path }P} \vec{F}\,^{T} g \, d\vec{r}$$ where $$g = (g_{ij})$$ is any $$3\times 3$$ non-degenerate symmetric matrix (giving a non-degenerate bilinear form), $$g^{-1} = (g^{ij})$$, and $$\nabla_{g}U := g^{ij}\frac{\partial U}{\partial x^{i}} \, \vec{e}_{j}.$$ As far as I have checked, analogues of the conservation of energy and the work-energy theorem still hold.

In Newtonian mechanics, Newton's three laws make no reference to any metric. Newton's first law uses the notion of straight lines and constant speeds, but that's not enough to define a metric. Hence the new notion of kinetic energy doesn't contradict Newtonian mechanics.

In Lagrangian mechanics, there can be more than one Lagrangian describing the same system, so again our different notion of kinetic energy also won't run into problems there, assuming an appropriate potential energy function is chosen.

Questions.

• Is there anything special about using $$g = \text{diag}(1, 1, 1)$$ in defining kinetic energy, potential energy, and work instead of any other $$g$$?
• What observable, experimental difference does this make? Is there any experiment that rules these notions out as having the same validity as ordinary kinetic and potential energy?
• Do we need to appeal to other theories like thermodynamics and special relativity to make the determination of the "correct" notion of kinetic energy? If so, how?

A related post here asks about the necessity of kinetic energy and work in kinematics and dynamics. It seems that one reason for the possibility/ambiguity of non-standard kinetic energy is that they are not unavoidable. If one had the persistence, one could in principle work with forces only. Likewise, one could work with a non-standard notion of energy. However, I suspect this is true in dynamics only (hence my question about other theories).

• What is special about $g=\mathrm{diag}(1,1,1)$ is that it is, up to an overall factor, the only bilinear form that is invariant under the SO(3) group of spatial rotations. As long as you want the laws of nature to be isotropic, this is the only possible choice. Dec 26, 2021 at 19:10
• The metric g is function of x ?
– Eli
Dec 26, 2021 at 21:30
• @Eli In my post, no. I'm only thinking about a constant g. Dec 26, 2021 at 21:34

1. The standard Euclidean metric $$g=\mathrm{diag}(1,1,1)$$ is special in that it does not pick out any particular direction as special (it is isotropic). Using something else will make the spatial directions behave differently w.r.t the kinetic energy, why would you want that? We do not observe e.g. that it is harder to accelerate something in the x-direction than in the y-direction.

2. Changing the metric changes the Euler-Lagrange equations you obtain from the Lagrangian $$T-V$$ (or simply the forces $$-\nabla_gU$$ if you're doing Newtonian mechanics), i.e. you're not playing around just with abstract definitions of energy here. I'm puzzled why you would want to appeal to special relativity before appealing to experiment: You can just do experiments and observe that the standard Euclidean metric leads to the correct equations of motion.

• I edited my question to accommodate this post (the questions are now slightly different though). The Euler-Lagrange equations don't need to be changed actually. Sure, you can appeal to the fact that the Euclidean metric is isotropic, but even with the alternative notions of potential and kinetic energies, experimental predictions are exactly the same, as far as kinematics goes. The entire point of my post is that I don't immediately know of any experiment that rules these notions out. Dec 26, 2021 at 20:17
• @MaximalIdeal I don't think you showed what you think you showed. Your equations of motion are $m\ddot{x} = -\nabla_gU$, which is not the standard equation of motion when $g$ is not the Euclidean metric. That's exactly what I mean - a non-standard metric leads to non-standard equations of motion. Dec 26, 2021 at 20:22
• @MaximalIdeal Here is an experiment that can help you rule out a non-isotropic kinetic energy. Consider an elastic head-on collision of two balls with equal speeds. After the collision, the velocities of the balls will again be equal in magnitude and opposite in direction, but the direction itself is not fixed by conservation laws. If the kinetic energy were not isotropic as you suggest, then the speed of the balls after the collision would depend on the direction in which they move. You can actually check this at home. Dec 26, 2021 at 21:15
• @TomášBrauner I've thought about your comment and I found it tremendously helpful. Now you said "head-on" collisions which made me think you were thinking of 1D collisions, but I don't think 1D collisions could rule out non-isotropic KE, because, well, in 1D the kinetic energy can only be superficially rescaled with no physical difference. However, I found out that in 2D elastic collisions non-isotropic KE does not get conserved. Only the usual notion of KE is conserved. I take this as a valid and a strong argument in favor of standard KE, so this makes sense. Jan 7 at 5:16
• @MaximalIdeal Yes, I meant 2d collisions, perhaps just used a wrong term. "Head-on" was supposed to indicate that before the collision the two balls have momenta of equal magnitude and opposite direction. Otherwise, I find your argumentation a bit strange. In the absence of a potential, any kinetic energy bilinear in velocities (regardless of the choice of $g$) will be conserved as a consequence of invariance under time translations. The energy of the in- and out-states of the collision will therefore be independent of the force law as long as this is of sufficiently short range. Jan 7 at 21:54

It turns out I had an implicit assumption in my post that I falsely took for granted. This false assumption is that the potential energy function $$U$$ for the alternative notion of energy exists at all to begin with.

Let $$e$$ be the usual Euclidean inner-product, and let $$g$$ be any real inner-product that is not a scalar multiple of $$e$$. The main lesson is that just because $$\vec{F} = -\nabla_{e}U_{1}$$ for some function $$U_{1} = U_{1}(\vec{x})$$, it does not mean that $$\vec{F} = -\nabla_{g}U_{2}$$ for some function $$U_{2} = U_{2}(\vec{x})$$. Indeed, this seems to be the main reason why we favor $$e$$-kinetic energy (based on the Euclidean metric $$e$$) rather than $$g$$-kinetic energy (based on $$g$$).

Remember that in order for the potential energy $$U = U(\vec{x})$$ to be well-defined, the force field $$\vec{F}$$ must be conservative (i.e. the work must be path-independent). According to this post here, the force field is conservative if and only if $$\nabla\times\vec{F} = 0$$, and it turns out that almost all natural laws have zero curl with respect to metric $$e$$ but not any other metric $$g$$.

Now I also asked, how come Newton's three laws make no reference to energy, yet the usual notion of kinetic energy is special? The answer lies in the force laws that we use to model systems. Those force laws for which $$\nabla\times_{e}\vec{F} = 0$$ are usually not the force laws for which $$\nabla\times_{g}\vec{F} = 0$$. Think about Newton's law of gravitation or Coulomb's law as examples. Their curl is zero only with respect to the Euclidean metric $$e$$.

In fact, we can even consider all possible force laws for which $$\nabla\times_{e}\vec{F} = 0$$ and $$\nabla\times_{g}\vec{F} = 0$$ simultaneously both hold, and we find that the set of such possible forces is very limiting. According to my answer in this MSE question of mine, the matrix of $$g$$ must have three eigenvectors $$\vec{p}_{1}, \vec{p}_{2}, \vec{p}_{3}$$ by the spectral theorem for symmetric matrices, and any vector field $$\vec{F}(\vec{x})$$ for which $$\nabla\times_{e}\vec{F} = 0$$ and $$\nabla\times_{g}\vec{F} = 0$$ both hold must be of the form \begin{align}\tag{1} \vec{F}(\vec{x}) = f_{1}(\vec{x}\cdot\vec{p}_{1})\vec{p}_{1} + f_{2}(\vec{x}\cdot\vec{p}_{2})\vec{p}_{2} + f_{3}(\vec{x}\cdot\vec{p}_{3})\vec{p}_{3} \end{align} or \begin{align}\tag{2} \vec{F}(\vec{x}) = f_{1}(\text{proj }\vec{x})\vec{p}_{1} + f_{2}(\text{proj }\vec{x})\vec{p}_{2} + f_{3}(\vec{x}\cdot\vec{p}_{3})\vec{p}_{3} \end{align} where $$\text{proj }\vec{x}$$ projects $$\vec{x}$$ to $$\text{Span}(\vec{p}_{1}, \vec{p}_{2})$$ according to $$e$$-orthogonality or \begin{align}\tag{3} \vec{F}(\vec{x}) = f_{1}(\text{proj }\vec{x})\vec{p}_{1} + f_{2}(\vec{x}\cdot\vec{p}_{2})\vec{p}_{2} + f_{3}(\text{proj }\vec{x})\vec{p}_{3} \end{align} where $$\text{proj }\vec{x}$$ projects $$\vec{x}$$ to $$\text{Span}(\vec{p}_{1}, \vec{p}_{3})$$ according to $$e$$-orthogonality or \begin{align}\tag{4} \vec{F}(\vec{x}) = f_{1}(\vec{x}\cdot\vec{p}_{1})\vec{p}_{1} + f_{2}(\text{proj }\vec{x})\vec{p}_{2} + f_{3}(\text{proj }\vec{x})\vec{p}_{3} \end{align} where $$\text{proj }\vec{x}$$ projects $$\vec{x}$$ to $$\text{Span}(\vec{p}_{2}, \vec{p}_{3})$$ according to $$e$$-orthogonality.

The point is, so long as $$g$$ is not a scalar multiple of $$e$$, the vector field $$\vec{F}$$ in question must be separable in the way written above and (as far as I understand) no natural force law is like this. Every force law we know of decreases in magnitude with distance, yet $$\vec{F}(\vec{x})$$ in $$(1)$$ and $$(2)$$ can have constant $$\vec{p}_{1}$$ and $$\vec{p}_{2}$$ components no matter where you are at along the $$\vec{p}_{3}$$-axis. Forces of the form $$(3)$$ and $$(4)$$ display similar kind of behavior. Thus, any force law for which $$\nabla\times_{e}\vec{F} = 0$$ and $$\nabla\times_{g}\vec{F} = 0$$ both hold is simply not found in nature. Any natural force law must have zero curl according to at most one metric (up to scaling).

Without any potential energy function whose $$g$$-gradient is $$\vec{F}$$, we cannot talk about conservation of energy with respect to metric $$g$$. Hence the $$g$$-kinetic energy is meaningless.

Tomáš Brauner pointed out that elastic collisions involve kinetic energy conservation. Now we can't discriminate between $$e$$-kinetic energy and $$g$$-kinetic energy in 1D collisions, because in 1D $$e$$- and $$g$$-kinetic energies differ only by a scalar factor. However, I found out that in 2D elastic collisions, only the $$e$$-kinetic energy is conserved and not any other. I take this a valid empirical reason for taking $$e$$-kinetic energy to be the kinetic energy as opposed to any other $$g$$-kinetic energy, but this made me wonder why this was the case when Newton's laws made no reference to kinetic energy.

Note: I'm passing from 3D to 2D here to simplify things. Everything said above carries over to 2D if you adjust various statements accordingly. We define the 2D-curl by $$\nabla\times_{e}\vec{v} := \partial_{1}v_{2} - \partial_{2}v_{1}$$ and the more general 2D-$$g$$-curl by $$\nabla\times_{g}\vec{v} := \frac{1}{\sqrt{|g|}} \nabla\times_{e} g\vec{v}.$$ Note 2: For reference, this excellent post by cromod gives the formulas for the final velocities in a 2D elastic collision. Compactifying the notation a little, we have $$\vec{v}_{1}\,' = \vec{v}_{1} - \frac{2m_{2}}{m_{1}+m_{2}} \frac{\vec{v}_{12}\cdot\vec{x}_{12}}{\vec{x}_{12}^{2}} \vec{x}_{12} \quad\text{ and }\quad \vec{v}_{2}\,' = \vec{v}_{2} - \frac{2m_{1}}{m_{1}+m_{2}} \frac{\vec{v}_{21}\cdot\vec{x}_{21}}{\vec{x}_{21}^{2}} \vec{x}_{21}$$ where the $$\vec{v}_{i}$$'s are initial velocities, $$\vec{v}_{i}\,'$$'s are final velocities, $$\vec{x}_{i}$$'s are the positions of balls at the collision, $$\vec{x}_{ij} = \vec{x}_{j} - \vec{x}_{i}$$, and $$\vec{v}_{ij} = \vec{v}_{j} - \vec{v}_{i}$$.

As far as I understand, I think there is a way to derive the 2D collision formulas using only Newton's laws. Imagine two particles with forces $$\vec{F}_{1} = \begin{cases} \beta\,\vec{e}_{21} &\text{ if } |\vec{x}_{1}-\vec{x}_{2}| < D, \\ 0 &\text{ otherwise} \end{cases}$$ on particle $$1$$ and $$\vec{F}_{2} = \begin{cases} -\beta\,\vec{e}_{21} &\text{ if } |\vec{x}_{1}-\vec{x}_{2}| < D, \\ 0 &\text{ otherwise} \end{cases}$$ on particle $$2$$ where $$\vec{e}_{21} = (\vec{x}_{1}-\vec{x}_{2})/|\vec{x}_{1}-\vec{x}_{2}|$$. In principle, one can consider the velocities as a function of time and then send $$\beta\rightarrow\infty$$. I don't have any proof for this, but I think this leads to the same 2D elastic collision formulas as written above.

Now one can check that for finite $$\beta$$ there is an $$e$$-potential energy function $$U = U(\vec{x}_{1}, \vec{x}_{2})$$ such that $$\vec{F}_{i} = -\nabla_{e, i}U\;\; (i=1, 2)$$, but there is no $$g$$-potential energy function if $$g$$ is not a scalar multiple of $$e$$.

In fact, it's even worse. Keeping $$\beta$$ finite, we can imagine the two particles with velocities parallel to the $$x$$-axis and coming into each other's interaction distance $$D$$. Then by the force law defined above, they push each other in opposite directions, and if their lines of motion are slightly off, it is reasonable to assume they will be outgoing in new directions as depicted below.

Since the outgoing particles go in different directions, it is not guaranteed that $$T = \frac{1}{2}m_{1}\vec{v}_{1}\,^{T}g\vec{v}_{1} + \frac{1}{2}m_{2}\vec{v}_{2}\,^{T}g\vec{v}_{2}$$ is conserved. I can't provide an analytic solution to the above scenario, so instead I turned to Python to give numerical examples.

After doing at least three numerical examples with various $$\beta$$ and $$D$$ parameters and initial conditions, one can use linear algebra show that the only symmetric matrices $$g$$ for which $$T = \frac{1}{2}m_{1}\vec{v}_{1}\,^{T}g\vec{v}_{1} + \frac{1}{2}m_{2}\vec{v}_{2}\,^{T}g\vec{v}_{2}$$ is conserved are of the form $$g = \text{diag}(\lambda, \lambda) = \lambda e$$. Thus, $$g$$-kinetic energy is not conserved if $$g$$ is not a scalar multiple of $$e$$. And of course, there's no corresponding potential energy that we could blame as having provided this change in $$g$$-kinetic energy.

Since the $$g$$-kinetic energy is not conserved here, it is no surprise that it is not conserved as we send $$\beta\rightarrow\infty$$, and in this scenario it is in fact expected. Thus, we can definitively say only the $$e$$-kinetic energy is meaningful (up to a scalar factor).

Python Code

In case anyone is interested, this is the code I used to numerically solve the 2D problem I set up above.

import random
import numpy as np
import scipy as sp
import matplotlib.pyplot as plt
from scipy.integrate import solve_ivp

# define parameters and functions
interaction_dist = 2.0
beta = 1.0
mass1, mass2 = 1.0, 1.0

def function_force2on1(x1, y1, x2, y2):
sqdist = (x1-x2)**2 + (y1-y2)**2
if (0 < sqdist < interaction_dist * interaction_dist):
return beta*(x1 - x2)/np.sqrt(sqdist), beta*(y1 - y2)/np.sqrt(sqdist)
else:
return 0, 0

# initial conditions
initial_X1, initial_Y1 = -5.0, -0.5
initial_X2, initial_Y2 = 5.0, 0.5
initial_velX1, initial_velY1 = 1.0, 0.0
initial_velX2, initial_velY2 = -1.0, 0.0

# ODE
# vect[0] = X1, vect[1] = Y1, vect[2] = X2, vect[3] = Y2
# vect[4] = velX1, vect[5] = velY1, vect[6] = velX2, vect[7] = velY2
def function_ODE(t, vect):
forceX, forceY = function_force2on1(vect[0], vect[1], vect[2], vect[3])
return [vect[4], vect[5], vect[6], vect[7], forceX/mass1, forceY/mass1,
-forceX/mass2, -forceY/mass2]

sol = solve_ivp(function_ODE, [0.0, 10.0],
[initial_X1, initial_Y1, initial_X2, initial_Y2,
initial_velX1, initial_velY1, initial_velX2, initial_velY2],
rtol=1e-6)

# plot trajectories
plt.plot(sol.y[0], sol.y[1], color='b')
plt.plot(sol.y[2], sol.y[3], color='r')

plt.xlim(-6, 6), plt.ylim(-6, 6)
plt.xticks(np.arange(-6, 6, 1.0)), plt.yticks(np.arange(-6, 6, 1.0))